🧠:[<think>Okay, so I need to find all functions f from the real numbers to the real numbers that satisfy the equation f(x + y f(x + y)) = y² + f(x)f(y) for all real x and y. Hmm, functional equations can be tricky, but let's take it step by step.First, maybe I can try plugging in some specific values for x or y to simplify the equation. Let's see. Let me start by setting x = 0. Then the equation becomes:f(0 + y f(0 + y)) = y² + f(0)f(y)Simplifying the left side: f(y f(y)) = y² + f(0)f(y)Hmm, interesting. So for any y, f evaluated at y times f(y) is equal to y squared plus f(0) times f(y). Maybe that's a useful identity. Let me note that down: f(y f(y)) = y² + f(0)f(y) for all y.Another idea: set y = 0. Then the original equation becomes:f(x + 0 * f(x + 0)) = 0² + f(x)f(0)Simplifying the left side: f(x) = f(x)f(0)So, f(x) = f(x)f(0) for all x. That implies that either f(0) = 1 or f(x) = 0 for all x. Wait, but if f(0) ≠ 1, then we must have f(x) = 0 for all x. Let me check if the zero function is a solution.Suppose f(x) = 0 for all x. Then the left side of the original equation becomes f(x + y*0) = f(x) = 0. The right side becomes y² + 0*0 = y². So 0 = y² for all y, which is only true if y = 0. But y can be any real number, so the zero function is not a solution. Therefore, f(0) must be 1. So f(0) = 1.Okay, so we have f(0) = 1. That's a useful piece of information. Let me write that down.Now, let's try to find more information. Maybe set x = -y. Let's try x = -y. Then the original equation becomes:f(-y + y f(-y + y)) = y² + f(-y)f(y)Simplifying the left side: f(-y + y f(0)) since x + y = -y + y = 0. Then, f(-y + y f(0)) = f(-y + y*1) because f(0)=1. That simplifies to f(-y + y) = f(0) = 1. So the left side becomes 1. Therefore, 1 = y² + f(-y)f(y) for all y.So we have y² + f(-y)f(y) = 1 for all y. Rearranging, f(-y)f(y) = 1 - y². Hmm, that's another useful equation. So the product of f(-y) and f(y) is 1 - y². That suggests that f is even or maybe related to sqrt(1 - y²), but not sure yet. Wait, but 1 - y² must be non-negative if we're dealing with real functions, because the product of two real numbers is real. So 1 - y² ≥ 0, which implies that |y| ≤ 1. But this must hold for all real y. Wait, that's a problem. If y is any real number, then 1 - y² can be negative, but the right-hand side f(-y)f(y) must still equal 1 - y². However, if f maps real numbers to real numbers, then f(-y)f(y) is real, but 1 - y² can be negative. So for |y| > 1, 1 - y² is negative, which means f(-y)f(y) must be negative. So for |y| > 1, f(-y) and f(y) must have opposite signs. Hmm, but this is okay, as long as the product is 1 - y². Maybe f(y) = 1 - y²? Wait, let's test that.Wait, let's test the function f(y) = 1 - y². Let's check if this satisfies the original equation. Let's substitute f(x) = 1 - x² into the equation:Left side: f(x + y f(x + y)) = f(x + y(1 - (x + y)^2)) = 1 - [x + y(1 - (x + y)^2)]²Right side: y² + f(x)f(y) = y² + (1 - x²)(1 - y²) = y² + 1 - x² - y² + x² y² = 1 - x² + x² y²So, we need to check if 1 - [x + y(1 - (x + y)^2)]² equals 1 - x² + x² y².This seems complicated. Let me compute the left side:First, compute the argument of f on the left side: x + y(1 - (x + y)^2) = x + y - y(x + y)^2Then, square that expression and subtract from 1:1 - [x + y - y(x + y)^2]^2That's going to be messy. Let's see if perhaps f(y) = 1 - y² is a solution. Alternatively, maybe another function.Alternatively, perhaps f is linear. Let's assume f is linear. Suppose f(x) = ax + b. Let's see if such a function can satisfy the equation. Let's substitute into the original equation:Left side: f(x + y f(x + y)) = a(x + y f(x + y)) + b = a x + a y f(x + y) + bBut f(x + y) = a(x + y) + b, so substituting:Left side: a x + a y [a(x + y) + b] + b = a x + a y (a x + a y + b) + bRight side: y² + f(x) f(y) = y² + (a x + b)(a y + b)Let's expand both sides:Left side:a x + a y (a x + a y + b) + b= a x + a^2 x y + a^2 y² + a b y + bRight side:y² + (a x + b)(a y + b)= y² + a^2 x y + a b x + a b y + b²Now, set left side equal to right side:a x + a^2 x y + a^2 y² + a b y + b = y² + a^2 x y + a b x + a b y + b²Compare coefficients of like terms.First, coefficients of y²:Left side: a^2Right side: 1So a^2 = 1 => a = 1 or a = -1.Coefficients of x:Left side: aRight side: a bTherefore, a = a b. If a ≠ 0, then 1 = b. So if a = 1 or -1, then b = 1.Coefficients of y:Left side: a bRight side: a bSo same on both sides.Constants:Left side: bRight side: b²So b = b². Since we already have b = 1 if a ≠ 0, then 1 = 1², which is true. So if a = 1 or -1, and b = 1, then we can check if the other terms match.Now, check the remaining terms. For example, the coefficient of x y:Left side: a^2Right side: a^2So that's same on both sides. So if a^2 = 1, then okay.But let's check the coefficients of y² again. Left side: a^2 = 1, right side: 1. So that's okay. Wait, but if a = -1, then a^2 is still 1, so that's okay.But then, let's check the entire equation with a = 1, b = 1:Left side: 1*x + 1*1*x y + 1*1 y² + 1*1 y + 1 = x + x y + y² + y + 1Right side: y² + (1*x +1)(1*y +1) = y² + (x +1)(y +1) = y² + x y + x + y +1Comparing left and right:Left: x + x y + y² + y + 1Right: y² + x y + x + y +1They are the same. So f(x) = x + 1 is a solution? Wait, but if a =1, b=1, then f(x) = x +1. Let's check if that works in the original equation.Wait, let's test f(x) = x +1. So substitute into the original equation:Left side: f(x + y f(x + y)) = (x + y f(x + y)) +1But f(x + y) = (x + y) +1, so substitute that in:Left side = x + y*(x + y +1) +1 = x + x y + y² + y +1Right side: y² + f(x)f(y) = y² + (x +1)(y +1) = y² + x y + x + y +1Comparing both sides: x + x y + y² + y +1 vs. y² + x y + x + y +1. They are equal. So yes, f(x) = x +1 is a solution.What about a = -1, b=1? Let's check f(x) = -x +1.Left side: f(x + y f(x + y)) = - [x + y f(x + y)] +1But f(x + y) = - (x + y) +1, so substitute:Left side = - [x + y*(-x - y +1)] +1 = - [x - x y - y² + y] +1 = -x + x y + y² - y +1Right side: y² + f(x)f(y) = y² + (-x +1)(-y +1) = y² + (x y - x - y +1)So right side = y² + x y - x - y +1Compare left and right:Left: -x + x y + y² - y +1Right: y² + x y - x - y +1They are the same. So f(x) = -x +1 is also a solution. So both f(x) = x +1 and f(x) = -x +1 satisfy the equation. But wait, we have f(0) =1, which is satisfied by both: f(0) = 0 +1 =1 and f(0)= -0 +1=1.But earlier, when we set x=0, we had f(y f(y)) = y² + f(0)f(y) = y² + f(y). Let's check if these functions satisfy that.Take f(y) = y +1:Left side: f(y f(y)) = f(y(y +1)) = f(y² + y) = (y² + y) +1 = y² + y +1Right side: y² + f(y) = y² + (y +1) = y² + y +1. So yes, equal.For f(y) = -y +1:Left side: f(y f(y)) = f(y*(-y +1)) = f(-y² + y) = -(-y² + y) +1 = y² - y +1Right side: y² + f(y) = y² + (-y +1) = y² - y +1. Also equal. So both functions satisfy the equation when x=0.Now, but we need to check if there are any other solutions. Let's see. So far, we have two linear solutions. Could there be non-linear solutions?Earlier, when we set x = -y, we found that f(-y)f(y) =1 - y². Let's check this for our solutions.For f(y) = y +1: f(-y)f(y) = (-y +1)(y +1) = (-y)(y) + (-y)(1) +1(y) +1(1) = -y² - y + y +1 = -y² +1 =1 - y². Correct.For f(y) = -y +1: f(-y)f(y) = (y +1)(-y +1) = -y² +1 + y - y = -y² +1. Also correct.So both functions satisfy the relation f(-y)f(y) =1 - y².Now, the question is, are there any other functions that satisfy the original equation?Suppose there is a quadratic function. Let's suppose f(y) = a y² + b y + c. Let's see if such a function can satisfy the equation. But this might get complicated. Let's think.Alternatively, maybe f is an even function? Because f(-y)f(y) =1 - y², which is even. So if f is even, then f(-y) = f(y), so [f(y)]² =1 - y². Thus, f(y) = ±√(1 - y²). But 1 - y² must be non-negative, so |y| ≤1. But the domain is all real numbers, so if |y| >1, then 1 - y² is negative, and square root is not real. Therefore, such a function cannot be real for all real numbers y. Hence, f cannot be even, except if we restrict y, but the problem states that f is defined for all real numbers. So f(y) = sqrt(1 - y²) is not possible. Thus, such a function is invalid. Therefore, the only possible solutions are the linear ones we found earlier.Wait, but what if f is neither linear nor even? Let's see.Alternatively, suppose f is a quadratic function. Let's try f(y) =1 - y². Wait, but earlier, when I tried substituting f(y) =1 - y² into the original equation, it didn't seem to work, but let me check again more carefully.Let me try f(x) =1 - x². Then let's compute both sides.Left side: f(x + y f(x + y)).First compute f(x + y): f(x + y) =1 - (x + y)^2. Then, multiply by y: y*(1 - (x + y)^2) = y - y(x + y)^2. Then add x: x + y - y(x + y)^2. Then apply f: 1 - [x + y - y(x + y)^2]^2.Right side: y² + f(x)f(y) = y² + (1 - x²)(1 - y²) = y² +1 - x² - y² + x² y² =1 - x² + x² y².So left side is 1 - [x + y - y(x + y)^2]^2, and right side is 1 - x² + x² y². Let's compute the left side:Let me denote A = x + y - y(x + y)^2. Then left side is 1 - A².Expand A: x + y - y(x² + 2x y + y²) =x + y - y x² -2x y² - y³.So A = x + y - y x² -2x y² - y³. Then A² is going to be a big expression, and 1 - A² is going to be very complicated. It's unlikely that this would simplify to 1 - x² +x² y². Let's check for specific values. Let me set x=0, y=1.Left side: f(0 +1*f(0 +1)) =f(0 +1*f(1)). Since f(1)=1 -1=0, so left side is f(0 +0)=f(0)=1.Right side:1² +f(0)f(1)=1 +1*0=1. So equality holds here.Another test: x=1, y=0.Left side: f(1 +0*f(1+0))=f(1)=0.Right side:0² +f(1)f(0)=0 +0*1=0. Okay, equality holds.Another test: x=1, y=1.Left side: f(1 +1*f(2)). f(2)=1 -4= -3, so f(1 +1*(-3))=f(-2)=1 -4= -3.Right side:1² +f(1)f(1)=1 +0*0=1. But left side is -3, which is not equal to 1. So f(x)=1 -x² is not a solution. Therefore, f(x) =1 -x² doesn't work. So the quadratic function is invalid.Therefore, so far, only the linear functions f(x)=x+1 and f(x)=-x +1 work. Let me see if there are any other solutions.Another approach: suppose f is injective or surjective. Let's check if f is injective. Suppose f(a) = f(b). Let's see if this implies a = b. Not sure yet.Alternatively, let's check if f is surjective. For example, take f(x) =x +1. It's surjective because for any real number z, z = f(z -1). Similarly, f(x)=-x +1 is also surjective, since for any z, z = f(1 - z). So both linear solutions are surjective.Alternatively, maybe we can prove that f is linear. Let's suppose that f is a polynomial function. Suppose f is a polynomial. Let's see. Then, since f(0) =1, the constant term is 1. Let’s assume f is a quadratic polynomial: f(x) = a x² + b x +1. Let's see if that can work.But earlier, when we tried f(x)=1 -x², it didn't work. So maybe no quadratic polynomial works. Let's suppose f is linear, so degree 1. Then we already found the solutions. If f is a constant function, but f(0)=1, then f(x)=1 for all x. Let's check if that works. f(x + y*1) =1= y² +1*1. So 1 = y² +1 => y²=0 for all y, which is false. Hence, constant function is invalid.Alternatively, maybe f is a cubic polynomial? That seems more complicated, but let me try.Suppose f(x)=a x³ +b x² +c x +d. But this might get too messy. Alternatively, maybe f is linear, given that we found two linear solutions. Let's see.Let’s suppose that f is additive or multiplicative, but given the original equation, it's not straightforward.Alternatively, let's try to find f(1). Let me compute f(1). Let's set x=1 and y=1 in the original equation.Left side: f(1 +1*f(2)). But we don't know f(2) yet. Alternatively, maybe set y=1.Set y=1. Then the equation becomes:f(x +1*f(x +1)) =1 + f(x)f(1)Let’s denote z = x + f(x +1). Then f(z) =1 + f(x)f(1). Hmm, but not sure how to use that.Alternatively, set x=1 and y=1. Then:Left side: f(1 +1*f(2)) = f(1 +f(2))Right side:1 + f(1)f(1) =1 + [f(1)]²So f(1 +f(2)) =1 + [f(1)]²But without knowing f(1) or f(2), this is not helpful yet.Wait, for our solutions f(x)=x +1 and f(x)=-x +1, let's compute f(1). For f(x)=x +1, f(1)=2. For f(x)=-x +1, f(1)=0.Wait, but earlier, when we set x=1, y=1 in f(x)=1 -x², which isn't a solution, but in the valid solutions, f(1)=2 or 0.Wait, maybe we can find f(1). Let me set x=0, y=1 in the original equation.Left side: f(0 +1*f(1)) =f(f(1))Right side:1² +f(0)f(1)=1 +1*f(1)=1 +f(1)Therefore, f(f(1))=1 +f(1). Let's call A =f(1). Then f(A)=1 +A. For our solutions:For f(x)=x +1: f(1)=2, then f(2)=3, and 1 +2=3, which holds.For f(x)=-x +1: f(1)=0, then f(0)=1, and 1 +0=1, which holds. So the equation f(f(1))=1 +f(1) is satisfied by both solutions.But how does this help us? Maybe consider that if f is linear, then this holds. If f is non-linear, maybe it can also hold. Not sure.Another idea: let's suppose that f is invertible. Then maybe we can find an expression for its inverse. But since f(0)=1, unless f is bijective, we might not have an inverse. However, the linear functions we found are bijective (since they have non-zero slope), so they have inverses.Alternatively, let's try to set y=1 in the original equation. Then:f(x +1*f(x +1))=1 +f(x)f(1)Let’s denote u =x +f(x +1). Then f(u)=1 +f(x)f(1). If we can express x in terms of u, maybe we can find a relation.But this seems challenging. Alternatively, if f is linear, then f(x +1)=a(x +1)+b. Let’s plug into the equation for y=1:For f(x)=ax +b, we have:Left side: f(x +1*(a(x +1) +b))=a(x + a(x +1) +b) +b =a x +a²(x +1) +a b +bRight side:1 + (a x +b)(a*1 +b) =1 + (a x +b)(a +b)If f is linear, a and b must satisfy the equation:a x +a² x +a² +a b +b =1 + (a x +b)(a +b)Expanding the right side:1 + a(a +b)x +b(a +b)Comparing coefficients:Coefficient of x:Left: a +a²Right: a(a +b)So a +a² =a(a +b) => a +a² =a² +a b => a =a b => If a ≠0, then 1 =b. Which is consistent with previous results.Constant terms:Left: a² +a b +bRight:1 +b(a +b)So setting them equal:a² +a b +b =1 +a b +b²Simplify: a² +a b +b -1 -a b -b² =0 => a² - b² +b -1=0But from previous, if a ≠0, then b=1. So substitute b=1:a² -1 +1 -1=0 =>a² -1=0 =>a²=1 =>a=1 or -1, which matches our solutions. Therefore, the only linear solutions are f(x)=x +1 and f(x)=-x +1.But could there be non-linear solutions?Suppose there exists a non-linear solution. Let's assume that f is non-linear. Let's see if we can derive a contradiction.From the equation f(-y)f(y)=1 -y², if we suppose that f is a quadratic function, as before, but that didn't work. Alternatively, perhaps f is a cubic function. Let me try a cubic function.But this might be too time-consuming. Alternatively, let's try to find f(y) for specific y.For example, set y=1. Then f(-1)f(1)=1 -1=0. So either f(-1)=0 or f(1)=0. For our solutions:For f(x)=x +1: f(1)=2, f(-1)=0. So 0*2=0, which matches 1 -1=0.For f(x)=-x +1: f(1)=0, f(-1)=2. So 2*0=0, which also matches.So in both cases, the product is zero. So if there is another solution, perhaps f(1)=0 or f(-1)=0. Let's suppose f(1)=0. Then, from the equation f(f(1))=1 +f(1), we have f(0)=1 +0=1, which is consistent with f(0)=1.Similarly, if f(-1)=0, then from the equation f(-y)f(y)=1 -y² with y=1, we have f(-1)f(1)=0, but since f(-1)=0, then f(1) can be arbitrary? Wait, no. Because f(-1)f(1)=1 -1=0, which holds regardless. So perhaps f(1) can be zero or not. But for the known solutions, one has f(1)=0 and the other f(1)=2.Alternatively, let's suppose that f(1)=0. Let me see if I can derive more properties. Suppose f(1)=0.From the original equation, set x=1, y=1:Left side: f(1 +1*f(2))=f(1 +f(2))=f(1 +f(2))Right side:1 + f(1)f(1)=1 +0=1So f(1 +f(2))=1But if f(1)=0, and we need to find f(2). From the equation f(-y)f(y)=1 -y², set y=2:f(-2)f(2)=1 -4= -3. So f(-2)f(2)=-3.But without knowing more about f(2), this might not help. However, suppose we also set x=1, y=-1 in the original equation:Left side: f(1 + (-1)f(0))=f(1 -1*1)=f(0)=1Right side: (-1)^2 +f(1)f(-1)=1 +0*f(-1)=1So that holds regardless. Not helpful.Alternatively, set x=1, y=2:Left side: f(1 +2 f(3))Right side:4 +f(1)f(2)=4 +0*f(2)=4So f(1 +2 f(3))=4.But unless we know f(3), this is not helpful.Alternatively, set y=1 in the original equation:f(x +1*f(x +1))=1 +f(x)f(1)=1 +0=1.So f(x +f(x +1))=1 for all x.But if f(x +f(x +1))=1 for all x, then since f(0)=1, this suggests that x +f(x +1) is a root of f(z)=1. So if f(z)=1, then z=0 or any other roots? Wait, but for our solution f(x)= -x +1, f(z)=1 implies -z +1=1 => z=0. So in this case, x +f(x +1)=0 for all x. Let's check:For f(x)= -x +1: f(x +1)= - (x +1) +1= -x. So x +f(x +1)=x -x=0. So indeed, f(0)=1.But for f(x)=x +1: f(x +1)=x +1 +1=x +2. Then x +f(x +1)=x +x +2=2x +2. So f(2x +2)=1. But f(2x +2)= (2x +2) +1=2x +3. Setting this equal to1: 2x +3=1 => x= -1. But the equation f(x +f(x +1))=1 must hold for all x, but in this case, it only holds when x=-1. Wait, but this contradicts. Wait, but earlier, when we considered f(x)=x +1, and set y=1 in the original equation, we had:f(x +1*f(x +1))=1 +f(x)f(1)For f(x)=x +1, f(1)=2, so the right side is1 + (x +1)*2=1 +2x +2=2x +3. The left side is f(x +1*(x +2))=f(x +x +2)=f(2x +2)=2x +2 +1=2x +3. So equality holds. But when we set y=1, the equation becomes f(x +f(x +1))=1 +2(x +1). Wait, but earlier, I thought that f(x +f(x +1))=1 for all x if f(1)=0, but if f(1)=2, then f(x +f(x +1))=2x +3, which is not a constant. So that approach was incorrect.Wait, in the case where f(1)=0, we have from setting y=1: f(x +f(x +1))=1 for all x. But if f(1)=0, then perhaps this gives us a functional equation for f. Let me explore that.Suppose f(x +f(x +1))=1 for all x. Let me set z =x +1. Then x =z -1. So the equation becomes f(z -1 +f(z))=1 for all z. Therefore, f(z -1 +f(z))=1 for all z.But we know f(0)=1, so this implies that z -1 +f(z)=0, or z -1 +f(z)=c where f(c)=1. But since f(0)=1, perhaps c=0. So z -1 +f(z)=0 => f(z)=1 -z for all z. Which is one of our solutions, f(z)= -z +1.So if f(z)= -z +1, then indeed f(z -1 +f(z))=f(z -1 + (-z +1))=f(0)=1. So that works.Alternatively, if f(z -1 +f(z))=1 and we don't assume that z -1 +f(z)=0, but maybe there are other solutions where z -1 +f(z) is some other root of f. However, since f(0)=1 and if there are other roots, then f would take the value 1 at multiple points. But for our other solution f(z)=z +1, f(z)=1 implies z +1=1 =>z=0. So in that case, the only root is z=0. Therefore, if f(z -1 +f(z))=1 for all z, then for the solution f(z)=z +1, this would require z -1 +f(z)=0 for all z. But z -1 +z +1=2z=0 =>z=0. Which is not for all z. Contradiction. Wait, but earlier when we checked f(z)=z +1, the equation f(x +f(x +1))=2x +3=1 +2x +2, which equals the right side. But if we set f(z -1 +f(z))=1, but for f(z)=z +1, this is f(z -1 +z +1)=f(2z)=2z +1=1 =>2z +1=1 =>z=0. Which only holds for z=0, not for all z. Hence, in this case, when f(1)=2, the equation f(x +f(x +1))=1 +f(x)f(1) results in f(2x +2)=2x +3, which holds for all x, which is valid.But if we have another function where f(z -1 +f(z))=1 for all z, then z -1 +f(z)=0 for all z, leading to f(z)=1 -z. Hence, this gives uniqueness for f(z)= -z +1.Similarly, for the case when f(1)=0, we derive f(z)= -z +1.Thus, combining these observations, it seems that the only solutions are f(z)=z +1 and f(z)=-z +1.Another angle: the original equation must hold for all x and y. Let's suppose that f is differentiable. Although the problem doesn't specify differentiability, but assuming that could help. Let's try differentiating both sides with respect to y and set y=0. However, this might be complicated.Alternatively, notice that the equation resembles a quadratic in y. Let's fix x and see the equation as f(x + y f(x + y)) - f(x)f(y) = y². Maybe expanding in terms of y.But this is vague. Alternatively, for fixed x, let's consider the behavior as y approaches 0.But not sure. Alternatively, set y=1 and vary x.Alternatively, use the relation f(-y)f(y)=1 -y². Let me write this as f(y)f(-y)=1 -y². Suppose we let y=z, then f(z)f(-z)=1 - z². If we let y=-z, then f(-z)f(z)=1 -z², which is the same. So this is symmetric.Suppose we set y=1, then f(1)f(-1)=0. So either f(1)=0 or f(-1)=0. For our solutions:- f(x)=x +1: f(1)=2, f(-1)=0. So f(-1)=0.- f(x)=-x +1: f(1)=0, f(-1)=2.Thus, in both cases, either f(1)=0 or f(-1)=0. So if we have another solution, it must satisfy one of these. Let's explore f(1)=0 case.Assume f(1)=0. Then, from f(-y)f(y)=1 -y², set y=1: f(-1)*0=1 -1=0, which holds for any f(-1). But from the equation f(-1)f(1)=0, which is 0=0. So f(-1) is arbitrary? Wait, but in the relation f(-y)f(y)=1 -y², if y=1, then f(-1)f(1)=0=1 -1=0, which holds regardless of f(-1). So if f(1)=0, then f(-1) can be anything. But from the original functional equation, there may be constraints.But in our known solutions, when f(1)=0, f(-1)=2. Let's see if we can derive that.From the original equation, set x= -1, y=1:f(-1 +1*f(-1 +1))=f(-1 +1*f(0))=f(-1 +1*1)=f(0)=1Right side:1² +f(-1)f(1)=1 +f(-1)*0=1So it holds regardless. Not helpful.Set x= -1, y= -1:Left side: f(-1 + (-1)f(-2))=f(-1 -f(-2))Right side: (-1)^2 +f(-1)f(-1)=1 +[f(-1)]²So f(-1 -f(-2))=1 +[f(-1)]²But this seems complicated without knowing f(-2) or f(-1).Alternatively, set x=1, y=-1:Left side: f(1 + (-1)f(0))=f(1 -1)=f(0)=1Right side: (-1)^2 +f(1)f(-1)=1 +0*f(-1)=1Again, holds regardless.Perhaps try to find f(2). From y=2 in f(-2)f(2)=1 -4=-3. So f(-2)f(2)=-3. If we can express f(2) in terms of other values.From the original equation, set x=0, y=2:f(0 +2 f(2))=2² +f(0)f(2)=4 +1*f(2)=4 +f(2)Left side: f(2 f(2))=4 +f(2)But we also have from the relation f(-2)f(2)=-3. So if we let f(2)=a, then f(-2)= -3/a.Therefore, f(2a)=4 +a.For our solutions:- For f(x)=x +1: f(2)=3, so a=3. Then f(2a)=f(6)=7, and 4 +a=7. Holds.- For f(x)=-x +1: f(2)= -2 +1= -1, so a=-1. Then f(2a)=f(-2)=3, and 4 +a=3. 4 +(-1)=3. Holds.So this equation f(2a)=4 +a is satisfied by both solutions. Suppose there is another solution where f(2)=a, then f(2a)=4 +a and f(-2)= -3/a. Let me see if this leads to a unique solution.Assume f(2)=a, then f(2a)=4 +a. Also, from the original equation, set x=2, y=1:Left side: f(2 +1*f(3))=f(2 +f(3))Right side:1 +f(2)f(1)=1 +a*0=1So f(2 +f(3))=1.But for f(x)=x +1: f(3)=4, so 2 +4=6, f(6)=7≠1. Wait, but this contradicts. Wait, no, for f(x)=x +1, setting x=2, y=1:Left side: f(2 +1*f(3))=f(2 +4)=f(6)=7Right side:1 +f(2)f(1)=1 +3*2=7. So yes, equality holds. Wait, but f(2 +f(3))=7 and 1 +f(2)f(1)=7. So 7=7.But if in another solution where f(1)=0 and f(2)=a, then f(2 +f(3))=1. But if f(1)=0, then f(3) can be determined?Wait, if f(1)=0, then from f(-1)f(1)=0, which holds, but how to find f(3)?From f(-3)f(3)=1 -9= -8. So f(-3)f(3)= -8.If we set x=3, y=0: f(3 +0)=0² +f(3)f(0) =>f(3)=1 +f(3)*1. Wait, but f(0)=1. So f(3)=1 +f(3). Subtract f(3): 0=1. Contradiction. Therefore, there is no solution with f(1)=0 except f(x)=-x +1.Wait, but for f(x)=-x +1, f(3)=-3 +1=-2. Then f(3)= -2. Let's check x=3, y=0:Left side: f(3 +0)=f(3)= -2Right side:0 +f(3)f(0)= -2*1= -2. So equality holds. Also, for x=2, y=1:Left side: f(2 +1*f(3))=f(2 +1*(-2))=f(0)=1Right side:1 +f(2)f(1)=1 +(-1)*0=1. So equality holds. So even though f(3)= -2, which is not 1, the equation works.But if there was another function with f(1)=0, then x=3, y=0 would require f(3)=1 +f(3), leading to 0=1, which is impossible. Therefore, the only possible function with f(1)=0 is f(x)=-x +1.Therefore, combining all these steps and checks, it seems that the only solutions are f(x)=x +1 and f(x)=-x +1. No other functions satisfy the given equation.Final AnswerThe solutions are boxed{f(x) = x + 1} and boxed{f(x) = -x + 1}.</think>To find all functions ( f: mathbb{R} rightarrow mathbb{R} ) such that[f(x + y f(x + y)) = y^2 + f(x) f(y)]for all ( x, y in mathbb{R} ), we proceed as follows:1. Setting ( x = 0 ): [ f(y f(y)) = y^2 + f(0) f(y) ] This implies ( f(y f(y)) = y^2 + f(0) f(y) ).2. Setting ( y = 0 ): [ f(x) = f(x) f(0) ] This implies either ( f(0) = 1 ) or ( f(x) = 0 ) for all ( x ). The zero function is not a solution, so ( f(0) = 1 ).3. Setting ( x = -y ): [ f(-y + y f(0)) = y^2 + f(-y) f(y) ] Simplifying, we get: [ f(0) = 1 implies 1 = y^2 + f(-y) f(y) ] Thus, ( f(-y) f(y) = 1 - y^2 ).4. Testing linear functions: - Assume ( f(x) = ax + b ). - Substituting into the original equation and comparing coefficients, we find ( a = 1 ) or ( a = -1 ) with ( b = 1 ).5. Verifying solutions: - For ( f(x) = x + 1 ): [ f(x + y f(x + y)) = x + y + y(x + y + 1) + 1 = x + xy + y^2 + y + 1 ] Right side: [ y^2 + (x + 1)(y + 1) = y^2 + xy + x + y + 1 ] Both sides match. - For ( f(x) = -x + 1 ): [ f(x + y f(x + y)) = - (x + y(-x - y + 1)) + 1 = -x + xy + y^2 - y + 1 ] Right side: [ y^2 + (-x + 1)(-y + 1) = y^2 + xy - x - y + 1 ] Both sides match.6. Checking non-linear solutions: - Testing quadratic functions like ( f(x) = 1 - x^2 ) leads to contradictions. - Polynomial and other non-linear functions do not satisfy the given functional equation.7. Conclusion: The only solutions are the linear functions ( f(x) = x + 1 ) and ( f(x) = -x + 1 ).[boxed{f(x) = x + 1} quad text{and} quad boxed{f(x) = -x + 1}],<think>Okay, let's tackle this functional equation problem. The goal is to find all functions ( f: mathbb{R} rightarrow mathbb{R} ) such that for all real numbers ( x ) and ( y ), the equation[f(x + y f(x + y)) = y^2 + f(x) f(y)]holds true. Hmm, functional equations can be tricky, but maybe by plugging in specific values for ( x ) and ( y ), I can figure out some properties of ( f ).First, let's try plugging in ( x = 0 ). That often simplifies things. So, set ( x = 0 ):[f(0 + y f(0 + y)) = y^2 + f(0) f(y)]Simplifies to:[f(y f(y)) = y^2 + f(0) f(y)]Interesting. So for any real ( y ), ( f(y f(y)) ) is equal to ( y^2 + f(0) f(y) ). Maybe this can help us find ( f(0) ). Let's try plugging in ( y = 0 ) into the original equation. Let's see:Set ( y = 0 ):[f(x + 0 cdot f(x + 0)) = 0^2 + f(x) f(0)]Simplifies to:[f(x) = f(x) f(0)]So, ( f(x) = f(x) f(0) ) for all ( x ). This equation tells us that either ( f(0) = 1 ) or ( f(x) = 0 ) for all ( x ). Let's check if ( f(x) = 0 ) is a solution.If ( f(x) = 0 ) for all ( x ), substitute into the original equation:Left-hand side (LHS): ( f(x + y cdot 0) = f(x) = 0 )Right-hand side (RHS): ( y^2 + 0 cdot 0 = y^2 )So, ( 0 = y^2 ) for all ( y ), which is only true if ( y = 0 ), but since this must hold for all real ( y ), the zero function is not a solution. Therefore, ( f(0) ) must be equal to 1. So, ( f(0) = 1 ).Now, knowing that ( f(0) = 1 ), let's revisit the equation we had when ( x = 0 ):[f(y f(y)) = y^2 + f(0) f(y) = y^2 + f(y)]So, ( f(y f(y)) = y^2 + f(y) ). Hmm, maybe this can help us figure out the form of ( f ).Another approach: let's try to see if ( f ) is a quadratic function. Sometimes functional equations like this have polynomial solutions. Let's assume ( f(x) ) is a quadratic function, say ( f(x) = ax^2 + bx + c ). Let's test this assumption.But before that, maybe try other specific values. Let's set ( y = 1 ). Then the original equation becomes:[f(x + 1 cdot f(x + 1)) = 1^2 + f(x) f(1) implies f(x + f(x + 1)) = 1 + f(x) f(1)]This might not be immediately helpful, but if we can find a relationship between ( f(x + 1) ) and other terms, maybe.Alternatively, let's see if ( f ) is injective or surjective. If we can prove that ( f ) is injective or surjective, that might help. But how?Wait, let's see if we can find ( f(0) ), which we already did. ( f(0) = 1 ). Also, perhaps setting ( y = 1 ) and then varying ( x ). Let me try ( y = 1 ):Original equation:[f(x + 1 cdot f(x + 1)) = 1 + f(x) f(1)]Suppose we let ( x = -f(x + 1) ). Wait, but that would require ( x ) to be expressed in terms of ( f(x + 1) ), which is a bit circular. Maybe not helpful.Alternatively, let's set ( x = -y ). Then the equation becomes:[f(-y + y f(-y + y)) = y^2 + f(-y) f(y)]Simplify the argument of ( f ):First, compute ( -y + y f(-y + y) ). The inner function is ( f(0) ), which is 1. So:[f(-y + y cdot 1) = f(0) = 1]So the left-hand side is ( f(0) = 1 ). Therefore:[1 = y^2 + f(-y) f(y)]Therefore, for all real ( y ):[f(-y) f(y) = 1 - y^2]Interesting! So ( f(-y) f(y) = 1 - y^2 ). This is a useful relation. Let's note this down:For all ( y in mathbb{R} ), ( f(-y) f(y) = 1 - y^2 ).This suggests that ( f ) is even or has some symmetric properties? Wait, not necessarily, unless ( f(-y) = f(y) ), but if that were the case, then ( f(y)^2 = 1 - y^2 ), which would imply ( f(y) = pm sqrt{1 - y^2} ), but that's only real if ( |y| leq 1 ), which contradicts the domain being all real numbers. Therefore, ( f(-y) ) isn't necessarily equal to ( f(y) ).But the product ( f(y) f(-y) ) is ( 1 - y^2 ). This seems similar to the equation for reciprocal functions but with a quadratic term. Maybe if we consider ( f(y) = 1 - y ), but let's test this.Wait, if we suppose ( f(y) = 1 - y ), then ( f(-y) = 1 + y ), so the product is ( (1 - y)(1 + y) = 1 - y^2 ), which matches the equation ( f(y) f(-y) = 1 - y^2 ). So that works! So maybe ( f(y) = 1 - y ) is a solution? Let's check.Let me test ( f(y) = 1 - y ).Substitute into the original equation:Left-hand side (LHS): ( f(x + y f(x + y)) )First compute ( f(x + y) = 1 - (x + y) ). Then, ( y f(x + y) = y(1 - x - y) ). Therefore, the argument of the outer ( f ) is ( x + y(1 - x - y) = x + y - xy - y^2 ). Then, apply ( f ):( f(x + y - xy - y^2) = 1 - (x + y - xy - y^2) = 1 - x - y + xy + y^2 ).Right-hand side (RHS): ( y^2 + f(x) f(y) ). Since ( f(x) = 1 - x ), ( f(y) = 1 - y ), so:( y^2 + (1 - x)(1 - y) = y^2 + 1 - x - y + xy ).Compare LHS and RHS:LHS: ( 1 - x - y + xy + y^2 )RHS: ( 1 - x - y + xy + y^2 )They are equal! So ( f(y) = 1 - y ) is indeed a solution. Nice!But the problem says "Find all functions", so we need to check if there are other solutions as well. Let's see.Earlier, we found that ( f(-y) f(y) = 1 - y^2 ). Suppose there is another function ( f ) satisfying this, perhaps quadratic? Let's think.Assume ( f ) is a quadratic function. Let ( f(y) = ay^2 + by + c ). Then, ( f(-y) = ay^2 - by + c ). Their product is:( (ay^2 + by + c)(ay^2 - by + c) = a^2 y^4 + (2ac - b^2) y^2 + c^2 ).But according to the equation, this product should equal ( 1 - y^2 ). Therefore:Equate coefficients:- Coefficient of ( y^4 ): ( a^2 = 0 ) ⇒ ( a = 0 )- Coefficient of ( y^2 ): ( 2ac - b^2 = -1 )- Coefficient of ( y^0 ): ( c^2 = 1 )Since ( a = 0 ), the quadratic term disappears. So the function is linear: ( f(y) = by + c ).Then, ( f(-y) = -by + c ). Their product:( (by + c)(-by + c) = -b^2 y^2 + c^2 ).Set equal to ( 1 - y^2 ):Thus,- Coefficient of ( y^2 ): ( -b^2 = -1 ) ⇒ ( b^2 = 1 ) ⇒ ( b = 1 ) or ( b = -1 )- Constant term: ( c^2 = 1 ) ⇒ ( c = 1 ) or ( c = -1 )But earlier, we found that ( f(0) = c = 1 ). So ( c = 1 ). Therefore, ( f(y) = by + 1 ), where ( b = pm 1 ). So two possibilities: ( f(y) = y + 1 ) or ( f(y) = -y + 1 ).Wait, but we previously tested ( f(y) = 1 - y ), which is ( -y + 1 ), and it worked. What about ( f(y) = y + 1 )? Let's test that.Let me check ( f(y) = y + 1 ).Substitute into the original equation:LHS: ( f(x + y f(x + y)) )First, compute ( f(x + y) = (x + y) + 1 = x + y + 1 ). Then, ( y f(x + y) = y(x + y + 1) = xy + y^2 + y ). So the argument of the outer ( f ) is ( x + xy + y^2 + y = x(1 + y) + y^2 + y ). Then, apply ( f ):( f(x(1 + y) + y^2 + y) = [x(1 + y) + y^2 + y] + 1 = x(1 + y) + y^2 + y + 1 ).RHS: ( y^2 + f(x) f(y) ). Since ( f(x) = x + 1 ), ( f(y) = y + 1 ), so:( y^2 + (x + 1)(y + 1) = y^2 + xy + x + y + 1 ).Compare LHS and RHS:LHS: ( x(1 + y) + y^2 + y + 1 = x + xy + y^2 + y + 1 )RHS: ( y^2 + xy + x + y + 1 )They are the same! So ( f(y) = y + 1 ) also satisfies the equation. Wait, but earlier when we considered ( f(-y) f(y) = 1 - y^2 ), if ( f(y) = y + 1 ), then ( f(-y) = -y + 1 ), and their product is ( (y + 1)(-y + 1) = -y^2 + 1 ), which is indeed ( 1 - y^2 ). So that's consistent.But hold on, we have two possible linear functions: ( f(y) = 1 - y ) and ( f(y) = 1 + y ). But both satisfy the equation. Are both valid solutions?Wait, earlier when we set ( f(y) = 1 - y ), it worked, and now ( f(y) = 1 + y ) also works. Let me verify with ( f(y) = 1 + y ).Take ( f(y) = 1 + y ). Let's check the original equation:LHS: ( f(x + y f(x + y)) )Compute ( f(x + y) = 1 + x + y ), so ( y f(x + y) = y(1 + x + y) = y + xy + y^2 ). Therefore, the argument of the outer function is ( x + y + xy + y^2 ). Then, applying ( f ):( f(x + y + xy + y^2) = 1 + x + y + xy + y^2 )RHS: ( y^2 + f(x) f(y) ). ( f(x) = 1 + x ), ( f(y) = 1 + y ), so:( y^2 + (1 + x)(1 + y) = y^2 + 1 + x + y + xy )Compare LHS and RHS:LHS: ( 1 + x + y + xy + y^2 )RHS: ( 1 + x + y + xy + y^2 )They match! So yes, ( f(y) = 1 + y ) is also a solution. So both ( f(y) = 1 + y ) and ( f(y) = 1 - y ) satisfy the equation. Wait, but according to our earlier analysis when assuming a linear function, we have ( b = pm 1 ), so both possibilities are allowed. But does this mean both are solutions?Wait, but let's check if there are any contradictions. For example, if ( f(y) = 1 + y ), then ( f(-y) = 1 - y ), and their product is ( (1 + y)(1 - y) = 1 - y^2 ), which matches our earlier equation. Similarly, for ( f(y) = 1 - y ), the product ( f(y) f(-y) = (1 - y)(1 + y) = 1 - y^2 ), which is also correct. So both functions satisfy the necessary condition ( f(y) f(-y) = 1 - y^2 ).But wait, when we tested the original equation, both functions worked. So both are solutions. Therefore, the problem might have two solutions? But let's check again.Wait, hold on. Let me test ( f(y) = 1 + y ) with another case. Let's set ( x = 0 ), then as before:( f(y f(y)) = y^2 + f(0) f(y) )With ( f(0) = 1 ), and ( f(y) = 1 + y ):Left-hand side: ( f(y(1 + y)) = f(y + y^2) = 1 + y + y^2 )Right-hand side: ( y^2 + 1 cdot (1 + y) = y^2 + 1 + y )Which matches. Similarly for ( f(y) = 1 - y ):Left-hand side: ( f(y(1 - y)) = 1 - y(1 - y) = 1 - y + y^2 )Right-hand side: ( y^2 + 1 cdot (1 - y) = y^2 + 1 - y )Which also matches. So both functions satisfy the equation when ( x = 0 ).Wait, but how can there be two solutions? Let me check another value. Let's take ( x = 1 ), ( y = 1 ).For ( f(y) = 1 + y ):LHS: ( f(1 + 1 cdot f(2)) ). ( f(2) = 1 + 2 = 3 ). So, argument is ( 1 + 3 = 4 ). Then, ( f(4) = 1 + 4 = 5 ).RHS: ( 1^2 + f(1) f(1) = 1 + (2)(2) = 5 ). So it works.For ( f(y) = 1 - y ):LHS: ( f(1 + 1 cdot f(2)) ). ( f(2) = 1 - 2 = -1 ). So, argument is ( 1 + (-1) = 0 ). Then, ( f(0) = 1 ).RHS: ( 1^2 + f(1) f(1) = 1 + (0)(0) = 1 ). So LHS = 1, RHS = 1. It also works.Wait a minute, so both functions work here as well. So perhaps both ( f(y) = 1 + y ) and ( f(y) = 1 - y ) are solutions.But this seems strange. Let me check another value. Let's take ( x = 1 ), ( y = 2 ).For ( f(y) = 1 + y ):LHS: ( f(1 + 2 f(3)) ). ( f(3) = 4 ), so argument is ( 1 + 2 cdot 4 = 9 ). ( f(9) = 10 ).RHS: ( 2^2 + f(1) f(2) = 4 + 2 cdot 3 = 10 ). So matches.For ( f(y) = 1 - y ):LHS: ( f(1 + 2 f(3)) ). ( f(3) = 1 - 3 = -2 ), so argument is ( 1 + 2(-2) = 1 - 4 = -3 ). Then ( f(-3) = 1 - (-3) = 4 ).RHS: ( 2^2 + f(1) f(2) = 4 + 0 cdot (-1) = 4 + 0 = 4 ). So LHS = 4, RHS = 4. Also matches.Hmm, so both functions work in this case as well. Maybe both are indeed solutions.Wait, but how come both are solutions? Let me check if there's a mistake in my earlier reasoning. When I assumed the function was linear, we found that ( f(y) = 1 + y ) and ( f(y) = 1 - y ) both satisfy the equation. But perhaps there are other non-linear solutions as well? Wait, but the equation ( f(-y) f(y) = 1 - y^2 ) must hold for all ( y ), which constrains the function significantly.Suppose there exists another solution that isn't linear. Let's suppose there is a quadratic function. But earlier, when assuming quadratic, we found that the coefficient of ( y^4 ) must be zero, so the function must be linear. Therefore, perhaps the only solutions are the two linear functions ( f(y) = 1 + y ) and ( f(y) = 1 - y ).Wait, but let me verify with another test. Suppose ( f(y) = 1 + y ). Then ( f(x + y f(x + y)) = 1 + x + y + xy + y^2 ), which is equal to ( y^2 + (1 + x)(1 + y) ), so that's correct.Similarly, for ( f(y) = 1 - y ), ( f(x + y f(x + y)) = 1 - x - y + xy + y^2 ), which is equal to ( y^2 + (1 - x)(1 - y) ), which is correct.Therefore, both functions satisfy the equation for all ( x, y ). Therefore, these are two solutions. Are there any other solutions?Wait, let's suppose there exists a function ( f ) such that ( f(y) = 1 + y ) for some ( y ) and ( f(y) = 1 - y ) for others. But since functional equations usually require consistency across all real numbers, such a piecewise function might not work unless it's consistent.But to check, suppose for example ( f(y) = 1 + y ) for ( y geq 0 ) and ( f(y) = 1 - y ) for ( y < 0 ). Let's see if this works. Take ( y = 1 ), ( x = 0 ):LHS: ( f(0 + 1 cdot f(1)) = f(f(1)) = f(2) = 1 + 2 = 3 )RHS: ( 1^2 + f(0) f(1) = 1 + 1 cdot 2 = 3 ). So works.Take ( y = -1 ), ( x = 0 ):LHS: ( f(0 + (-1) cdot f(-1)) = f(-f(-1)) ). Since ( f(-1) = 1 - (-1) = 2 ), so ( f(-2) ). If ( -2 < 0 ), ( f(-2) = 1 - (-2) = 3 ).RHS: ( (-1)^2 + f(0) f(-1) = 1 + 1 cdot 2 = 3 ). So works.But let's check another case. Take ( x = 1 ), ( y = -1 ):LHS: ( f(1 + (-1) f(1 + (-1))) = f(1 + (-1) f(0)) = f(1 + (-1)(1)) = f(0) = 1 )RHS: ( (-1)^2 + f(1) f(-1) = 1 + 2 cdot 2 = 5 ). Wait, LHS is 1, RHS is 5. That doesn't match. So this piecewise function doesn't work.Therefore, such a piecewise function isn't a solution. Therefore, the only possible solutions are the two linear functions ( f(y) = 1 + y ) and ( f(y) = 1 - y ).But wait, let's check if these are the only solutions. Suppose there exists another function that isn't linear but satisfies the equation. However, from our earlier analysis, when we assumed a quadratic function, we ended up forcing it to be linear. Therefore, maybe only linear functions satisfy the equation.Alternatively, maybe ( f ) has to be linear. Let's try to show that.Suppose ( f ) is not linear. But given that ( f(-y) f(y) = 1 - y^2 ), which is similar to the product of a function and its reflection. If ( f ) is linear, then it can be expressed as ( f(y) = 1 + ky ), where ( k = pm 1 ). So let's see if there are other possibilities.Suppose ( f ) is affine linear, i.e., ( f(y) = ay + b ). Then, as before:( f(-y) f(y) = ( -ay + b )( ay + b ) = -a^2 y^2 + b^2 ). This must equal ( 1 - y^2 ), so:- ( -a^2 = -1 ) ⇒ ( a^2 = 1 ) ⇒ ( a = pm 1 )- ( b^2 = 1 ) ⇒ ( b = pm 1 )But we already know ( f(0) = b = 1 ), so ( b = 1 ). Therefore, ( f(y) = pm y + 1 ). So these are the only possible linear solutions. Therefore, the only solutions are ( f(y) = y + 1 ) and ( f(y) = -y + 1 ).But earlier, when we considered quadratic functions, we saw that the only possible solutions are linear. Therefore, there are no quadratic or higher-degree polynomial solutions. What about non-polynomial solutions?Suppose there exists a non-linear solution. Let me try to assume such a function exists. However, the equation ( f(-y) f(y) = 1 - y^2 ) is quite restrictive. For example, if we write ( f(y) = sqrt{1 - y^2} ), but that's only real for ( |y| leq 1 ), which isn't the case here. Also, such a function wouldn't satisfy the original functional equation.Alternatively, maybe ( f(y) = 1 - y^2 / f(-y) ), but this seems recursive and difficult to resolve without more information.Wait, but since ( f(-y) f(y) = 1 - y^2 ), we can express ( f(-y) = (1 - y^2)/f(y) ). So this gives a relation between ( f(-y) ) and ( f(y) ). If we suppose ( f ) is differentiable or has some other properties, maybe we can derive more, but the problem doesn't specify any regularity conditions.Alternatively, let's try to express ( f(x + y f(x + y)) ) in terms of the original equation. Maybe set ( z = x + y ), but I'm not sure. Alternatively, suppose we let ( x = t - y f(t) ), where ( t = x + y f(x + y) ). Wait, this might complicate things.Wait, another idea: set ( y = 1 ). Then the equation becomes:( f(x + f(x + 1)) = 1 + f(x) f(1) )But since we have two possibilities for ( f(1) ):If ( f(y) = 1 + y ), then ( f(1) = 2 ), so the equation becomes:( f(x + f(x + 1)) = 1 + 2 f(x) )But ( f(x + 1) = 1 + x + 1 = x + 2 ), so left-hand side is ( f(x + x + 2) = f(2x + 2) = 1 + 2x + 2 = 2x + 3 )Right-hand side: ( 1 + 2 f(x) = 1 + 2(1 + x) = 1 + 2 + 2x = 2x + 3 ). So equality holds.For ( f(y) = 1 - y ), ( f(1) = 0 ), so the equation becomes:( f(x + f(x + 1)) = 1 + 0 cdot f(x) = 1 )But ( f(x + 1) = 1 - (x + 1) = -x ), so left-hand side is ( f(x + (-x)) = f(0) = 1 ). So equality holds. So both functions satisfy this case.Alternatively, let's consider the case where ( y = 1 ) and ( x = t ), leading to ( f(t + f(t + 1)) = 1 + f(t) f(1) ). For the linear functions, this holds, but for a non-linear function, this might impose additional constraints.Alternatively, suppose we set ( y = 1 ) and iterate. For example, starting with ( x = 0 ):( f(0 + f(1)) = 1 + f(0) f(1) implies f(f(1)) = 1 + f(1) )For ( f(y) = 1 + y ), ( f(f(1)) = f(2) = 3 ), and ( 1 + f(1) = 1 + 2 = 3 ), which holds.For ( f(y) = 1 - y ), ( f(f(1)) = f(0) = 1 ), and ( 1 + f(1) = 1 + 0 = 1 ), which also holds. So again, no conflict.But this might not help us rule out other solutions. Let me think if there's a way to show that ( f ) must be linear.From the equation ( f(-y) f(y) = 1 - y^2 ), if we suppose that ( f ) is linear, then the only possibilities are ( f(y) = 1 + y ) or ( f(y) = 1 - y ). If ( f ) is non-linear, then ( f(-y) = (1 - y^2)/f(y) ). Let's see if such a function can satisfy the original equation.Suppose ( f ) is non-linear. Let's attempt to construct an example. Let me pick a specific ( y ) and see if it leads to a contradiction.Take ( y = 1 ). Then ( f(-1) f(1) = 1 - 1 = 0 ). So either ( f(-1) = 0 ) or ( f(1) = 0 ). But in our linear solutions, for ( f(y) = 1 + y ), ( f(-1) = 0 ), and for ( f(y) = 1 - y ), ( f(1) = 0 ). So in general, one of them is zero. For a non-linear function, suppose ( f(1) = 0 ), then ( f(-1) ) is undefined (since ( f(-1) = (1 - 1)/f(1) )), but ( f(1) = 0 ) would make ( f(-1) ) undefined, which is not allowed because the function is supposed to be defined for all real numbers. Similarly, if ( f(-1) = 0 ), then ( f(1) = (1 - 1)/f(-1) ), which is undefined. Hence, any non-linear function would result in a division by zero in the relation ( f(-y) = (1 - y^2)/f(y) ) when either ( f(y) ) or ( f(-y) ) is zero. However, in our linear solutions, when ( f(1) = 0 ) (for ( f(y) = 1 - y )), ( f(-1) = 2 ), but according to the relation ( f(-1) f(1) = 1 - 1 = 0 ), which holds because ( f(1) = 0 ). Similarly, for ( f(y) = 1 + y ), ( f(-1) = 0 ), so ( f(-1) f(1) = 0 times 2 = 0 = 1 - 1 = 0 ). Therefore, in these cases, the problematic division by zero is avoided because the other term is zero. However, in a non-linear function, if there exists a ( y neq pm 1 ) such that ( f(y) = 0 ), then ( f(-y) f(y) = 0 ), but according to the equation ( f(-y) f(y) = 1 - y^2 ), which would require ( 1 - y^2 = 0 ), so ( y = pm 1 ). Therefore, the only zeros of ( f ) can be at ( y = 1 ) or ( y = -1 ). So if ( f ) is non-linear, it cannot have zeros elsewhere. However, the linear solutions only have zeros at ( y = -1 ) (for ( f(y) = 1 + y )) or at ( y = 1 ) (for ( f(y) = 1 - y )). Therefore, any non-linear function would have to avoid zeros except at ( y = pm 1 ), but it's not clear how to construct such a function.Alternatively, suppose we consider the function ( f ) to be injective. If ( f ) is injective, then perhaps we can manipulate the original equation. Let's check if the known solutions are injective.For ( f(y) = 1 + y ), it's injective because it's linear with slope 1. Similarly, ( f(y) = 1 - y ) is also injective (slope -1). So injective.Assume ( f ) is injective. Then, perhaps from the original equation, we can express ( x + y f(x + y) ) in terms of injectivity. However, I'm not sure how to proceed.Alternatively, let's set ( x = z - y f(x + y) ). Wait, but ( x ) is on both sides. Maybe another substitution.Alternatively, consider fixing ( x + y ) as a constant. Let me set ( s = x + y ). Then ( x = s - y ). Substitute into the original equation:( f( (s - y) + y f(s) ) = y^2 + f(s - y) f(y) )Simplifies to:( f( s - y + y f(s) ) = y^2 + f(s - y) f(y) )This might not be helpful unless we choose specific values for ( s ).Alternatively, set ( s = 0 ), then ( x = -y ). Wait, we already did that earlier.Alternatively, set ( y = 1 ). Then the equation becomes:( f(x + f(x + 1)) = 1 + f(x) f(1) )If ( f(1) = 0 ), then RHS is 1. But in this case, ( f(x + f(x + 1)) = 1 ). Since ( f(0) = 1 ), this implies ( x + f(x + 1) = 0 ), so ( f(x + 1) = -x ). Therefore, ( f(z) = - (z - 1) = 1 - z ). So this leads us to ( f(z) = 1 - z ), which is one of our solutions.Similarly, if ( f(1) = 2 ), then the equation becomes ( f(x + f(x + 1)) = 1 + 2 f(x) ). For ( f(x) = 1 + x ), ( f(x + 1) = x + 2 ), so LHS is ( f(x + x + 2) = f(2x + 2) = 2x + 3 ), and RHS is ( 1 + 2(1 + x) = 2x + 3 ), which matches. So this leads us to ( f(z) = 1 + z ).Therefore, regardless of the value of ( f(1) ), we can derive the form of ( f ).Alternatively, suppose ( f ) is linear. Then as shown earlier, ( f(y) = 1 pm y ). Since these are the only linear solutions and we can't find any non-linear solutions due to the constraints from ( f(-y) f(y) = 1 - y^2 ), it's likely that these are the only solutions.Therefore, the solutions are ( f(y) = 1 + y ) and ( f(y) = 1 - y ).But wait, let's check one more thing. Suppose there's a function that equals ( 1 + y ) for some ( y ) and ( 1 - y ) for others. But as shown earlier, such a piecewise function would lead to contradictions unless it's entirely one or the other. For example, if ( f(1) = 0 ) (from ( f(y) = 1 - y )), then ( f(-1) = 2 ), but if we switch to ( f(y) = 1 + y ) at some point, it would cause inconsistencies. Therefore, the function must be entirely ( 1 + y ) or ( 1 - y ).Hence, the only solutions are ( f(y) = 1 + y ) and ( f(y) = 1 - y ).But let me verify once more with a different value. Take ( x = 2 ), ( y = 1 ).For ( f(y) = 1 + y ):LHS: ( f(2 + 1 cdot f(3)) = f(2 + 4) = f(6) = 7 )RHS: ( 1^2 + f(2) f(1) = 1 + 3 cdot 2 = 7 ). Correct.For ( f(y) = 1 - y ):LHS: ( f(2 + 1 cdot f(3)) = f(2 + (-2)) = f(0) = 1 )RHS: ( 1^2 + f(2) f(1) = 1 + (-1)(0) = 1 ). Correct.Another test: ( x = -1 ), ( y = 2 ).For ( f(y) = 1 + y ):LHS: ( f(-1 + 2 f(1)) = f(-1 + 2 cdot 2) = f(3) = 4 )RHS: ( 2^2 + f(-1) f(2) = 4 + 0 cdot 3 = 4 ). Correct.For ( f(y) = 1 - y ):LHS: ( f(-1 + 2 f(1)) = f(-1 + 2 cdot 0) = f(-1) = 2 )RHS: ( 2^2 + f(-1) f(2) = 4 + 2 cdot (-1) = 4 - 2 = 2 ). Correct.Alright, after multiple tests and checks, both functions ( f(y) = 1 + y ) and ( f(y) = 1 - y ) satisfy the equation for all tested values, and there doesn't seem to be any contradictions. Additionally, the earlier analysis suggests that these are the only possible solutions given the constraints derived from the functional equation.Therefore, the solutions are ( f(x) = 1 + x ) and ( f(x) = 1 - x ).Final AnswerThe functions satisfying the equation are (boxed{f(x) = x + 1}) and (boxed{f(x) = -x + 1}).</think>To find all functions ( f: mathbb{R} rightarrow mathbb{R} ) such that[f(x + y f(x + y)) = y^2 + f(x) f(y)]for all ( x, y in mathbb{R} ), we start by analyzing specific values and properties of ( f ).1. Setting ( x = 0 ): [ f(y f(y)) = y^2 + f(0) f(y) ] This equation suggests a relationship between ( f(y f(y)) ) and ( y^2 ).2. Setting ( y = 0 ): [ f(x) = f(x) f(0) ] This implies ( f(0) = 1 ) since ( f(x) = 0 ) for all ( x ) leads to a contradiction.3. Setting ( x = -y ): [ f(0) = y^2 + f(-y) f(y) ] Since ( f(0) = 1 ), we get: [ f(-y) f(y) = 1 - y^2 ]4. Assuming linear functions: We test linear functions ( f(y) = ay + b ). Given ( f(0) = 1 ), we have ( b = 1 ). The product ( f(-y) f(y) = 1 - y^2 ) leads to: [ (ay + 1)(-ay + 1) = 1 - a^2 y^2 ] Comparing coefficients, we find ( a^2 = 1 ), giving ( a = 1 ) or ( a = -1 ). Thus, the linear functions are ( f(y) = y + 1 ) and ( f(y) = -y + 1 ).5. Verification: Both functions ( f(y) = y + 1 ) and ( f(y) = -y + 1 ) satisfy the original functional equation. Testing these functions with various values of ( x ) and ( y ) confirms they are solutions.After thorough analysis and verification, the functions satisfying the given equation are:[boxed{f(x) = x + 1} quad text{and} quad boxed{f(x) = -x + 1}]]

🔑:Note that the function f(x)=0 for all x does not satisfy the condition. Therefore, there is an a such that f(a) neq 0. Substitute x=a and y=0, which gives f(a)=f(a) f(0), so f(0)=1. Now substitute x=1 and y=-1, which gives f(1-f(0))=1+f(1) f(-1). Since f(0)=1, this becomes 1=1+f(1) f(-1), so f(1)=0 or f(-1)=0. We will consider these two cases separately.First, assume that f(1)=0. Substitute x=t and y=1-t, and then x=1-t and y=t, which gives the two equationsbegin{aligned}f(t) & =(1-t)^{2}+f(t) f(1-t) f(1-t) & =t^{2}+f(t) f(1-t)end{aligned}Subtracting these equations gives f(t)-f(1-t)=(1-t)^{2}-t^{2}=1-2 t, so f(1-t)=f(t)+2 t-1. Substituting this into the first of the above two equations givesf(t)=(1-t)^{2}+f(t)^{2}+(2 t-1) f(t)which can be rewritten asf(t)^{2}+(2 t-2) f(t)+(1-t)^{2}=0or (f(t)-(1-t))^{2}=0. We conclude that f(t)=1-t. Checking this function in the original functional equation gives on the left 1-(x+y(1-x-y))=1-left(x+y-x y-y^{2}right)=1-x-y+x y+y^{2} and on the right y^{2}+(1-x)(1-y)=y^{2}+1-x-y+x y and that is the same, so the function satisfies the equation.Now assume f(-1)=0. Substitute x=t and y=-1-t, and then x=-1-t and y=t, which gives the two equationsbegin{aligned}f(t) & =(-1-t)^{2}+f(t) f(-1-t), f(-1-t) & =t^{2}+f(t) f(-1-t)end{aligned}Subtracting these equations gives f(t)-f(-1-t)=(-1-t)^{2}-t^{2}=1+2 t, so f(-1-t)=f(t)-2 t-1. Substituting this into the first of the above two equations givesf(t)=(-1-t)^{2}+f(t)^{2}+(-2 t-1) f(t)which can be rewritten asf(t)^{2}-(2 t+2) f(t)+(t+1)^{2}=0or (f(t)-(t+1))^{2}=0. We conclude that f(t)=t+1. Checking this function in the original functional equation gives on the left x+y(x+y+1)+1=x+x y+y^{2}+y+1 and on the right y^{2}+(x+1)(y+1)=y^{2}+x y+x+y+1 and that is the same, so the function satisfies the equation.We conclude that there are exactly two solutions: f(x)=1-x for all x and f(x)=x+1 for all x.

🧠:[<think>Okay, let's see. I need to solve this inequality involving the function f(x) and its derivative. The problem gives me that f(x) is differentiable on (-∞, 0), and it satisfies 2f(x) + x f’(x) > x². Then, I have to find the solution set for (x + 2014)² f(x + 2014) - 4f(-2) > 0. The options are A to D, which are intervals. First, let's parse the given information. The function is defined for all x less than 0, so the domain is negative numbers. The derivative f’(x) exists there. The inequality 2f(x) + x f’(x) > x² must hold for all x in the domain. Then, the inequality we need to solve involves shifting x by 2014, so x + 2014 must still be in the domain, which is (-∞, 0). Therefore, x must be less than -2014, right? Because if x + 2014 < 0, then x < -2014. So the domain for the inequality we need to solve is x < -2014. But the answer options include intervals that go up to 0, but maybe the solution is a subset of that domain?Wait, but the answer options are A: (-∞, -2012), B: (-2012, 0), C: (-∞, -2016), D: (-2016, 0). So these are all intervals to the left of 0, but shifted by 2014. Hmm. Let me think step by step.First, the given inequality 2f(x) + x f’(x) > x². That looks like a differential inequality. Maybe I can write it as a derivative of some product. Let me see. If I have 2f(x) + x f’(x), does that remind me of the derivative of something? Let's think: the derivative of x² f(x) would be 2x f(x) + x² f’(x). Hmm, that's similar but not quite the same. Wait, if I divide that by x, I get (2x f(x) + x² f’(x))/x = 2f(x) + x f’(x). Oh, right! So the expression 2f(x) + x f’(x) is equal to the derivative of x² f(x) divided by x. Wait, but actually, if I let’s consider the derivative of x² f(x):d/dx [x² f(x)] = 2x f(x) + x² f’(x). So if I divide both sides by x, assuming x ≠ 0, then I get (1/x) d/dx [x² f(x)] = 2f(x) + x f’(x). So the given inequality is (1/x) d/dx [x² f(x)] > x². Wait, but the left-hand side is 2f(x) + x f’(x) > x². So actually, the given inequality can be written as:(1/x) * d/dx [x² f(x)] > x².But since x is in (-∞, 0), x is negative. So multiplying both sides by x (which is negative) would reverse the inequality:d/dx [x² f(x)] < x³.Hmm, not sure if that helps. Alternatively, maybe think of the expression 2f(x) + x f’(x) as part of a differential equation. Let's rearrange the inequality:2f(x) + x f’(x) > x².Let me divide both sides by x², assuming x ≠ 0:(2f(x))/x² + f’(x)/x > 1.Hmm, maybe that can be written as the derivative of something. Let me check. Suppose I have d/dx [f(x)/x²]. Let's compute that:Using the quotient rule: [f’(x) * x² - f(x) * 2x] / x⁴ = [x² f’(x) - 2x f(x)] / x⁴ = [x f’(x) - 2f(x)] / x³.Wait, that's similar but not exactly the same. Let's see:Original inequality: 2f(x) + x f’(x) > x².If I rearrange terms: x f’(x) + 2f(x) > x². Let me divide both sides by x³, assuming x ≠ 0:(x f’(x) + 2f(x))/x³ > 1/x.Hmm. Then, the left side is [x f’(x) + 2f(x)] / x³. Let's see if that's the derivative of something. Let me think about f(x)/x². The derivative is [f’(x) x² - 2x f(x)] / x⁴ = [x f’(x) - 2f(x)] / x³. That's similar but with a negative sign. So the derivative of f(x)/x² is [x f’(x) - 2f(x)] / x³ = (x f’(x) - 2f(x))/x³. Comparing with our left-hand side [x f’(x) + 2f(x)] / x³, which is different. Hmm, not quite.Wait, maybe multiply through by an integrating factor. Let's consider the inequality as a linear differential inequality:x f’(x) + 2f(x) > x².If I write this as f’(x) + (2/x) f(x) > x.This is a linear differential inequality. The integrating factor would be e^{∫ (2/x) dx} = e^{2 ln |x|} = x². Since x is negative, |x| = -x, so e^{2 ln |x|} = |x|² = x². So the integrating factor is x². Multiply both sides by x²:x² f’(x) + 2x f(x) > x³.But the left-hand side is the derivative of x² f(x). So:d/dx [x² f(x)] > x³.Then, integrating both sides from some a to b, but since we have an inequality, perhaps we can consider the antiderivative. Let me integrate both sides from x = c to x = d, where c < d < 0.Integrate the left side: ∫_c^d d/dx [x² f(x)] dx = x² f(x) |_c^d = d² f(d) - c² f(c).Integrate the right side: ∫_c^d x³ dx = [x⁴ /4]_c^d = (d⁴ - c⁴)/4.So the inequality becomes:d² f(d) - c² f(c) > (d⁴ - c⁴)/4.But I'm not sure if that helps directly. Alternatively, maybe consider solving the differential inequality as if it were an equation. Let's consider the differential equation:x f’(x) + 2f(x) = x² + g(x), where g(x) > 0.But perhaps solving the equality first. Let me rewrite the equation:f’(x) + (2/x) f(x) = x.The integrating factor is x², as before. Multiply both sides:x² f’(x) + 2x f(x) = x³.Left side is d/dx [x² f(x)] = x³.Integrate both sides:x² f(x) = ∫ x³ dx + C = x⁴ /4 + C.So, f(x) = x² /4 + C / x².But since this is the solution to the equality, but our inequality is f’(x) + (2/x) f(x) > x. So the solution to the inequality would be f(x) > x² /4 + C / x² if the integrating factor is positive. Wait, but the integrating factor is x², which is positive even when x is negative, since squared. So the direction of the inequality is preserved when multiplying by the integrating factor. Therefore, the general solution to the inequality f’(x) + (2/x) f(x) > x is f(x) > x²/4 + C / x² for some constant C. Hmm.But wait, actually, when solving the differential inequality, the solution would be f(x) > particular solution + homogeneous solution. But perhaps I need to consider the inequality:After multiplying by the integrating factor x², we have d/dx [x² f(x)] > x³.Then, integrating both sides from some lower bound to x. Let's pick a lower bound, say a, which is less than x (since x is in (-∞, 0)). Let me integrate from a to x:∫_a^x d/dt [t² f(t)] dt > ∫_a^x t³ dt.This gives x² f(x) - a² f(a) > [x⁴ /4 - a⁴ /4].Then, rearranging:x² f(x) > a² f(a) + (x⁴ - a⁴)/4.Therefore,f(x) > [a² f(a) + (x⁴ - a⁴)/4] / x².But since this must hold for all a < x (since we can choose any a as the lower limit), I think this might not directly help. Maybe instead, suppose we pick a specific value for a? For example, take a approaching -∞. But if x is fixed and a approaches -∞, then a² f(a) term might dominate unless f(a) behaves nicely. Hmm, perhaps this approach is getting too complicated.Alternatively, maybe we can use the given inequality to compare f(x) with some function. Since we have f(x) > x²/4 + C/x², but without knowing C, maybe not helpful. Wait, but if we can find an expression for f(x), maybe we can relate f(x + 2014) and f(-2). But since the problem gives us an inequality involving (x + 2014)^2 f(x + 2014) - 4f(-2) > 0, perhaps we need to find a relationship between f(x) at different points.Alternatively, let's look at the expression (x + 2014)^2 f(x + 2014) - 4f(-2) > 0. Let me denote y = x + 2014. Then, the inequality becomes y² f(y) - 4f(-2) > 0, where y = x + 2014. Since x must be in (-∞, 0), then y = x + 2014 is in (-∞, 2014). But the original domain of f is (-∞, 0), so y must also be in (-∞, 0). Therefore, x + 2014 < 0 ⇒ x < -2014. So the domain for the original inequality is x < -2014, so y < 0. Therefore, we can rewrite the inequality as y² f(y) > 4f(-2), with y < 0. So we need to find y such that y² f(y) > 4f(-2). Then, once we solve for y, we can relate back to x via y = x + 2014, so x = y - 2014. So if we find y < 0 such that y² f(y) > 4f(-2), then x = y - 2014 would be the solution. Then, the solution set for x is y - 2014 where y satisfies y² f(y) > 4f(-2). So the answer will depend on how y² f(y) compares to 4f(-2). To analyze y² f(y), perhaps use the given differential inequality. Recall that 2f(x) + x f’(x) > x². Let's manipulate this. Let's consider the expression x² f(x). Let’s compute its derivative:d/dx [x² f(x)] = 2x f(x) + x² f’(x) = x [2f(x) + x f’(x)]. But from the given inequality, 2f(x) + x f’(x) > x². Therefore, substituting:d/dx [x² f(x)] = x [2f(x) + x f’(x)] > x * x² = x³.Therefore, the derivative of x² f(x) is greater than x³. So, integrating this inequality. Let's integrate both sides from some point a to b, where a < b < 0. ∫_a^b d/dt [t² f(t)] dt > ∫_a^b t³ dt.This gives:b² f(b) - a² f(a) > (b⁴ - a⁴)/4.So, rearranged:b² f(b) > a² f(a) + (b⁴ - a⁴)/4.If we set a = -2, then for b < -2, we have:b² f(b) > (-2)² f(-2) + (b⁴ - (-2)^4)/4 = 4f(-2) + (b⁴ - 16)/4.Therefore,b² f(b) > 4f(-2) + (b⁴ - 16)/4.Therefore,b² f(b) - 4f(-2) > (b⁴ - 16)/4.So, the left-hand side is exactly the expression in our inequality: y² f(y) - 4f(-2) > 0. Therefore, substituting y = b, we have:y² f(y) - 4f(-2) > (y⁴ - 16)/4.Therefore, if (y⁴ - 16)/4 ≥ 0, then the left-hand side would be positive. But we need y² f(y) - 4f(-2) > 0. So, even if (y⁴ -16)/4 is positive, we can guarantee the inequality. But (y⁴ -16)/4 is positive when y⁴ > 16, i.e., |y| > 2. Since y < 0, this would be y < -2. Therefore, when y < -2, (y⁴ -16)/4 > 0, so y² f(y) -4f(-2) > (positive number). Therefore, for y < -2, the inequality y² f(y) -4f(-2) > 0 holds. However, when y ≥ -2 (but still y < 0), (y⁴ -16)/4 is negative, so we can't be sure if the left-hand side is positive. But maybe even in that region, the inequality could still hold. Wait, but we need to check.Alternatively, perhaps the inequality y² f(y) -4f(-2) >0 is equivalent to y² f(y) >4f(-2). From the previous result, for any b < -2, we have:b² f(b) > 4f(-2) + (b⁴ -16)/4.So, for y < -2, we can write:y² f(y) > 4f(-2) + (y⁴ -16)/4.Thus, y² f(y) -4f(-2) > (y⁴ -16)/4.So, if (y⁴ -16)/4 >0, which is y⁴ >16 ⇒ |y|>2 ⇒ y < -2 (since y <0), then the right-hand side is positive. Therefore, for y < -2, y² f(y) -4f(-2) > (positive number), which implies the inequality holds. So, for y < -2, the inequality is true. But what about when -2 ≤ y <0? Then, (y⁴ -16)/4 <0, so we can't conclude from the inequality y² f(y) -4f(-2) > (negative number). Therefore, in this region, it's possible that y² f(y) -4f(-2) is positive or negative. Therefore, we need to check whether the inequality holds for y between -2 and 0. However, since our original problem requires solving (x +2014)^2 f(x +2014) -4f(-2) >0, and since x +2014 = y, which must be less than 0, but x must be less than -2014 to make y <0, but the answer options include intervals like (-2016,0), which would correspond to y =x +2014 between -2 and 0 (since x between -2016 and 0 would make y between -2 and 2014, but since y must be <0, x must be < -2014). Wait, actually, if x is between -2016 and -2014, then y =x +2014 is between -2 and 0. If x is less than -2016, then y is less than -2.Wait, hold on. Let me check:If x < -2014, then y = x +2014 < 0. If x is between -2016 and -2014, then y is between -2 and 0. If x < -2016, then y < -2. So the regions are:x < -2016 ⇒ y < -2-2016 <x < -2014 ⇒ -2 < y <0But since the original domain for the inequality is x < -2014 (since y must be <0), the possible x values are split into two intervals: x < -2016 (y < -2) and -2016 <x < -2014 ( -2 < y <0). From the previous analysis, when y < -2 (i.e., x < -2016), the inequality y² f(y) -4f(-2) >0 holds. But when y is between -2 and 0 (i.e., x between -2016 and -2014), we don't know for sure. So maybe the solution set is x < -2016, which is option C: (-∞, -2016). But let's verify.Wait, but how do we know that for y between -2 and 0, the inequality doesn't hold? Let me think. Maybe we can use the given differential inequality to analyze the behavior of f(y) in that interval.Suppose y is between -2 and 0, so y is greater than -2. Let's pick a = y and b = -2. Then, integrating the inequality from a to b (since a < b <0, but here a is y and b is -2, so y < -2? Wait, no. If y is between -2 and 0, then a = y > -2, and b = -2. But in the previous integral, we had a < b. So maybe we need to reverse the limits. Let me try.From the differential inequality, d/dt [t² f(t)] > t³.Integrate from y to -2:∫_y^{-2} d/dt [t² f(t)] dt > ∫_y^{-2} t³ dt.Left side: (-2)² f(-2) - y² f(y) =4f(-2) - y² f(y).Right side: [(-2)^4 /4 - y⁴ /4] = (16/4 - y⁴ /4) = 4 - y⁴ /4.Therefore, the inequality becomes:4f(-2) - y² f(y) >4 - y⁴ /4.Rearranged:- y² f(y) >4 - y⁴ /4 -4f(-2).Multiply both sides by -1 (inequality reverses):y² f(y) < -4 + y⁴ /4 +4f(-2).But our original inequality is y² f(y) >4f(-2). So combining these two:4f(-2) < y² f(y) < -4 + y⁴ /4 +4f(-2).So this implies that 4f(-2) < -4 + y⁴ /4 +4f(-2).Subtract 4f(-2) from both sides:0 < -4 + y⁴ /4.Thus,y⁴ /4 >4 ⇒ y⁴ >16 ⇒ |y| >2 ⇒ y < -2 or y >2. But since y is between -2 and 0 here, this is impossible. Therefore, this leads to a contradiction, which suggests that in the interval -2 < y <0, the inequality y² f(y) >4f(-2) cannot hold. Therefore, the only solutions are when y < -2, i.e., x < -2016. Therefore, the solution set is (-∞, -2016), which is option C.Wait, but let me confirm. The key step was integrating from y to -2 when y is in (-2,0). The result of that integration led us to an impossible inequality, meaning that our assumption that y is in (-2,0) and y² f(y) >4f(-2) leads to a contradiction. Therefore, there are no solutions in (-2016, -2014), which corresponds to y in (-2,0). Therefore, the only solutions are when y < -2, i.e., x < -2016.Therefore, the answer should be option C: (-∞, -2016).Final Answerboxed{C}</think>Given the function ( f(x) ) defined on ( (-infty, 0) ) with its derivative ( f'(x) ), and the inequality ( 2f(x) + x f'(x) > x^2 ), we need to solve the inequality ( (x + 2014)^2 f(x + 2014) - 4f(-2) > 0 ).First, we recognize that the given differential inequality ( 2f(x) + x f'(x) > x^2 ) can be rewritten using the integrating factor method. Multiplying through by ( x^2 ), we get:[frac{d}{dx} [x^2 f(x)] > x^3]Integrating this inequality from ( a ) to ( b ) (where ( a < b < 0 )), we find:[b^2 f(b) - a^2 f(a) > frac{b^4 - a^4}{4}]Setting ( a = -2 ) and ( b = y ), we analyze the inequality:[y^2 f(y) > 4f(-2) + frac{y^4 - 16}{4}]This implies:[y^2 f(y) - 4f(-2) > frac{y^4 - 16}{4}]For ( y < -2 ), the right-hand side ( frac{y^4 - 16}{4} ) is positive, ensuring the inequality holds. For ( -2 leq y < 0 ), the right-hand side is negative, leading to a contradiction when checked against the original inequality.Thus, the solution set for ( y ) is ( y < -2 ), translating back to ( x ) as ( x + 2014 < -2 ), which simplifies to ( x < -2016 ).The final answer is (boxed{C}).,<think>Okay, so I need to solve this inequality involving the function f(x), which is differentiable on (-∞, 0). The problem gives me that 2f(x) + x f’(x) > x², and then asks about the solution set for (x + 2014)² f(x + 2014) - 4 f(-2) > 0. The answer choices are intervals, so I need to figure out for which x values this inequality holds.First, let me parse the given information. The function f(x) is defined on (-∞, 0), which means that x + 2014 must also be in that domain for the inequality to make sense. So x + 2014 < 0, which implies x < -2014. That might narrow down the possible answer choices. Let's check the options:A: (-∞, -2012)B: (-2012, 0)C: (-∞, -2016)D: (-2016, 0)But since x must be less than -2014, the valid interval must be a subset of (-∞, -2014). Looking at the options, A is (-∞, -2012), which includes some values greater than -2014, but B and D are intervals that start at -2012 and -2016 respectively, and go to 0. C is (-∞, -2016), which is entirely less than -2014. So maybe C is a candidate, but let me think further.Wait, the original problem says the function is defined on (-∞, 0), so x + 2014 has to be in that interval. Therefore, x < -2014. So the domain of the inequality is x < -2014. Therefore, the solution set must be a subset of (-∞, -2014). So looking at the answer choices, only option C is entirely within (-∞, -2014), and option A includes (-∞, -2012), which includes numbers between -2014 and -2012, which are not in the domain. So maybe the answer is C? But let me not jump to conclusions yet.Now, the inequality given is 2f(x) + x f’(x) > x². That looks like a differential inequality. Maybe I can relate this to the derivative of some product. Let me see. Let's consider if the left-hand side is the derivative of something.Let me think: if I have x² f(x), then the derivative would be 2x f(x) + x² f’(x). Hmm, but the given expression is 2f(x) + x f’(x). That's similar but not the same. Wait, if I factor x from the derivative term, maybe?Alternatively, maybe if I divide x² f(x) by x, but that complicates things. Let me check:Suppose I let g(x) = x² f(x). Then g’(x) = 2x f(x) + x² f’(x). That's 2x f(x) + x² f’(x). But the given inequality is 2f(x) + x f’(x) > x². If I factor out an x from g’(x), I get x(2f(x) + x f’(x)) = g’(x). Wait, so the given inequality is (g’(x))/x > x². Since x is negative (domain is (-∞, 0)), so multiplying both sides by x (which is negative) would reverse the inequality.Wait, let me try:Given 2f(x) + x f’(x) > x². Let's note that g’(x) = 2x f(x) + x² f’(x) = x(2f(x) + x f’(x)). So, g’(x) = x*(left-hand side of given inequality). Therefore, g’(x) > x * x² = x³. So, g’(x) > x³.But x is negative here. So perhaps integrating this inequality? Let's see.But maybe we can write the original inequality in terms of g(x). Since g(x) = x² f(x), then f(x) = g(x)/x². Let's substitute back into the inequality:2*(g(x)/x²) + x*(d/dx [g(x)/x²]) > x²Compute the derivative term:d/dx [g(x)/x²] = [g’(x) * x² - g(x) * 2x] / x⁴But this might get complicated. Alternatively, perhaps using the expression for g’(x):Given that g’(x) = x*(2f(x) + x f’(x)) > x³.So, g’(x) > x³. Then, integrating both sides from some a to b in the domain.But maybe instead of integrating, we can write this as a differential inequality for g(x):g’(x) > x³.If we can solve this differential inequality, perhaps we can find an expression for g(x), which is x² f(x). Let me try integrating both sides. Suppose we have g’(x) > x³. Then, integrating from some point c to x:g(x) - g(c) > ∫_c^x t³ dt = [t⁴/4]_c^x = x⁴/4 - c⁴/4.But since the inequality is for all x in the domain, perhaps we can write g(x) > x⁴/4 + K, where K is a constant. However, since the inequality is just g’(x) > x³, the general solution would be g(x) > x⁴/4 + C, where C is a constant of integration. But we need to verify if this is valid.Alternatively, consider the differential inequality g’(x) > x³. To solve this, we can consider the related equation g’(x) = x³ + ε(x), where ε(x) > 0. Integrating gives g(x) = x⁴/4 + ∫ ε(x) dx + C. Since ε(x) > 0, the integral ∫ ε(x) dx is an increasing function. Therefore, the general solution is g(x) > x⁴/4 + C. However, without an initial condition, we can't determine C. Maybe there's another approach.Alternatively, let's think about the inequality (x + 2014)² f(x + 2014) - 4 f(-2) > 0. Let me make a substitution to simplify this. Let y = x + 2014. Then, since x < -2014 (as established earlier), y < 0. The inequality becomes y² f(y) - 4 f(-2) > 0. So, y² f(y) > 4 f(-2). Therefore, y² f(y) > 4 f(-2). So, the inequality is equivalent to y² f(y) > 4 f(-2). Let me write that as y² f(y) > 4 f(-2). Let me divide both sides by y² (since y < 0, y² is positive, so inequality remains the same):f(y) > 4 f(-2) / y².But I don't know if that's helpful. Alternatively, maybe relate this to the function g(y) = y² f(y). Then the inequality becomes g(y) > 4 f(-2). So we need g(y) > 4 f(-2). So if I can find the values of y where g(y) > 4 f(-2), then x + 2014 = y, so x = y - 2014. Therefore, once I find y, I can get x.But how to relate g(y) to the given differential inequality? Earlier, we saw that g’(y) = 2y f(y) + y² f’(y). Wait, but the given differential inequality is 2f(x) + x f’(x) > x². Let's check for x = y:2f(y) + y f’(y) > y². Multiply both sides by y (but y is negative, so inequality reverses):y*(2f(y) + y f’(y)) < y³. But y*(2f(y) + y f’(y)) is exactly g’(y). Therefore, g’(y) < y³.Wait, this contradicts my previous conclusion. Wait, let's check again:Given 2f(x) + x f’(x) > x². If we set x = y, then 2f(y) + y f’(y) > y². Then, multiplying both sides by y (which is negative, since y < 0), we get:y*(2f(y) + y f’(y)) < y³. But the left-hand side is y*2f(y) + y² f’(y) = 2y f(y) + y² f’(y) = g’(y). Therefore, we have g’(y) < y³.Wait, so earlier, I thought that g’(x) = x*(2f(x) + x f’(x)), which is correct. So since x is negative, multiplying the inequality 2f(x) + x f’(x) > x² by x (negative) reverses the inequality: x*(2f(x) + x f’(x)) < x³. Therefore, g’(x) < x³.Therefore, the derivative of g(x) is less than x³. So, integrating this might give us an upper bound on g(x).But how does this help us with the inequality g(y) > 4 f(-2)? Hmm.Alternatively, maybe we can consider the function h(y) = g(y) - (y⁴)/4. Then, h’(y) = g’(y) - y³. But we know that g’(y) < y³, so h’(y) = g’(y) - y³ < 0. Therefore, h(y) is decreasing.So h(y) is a decreasing function. Therefore, if y1 < y2, then h(y1) > h(y2). Therefore, for y < -2, since h is decreasing, h(y) > h(-2). Therefore, g(y) - y⁴/4 > g(-2) - (-2)^4 /4 = g(-2) - 16/4 = g(-2) - 4.But g(-2) = (-2)^2 f(-2) = 4 f(-2). Therefore, h(-2) = 4 f(-2) - 4. So h(y) > h(-2) when y < -2. Therefore, for y < -2:g(y) - y⁴/4 > 4 f(-2) - 4Therefore, g(y) > y⁴/4 + 4 f(-2) - 4But I don't see how this directly helps. Wait, our target is to find when g(y) > 4 f(-2). So if we can show that g(y) > 4 f(-2), then y must satisfy some inequality.But perhaps we can use the fact that h(y) is decreasing. Let me see:Since h(y) = g(y) - y⁴/4 is decreasing, then for y < -2, h(y) > h(-2). So:g(y) - y⁴/4 > g(-2) - (-2)^4 /4=> g(y) > y⁴/4 + g(-2) - 4But g(-2) = 4 f(-2), so:g(y) > y⁴/4 + 4 f(-2) - 4So if we want g(y) > 4 f(-2), then:y⁴/4 + 4 f(-2) - 4 > 4 f(-2)Subtracting 4 f(-2) from both sides:y⁴/4 - 4 > 0=> y⁴/4 > 4=> y⁴ > 16=> |y| > 2Since y is negative (because y < 0), this implies y < -2.Therefore, if y < -2, then the inequality g(y) > 4 f(-2) would hold? Wait, let me check:From h(y) decreasing, and h(y) > h(-2) for y < -2, which gives:g(y) - y⁴/4 > 4 f(-2) - 4So rearranged:g(y) > y⁴/4 + 4 f(-2) - 4Therefore, to have g(y) > 4 f(-2), it's sufficient that y⁴/4 - 4 > 0, which is y⁴ > 16, i.e., |y| > 2. Since y < 0, this is y < -2.Therefore, if y < -2, then g(y) > 4 f(-2). Therefore, the inequality y² f(y) > 4 f(-2) holds when y < -2. Therefore, substituting back to x:y = x + 2014 < -2=> x + 2014 < -2=> x < -2016Therefore, the solution set is x < -2016, which is (-∞, -2016), which is option C.Wait, but let me verify this logic again. So h(y) is decreasing, so if y < -2, then h(y) > h(-2). Then, from h(y) = g(y) - y⁴/4, so:h(y) > h(-2) => g(y) - y⁴/4 > g(-2) - (-2)^4 /4Compute (-2)^4 /4: 16/4 = 4g(-2) = 4 f(-2)Thus:g(y) - y⁴/4 > 4 f(-2) - 4So,g(y) > y⁴/4 + 4 f(-2) - 4We want g(y) > 4 f(-2), so:y⁴/4 + 4 f(-2) - 4 > 4 f(-2)Subtract 4 f(-2):y⁴/4 - 4 > 0 => y⁴ > 16 => |y| > 2. Since y < 0, this is y < -2. Therefore, when y < -2, the inequality g(y) > 4 f(-2) holds. Therefore, x + 2014 < -2 => x < -2016.Thus, the solution set is x < -2016, which is option C.But let me check if there could be a case where even if y >= -2, maybe g(y) > 4 f(-2). But since h(y) is decreasing, when y > -2, h(y) < h(-2), so:g(y) - y⁴/4 < 4 f(-2) - 4Therefore,g(y) < y⁴/4 + 4 f(-2) - 4But for y > -2, y is negative but closer to zero. Let's take y = -1. Then y⁴/4 = 1/4. So:g(-1) < 1/4 + 4 f(-2) - 4But unless 1/4 - 4 is positive, which it isn't, we can't say. But 4 f(-2) is a term we don't know the value of. Wait, maybe this approach is missing something.Alternatively, maybe we can analyze the differential inequality more directly. Let's consider the original inequality:2f(x) + x f’(x) > x²Let me divide both sides by x² (since x ≠ 0, and x < 0):(2f(x))/x² + f’(x)/x > 1Wait, maybe this is a stretch. Let me try to write this as a derivative of some function. Let's think of the left-hand side.Let me consider the function f(x)/x². Let's compute its derivative:d/dx [f(x)/x²] = (f’(x) x² - 2x f(x)) / x⁴ = (x f’(x) - 2 f(x)) / x³But our inequality is 2f(x) + x f’(x) > x². Let me rearrange the inequality:x f’(x) + 2f(x) > x²Dividing both sides by x³ (since x < 0, x³ is negative):(x f’(x) + 2f(x))/x³ > x²/x³ = 1/xBut (x f’(x) + 2 f(x))/x³ can be written as (x f’(x) + 2 f(x))/x³ = (f’(x)/x² + 2 f(x)/x³). Alternatively, maybe integrating factor.Wait, let's consider the differential inequality:x f’(x) + 2 f(x) > x²Divide both sides by x (x ≠ 0):f’(x) + (2/x) f(x) > xThis is a linear differential inequality. Let's write it as:f’(x) + (2/x) f(x) > xThe integrating factor is e^{∫ (2/x) dx} = e^{2 ln |x|} = |x|² = x² (since x ≠ 0). But since x is in (-∞, 0), we can write |x| = -x, so |x|² = x².Multiply both sides by the integrating factor x²:x² f’(x) + 2x f(x) > x³Notice that the left-hand side is the derivative of x² f(x):d/dx [x² f(x)] = 2x f(x) + x² f’(x) = left-hand sideTherefore, the inequality becomes:d/dx [x² f(x)] > x³So, integrating both sides from a to b, where a < b < 0.Integrate from a to b:∫_a^b d/dt [t² f(t)] dt > ∫_a^b t³ dtWhich gives:x² f(x) | from a to b > [t⁴ /4]_a^bTherefore,b² f(b) - a² f(a) > (b⁴ - a⁴)/4But perhaps choosing a specific a would help. Let me pick a = -2, since we have f(-2) in the target inequality. Let me set a = -2 and b = y, where y < -2 (since we need to relate to the inequality g(y) > 4 f(-2)).So, integrating from -2 to y:y² f(y) - (-2)² f(-2) > (y⁴ - (-2)^4)/4=> y² f(y) - 4 f(-2) > (y⁴ - 16)/4Multiply both sides by 4:4 y² f(y) - 16 f(-2) > y⁴ - 16Bring all terms to the left side:4 y² f(y) - 16 f(-2) - y⁴ + 16 > 0Factor 16:4 y² f(y) - y⁴ -16 f(-2) +16 >0Hmm, not sure if that helps. Wait, the original inequality we need to solve is y² f(y) -4 f(-2) >0, which is equivalent to y² f(y) >4 f(-2). From the above inequality:From the integration result:y² f(y) -4 f(-2) > (y⁴ -16)/4Therefore,y² f(y) -4 f(-2) > (y⁴ -16)/4But we want y² f(y) -4 f(-2) >0. Therefore, it's sufficient that (y⁴ -16)/4 >=0, but actually, the inequality from integration is:y² f(y) -4 f(-2) > (y⁴ -16)/4Therefore, if (y⁴ -16)/4 >=0, then y² f(y) -4 f(-2) >0. But (y⁴ -16)/4 >=0 implies y⁴ >=16, which is |y| >=2. Since y <0, this means y <=-2. Therefore, for y <=-2, we have that (y⁴ -16)/4 >=0, so y² f(y) -4 f(-2) > something >=0. But wait, not exactly. Let me think.Suppose y <= -2. Then (y⁴ -16)/4 >=0 because y⁴ >=16. Therefore, the right-hand side is non-negative, so the left-hand side y² f(y) -4 f(-2) is greater than a non-negative number. Therefore, the left-hand side is positive. Therefore, y² f(y) -4 f(-2) >0 for y <=-2. Therefore, x +2014 <= -2 implies x <= -2016. Therefore, the solution set is (-∞, -2016), which is option C.Alternatively, if y > -2 (but y <0), then (y⁴ -16)/4 <0, so y² f(y) -4 f(-2) is greater than a negative number. So it's possible that y² f(y) -4 f(-2) is positive or negative. But we can't conclude from this inequality. Therefore, the only interval where we can be sure that y² f(y) -4 f(-2) >0 is when y <= -2, which translates to x <= -2016. Therefore, the answer is C.So after going through this differential inequality approach and integrating, it does seem like option C is the correct answer. Let me just check once more for any possible mistakes.Wait, integrating the inequality d/dt [t² f(t)] > t³ from -2 to y gives:y² f(y) -4 f(-2) > (y⁴ -16)/4Therefore, rearranged:y² f(y) >4 f(-2) + (y⁴ -16)/4We want y² f(y) >4 f(-2). So:4 f(-2) + (y⁴ -16)/4 >4 f(-2)Subtract 4 f(-2):(y⁴ -16)/4 >0Which implies y⁴ >16, so |y|>2, hence y < -2 (since y <0). Therefore, when y < -2, we have y² f(y) >4 f(-2). Therefore, x +2014 < -2 => x < -2016. Therefore, solution set is (-∞, -2016), which is option C. Therefore, the correct answer is C.Final Answerboxed{C}</think>Given the function ( f(x) ) defined on ( (-infty, 0) ) with its derivative ( f'(x) ), we know that it satisfies the inequality ( 2f(x) + x f'(x) > x^2 ). We need to solve the inequality ( (x + 2014)^2 f(x + 2014) - 4f(-2) > 0 ).First, we note that ( x + 2014 ) must be in the domain ( (-infty, 0) ), implying ( x < -2014 ). This narrows down the solution set to a subset of ( (-infty, -2014) ).Next, we consider the given differential inequality ( 2f(x) + x f'(x) > x^2 ). By multiplying both sides by ( x ) (which is negative), we reverse the inequality and recognize it as the derivative of ( x^2 f(x) ). This leads us to consider the function ( g(x) = x^2 f(x) ), whose derivative ( g'(x) = x(2f(x) + x f'(x)) ) is less than ( x^3 ).We integrate this inequality from ( -2 ) to ( y ) (where ( y < -2 )) and find that ( g(y) - 4f(-2) > frac{y^4 - 16}{4} ). For ( y leq -2 ), this implies ( y^4 geq 16 ), leading to ( y leq -2 ).Substituting back ( y = x + 2014 ), we get ( x + 2014 leq -2 ), which simplifies to ( x leq -2016 ). Therefore, the solution set is ( (-infty, -2016) ).Thus, the correct answer is (boxed{C}).]

🔑:AnalysisThis problem examines the use of monotonicity to solve inequalities. Constructing a function from the given conditions and utilizing the relationship between function monotonicity and derivatives is key to solving this problem. Transforming the inequality allows us to reach a conclusion.SolutionGiven 2f(x)+xf′(x) > x^{2}, (x 0,Since F(x) is a decreasing function on (-infty,0),From F(x+2014) > F(-2),We get x+2014 < -2,Which means x < -2016.Therefore, the correct choice is boxed{text{C}}.

🧠:[<think>Okay, so I have this problem here: If the tangent line of the curve y = ln x at point P(x₁, y₁) is tangent to the curve y = eˣ at point Q(x₂, y₂), then find the value of (2)/(x₁ - 1) + x₂. Hmm, interesting. Let me try to figure this out step by step.First, I know that a tangent line to a curve at a point has two main properties: it passes through that point, and its slope is equal to the derivative of the curve at that point. So, for point P on y = ln x, the slope of the tangent line will be the derivative of ln x evaluated at x = x₁. Similarly, the same tangent line is also tangent to y = eˣ at point Q, so the slope there should be the derivative of eˣ at x = x₂. Since it's the same line, these two slopes must be equal. Also, the tangent line must touch both curves at their respective points. So, I need to find x₁ and x₂ such that the tangent line at P on ln x is the same as the tangent line at Q on eˣ. Then compute 2/(x₁ - 1) + x₂.Alright, let's break it down. Let's start by finding the equation of the tangent line at point P on y = ln x.The derivative of ln x is 1/x, so at x = x₁, the slope m = 1/x₁. The point P is (x₁, ln x₁). So, the equation of the tangent line at P is:y - ln x₁ = (1/x₁)(x - x₁)Simplifying that:y = (1/x₁)x - (1/x₁)x₁ + ln x₁Which simplifies further to:y = (1/x₁)x - 1 + ln x₁So, the tangent line equation is y = (1/x₁)x + (ln x₁ - 1)Now, this same line is supposed to be tangent to y = eˣ at point Q(x₂, y₂). So, let's write the equation of the tangent line to y = eˣ at Q.The derivative of eˣ is eˣ, so the slope at x = x₂ is m = e^{x₂}. The point Q is (x₂, e^{x₂}). So, the equation of the tangent line is:y - e^{x₂} = e^{x₂}(x - x₂)Simplifying:y = e^{x₂}x - e^{x₂}x₂ + e^{x₂}Which becomes:y = e^{x₂}x + e^{x₂}(1 - x₂)Now, since these two tangent lines are the same line, their slopes must be equal, and their y-intercepts must be equal. So, set the two equations equal:First, compare the slopes:1/x₁ = e^{x₂} ...(1)Then, compare the y-intercepts:ln x₁ - 1 = e^{x₂}(1 - x₂) ...(2)So now, we have a system of two equations with two variables x₁ and x₂. Our goal is to solve for x₁ and x₂ and then compute 2/(x₁ - 1) + x₂.From equation (1), we can express e^{x₂} as 1/x₁. Let's substitute that into equation (2):ln x₁ - 1 = (1/x₁)(1 - x₂)Multiply both sides by x₁ to eliminate the denominator:x₁(ln x₁ - 1) = 1 - x₂So, rearranged:x₂ = 1 - x₁(ln x₁ - 1)Which simplifies to:x₂ = 1 - x₁ ln x₁ + x₁So, x₂ = 1 + x₁ - x₁ ln x₁Hmm. So, if I can find x₁, then I can compute x₂. But we need another equation to solve for x₁. Wait, we already used both equations. Equation (1) gives e^{x₂} = 1/x₁, so x₂ = ln(1/x₁) = -ln x₁. Wait, that might be helpful. Wait, hold on: If e^{x₂} = 1/x₁, then take natural logarithm of both sides: x₂ = ln(1/x₁) = -ln x₁. Therefore, x₂ = -ln x₁. So that gives another expression for x₂.Wait, so previously, from equation (2), we had x₂ = 1 + x₁ - x₁ ln x₁. But from equation (1), x₂ = -ln x₁. Therefore, setting these equal:- ln x₁ = 1 + x₁ - x₁ ln x₁So, -ln x₁ = 1 + x₁ - x₁ ln x₁Hmm. Let's rearrange terms. Let me bring all terms to one side:- ln x₁ - 1 - x₁ + x₁ ln x₁ = 0Factor terms where possible:x₁ ln x₁ - ln x₁ - x₁ - 1 = 0Factor ln x₁ from the first two terms:ln x₁ (x₁ - 1) - x₁ - 1 = 0Hmm, that's an equation in terms of x₁. Let me write it as:ln x₁ (x₁ - 1) - (x₁ + 1) = 0This seems a bit complicated. Maybe there's a substitution here. Let's see.Let me denote t = x₁ - 1. Then, x₁ = t + 1. Maybe substituting t will help. Let's try.First, ln x₁ becomes ln(t + 1). Then:ln(t + 1) * t - (t + 1 + 1) = 0Wait, wait. Original expression after substitution:ln(t + 1) * t - ( (t + 1) + 1 ) = 0Which is:t ln(t + 1) - t - 2 = 0Hmm, not sure if that helps. Alternatively, maybe trying to see if x₁ is a value that satisfies this equation.Alternatively, perhaps we can use the expression x₂ = -ln x₁ from equation (1), and plug that into equation (2).Wait, let's recall equation (2):ln x₁ - 1 = (1/x₁)(1 - x₂)But if x₂ = -ln x₁, then substituting into equation (2):ln x₁ - 1 = (1/x₁)(1 - (-ln x₁)) = (1/x₁)(1 + ln x₁)Multiply both sides by x₁:x₁ (ln x₁ - 1) = 1 + ln x₁Bring all terms to left:x₁ ln x₁ - x₁ - 1 - ln x₁ = 0Factor ln x₁:ln x₁ (x₁ - 1) - (x₁ + 1) = 0Same as before. So, same equation. So, we need to solve ln x₁ (x₁ - 1) - (x₁ + 1) = 0.This seems like a transcendental equation, which might not have a closed-form solution. Hmm. Maybe we can guess a value for x₁. Let's try x₁ = 2.Plugging x₁ = 2:ln 2 (2 - 1) - (2 + 1) = ln 2 - 3 ≈ 0.6931 - 3 ≈ -2.3069 ≠ 0x₁ = 1: Well, x₁ can't be 1 because ln x₁ would be 0, but then x₁ -1 would be 0, and in the original problem, we have 2/(x₁ - 1), which would be undefined. So x₁ cannot be 1.x₁ = e: Let's try x₁ = e ≈ 2.718.ln e (e -1) - (e +1) = 1*(e -1) - (e +1) = e -1 - e -1 = -2 ≠ 0Not zero. Hmm.x₁ = 3:ln 3 (3 -1) - (3 +1) = ln 3 * 2 - 4 ≈ 2*1.0986 - 4 ≈ 2.1972 -4 ≈ -1.8028Still negative.x₁ = 4:ln4 *3 -5 ≈ 1.3863*3 -5 ≈4.1589 -5≈-0.8411x₁=5:ln5 *4 -6≈1.6094*4 -6≈6.4376 -6≈0.4376So between x₁=4 and x₁=5, the expression crosses from negative to positive. Let's check x₁=4.5:ln(4.5)*3.5 -5.5≈1.5041*3.5≈5.2643 -5.5≈-0.2357Still negative.x₁=4.75:ln(4.75)*3.75 -5.75≈1.5581*3.75≈5.8429 -5.75≈0.0929So between 4.5 and 4.75. Let's try 4.6:ln(4.6)*3.6 -5.6≈1.5261*3.6≈5.494 -5.6≈-0.106x₁=4.7:ln(4.7)*3.7 -5.7≈1.5476*3.7≈5.726 -5.7≈0.026x₁=4.65:ln(4.65)*3.65 -5.65≈1.5369*3.65≈5.6097 -5.65≈-0.0403x₁=4.68:ln(4.68)*3.68 -5.68≈1.5444*3.68≈5.684 -5.68≈0.004Close to zero. So around x₁≈4.68.But this is getting tedious. Wait, maybe there's a smarter way here. Let me think.Alternatively, maybe there is a substitution. Let's let t = x₁ -1. Then x₁ = t +1. Then our equation is:ln(x₁)(x₁ -1) - (x₁ +1) = 0Substituting t:ln(t + 1) * t - (t +1 +1) = 0So:t ln(t +1) - t -2 =0Or:t (ln(t +1) -1 ) -2 =0Hmm. Still not obvious. Maybe try another substitution. Let me set z = t +1 = x₁. Wait, no, that's going back. Alternatively, let me write the equation as:ln(x₁) = (x₁ +1)/(x₁ -1)Because from ln x₁ (x₁ -1) - (x₁ +1) =0, we can rearrange:ln x₁ = (x₁ +1)/(x₁ -1)So, ln x = (x +1)/(x -1). Let's see if this equation has a solution. Let me graph ln x and (x +1)/(x -1). Hmm. For x >1, since x₁ must be greater than 0 (because ln x is defined for x>0), and in the original expression, x₁ cannot be 1. Also, from the tangent line on ln x, the slope is 1/x₁. If x₁ ≤0, it's invalid. So x₁>0. But the tangent line is also tangent to e^x, which is defined for all real numbers. But x₂ can be any real number.But from equation (1), e^{x₂} =1/x₁. Since e^{x₂} is always positive, 1/x₁ must be positive, so x₁>0, which we already knew.So, back to ln x = (x +1)/(x -1). Let's see. Let me check x=2:ln 2 ≈0.6931, (2 +1)/(2 -1)=3/1=3. Not equal.x=3: ln3≈1.0986, (4)/2=2. Not equal.x=1.5: ln1.5≈0.4055, (2.5)/0.5=5. Not equal.x approaches 1 from the right: ln x approaches 0, (x +1)/(x -1) approaches (2)/0+ = +infty. So not equal.x approaches infty: ln x grows to infty, (x +1)/(x -1) approaches 1. So, no solution for x>1 except maybe somewhere.Wait, when x=4.68, as we saw earlier, ln x ≈1.544, (x +1)/(x -1)=5.68/3.68≈1.543. So that's very close. So x≈4.68 gives ln x ≈1.544 and (x +1)/(x -1)≈1.543. So approximately, x≈4.68 is a solution.But this seems like it's not an exact value. Maybe the problem expects an exact value? Wait, maybe there's a trick here.Wait, the problem asks for 2/(x₁ -1) + x₂. Maybe we don't need to find x₁ and x₂ explicitly. Maybe there's a relationship between x₁ and x₂ that allows us to compute this expression without knowing their exact values.Let me recall:From equation (1): e^{x₂} =1/x₁ => x₂ = -ln x₁From equation (2): ln x₁ -1 = (1/x₁)(1 - x₂)But since x₂ = -ln x₁, substitute into equation (2):ln x₁ -1 = (1/x₁)(1 - (-ln x₁)) = (1 + ln x₁)/x₁Multiply both sides by x₁:x₁ ln x₁ - x₁ = 1 + ln x₁Bring all terms to left:x₁ ln x₁ - x₁ -1 - ln x₁ =0Factor ln x₁:ln x₁ (x₁ -1) - (x₁ +1)=0Same equation as before.But perhaps instead of solving for x₁, we can find an expression for 2/(x₁ -1) + x₂ in terms of x₁.Given that x₂ = -ln x₁, then:2/(x₁ -1) + x₂ = 2/(x₁ -1) - ln x₁So, if we can express this in terms of the equation we have.From the equation ln x₁ = (x₁ +1)/(x₁ -1)Wait, from the equation:ln x₁ (x₁ -1) - (x₁ +1) =0 => ln x₁ = (x₁ +1)/(x₁ -1)So, substituting into 2/(x₁ -1) - ln x₁:2/(x₁ -1) - (x₁ +1)/(x₁ -1) = [2 - (x₁ +1)]/(x₁ -1) = (1 - x₁)/(x₁ -1) = - (x₁ -1)/(x₁ -1) = -1Oh! Wait, that's clever. Let me verify:Given that ln x₁ = (x₁ +1)/(x₁ -1), so substituting that into 2/(x₁ -1) - ln x₁ gives:2/(x₁ -1) - (x₁ +1)/(x₁ -1) = [2 - (x₁ +1)]/(x₁ -1) = (1 - x₁)/(x₁ -1) = - (x₁ -1)/(x₁ -1) = -1So, 2/(x₁ -1) + x₂ = -1Therefore, the answer is -1. So, boxed{-1}Wow, that's neat. I was going down the path of trying to solve for x₁ numerically, but it turns out that by manipulating the equations and recognizing the expression needed, we can substitute and simplify to find the answer directly. So, the key was to express 2/(x₁ -1) + x₂ in terms of the given equations, substitute using the relationships we found, and simplify. That's pretty cool. I need to remember to look for such substitutions instead of getting bogged down in solving transcendental equations.Final Answerboxed{-1}</think>To solve the problem, we need to find the value of (frac{2}{x_1 - 1} + x_2) where the tangent line to (y = ln x) at point (P(x_1, y_1)) is also tangent to (y = e^x) at point (Q(x_2, y_2)).1. Equation of the tangent line at (P(x_1, y_1)) on (y = ln x): - The slope at (x_1) is (frac{1}{x_1}). - The equation of the tangent line is (y = frac{1}{x_1}x + (ln x_1 - 1)).2. Equation of the tangent line at (Q(x_2, y_2)) on (y = e^x): - The slope at (x_2) is (e^{x_2}). - The equation of the tangent line is (y = e^{x_2}x + e^{x_2}(1 - x_2)).3. Equating the slopes and y-intercepts: - The slopes must be equal: (frac{1}{x_1} = e^{x_2}). - The y-intercepts must be equal: (ln x_1 - 1 = e^{x_2}(1 - x_2)).4. Substituting (e^{x_2} = frac{1}{x_1}) into the y-intercept equation: - (ln x_1 - 1 = frac{1}{x_1}(1 - x_2)). - Multiplying both sides by (x_1): (x_1 ln x_1 - x_1 = 1 - x_2). - Rearranging gives: (x_2 = 1 + x_1 - x_1 ln x_1).5. Expressing (x_2) in terms of (x_1): - From (frac{1}{x_1} = e^{x_2}), we get (x_2 = -ln x_1).6. Substituting (x_2 = -ln x_1) into the expression: - The expression to find is (frac{2}{x_1 - 1} + x_2). - Substitute (x_2): (frac{2}{x_1 - 1} - ln x_1).7. Using the equation (ln x_1 = frac{x_1 + 1}{x_1 - 1}): - Substitute (ln x_1) into the expression: (frac{2}{x_1 - 1} - frac{x_1 + 1}{x_1 - 1}). - Simplify: (frac{2 - (x_1 + 1)}{x_1 - 1} = frac{1 - x_1}{x_1 - 1} = -1).Thus, the value of (frac{2}{x_1 - 1} + x_2) is (boxed{-1}).,<think>Okay, let me try to figure this problem out. So, we have two curves: y = ln x and y = e^x. We need to find a point P(x1, y1) on the ln x curve where the tangent line at P is also tangent to the e^x curve at another point Q(x2, y2). Then, compute 2/(x1 - 1) + x2. Hmm, interesting. Let's break this down step by step.First, I remember that the equation of a tangent line at a point (a, f(a)) is given by y = f'(a)(x - a) + f(a). So, for both curves, we can write their tangent lines and set them equal since they're the same line. That should give us some equations to solve for x1 and x2.Starting with the first curve, y = ln x at point P(x1, y1). The derivative of ln x is 1/x, so the slope at x1 is 1/x1. Therefore, the tangent line at P is:y = (1/x1)(x - x1) + ln x1.Simplify that equation:y = (1/x1)x - 1 + ln x1.So, the equation is y = (1/x1)x + (ln x1 - 1).Now, for the second curve, y = e^x at point Q(x2, y2). The derivative of e^x is e^x, so the slope at x2 is e^{x2}. Therefore, the tangent line at Q is:y = e^{x2}(x - x2) + e^{x2}.Simplify that:y = e^{x2}x - e^{x2}x2 + e^{x2} = e^{x2}x + e^{x2}(1 - x2).Since these two tangent lines are the same line, their slopes and y-intercepts must be equal. So, setting the slopes equal:1/x1 = e^{x2}. ...(1)And setting the y-intercepts equal:ln x1 - 1 = e^{x2}(1 - x2). ...(2)So now we have two equations with two variables x1 and x2. Our goal is to solve these equations and then compute 2/(x1 - 1) + x2.Let me write down equations (1) and (2):1. 1/x1 = e^{x2}2. ln x1 - 1 = e^{x2}(1 - x2)From equation (1), we can express e^{x2} as 1/x1. Let's substitute that into equation (2):ln x1 - 1 = (1/x1)(1 - x2).So, equation (2) becomes:ln x1 - 1 = (1 - x2)/x1.Multiply both sides by x1 to eliminate the denominator:x1(ln x1 - 1) = 1 - x2.So, rearranged:x2 = 1 - x1(ln x1 - 1).Simplify that:x2 = 1 - x1 ln x1 + x1.So, x2 = 1 + x1 - x1 ln x1.Hmm. So, now we have expressions for x2 in terms of x1. But we still need another relation to solve for x1 and x2. But wait, equation (1) is 1/x1 = e^{x2}, so x2 = ln(1/x1) = -ln x1. Because e^{x2} = 1/x1 implies x2 = ln(1/x1) = -ln x1.Wait a minute, that's another expression for x2. So, we have:From equation (1): x2 = -ln x1.But we also have from equation (2):x2 = 1 + x1 - x1 ln x1.Therefore, setting these two expressions for x2 equal:- ln x1 = 1 + x1 - x1 ln x1.So, let's write that:- ln x1 = 1 + x1 - x1 ln x1.Bring all terms to one side:- ln x1 - 1 - x1 + x1 ln x1 = 0.Factor terms if possible. Let's see:x1 ln x1 - ln x1 - x1 -1 = 0.Factor ln x1 from the first two terms:ln x1(x1 - 1) - x1 -1 = 0.Hmm. So, ln x1 (x1 -1) - (x1 +1) =0.Hmm, this seems a bit complicated. Maybe there's a substitution here.Let me let t = x1 -1, but not sure. Alternatively, maybe try to rearrange terms:ln x1 (x1 -1) = x1 +1.So,ln x1 = (x1 +1)/(x1 -1).Hmm. So, this equation is ln x1 = (x1 +1)/(x1 -1).This seems like a transcendental equation, which might not have a closed-form solution. Maybe we can solve it numerically or see if there's an obvious solution.Let me check x1 = 2. Let's test x1 =2.Left side: ln 2 ≈0.693.Right side: (2+1)/(2-1)= 3/1=3. Not equal.x1=1. Let's check x1 approaching 1 from the right. Then (x1 +1)/(x1 -1) tends to infinity. But ln 1=0, so not helpful.x1=3. ln3≈1.098. (3+1)/(3-1)=4/2=2. Not equal.x1=0.5. ln(0.5)= -0.693. (0.5 +1)/(0.5 -1)=1.5/(-0.5)= -3. Not equal.x1= e. Let's see, ln e =1. (e +1)/(e -1)≈(2.718+1)/(1.718)=3.718/1.718≈2.164. Not equal to 1.Hmm. Maybe x1= something else. Let me try x1=1. Let's see, but x1=1 is on the curve y=ln x, but x1 can't be 1 because then denominator in (x1 -1) would be zero. So, x1 must be greater than 0, and not equal to 1.Alternatively, maybe there is a solution when x1=2? Wait, we tried x1=2, didn't work.Alternatively, perhaps x1= something else. Let's try x1= (1 + sqrt(5))/2 or something, but this is just guessing.Alternatively, maybe we can rearrange the equation:Let me write ln x1 = (x1 +1)/(x1 -1).Let me denote t = x1 -1, so x1 = t +1. Then:ln(t +1) = (t +1 +1)/t = (t +2)/t =1 + 2/t.So, equation becomes ln(t +1) =1 + 2/t.Hmm. Maybe now with substitution t, but still not straightforward. Let's see if t=1:ln(2)=1 +2/1=3. ln2≈0.693≈3? No. t=2: ln(3)=1 +2/2=2. ln3≈1.098≈2? No. t=0.5: ln(1.5)=1 +2/0.5=1+4=5. ln1.5≈0.405≠5.t= -0.5: x1=0.5, which we checked earlier. Not helpful.t= -1: x1=0, which is invalid since ln x is undefined.Hmm, maybe another approach. Let me consider the function f(x) = ln x - (x +1)/(x -1). We need to find x where f(x)=0.Compute f(2)= ln2 -3/1≈0.693 -3≈-2.307.f(3)=ln3 -4/2≈1.098 -2≈-0.902.f(4)=ln4 -5/3≈1.386 -1.666≈-0.28.f(5)=ln5 -6/4≈1.609 -1.5≈0.109.So, f(5)≈0.109, which is positive. So between x=4 and x=5, f(x) crosses zero. Wait, but the original problem must have a unique solution, perhaps? Maybe not. Wait, but the problem states that the tangent line is tangent to both curves, so maybe such x1 and x2 exist uniquely. But according to the values, f(4) is negative, f(5) positive, so by Intermediate Value Theorem, there is a root between 4 and 5. Similarly, let's check f(1.5):f(1.5)=ln1.5 - (1.5 +1)/(1.5 -1)=0.405 -2.5/0.5=0.405 -5≈-4.595.Negative. So maybe only one root between 4 and 5. Hmm. But the problem is presented as a problem that likely has an integer or simple fractional solution, so maybe I made a mistake in my approach.Wait, maybe there's another way to think about it. Let's recall that the tangent line to both curves must touch both curves at points P and Q, so the line is tangent to both. Therefore, there's a system of equations here.Wait, let's recap:We have the tangent line to y=ln x at x1 is the same as the tangent line to y=e^x at x2.We derived that x2 = -ln x1 from equation (1): e^{x2}=1/x1.And from equation (2), we have x2 =1 +x1 -x1 ln x1.Therefore, substituting x2 = -ln x1 into the second equation:- ln x1 =1 +x1 -x1 ln x1.So, bringing all terms to left-hand side:- ln x1 -1 -x1 +x1 ln x1 =0.Factor:x1 ln x1 - ln x1 -x1 -1=0.Factor ln x1 (x1 -1) - (x1 +1)=0.Hmm. So ln x1 = (x1 +1)/(x1 -1).This is the equation we need to solve for x1.Alternatively, maybe consider substituting t = x1 -1, so x1 = t +1. Then:ln(t +1) = (t +1 +1)/t = (t +2)/t =1 + 2/t.So, ln(t +1) -1 = 2/t.Hmm. Not sure. Maybe try plotting these functions or using numerical methods.Alternatively, maybe assume that x1 is 2, but we saw that didn't work. If x1 is 2, then ln2 ≈0.693, and (2 +1)/(2 -1)=3. Not equal.Alternatively, let's suppose x1= e, then ln e=1, and (e +1)/(e -1)≈(2.718+1)/1.718≈3.718/1.718≈2.164. Not equal.Alternatively, maybe x1=3. Let's compute ln3≈1.0986. (3+1)/(3-1)=4/2=2. So ln3≈1.0986 vs 2. Not equal.x1=4: ln4≈1.386, (4+1)/(4-1)=5/3≈1.666. Still not equal.x1=5: ln5≈1.609, (5+1)/4=6/4=1.5. So ln5≈1.609 vs 1.5.Wait, so between x1=4 and x1=5, the left side (ln x1) increases from 1.386 to 1.609, and the right side (x1 +1)/(x1 -1) decreases from 5/3≈1.666 to 6/4=1.5. So, they cross somewhere between x1=4 and x1=5.But this seems messy. Maybe there's a smarter substitution or recognizing that this equation might relate to the Lambert W function? Let me think.Starting from ln x1 = (x1 +1)/(x1 -1).Multiply both sides by (x1 -1):(x1 -1) ln x1 =x1 +1.Hmm. Let me rearrange:(x1 -1) ln x1 -x1 -1=0.Let me denote u = x1 -1. Then x1 = u +1. Substitute:u ln(u +1) - (u +1) -1=0 → u ln(u +1) -u -2=0.So, u ln(u +1) =u +2.Divide both sides by u (assuming u ≠0):ln(u +1) =1 + 2/u.Hmm, similar to before. Not sure. This seems like an equation that might require Lambert W, but perhaps not straightforward.Alternatively, maybe make substitution v = u +1, so u = v -1. Then:(v -1) ln v - (v -1) -2=0 → (v -1)(ln v -1) -2=0.But this might not help.Alternatively, set z = ln(u +1), then u +1 = e^z. Then, from the equation:z =1 +2/u.But u = e^z -1. Substitute into z:z =1 +2/(e^z -1).So,z -1 = 2/(e^z -1).Multiply both sides by (e^z -1):(z -1)(e^z -1)=2.This is getting complicated. Maybe not helpful.Alternatively, think numerically. Let's attempt to approximate the solution.We know that for x1 between 4 and 5, ln x1 is between 1.386 and 1.609, and (x1 +1)/(x1 -1) decreases from 1.666 to 1.5. So, the two functions cross somewhere in that interval.Let's try x1=4.5:ln4.5≈1.504.(4.5 +1)/(4.5 -1)=5.5/3.5≈1.571.So, ln4.5≈1.504 <1.571. So, lnx1 < (x1 +1)/(x1 -1).We need to find x1 where lnx1=(x1 +1)/(x1 -1). Let's try x1=4.3:ln4.3≈1.458.(4.3 +1)/3.3≈5.3/3.3≈1.606. Still lnx1 < RHS.x1=4.8:ln4.8≈1.568.(4.8 +1)/3.8≈5.8/3.8≈1.526.Now, ln4.8≈1.568 >1.526. So, somewhere between x1=4.5 and x1=4.8.Let's do a table:x1 | lnx1 | (x1 +1)/(x1 -1)4.5 |1.504 |1.5714.6 |1.526 |(5.6)/3.6≈1.5554.7 |1.547 |(5.7)/3.7≈1.5404.75|1.558 |(5.75)/3.75≈1.5334.8 |1.568 |1.526So, at x1=4.7, lnx1≈1.547 vs RHS≈1.540. So, lnx1 > RHS here.At x1=4.6: ln≈1.526 vs RHS≈1.555. So, lnx1 < RHS.So crossing between 4.6 and 4.7.At x1=4.65:ln4.65≈1.536.(4.65 +1)/(4.65 -1)=5.65/3.65≈1.548.So, ln≈1.536 <1.548. So, still lnx1 < RHS.At x1=4.67:ln≈1.542.(4.67+1)/3.67≈5.67/3.67≈1.544.So, close. ln≈1.542 vs 1.544. Almost equal.x1=4.675:ln≈1.543.(4.675 +1)/3.675≈5.675/3.675≈1.544.So, lnx1=1.543 vs RHS≈1.544. Very close.x1=4.678:ln≈1.544.(4.678+1)/3.678≈5.678/3.678≈1.543.So, now lnx1=1.544 vs RHS≈1.543. So, crossing around x1≈4.677.Therefore, approximate solution x1≈4.677, then x2=-lnx1≈-1.544.But then we need to compute 2/(x1 -1) +x2≈2/(4.677 -1) +(-1.544)=2/3.677≈0.543 -1.544≈-1.001.Wait, approximately -1? That seems interesting. Maybe the exact answer is -1? Let me check.Wait, if 2/(x1 -1) +x2 = -1, then let's see if that's possible.Suppose 2/(x1 -1) +x2 = -1.But from earlier, we had x2 = -ln x1.Therefore, 2/(x1 -1) - ln x1 = -1.So,2/(x1 -1) = ln x1 -1.But from equation (2):ln x1 -1 = (1 -x2)/x1.But x2 = -ln x1, so:ln x1 -1 = (1 - (-ln x1))/x1 = (1 + ln x1)/x1.So,2/(x1 -1) = (1 + ln x1)/x1.Cross-multiplying:2x1 = (1 + ln x1)(x1 -1).Expand the right side:(1)(x1 -1) + (ln x1)(x1 -1) =x1 -1 +x1 ln x1 - ln x1.Therefore,2x1 =x1 -1 +x1 ln x1 - ln x1.Bring all terms to left side:2x1 -x1 +1 -x1 ln x1 + ln x1=0.Simplify:x1 +1 -x1 ln x1 + ln x1=0.Factor:x1(1 - ln x1) +1 + ln x1=0.Hmm, not sure. Wait, perhaps this is the same equation we had before. Let's check.Original equation after substituting x2 = -ln x1 into equation (2):- ln x1 =1 +x1 -x1 ln x1.Bring all terms to left:- ln x1 -1 -x1 +x1 ln x1=0.Which is same as:x1 ln x1 - ln x1 -x1 -1=0.Which factors as:ln x1 (x1 -1) - (x1 +1)=0.Yes, same equation. Therefore, if we assume that 2/(x1 -1) +x2 = -1, and given that x2 = -ln x1, then:2/(x1 -1) - ln x1 = -1.But when we derived from equation (2), we saw that ln x1 -1 = (1 -x2)/x1.But substituting x2 = -ln x1 gives:ln x1 -1 = (1 + ln x1)/x1.Therefore, 2/(x1 -1) = (1 + ln x1)/x1.So, if these two expressions are equal, then:2/(x1 -1) = (1 + ln x1)/x1.But how does this relate to the original equation?Let me see. From the original equation:ln x1 = (x1 +1)/(x1 -1).So, substitute ln x1 into the above equation:2/(x1 -1) = (1 + (x1 +1)/(x1 -1))/x1.Compute numerator:1 + (x1 +1)/(x1 -1) = [ (x1 -1) +x1 +1 ] / (x1 -1) )= [2x1]/(x1 -1).Therefore,2/(x1 -1) = [2x1/(x1 -1)] /x1 = 2/(x1 -1).Thus, it's an identity. Therefore, the equation 2/(x1 -1) +x2 = -1 is indeed valid given the original equations.Therefore, even without knowing the exact value of x1 and x2, we can conclude that 2/(x1 -1) +x2 = -1.Therefore, the answer is boxed{-1}.Final Answerboxed{-1}</think>Given the problem of finding the value of (frac{2}{x_1 - 1} + x_2) where the tangent line to the curve (y = ln x) at point (P(x_1, y_1)) is also tangent to the curve (y = e^x) at point (Q(x_2, y_2)), we start by determining the equations of the tangent lines for both curves.1. Tangent line at (P(x_1, y_1)) on (y = ln x): - The derivative of (ln x) is (frac{1}{x}), so the slope at (x_1) is (frac{1}{x_1}). - The equation of the tangent line is (y = frac{1}{x_1}(x - x_1) + ln x_1), which simplifies to (y = frac{1}{x_1}x + (ln x_1 - 1)).2. Tangent line at (Q(x_2, y_2)) on (y = e^x): - The derivative of (e^x) is (e^x), so the slope at (x_2) is (e^{x_2}). - The equation of the tangent line is (y = e^{x_2}(x - x_2) + e^{x_2}), which simplifies to (y = e^{x_2}x + e^{x_2}(1 - x_2)).Since these tangent lines are the same, their slopes and y-intercepts must be equal:- Equating the slopes: (frac{1}{x_1} = e^{x_2}).- Equating the y-intercepts: (ln x_1 - 1 = e^{x_2}(1 - x_2)).From the slope equation, we get (e^{x_2} = frac{1}{x_1}), which implies (x_2 = -ln x_1).Substituting (e^{x_2} = frac{1}{x_1}) into the y-intercept equation:[ln x_1 - 1 = frac{1}{x_1}(1 - x_2)]Multiplying both sides by (x_1):[x_1 ln x_1 - x_1 = 1 - x_2]Rearranging gives:[x_2 = 1 + x_1 - x_1 ln x_1]We also know (x_2 = -ln x_1). Equating the two expressions for (x_2):[-ln x_1 = 1 + x_1 - x_1 ln x_1]Rearranging terms:[x_1 ln x_1 - ln x_1 - x_1 - 1 = 0]Factoring out (ln x_1):[ln x_1 (x_1 - 1) - (x_1 + 1) = 0]This simplifies to:[ln x_1 = frac{x_1 + 1}{x_1 - 1}]To find (frac{2}{x_1 - 1} + x_2), we use (x_2 = -ln x_1) and the equation (ln x_1 = frac{x_1 + 1}{x_1 - 1}):[frac{2}{x_1 - 1} + x_2 = frac{2}{x_1 - 1} - ln x_1]Substituting (ln x_1 = frac{x_1 + 1}{x_1 - 1}):[frac{2}{x_1 - 1} - frac{x_1 + 1}{x_1 - 1} = frac{2 - (x_1 + 1)}{x_1 - 1} = frac{1 - x_1}{x_1 - 1} = -1]Thus, the final answer is (boxed{-1}).]

🔑:To solve the problem, we first need to find the equations of the tangent lines to the curves at the given points and then equate them to find the relationship between x_1 and x_2.1. Finding the tangent line to y=ln x at P(x_{1}, y_{1}):The derivative of y=ln x is y'=frac{1}{x}. Thus, the equation of the tangent line at point P(x_{1}, y_{1}) is given by:[y-y_{1}=frac{1}{x_{1}}(x-x_{1})]Since y_{1}=ln x_{1}, we can rewrite this as:[y-ln x_{1}=frac{1}{x_{1}}(x-x_{1})]2. Finding the tangent line to y=e^{x} at Q(x_{2}, y_{2}):The derivative of y=e^{x} is y'=e^{x}. Thus, the equation of the tangent line at point Q(x_{2}, y_{2}) is given by:[y-y_{2}=e^{x_{2}}(x-x_{2})]Since y_{2}=e^{x_{2}}, we can rewrite this as:[y-e^{x_{2}}=e^{x_{2}}(x-x_{2})]3. Equating the slopes of the tangent lines:Since the tangent lines are the same, their slopes must be equal. Therefore, we have:[frac{1}{x_{1}}=e^{x_{2}}]And their y-intercepts must also be equal, leading to:[ln x_{1}-1=e^{x_{2}}(1-x_{2})]4. Finding the relationship between x_1 and x_2:From frac{1}{x_{1}}=e^{x_{2}}, we can deduce that x_{2}=-ln x_{1}. Substituting this into the equation for the y-intercepts gives:[ln x_{1}-1=frac{1}{x_{1}}(1+ln x_{1})]Simplifying this equation, we find:[ln x_{1}-1=frac{2}{{x}_{1}-1}]5. Calculating frac{2}{{x_1}-1}+x_{2}:Given x_{2}=-ln x_{1}, we substitute this into the equation to find:[frac{2}{{x_1}-1}+x_{2}=ln x_{1}-1-ln x_{1}=-1]Therefore, the final answer is boxed{-1}.

🧠:[<think>Okay, so I need to find all positive integer solutions (x, y, z) to the equation arctan(1/x) + arctan(1/y) + arctan(1/z) = π/4. Hmm, arctangent functions adding up to π/4. I remember there are some addition formulas for arctangent. Let me recall.The formula for adding two arctangent terms: arctan(a) + arctan(b) = arctan((a + b)/(1 - ab)), but this is only true when ab < 1, right? Otherwise, you have to add or subtract π. But in our case, since we have three arctangent terms adding up to π/4, which is a positive angle less than π/2, maybe all the individual arctans are small enough such that their sum doesn't exceed π/4. So maybe we can apply the addition formula step by step.But with three terms, applying the addition formula twice. Let me try to set up the equation step by step.Let’s denote:A = arctan(1/x), B = arctan(1/y), C = arctan(1/z).So the equation is A + B + C = π/4.If I first add A and B, then add C to the result. Let's first compute arctan(1/x) + arctan(1/y). Using the formula:arctan(a) + arctan(b) = arctan((a + b)/(1 - ab)) + kπ, where k is an integer. Since we are dealing with positive x and y, 1/x and 1/y are positive and less than 1 (since x, y ≥ 1, but x, y are positive integers, so x, y ≥ 1, but actually x, y ≥ 1, but 1/x and 1/y ≤ 1. However, if x or y is 1, then 1/x or 1/y is 1. So the maximum value for arctan(1/x) is π/4 when x=1. So adding three π/4 terms would get 3π/4, which is way larger than π/4. So likely, x, y, z are larger than 1, but not necessarily. So need to check.But given that the sum is π/4, which is quite small, so each of the arctan terms must be quite small. So 1/x, 1/y, 1/z are small, so x, y, z are large? Wait, but maybe some of them can be small. For example, if one of them is 1, which gives arctan(1) = π/4, but then adding three of them would be too much. But if one is 1, then the other two would need to subtract π/4 - π/4 = 0, which would require the sum of the other two arctans to be zero, but arctan is always positive, so that can't happen. So maybe none of x, y, z can be 1. Let me check that.Suppose x=1: then arctan(1) = π/4. Then the equation becomes π/4 + arctan(1/y) + arctan(1/z) = π/4, which implies arctan(1/y) + arctan(1/z) = 0. But since y and z are positive integers, 1/y and 1/z are positive, so their arctans are positive. So sum cannot be zero. Hence x cannot be 1. Similarly, y and z cannot be 1. Therefore, x, y, z ≥ 2.So all variables start from 2. So arctan(1/x) ≤ arctan(1/2) ≈ 0.4636 radians, which is about 26.565 degrees. So three of these would add up to at most around 1.39 radians, which is about 79.7 degrees, but we need them to add up to π/4, which is 45 degrees. So each term needs to be smaller, on average about 15 degrees. So maybe x, y, z are larger.But let's see. Maybe first try small integers. Let's consider possible small x, y, z ≥2 and check if their arctans sum to π/4.Alternatively, perhaps we can use the tangent function on both sides. Let’s try that.Take tangent of both sides:tan(A + B + C) = tan(π/4) = 1.Using the tangent addition formula for three angles.The formula for tan(A + B + C) is [tan A + tan B + tan C - tan A tan B tan C] / [1 - tan A tan B - tan A tan C - tan B tan C].So, substituting tan A = 1/x, tan B = 1/y, tan C = 1/z, we get:[ (1/x + 1/y + 1/z) - (1/x)(1/y)(1/z) ] / [ 1 - (1/x)(1/y) - (1/x)(1/z) - (1/y)(1/z) ] = 1.Therefore,( (1/x + 1/y + 1/z) - 1/(xyz) ) = 1 - (1/(xy) + 1/(xz) + 1/(yz) )Let me write that equation again:(1/x + 1/y + 1/z) - 1/(xyz) = 1 - (1/(xy) + 1/(xz) + 1/(yz))Let me rearrange terms:Left side: (1/x + 1/y + 1/z) - 1/(xyz)Right side: 1 - (1/(xy) + 1/(xz) + 1/(yz))Bring all terms to the left side:(1/x + 1/y + 1/z) - 1/(xyz) - 1 + (1/(xy) + 1/(xz) + 1/(yz)) = 0Let me group the terms:(-1) + (1/x + 1/y + 1/z) + (1/(xy) + 1/(xz) + 1/(yz)) - 1/(xyz) = 0Hmm, this seems a bit messy, but maybe we can factor it. Let me see.Let’s consider the equation:-1 + (1/x + 1/y + 1/z) + (1/(xy) + 1/(xz) + 1/(yz)) - 1/(xyz) = 0Multiply both sides by xyz to eliminate denominators:-xyz + yz + xz + xy + z + y + x - 1 = 0So:-xyz + (xy + yz + zx) + (x + y + z) -1 = 0Rearranged:xyz - (xy + yz + zx) - (x + y + z) + 1 = 0Hmm, interesting. So the equation simplifies to:xyz - (xy + yz + zx) - (x + y + z) + 1 = 0This is a diophantine equation in positive integers x, y, z ≥2.So now we need to find all positive integers x, y, z ≥2 satisfying this equation.Let me write that equation again:xyz - xy - yz - zx - x - y - z + 1 = 0Wait, no: original was xyz - (xy + yz + zx) - (x + y + z) + 1 = 0Yes. So:xyz - xy - yz - zx - x - y - z + 1 = 0This seems like a cubic equation. Solving this in positive integers ≥2 might be challenging, but maybe we can factor it somehow.Alternatively, perhaps we can add 1 to both sides:xyz - xy - yz - zx - x - y - z + 1 = 0 => xyz - xy - yz - zx - x - y - z = -1Alternatively, perhaps factor terms.Looking at the equation:xyz - xy - yz - zx - x - y - z + 1 = 0Let me try to factor. Maybe factor by grouping.Let me group terms:xyz - xy - yz - zx - x - y - z + 1Let me group xyz - xy - xz - x: x(yz - y - z -1)Similarly, the remaining terms: - yz - y - z +1Wait, that might not be helpful.Alternatively, perhaps consider the equation as:(xy)(z - 1) - (x + y)(z + 1) - z +1 = 0Not sure. Let me try substituting small values.Since x, y, z ≥2, let's try z=2 first, see if we can find x and y.Let’s set z=2. Then the equation becomes:2xy - xy - 2y - 2x - x - y - 2 +1 = 0Wait, let me compute each term:xyz = x*y*2 = 2xyxy + yz + zx = xy + 2y + 2xx + y + z = x + y +2So substituting into the equation:2xy - (xy + 2y + 2x) - (x + y + 2) +1 = 0Compute step by step:2xy - xy - 2y - 2x - x - y -2 +1 = 0Simplify:(2xy - xy) + (-2y - y) + (-2x - x) + (-2 +1) = 0So:xy -3y -3x -1 =0So:xy -3x -3y =1Hmm, this is a diophantine equation in x and y. Let's rearrange:xy -3x -3y =1Add 9 to both sides:xy -3x -3y +9 =10Then factor:(x -3)(y -3) =10Ah, clever! So (x -3)(y -3)=10. Since x and y are integers ≥2, x-3 and y-3 are integers ≥-1. But since (x -3)(y -3)=10, which is positive, so both factors must be positive. Therefore, x -3 ≥1 and y -3 ≥1, so x ≥4 and y ≥4.So the positive integer solutions for (x-3, y-3) are the factor pairs of 10: (1,10),(2,5),(5,2),(10,1)Therefore, (x, y) = (4,13), (5,8), (8,5), (13,4)Thus, when z=2, the solutions (x,y,z) are (4,13,2), (5,8,2), (8,5,2), (13,4,2). Similarly, since the equation is symmetric in x, y, z, we can expect similar solutions when we fix x=2 or y=2.But wait, in the original equation, are x, y, z symmetric? Let me check.The original equation is arctan(1/x) + arctan(1/y) + arctan(1/z) = π/4. Since addition is commutative, yes, the equation is symmetric in x, y, z. Therefore, all permutations of solutions are also solutions. So if we fix z=2, we get those four solutions, and similarly for x=2 and y=2, which would give permutations. So need to check if those are distinct.But first, let's check for z=3. Let's see if there are solutions.Set z=3:xyz - xy - yz - zx - x - y - z +1 =0Compute:3xy -xy -3y -3x -x -y -3 +1=0Again, compute step by step:3xy -xy =2xy-3y - y = -4y-3x -x = -4x-3 +1 =-2So equation:2xy -4x -4y -2 =0Divide both sides by 2:xy -2x -2y -1 =0Rearranged:xy -2x -2y =1Add 4 to both sides:xy -2x -2y +4 =5Factor:(x -2)(y -2)=5Since 5 is prime, the factor pairs are (1,5),(5,1)Thus, (x -2, y -2)=(1,5) => x=3, y=7Or (5,1) => x=7, y=3Therefore, solutions (3,7,3) and (7,3,3). But since z=3, so the solutions are (3,7,3) and (7,3,3). However, we need to check if x,y,z ≥2. Yes, 3,7,3 are all ≥2.But wait, z was set to 3, so the solutions are (3,7,3) and (7,3,3). However, note that these solutions might be permutations of each other. But in this case, since z is fixed as 3, but x and y can be swapped. So (3,7,3) and (7,3,3) are distinct if we consider order. So depending on whether the problem considers ordered triples or unordered, but the problem says "the number of all positive integer solutions (x,y,z)", so ordered triples.So these are distinct solutions. So when z=3, we have two solutions: (3,7,3) and (7,3,3). Wait, but z is fixed as 3 here. Wait, no. Wait, when we set z=3, we solve for x and y, so x and y can vary, but z is 3. So for z=3, the solutions are (3,7,3) and (7,3,3). Similarly, if we set x=3, then y and z would be variables, leading to other solutions. But since the equation is symmetric, all permutations are allowed.But let's stick to fixing one variable at a time. Let's check for z=4.Set z=4:Equation becomes:4xy -xy -4y -4x -x -y -4 +1 =0Compute step by step:4xy -xy =3xy-4y -y =-5y-4x -x =-5x-4 +1 =-3So equation:3xy -5x -5y -3 =0This seems a bit more complicated. Let me try rearranging.3xy -5x -5y =3Hmm, maybe add 25/3 to both sides to complete the rectangle, but since we are dealing with integers, perhaps another approach.Multiply both sides by 3 to eliminate the coefficient of xy:9xy -15x -15y =9Now, add 25 to both sides:9xy -15x -15y +25 =34Hmm, can we factor this?Looking at 9xy -15x -15y +25. Let me check:Maybe factor as (3x -5)(3y -5) ?Compute (3x -5)(3y -5) =9xy -15x -15y +25. Yes, exactly.So:(3x -5)(3y -5) =34Now, 34 factors into 1*34, 2*17. Since 34 is positive, and x,y ≥2, so 3x -5 and 3y -5 must be positive integers.Thus:3x -5 ≥1 (since x ≥2: 3*2 -5=1), so possible.Factor pairs of 34: (1,34),(2,17),(17,2),(34,1)Thus,Case 1:3x -5=1 => 3x=6 =>x=23y -5=34 =>3y=39 =>y=13Case 2:3x -5=2 =>3x=7 =>x=7/3 Not integer. Reject.Case3:3x -5=17 =>3x=22 =>x=22/3 Not integer.Case4:3x -5=34 =>3x=39 =>x=133y -5=1 =>3y=6 =>y=2Thus, the solutions are (x,y)=(2,13) and (13,2). Therefore, when z=4, the solutions are (2,13,4) and (13,2,4).But wait, x and y have to be ≥2, which they are.So those are two more solutions. So for z=4, we get two solutions. Similarly, if we fix z=5, maybe we can find solutions.But this process might take a while. Let me see if there's a pattern here.For z=2: 4 solutionsFor z=3: 2 solutionsFor z=4: 2 solutionsWait, maybe the number of solutions decreases as z increases. Let's check z=5.Set z=5:Equation becomes:5xy -xy -5y -5x -x -y -5 +1=0Compute step by step:5xy -xy=4xy-5y -y= -6y-5x -x= -6x-5 +1= -4Thus, equation:4xy -6x -6y -4=0Divide both sides by 2:2xy -3x -3y -2=0Rearranged:2xy -3x -3y=2Hmm, let's try to manipulate this equation.Add 9/2 to both sides to complete the rectangle, but since we have integer coefficients, perhaps another approach.Multiply both sides by 2:4xy -6x -6y =4Add 9 to both sides:4xy -6x -6y +9 =13Factor:(2x -3)(2y -3)=13Since 13 is prime, factor pairs are (1,13),(13,1)Thus,Case1:2x -3=1 =>2x=4 =>x=22y -3=13 =>2y=16 =>y=8Case2:2x -3=13 =>2x=16 =>x=82y -3=1 =>2y=4 =>y=2Thus, solutions (x,y)=(2,8) and (8,2). Therefore, when z=5, the solutions are (2,8,5) and (8,2,5).So two solutions again. Hmm.Similarly, proceeding to z=6:Set z=6:Equation becomes:6xy -xy -6y -6x -x -y -6 +1=0Compute:6xy -xy=5xy-6y -y= -7y-6x -x= -7x-6 +1= -5Thus, equation:5xy -7x -7y -5=0This seems a bit more complex. Let's try rearranging.5xy -7x -7y =5Perhaps add 49/5 to both sides, but since we need integers, let's see:Multiply both sides by 5:25xy -35x -35y =25Add 49 to both sides:25xy -35x -35y +49 =74Hmm, can we factor this? Let's see.Looking for factors in the form (ax + b)(ay + c). Not straightforward.Alternatively, write equation as:5xy -7x -7y =5Let me rearrange:5xy -7x -7y =5Bring all terms to left:5xy -7x -7y -5=0Maybe factor by grouping.Group terms:(5xy -7x) - (7y +5)=0x(5y -7) - (7y +5)=0Hmm, not helpful. Alternatively, isolate x:5xy -7x =7y +5x(5y -7)=7y +5Thus,x=(7y +5)/(5y -7)Since x must be a positive integer, (7y +5) must be divisible by (5y -7). Let's denote d=5y -7.Then, x=(7y +5)/d. But d=5y -7, so:Express 7y +5 in terms of d:d=5y -7 => y=(d +7)/5Thus,7y +5=7*(d +7)/5 +5= (7d +49)/5 +25/5= (7d +74)/5Therefore,x=(7d +74)/5 divided by d= (7d +74)/(5d)This must be an integer. So (7d +74) must be divisible by 5d. Let me write:7d +74 ≡0 mod 5dWhich implies:7d +74 ≡0 mod5d =>7d ≡-74 mod5dBut 5d divides 7d +74, so 5d divides 7d +74. Therefore, 5d | (7d +74). Therefore, 5d |74. Because 5d divides 7d +74 -1*(5d)=2d +74. Then 5d divides 2d +74. Then 5d divides 2d +74 -k*(5d) where k=0: 5d divides 2d +74. Then 5d divides 74 -3d. Continuing this is getting complicated.Alternatively, since 5d divides 74 -3d, then 5d ≤ |74 -3d|. But since d=5y -7 and y ≥2, so d ≥5*2 -7=3. So d ≥3.Let me try small d values starting from 3:d=3: 5*3=15 divides 74 -3*3=74 -9=65? 65/15=4.333… No.d=4:5*4=20 divides 74 -12=62? 62/20=3.1 No.d=5:25 divides 74 -15=59? No.d=6:30 divides 74 -18=56? 56/30≈1.866 No.d=7:35 divides 74 -21=53? No.d=8:40 divides 74 -24=50? 50/40=1.25 No.d=9:45 divides 74 -27=47? No.d=10:50 divides 74 -30=44? 44/50=0.88 No.d=11:55 divides 74 -33=41? No.d=12:60 divides 74 -36=38? No.d=13:65 divides 74 -39=35? 35/65=0.538… No.d=14:70 divides 74 -42=32? No.d=15:75 divides 74 -45=29? No.d=16:80 divides 74 -48=26? No.d=17:85 divides 74 -51=23? No.d=18:90 divides 74 -54=20? 20/90≈0.222… No.d=19:95 divides 74 -57=17? No.d=20:100 divides 74 -60=14? No.Continuing further, d increases but 74 -3d decreases. So no solution here. Therefore, there are no solutions when z=6.Hmm. So maybe z=2,3,4,5 gives solutions, but z=6 gives none. Let's check z=7.Set z=7:Equation becomes:7xy -xy -7y -7x -x -y -7 +1=0Compute step by step:7xy -xy=6xy-7y -y= -8y-7x -x= -8x-7 +1= -6Thus, equation:6xy -8x -8y -6=0Divide both sides by 2:3xy -4x -4y -3=0Rearranged:3xy -4x -4y=3Let me try to isolate variables. Let's isolate x:3xy -4x =4y +3x(3y -4)=4y +3Thus,x=(4y +3)/(3y -4)x must be a positive integer. So (4y +3) must be divisible by (3y -4). Let me denote d=3y -4, so y=(d +4)/3.Then,4y +3=4*(d +4)/3 +3= (4d +16)/3 +9/3= (4d +25)/3Thus,x=(4d +25)/(3d)Since x must be an integer, (4d +25) must be divisible by 3d. So 3d divides 4d +25. Therefore, 3d divides 25. Since d=3y -4 and y ≥2, d ≥3*2 -4=2. So d ≥2.Possible divisors of 25 are 1,5,25. Since 3d divides 25, and 3d ≥6 (since d ≥2), but 25 is not divisible by 6, 15, or 75. Thus, no solutions here.Therefore, no solutions when z=7.Similarly, as z increases, the equation becomes more restrictive, so likely no solutions for z ≥6.Now, let's check z=2,3,4,5 give us solutions. For z=2: 4 solutions, z=3:2 solutions, z=4:2 solutions, z=5:2 solutions. So total solutions when we fix z=2,3,4,5: 4+2+2+2=10 solutions. But considering the symmetry, we need to check if these are unique when considering all permutations.Wait, for example, when z=2, we have solutions (4,13,2), (5,8,2), (8,5,2), (13,4,2). Similarly, if we fix x=2, we would get solutions where x=2 and the other variables are permuted. But since the equation is symmetric, all permutations of these solutions are also solutions.Therefore, in order to count all solutions, we need to consider all permutations of the triples found when fixing each variable.However, in the above approach, by fixing z=2,3,4,5, and solving for x and y, we might miss solutions where all three variables are different, or where two variables are equal.Wait, but when we fix z, we allow x and y to vary. However, if there are solutions where all three variables are different, they should be captured when permuting the variables. However, when we fix z=2,3,4,5, we get solutions where z is fixed, but x and y can take different values, possibly equal to each other or different.Wait, but in our previous solutions, when z=3, we got (3,7,3) and (7,3,3). So variables can be equal. Let's check if there are solutions with two variables equal.Wait, for example, let’s suppose x=y. Let me check if such solutions exist.Let’s set x=y. Then the original equation becomes 2 arctan(1/x) + arctan(1/z) = π/4.Using the tangent formula again:tan(2 arctan(1/x) + arctan(1/z)) = tan(π/4)=1First compute tan(2 arctan(1/x)):tan(2θ) = 2 tanθ / (1 - tan²θ). So here θ=arctan(1/x), so tanθ=1/x. Thus,tan(2θ) = 2*(1/x)/(1 - 1/x²) = (2/x)/( (x² -1)/x² ) = 2x/(x² -1)Then, tan(2θ + arctan(1/z)) = [ tan(2θ) + tan(arctan(1/z)) ] / [1 - tan(2θ) tan(arctan(1/z)) ] = [ (2x/(x² -1) ) + (1/z) ] / [1 - (2x/(x² -1))(1/z) ]Set this equal to 1:[ (2x/(x² -1) + 1/z ) ] = [1 - (2x/(x² -1))(1/z) ]Multiply both sides by denominator:2x/(x² -1) + 1/z =1 - (2x)/(z(x² -1))Bring all terms to left:2x/(x² -1) +1/z + (2x)/(z(x² -1)) -1=0Combine terms:[2x/(x² -1) + 2x/(z(x² -1))] + (1/z -1)=0Factor 2x/(x² -1):2x/(x² -1) [1 + 1/z] + (1/z -1)=0Let me multiply through by z(x² -1) to eliminate denominators:2xz [1 +1/z] + (1/z -1) z(x² -1)=0Simplify:2xz + 2x + (1 - z)(x² -1)=0Expand (1 - z)(x² -1):(1)(x² -1) - z(x² -1)=x² -1 - z x² + zThus, equation becomes:2xz +2x +x² -1 -z x² + z =0Rearranged:x² - z x² +2xz +2x + z -1=0Factor x² terms:x²(1 - z) +2xz +2x + z -1=0Hmm, this seems complicated, but let's try plugging in specific values. Suppose x=2, then:(4)(1 - z) +4z +4 + z -1=0Simplify:4 -4z +4z +4 +z -1=0Combine terms:(4 +4 -1) + (-4z +4z +z)=0 =>7 +z=0 =>z=-7, which is invalid. So no solution for x=2.Try x=3:9(1 -z) +6z +6 + z -1=09 -9z +6z +6 +z -1=0Combine like terms:(9 +6 -1) + (-9z +6z +z)=14 -2z=0 =>2z=14 =>z=7Thus, when x=y=3, z=7. Check if this is a solution.Original equation: arctan(1/3) + arctan(1/3) + arctan(1/7) = ?Compute arctan(1/3) ≈0.32175 radians. Two of them sum to ≈0.6435 radians. Then adding arctan(1/7)≈0.1429 radians. Total≈0.7864 radians≈45.02 degrees. Which is approximately π/4≈0.7854 radians. Close, but not exact. Wait, but maybe exact?Wait, perhaps we can check if the tangent is exactly 1.Using the formula:tan(2 arctan(1/3) + arctan(1/7)) =1?Earlier computation when x=3, z=7:tan(2θ + φ)=1, where θ=arctan(1/3), φ=arctan(1/7).Compute tan(2θ)=2*(1/3)/(1 -1/9)= (2/3)/(8/9)= (2/3)*(9/8)= 3/4.Then tan(2θ + φ)= (3/4 +1/7)/(1 - (3/4)(1/7))= (21/28 +4/28)/(1 -3/28)= (25/28)/(25/28)=1. So yes, exactly 1. Thus, 2 arctan(1/3) + arctan(1/7)=π/4. Therefore, (3,3,7) is a solution, as well as permutations.But when we set x=y=3, we get z=7. So the solution (3,3,7). However, when we fixed z=3 earlier, we found solutions (3,7,3) and (7,3,3). But (3,3,7) is a different permutation. So we need to check if this solution is already accounted for.Wait, when z=7, did we check for solutions? Earlier, when z=7, we tried solving and found no solutions. Because when z=7, the equation led to no integer solutions. But here, by setting x=y, we found a solution (3,3,7). So perhaps there are solutions where two variables are equal, which were not captured by fixing z=2,3,4,5.Therefore, our initial approach missed these solutions because we fixed z and solved for x and y, but if x=y, then unless we consider that case separately, we might miss those solutions.Therefore, we need a different approach. Perhaps instead of fixing one variable, we should consider all possible cases where variables are ordered or not.Alternatively, we can proceed to find all solutions by considering the equation:xyz - (xy + yz + zx) - (x + y + z) +1 =0and look for all positive integers x ≤ y ≤ z (to avoid permutations), then multiply by the number of permutations.But since the original problem asks for all positive integer solutions (x,y,z), considering order, we need to count all permutations.But to ensure we don't miss any solutions, let me try to list all possible solutions found so far:From z=2:(4,13,2), (5,8,2), (8,5,2), (13,4,2)From z=3:(3,7,3), (7,3,3)From z=4:(2,13,4), (13,2,4)From z=5:(2,8,5), (8,2,5)From x=y=3:(3,3,7) and permutations.Wait, so (3,3,7) is a solution. Similarly, permutations would be (3,7,3) and (7,3,3), which we already found when z=3. So actually, the solution (3,7,3) and (7,3,3) are permutations of (3,3,7). Therefore, these are not new solutions.But wait, when z=3, we found (3,7,3) and (7,3,3). But (3,3,7) has two 3s and one 7, so it's a different permutation. However, when we set z=3 and found (3,7,3), which has two 3s and one 7. So actually, (3,7,3) is same as (3,3,7) up to ordering. Wait, no. (3,7,3) has two 3s and one 7, regardless of order. So if we consider all permutations, then (3,3,7) can be permuted in 3 ways: (3,3,7), (3,7,3), (7,3,3). So those three permutations.But earlier, when we fixed z=3, we only found two solutions: (3,7,3) and (7,3,3). Wait, why didn't we get (3,3,7) when z=3?Because when we fixed z=3, we solved for x and y, but in that case, x and y were 3 and 7. So if x=3, y=7, then the solution is (3,7,3). If x=7, y=3, then (7,3,3). But the case where x=3 and y=3 wasn't captured because when we set z=3 and tried to solve for x and y, we had (x-3)(y-3)=10, which requires x and y to be 4 and 13, etc. Wait, no, when z=3, we had:For z=3, we set z=3 and found the equation led to (x -2)(y -2)=5, giving x=3, y=7 and x=7, y=3. So the solutions when z=3 are (3,7,3) and (7,3,3). There is no solution with x=y=3 when z=3 because plugging x=y=3 into the equation for z=3 would mean:From z=3 equation: xy -2x -2y -1 =0. If x=y=3, then 9 -6 -6 -1= -4≠0. So no solution. Therefore, the solution (3,3,7) comes from a different case where x=y and z=7.So when x=y=3, z=7. Let's verify:Plug into the equation:3*3*7 - (3*3 +3*7 +3*7) - (3 +3 +7) +1= 63 - (9 +21 +21) -13 +1=63 -51 -13 +1=0. Yes, 63-51=12, 12-13=-1, -1+1=0. So it works.So (3,3,7) is a valid solution. Similarly, permutations of this are (3,7,3), (7,3,3). So these are three distinct ordered triples.But when we set z=3, we found (3,7,3) and (7,3,3). So two of the permutations. But (3,3,7) is another permutation. How come we didn't find that?Because when z=3, we set z=3 and solved for x and y, but x and y couldn't both be 3. So this solution exists when z=7 instead.Therefore, to capture all solutions, we need to consider cases where two variables are equal, which might not be found by fixing a single variable.Therefore, our initial approach misses some solutions. Hence, we need a better strategy.Alternative approach: To find all solutions (x, y, z) with x ≤ y ≤ z, and then count all permutations.First, find all triples with x ≤ y ≤ z.We have the equation:xyz - xy - yz - zx - x - y - z +1 =0Let me attempt small values for x.Start with x=2:Then the equation becomes:2yz - 2y -2z - yz -2 - y - z +1=0Wait, no. Wait, original equation:xyz - xy - yz - zx - x - y - z +1 =0If x=2:2yz -2y - yz -2z -2 - y - z +1=0Simplify:(2yz - yz) + (-2y - y) + (-2z - z) + (-2 +1)=0Thus:yz -3y -3z -1=0Rearranged:yz -3y -3z =1Add 9 to both sides:yz -3y -3z +9=10Factor:(y -3)(z -3)=10So since x=2 and y ≥x=2, but we assumed x ≤ y ≤ z, so y ≥2, z ≥ y.Thus, (y -3)(z -3)=10. Since y ≥2, y -3 can be negative, zero, or positive. However, the product is 10, which is positive. Therefore, both factors must be positive. Hence, y -3 ≥1 and z -3 ≥1. Thus, y ≥4, z ≥4.Factor pairs of 10: (1,10),(2,5),(5,2),(10,1). But since y ≤ z, then y -3 ≤ z -3. So the valid factor pairs are (1,10),(2,5).Thus:1. y -3=1, z -3=10 => y=4, z=132. y -3=2, z -3=5 => y=5, z=8Therefore, when x=2, we have two triples: (2,4,13) and (2,5,8). But wait, but in our earlier approach when z=2, we found (4,13,2) etc. But here, we're considering x=2, so ordered triples (2,4,13) and (2,5,8). Then, considering x ≤ y ≤ z, we can permute these, but since we fixed x=2 and sorted y and z, these are the minimal ordered triples. Then permutations would be all rearrangements.Next, x=3:Assume x=3, and x ≤ y ≤ z. Substitute x=3 into the equation:3yz -3y - yz -3z -3 - y - z +1=0Simplify:(3yz - yz) + (-3y - y) + (-3z - z) + (-3 +1)=0Thus:2yz -4y -4z -2=0Divide by 2:yz -2y -2z -1=0Rearranged:yz -2y -2z =1Add 4 to both sides:yz -2y -2z +4=5Factor:(y -2)(z -2)=5Factor pairs of 5: (1,5),(5,1). Since y ≤ z, (1,5). Thus:y -2=1 => y=3z -2=5 => z=7Therefore, the solution is (3,3,7). Since y ≥x=3, y=3, z=7.Next, check x=4:x=4, so equation:4yz -4y - yz -4z -4 - y - z +1=0Simplify:(4yz - yz) + (-4y - y) + (-4z - z) + (-4 +1)=0Thus:3yz -5y -5z -3=0Rearranged:3yz -5y -5z =3This seems familiar. Let's try solving for y and z.But with x=4 and y ≥4, z ≥y.Let me rearrange the equation:3yz -5y -5z =3Let me add 25/3 to both sides to complete the product:3yz -5y -5z +25/3 =3 +25/3= 34/3But this is not helpful for integers. Alternatively, isolate y:3yz -5y =5z +3y(3z -5)=5z +3Thus,y=(5z +3)/(3z -5)y must be an integer ≥4. So:(5z +3) must be divisible by (3z -5). Let’s compute:Let’s denote d=3z -5, so z=(d +5)/3Then,5z +3=5*(d +5)/3 +3= (5d +25)/3 +9/3= (5d +34)/3Thus,y=(5d +34)/(3d)This must be integer. So 3d divides 5d +34. Thus, 3d divides 34. Since z ≥ y ≥4, d=3z -5 ≥3*4 -5=7. So d ≥7. Factors of 34 are 1,2,17,34. Since d ≥7, possible d=17,34.Check d=17:Then z=(17 +5)/3=22/3≈7.333… Not integer.d=34:z=(34 +5)/3=39/3=13. Thus, z=13.Then y=(5*34 +34)/ (3*34)= (170 +34)/102=204/102=2. But y=2 is less than x=4. Contradicts y ≥x=4. Therefore, no solutions when x=4.Similarly, x=5:Equation becomes:5yz -5y -yz -5z -5 -y -z +1=0Simplify:(5yz - yz) + (-5y - y) + (-5z - z) + (-5 +1)=0Thus:4yz -6y -6z -4=0Divide by 2:2yz -3y -3z -2=0Rearranged:2yz -3y -3z =2Add 9/2 to both sides: but not helpful. Alternatively, isolate y:2yz -3y =3z +2y(2z -3)=3z +2Thus,y=(3z +2)/(2z -3)y must be integer ≥5. Let's check for z ≥ y ≥5.Let’s denote d=2z -3, so z=(d +3)/2Then,3z +2=3*(d +3)/2 +2= (3d +9)/2 +4/2= (3d +13)/2Thus,y=(3d +13)/(2d)Must be integer. So 2d divides 3d +13. Thus, 2d divides 13. Since z ≥ y ≥5, z=(d +3)/2 ≥5 => d +3 ≥10 =>d ≥7. 13 is prime, factors are 1 and13. Since d ≥7, d=13:Check d=13:Then z=(13 +3)/2=16/2=8y=(3*13 +13)/(2*13)= (39 +13)/26=52/26=2. But y=2 <5. Not valid. Hence no solution.Thus, x=5 gives no solutions.x=6:Equation becomes:6yz -6y -yz -6z -6 -y -z +1=0Simplify:5yz -7y -7z -5=0This is similar to the z=6 case earlier. It might not have solutions. Let me check.5yz -7y -7z =5Isolate y:5yz -7y =7z +5y(5z -7)=7z +5Thus,y=(7z +5)/(5z -7)Must be integer ≥6. Let’s check:Let d=5z -7, so z=(d +7)/5Then,7z +5=7*(d +7)/5 +5= (7d +49)/5 +25/5= (7d +74)/5Thus,y=(7d +74)/(5d)Must be integer. So 5d divides7d +74. Thus, 5d divides74. Since z ≥ y ≥6,z=(d +7)/5 ≥6 =>d +7 ≥30 =>d ≥23.Factors of74 are1,2,37,74. Since d ≥23, possible d=37,74.Check d=37:z=(37 +7)/5=44/5=8.8, not integer.d=74:z=(74 +7)/5=81/5=16.2, not integer.Thus, no solution.Therefore, x=6 gives no solutions. Similarly, higher x would give larger terms, likely no solutions.Thus, the minimal triples with x ≤ y ≤ z are:(2,4,13), (2,5,8), (3,3,7)Now, let's check if there are any more solutions.Wait, let's check x=3, y=4.But wait, according to our previous step-by-step approach, x=3,y=4 would require solving the equation, but since we already found all solutions with x ≤ y ≤ z, maybe these are all.Therefore, the minimal solutions are:1. (2,4,13)2. (2,5,8)3. (3,3,7)Now, we need to generate all permutations of these triples to get all ordered solutions.For each minimal triple, count the number of distinct permutations:1. (2,4,13): All elements are distinct. Number of permutations: 3! =62. (2,5,8): All elements distinct. Permutations:63. (3,3,7): Two elements same, one different. Permutations:3 (positions where 7 can be)Thus, total solutions:6+6+3=15.But wait, let's check the solutions we found earlier:When z=2: (4,13,2), (5,8,2), (8,5,2), (13,4,2) → 4 solutions, which correspond to permutations of (2,4,13) and (2,5,8). Indeed, (4,13,2) and (13,4,2) are permutations of (2,4,13). Similarly, (5,8,2) and (8,5,2) are permutations of (2,5,8). So each of these contributes 2 permutations, totaling 4 solutions.Similarly, when z=3: (3,7,3) and (7,3,3) → 2 solutions, which are permutations of (3,3,7). The third permutation is (3,3,7) itself, which wasn't found when we fixed z=3, because in that case, we fixed z=3 and solved for x and y, but (3,3,7) has z=7.When z=4: (2,13,4), (13,2,4) → 2 permutations of (2,4,13)When z=5: (2,8,5), (8,2,5) → 2 permutations of (2,5,8)Additionally, the solution (3,3,7) and its permutations (3,7,3), (7,3,3). So total solutions:From (2,4,13): 6 permutationsFrom (2,5,8):6 permutationsFrom (3,3,7):3 permutationsTotal:6+6+3=15 solutions.But wait, earlier when fixing z=2,3,4,5, we found:z=2:4 solutionsz=3:2 solutionsz=4:2 solutionsz=5:2 solutionsTotal:10 solutionsBut according to the permutation count, there should be 15 solutions. So discrepancy here. The missing solutions are the permutations where the smallest element is not fixed as z. For example, (4,2,13), (13,2,4), etc.But when we fixed z=2, we found all solutions where z=2, which includes permutations like (4,13,2), but when we set x=2 and found (2,4,13), which is a different permutation.Wait, but in our initial approach, by fixing z=2,3,4,5, we found 10 solutions. But according to the permutation count, there should be 15. Thus, where are the remaining 5 solutions?Ah, because when we fixed z=2,3,4,5, we only found solutions where the largest or other variable is fixed, but permutations where the fixed variable is in a different position are not captured.Wait, no. Because the equation is symmetric, all permutations are considered. For example, solution (4,13,2) is considered when z=2, but (4,2,13) would have z=13, which we didn't check because z=13 is too large. But in reality, if we fix z=13, would we find solutions?Wait, let's check. Suppose z=13:Equation becomes:13xy -xy -13y -13x -x -y -13 +1=0Simplify:12xy -14x -14y -12=0Divide by 2:6xy -7x -7y -6=0Rearranged:6xy -7x -7y =6Isolate x:6xy -7x =7y +6x(6y -7)=7y +6Thus,x=(7y +6)/(6y -7)Must be integer ≥2.Let’s denote d=6y -7, so y=(d +7)/6Then,7y +6=7*(d +7)/6 +6= (7d +49)/6 +36/6= (7d +85)/6Thus,x=(7d +85)/(6d)Must be integer. So 6d divides7d +85. Therefore,6d divides85. Since y≥2, d=6y -7≥6*2 -7=5. Factors of85 are1,5,17,85. Since d≥5, possible d=5,17,85.Check d=5:y=(5 +7)/6=12/6=2x=(7*5 +85)/(6*5)= (35 +85)/30=120/30=4Thus, solution (4,2,13). But since we fixed z=13, the solution is (4,2,13). Which is a permutation of (2,4,13). So indeed, this solution exists when z=13. However, in our initial approach, we didn't check z=13 because it's large, but by symmetry, it should be considered.Similarly, other permutations would be found when fixing other z values. However, manually checking all z up to 13 is impractical.Therefore, the correct total number of solutions is 15, accounting for all permutations of the minimal triples (2,4,13), (2,5,8), (3,3,7). Hence, the number of positive integer solutions is 15.But to confirm, let's list all permutations:For (2,4,13):All permutations are the 6 permutations of three distinct numbers:(2,4,13), (2,13,4), (4,2,13), (4,13,2), (13,2,4), (13,4,2)For (2,5,8):Similarly, 6 permutations:(2,5,8), (2,8,5), (5,2,8), (5,8,2), (8,2,5), (8,5,2)For (3,3,7):3 permutations:(3,3,7), (3,7,3), (7,3,3)Total:6+6+3=15 solutions.Therefore, the total number of positive integer solutions is 15.Final AnswerThe number of all positive integer solutions is boxed{15}.</think>To find the number of all positive integer solutions ((x, y, z)) to the equation (arctan frac{1}{x} + arctan frac{1}{y} + arctan frac{1}{z} = frac{pi}{4}), we start by converting the equation using the tangent function:[tanleft(arctan frac{1}{x} + arctan frac{1}{y} + arctan frac{1}{z}right) = tanleft(frac{pi}{4}right) = 1]Using the tangent addition formula for three angles, we derive the equation:[frac{frac{1}{x} + frac{1}{y} + frac{1}{z} - frac{1}{xyz}}{1 - left(frac{1}{xy} + frac{1}{xz} + frac{1}{yz}right)} = 1]Simplifying this, we get the diophantine equation:[xyz - xy - yz - zx - x - y - z + 1 = 0]We solve this equation by considering possible values for (x, y, z) and checking permutations. We find the following minimal solutions with (x leq y leq z):1. ((2, 4, 13))2. ((2, 5, 8))3. ((3, 3, 7))Next, we count all permutations of these solutions:- For ((2, 4, 13)) and ((2, 5, 8)), each with 3 distinct elements, there are (3! = 6) permutations each.- For ((3, 3, 7)) with two identical elements, there are (3) permutations.Summing these, we get:[6 + 6 + 3 = 15]Thus, the number of all positive integer solutions is (boxed{15}).,<think>Okay, so I need to find all positive integer solutions (x, y, z) to the equation arctan(1/x) + arctan(1/y) + arctan(1/z) = π/4. Hmm, arctangent functions... I remember there's a formula for adding arctans. Let me recall. If I have arctan(a) + arctan(b), it's equal to arctan((a + b)/(1 - ab)) plus some multiple of π depending on the quadrant. But since all the terms here are positive and we're adding three of them to get π/4, which is 45 degrees, each individual arctan must be less than π/4, right? Because arctan(1/x) is positive and if any of them were greater than π/4, the sum would exceed π/4. So, that means each 1/x must be less than 1, so x, y, z are all greater than 1? Wait, 1/x is less than 1 when x > 1. So x, y, z must be integers greater than or equal to 2? Wait, no, 1/1 is 1, arctan(1) is π/4. So if x, y, z are 1, then arctan(1) is π/4. But if we have three terms adding to π/4, each term must be less than π/4, so each 1/x must be less than 1. Therefore, x, y, z must be at least 2. Because 1/x < 1 implies x > 1, so x ≥ 2. So all variables are integers starting from 2. Okay, so maybe the variables start at 2. Now, how do I handle the addition of three arctans? I know the formula for two arctans, but three... Maybe I can apply the formula step by step. Let me try to first combine two of them and then combine the result with the third one. Suppose I let A = arctan(1/x) + arctan(1/y). Then, using the addition formula, A = arctan((1/x + 1/y)/(1 - 1/(xy))) = arctan((x + y)/(xy - 1)). Then, adding arctan(1/z) to A should give π/4. So:arctan((x + y)/(xy - 1)) + arctan(1/z) = π/4.Using the addition formula again, this sum equals arctan( [( (x + y)/(xy - 1) + 1/z ) / (1 - (x + y)/(xy - 1) * 1/z) ]) = π/4.But since the sum is π/4, which is a specific value, the argument of arctan must be 1, because arctan(1) = π/4. However, wait, arctan(a) + arctan(b) = π/4 only if their sum gives π/4, which might not necessarily require the argument to be 1. Let me check that.Actually, if arctan(α) + arctan(β) = π/4, then we can use the formula:tan(arctan(α) + arctan(β)) = (α + β)/(1 - αβ) = tan(π/4) = 1.So, (α + β)/(1 - αβ) = 1. Therefore, α + β = 1 - αβ. Which leads to α + β + αβ = 1. So, (α + β + αβ) = 1.Therefore, in our case, when we combine the first two arctans:Let’s denote α = (x + y)/(xy - 1) and β = 1/z. Then, according to the formula, we must have:α + β + αβ = 1.So, substituting α and β:[(x + y)/(xy - 1)] + (1/z) + [(x + y)/(xy - 1) * (1/z)] = 1.Let me write that out:(x + y)/(xy - 1) + 1/z + (x + y)/(z(xy - 1)) = 1.Combine the terms:First, factor out (x + y)/(xy - 1):[(x + y)/(xy - 1)]*(1 + 1/z) + 1/z = 1.Let’s compute (1 + 1/z) = (z + 1)/z. So,[(x + y)(z + 1)/ (z(xy - 1))] + 1/z = 1.Combine the two fractions over z(xy - 1):[(x + y)(z + 1) + (xy - 1)] / [z(xy - 1)] = 1.Multiply numerator and denominator:Numerator: (x + y)(z + 1) + (xy - 1)Denominator: z(xy - 1)Set equal to 1:(x + y)(z + 1) + (xy - 1) = z(xy - 1)Let me expand the numerator:First, expand (x + y)(z + 1):xz + x + yz + yThen add (xy - 1):xz + x + yz + y + xy - 1So, numerator becomes:xz + x + yz + y + xy - 1Denominator is z(xy - 1). So the equation is:xz + x + yz + y + xy - 1 = z(xy - 1)Expand the right-hand side:zxy - zSo, equation becomes:xz + x + yz + y + xy - 1 = xyz - zBring all terms to the left-hand side:xz + x + yz + y + xy - 1 - xyz + z = 0Let’s rearrange terms:-xyz + xz + yz + z + x + y + xy - 1 = 0Factor terms where possible:Let’s see:Looking at terms with xyz: -xyzTerms with xz: xzTerms with yz: yzTerms with z: zTerms with x: xTerms with y: yTerms with xy: xyConstant term: -1Hmm, this seems complicated. Maybe factor terms step by step.Let me group terms:-xyz + xz + yz + z = z(-xy + x + y + 1)Then, the remaining terms: x + y + xy -1So, total equation:z(-xy + x + y + 1) + (x + y + xy -1) = 0Factor (x + y + xy -1) can be written as (x + y + xy) -1.Wait, maybe factor terms:Let me factor out (x + y + xy -1):But the first term is z(-xy + x + y + 1) + (x + y + xy -1) = 0Notice that -xy + x + y + 1 = -(xy - x - y -1) = not sure. Maybe factor:Wait, -xy + x + y +1 = x(1 - y) + y + 1. Hmm, not helpful.Alternatively, perhaps notice that:x + y + xy -1 = (x + 1)(y + 1) - 2.Wait, (x + 1)(y + 1) = xy + x + y +1, so subtract 2: xy +x + y +1 -2 = xy +x + y -1. Close but not exactly. Hmm.Alternatively, maybe write x + y + xy -1 as (x +1)(y +1) -2. Let me check:(x +1)(y +1) = xy +x + y +1. So subtract 2: xy +x + y +1 -2 = xy +x + y -1. Yes, exactly. So x + y + xy -1 = (x +1)(y +1) -2.But not sure if that helps. Let's see.So the equation is:z*(-xy +x + y +1) + (x + y + xy -1) =0Let me factor terms:Let’s write it as:z*(x(1 - y) + y +1) + (x + y + xy -1) =0Alternatively, factor x(1 - y) + y +1 = x(1 - y) + (y +1). Hmm.Alternatively, maybe factor:Let’s note that x(1 - y) + (y +1) = x(1 - y) + (y +1) = (1 - y)x + (y +1). Maybe write as (y +1) - x(y -1). Not sure.Alternatively, maybe let’s see if we can factor variables.Alternatively, maybe express this equation in terms of variables.Let me consider variables x, y, z as positive integers ≥2.We need to solve:z*(-xy +x + y +1) + (x + y + xy -1) =0Let me rearrange terms:Bring all terms to the other side:z*(xy - x - y -1) = x + y + xy -1So,z = (x + y + xy -1)/(xy -x - y -1)Hmm, so z must be a positive integer. So the numerator divided by denominator must be integer. Let me denote:Let’s denote N = x + y + xy -1 = xy +x + y -1Denominator D = xy -x - y -1So, z = N/DBut z must be a positive integer, so D must divide N. Also, since z is positive, N and D must have the same sign.Given that x, y ≥2, let's compute D:D = xy -x - y -1. Let's see for x=2, y=2:D = 4 -2 -2 -1 = -1For x=2, y=3:D=6 -2 -3 -1=0For x=3, y=2: same as above.x=2, y=4:D=8 -2 -4 -1=1Hmm, interesting. So D can be negative, zero, or positive depending on x and y. Let's analyze.But since z must be a positive integer, N/D must be positive. So N and D must have the same sign.Case 1: D >0. Then N must be positive. If D>0, then:xy -x - y -1 >0 => xy -x - y >1 => (x-1)(y-1) >2. Since (x-1)(y-1) = xy -x -y +1, so xy -x -y = (x-1)(y-1) -1. So, (x-1)(y-1) -1 >1 => (x-1)(y-1) >2. So (x-1)(y-1) >2.Therefore, x-1 and y-1 are at least 1 (since x,y ≥2), so their product is at least 1*1=1. To be greater than 2, possible when (x-1, y-1) ≥ (2,2) or (3,1), (1,3). But x-1 and y-1 are integers ≥1.Wait, for example, x=3, y=3: (2)(2)=4>2. x=4, y=2: (3)(1)=3>2. x=2, y=4: (1)(3)=3>2. So D>0 when (x-1)(y-1) >2.Case 2: D=0. Then, denominator is zero. So D=0 implies xy -x - y -1=0. Let's solve for integers x, y ≥2:xy -x - y -1=0 => xy -x - y =1 => (x-1)(y-1)=2. Since x-1 and y-1 are positive integers, the possible factorizations of 2 are 1*2 or 2*1. Thus:(x-1, y-1)=(1,2) => (x,y)=(2,3)(x-1, y-1)=(2,1) => (x,y)=(3,2)So D=0 when (x,y)=(2,3) or (3,2). Then, z=N/D would be undefined (division by zero). Therefore, these cases are invalid, as z must be a positive integer. Therefore, these (x,y) pairs are excluded.Case3: D<0. Then, since N/D must be positive, N must be negative. Let's check N:N = xy +x + y -1. Since x,y ≥2, xy ≥4, x≥2, y≥2, so N ≥4 +2 +2 -1=7. So N is always positive. Therefore, if D<0, then N/D is negative, which contradicts z being a positive integer. Therefore, in D<0, no solution.Therefore, only possible case is D>0, i.e., (x-1)(y-1) >2, and z=N/D=(xy +x + y -1)/(xy -x - y -1) must be a positive integer.So now, the problem reduces to finding all pairs (x,y) with x,y ≥2 and (x-1)(y-1) >2, such that (xy +x + y -1) is divisible by (xy -x - y -1), and z=(xy +x + y -1)/(xy -x - y -1) is a positive integer.So we need to find all such x, y, compute z, and ensure z is also a positive integer.This seems challenging. Let's see if we can manipulate the expression for z.Let me write z=(xy +x + y -1)/(xy -x - y -1). Let's try to simplify this fraction.Let me denote numerator N = xy +x + y -1Denominator D = xy -x - y -1Compute N - D = (xy +x + y -1) - (xy -x - y -1) = xy +x + y -1 -xy +x + y +1= 2x + 2y.Therefore, N = D + 2x + 2ySo z = (D + 2x + 2y)/D = 1 + (2x + 2y)/DTherefore, z = 1 + 2(x + y)/DBut D = xy -x - y -1, so z = 1 + 2(x + y)/(xy -x - y -1)Therefore, in order for z to be an integer, (xy -x - y -1) must divide 2(x + y). Let's write this as:xy -x - y -1 | 2(x + y)Which means that D divides 2(x + y). So, D is a positive divisor of 2(x + y). Also, since D = xy -x - y -1, which is positive when (x-1)(y-1) >2, as established before.Therefore, for each x, y ≥2 such that (x-1)(y-1) >2, D divides 2(x + y), and z is given by 1 + 2(x + y)/D.But this is a bit abstract. Maybe we can make substitution for variables. Let me set a = x -1 and b = y -1. Then, since x, y ≥2, a, b ≥1. Then, D = (a +1)(b +1) - (a +1) - (b +1) -1 = ab +a + b +1 -a -1 -b -1 -1 = ab -1 -1 = ab -2. Wait, let's compute:Wait, original D = xy -x - y -1.x = a +1, y = b +1.Therefore, D = (a +1)(b +1) - (a +1) - (b +1) -1 = ab +a +b +1 -a -1 -b -1 -1 = ab +a +b +1 -a -b -2 = ab -1.Wait, let's check:(a +1)(b +1) = ab +a +b +1Subtract (a +1) + (b +1) +1:ab +a +b +1 -a -1 -b -1 -1 = ab -2. So D = ab - 2.So D = ab - 2. Since D must be positive, ab -2 >0 => ab >2. Which aligns with our previous conclusion that (x-1)(y-1) >2 since a = x -1, b = y -1, so ab >2.Also, z = 1 + 2(x + y)/D = 1 + 2((a +1) + (b +1))/(ab -2) = 1 + 2(a + b + 2)/(ab -2)So z = 1 + [2(a + b + 2)] / (ab -2)We need z to be a positive integer. Therefore, (ab -2) must divide 2(a + b + 2).Let me write this as:(ab - 2) divides 2(a + b + 2)So, ab -2 | 2(a + b + 2)Let me denote k = ab -2. Then, k divides 2(a + b + 2). Also, k = ab -2, so ab = k + 2.But since a and b are positive integers with ab >2, k is a positive integer.So, k | 2(a + b + 2). Also, since ab = k + 2, perhaps we can express a + b in terms of k.But this might not be straightforward. Alternatively, perhaps fix a and b and iterate through possible small values to find solutions.Given that a and b are positive integers with ab >2, let's start with small values of a and b and compute z.Let's start with a=1 (since a, b ≥1). If a=1:Then ab -2 = b -2. Since ab >2, and a=1, so b >2. So b ≥3.Then, z =1 + [2(1 + b + 2)]/(1*b -2)=1 + 2(b + 3)/(b -2)Simplify:2(b +3)/(b -2) = 2[(b -2) +5]/(b -2)=2 + 10/(b -2)Therefore, z=1 +2 +10/(b -2)=3 +10/(b -2)So z must be integer, so 10/(b -2) must be integer. Therefore, (b -2) divides 10. Since b ≥3, so b -2 ≥1. The divisors of 10 are 1,2,5,10. Thus, possible b -2=1,2,5,10 => b=3,4,7,12.Therefore, for a=1:b=3: z=3 +10/1=13Check if this works: a=1, b=3. So x=a+1=2, y=b+1=4. Then z=13. Let's check the original equation:arctan(1/2) + arctan(1/4) + arctan(1/13). Let's compute this sum.First, compute arctan(1/2) + arctan(1/4). Using the formula:tan(A + B) = (1/2 + 1/4)/(1 - (1/2)(1/4)) = (3/4)/(1 - 1/8) = (3/4)/(7/8) = 6/7. So arctan(6/7). Then add arctan(1/13). Compute tan(arctan(6/7) + arctan(1/13)) = (6/7 +1/13)/(1 - (6/7)(1/13)) = (78 +7)/ (91 -6) = 85/85 =1. So arctan(1) = π/4. Therefore, yes, this works. So (2,4,13) is a solution.Similarly, for b=4: z=3 +10/2=8Check x=2, y=5, z=8.Compute arctan(1/2) + arctan(1/5). Let's compute tan sum: (1/2 +1/5)/(1 -1/10)= (7/10)/(9/10)=7/9. Then add arctan(1/8). Compute tan(sum + arctan(1/8)) = (7/9 +1/8)/(1 -7/9*1/8)= (56 +9)/ (72 -7)=65/65=1. So again, π/4. So (2,5,8) is a solution.For b=7: z=3 +10/5=5Check x=2, y=9, z=5. Let's compute:arctan(1/2) + arctan(1/9). Tan sum: (1/2 +1/9)/(1 -1/18)= (11/18)/(17/18)=11/17. Then add arctan(1/5). Compute tan(sum + arctan(1/5))=(11/17 +1/5)/(1 -11/17*1/5)=(55 +17)/(85 -11)=72/74=36/37≈0.973. Hmm, arctan(36/37) is close to π/4 but not exactly. Wait, wait, maybe my calculation is wrong.Wait, let's redo:(11/17 +1/5) = (55 + 17)/85 =72/85Denominator: 1 - (11/17)*(1/5)=1 -11/85=74/85So tan= (72/85)/(74/85)=72/74=36/37. So arctan(36/37)≈0.767 radians≈43.8 degrees. But π/4 is 45 degrees, so this is close but not exactly. Wait, this contradicts our previous conclusion. Did we make a mistake?Wait, according to our formula, z=5. So the sum should be π/4, but the calculation shows it's arctan(36/37). Let me check the steps again.Wait, if a=1, b=7, then x=2, y=8. Wait, wait: a=1, b=7: x=a+1=2, y=b+1=8. Therefore, z=5. So the triplet is (2,8,5). Wait, but in my check above, I used y=9 by mistake. Let me correct that.Compute arctan(1/2) + arctan(1/8). Let's compute tan(sum): (1/2 +1/8)/(1 -1/16)= (5/8)/(15/16)= (5/8)*(16/15)= 2/3. Then add arctan(1/5). Compute tan(sum + arctan(1/5))=(2/3 +1/5)/(1 - (2/3)(1/5))=(13/15)/(13/15)=1. So arctan(1)=π/4. So that works. So (2,8,5) is a solution. My mistake earlier was using y=9 instead of y=8. So that's correct.Similarly, b=12: z=3 +10/10=4Check x=2, y=13, z=4. Compute arctan(1/2) + arctan(1/13). Let's compute:(1/2 +1/13)/(1 -1/26)= (15/26)/(25/26)=15/25=3/5. Then add arctan(1/4). Compute (3/5 +1/4)/(1 -3/5*1/4)= (17/20)/(17/20)=1. So arctan(1)=π/4. So (2,13,4) is a solution. So for a=1, we have four solutions: (2,4,13), (2,5,8), (2,8,5), (2,13,4). Note that (2,8,5) and (2,5,8) are different because z is different.Wait, but in the original problem, the variables x, y, z are ordered, so different permutations are considered distinct? The problem says "positive integer solutions (x, y, z)", so ordered triples. So these are all distinct solutions.Now, moving on to a=2:a=2, so b must satisfy ab >2 => 2b >2 => b≥2. Then D=ab -2=2b -2=2(b -1). Then, z=1 + [2(a + b +2)]/(ab -2)=1 + [2(2 + b +2)]/(2b -2)=1 + [2(b +4)]/[2(b -1)]=1 + (b +4)/(b -1)So z=1 + (b +4)/(b -1). Let's simplify:(b +4)/(b -1)= [ (b -1) +5 ] / (b -1)=1 +5/(b -1). So z=1 +1 +5/(b -1)=2 +5/(b -1). Therefore, 5/(b -1) must be integer. Therefore, (b -1) divides 5. Since b≥2, b -1 is a positive integer. The divisors of 5 are 1,5. Thus, b -1=1 => b=2, or b -1=5 => b=6.So for a=2:b=2: z=2 +5/1=7Check x=a+1=3, y=b+1=3, z=7. Compute arctan(1/3) + arctan(1/3) + arctan(1/7). First compute arctan(1/3) + arctan(1/3). Tan sum: (2/3)/(1 -1/9)= (2/3)/(8/9)= (2/3)*(9/8)=3/4. Then add arctan(1/7). Compute tan(sum + arctan(1/7))=(3/4 +1/7)/(1 -3/4*1/7)=(25/28)/(25/28)=1. So arctan(1)=π/4. So (3,3,7) is a solution.b=6: z=2 +5/5=3Check x=3, y=7, z=3. Compute arctan(1/3) + arctan(1/7). Tan sum: (1/3 +1/7)/(1 -1/21)= (10/21)/(20/21)=1/2. Then add arctan(1/3). Compute tan(sum + arctan(1/3))=(1/2 +1/3)/(1 -1/6)= (5/6)/(5/6)=1. So arctan(1)=π/4. Therefore, (3,7,3) is a solution.So for a=2, we get (3,3,7) and (3,7,3). Similarly, note that these are different ordered triples.Next, a=3:D=ab -2=3b -2. So z=1 + [2(a + b +2)]/(3b -2)=1 + [2(3 + b +2)]/(3b -2)=1 + [2(b +5)]/(3b -2)Simplify:z=1 + (2b +10)/(3b -2). Let's see if this can be simplified.Let me perform polynomial division or look for common factors.Let me write numerator as 2b +10 and denominator as 3b -2. Let me see:Express numerator as A*(3b -2) + B:2b +10 = A*(3b -2) + BSolve for A and B:3A b + (-2A) + B =2b +10Equate coefficients:3A=2 => A=2/3-2A + B=10 => -4/3 + B=10 => B=34/3But since we have fractions, this approach may not help. Alternatively, let's see if we can express this as a mixed number.Alternatively, set variable k =3b -2, then express 2b +10 in terms of k.k=3b -2 => b=(k +2)/3So 2b +10=2*(k +2)/3 +10= (2k +4)/3 +10= (2k +4 +30)/3= (2k +34)/3Therefore, (2b +10)/(3b -2)= (2k +34)/3 /k= (2k +34)/(3k)= (2 + 34/k)/3Thus, z=1 + (2 +34/k)/3= (3 +2 +34/k)/3= (5 +34/k)/3. Therefore, 34/k must be integer. So k must be a divisor of 34.But k=3b -2, which is a positive integer. Since a=3 and b≥1, and ab>2. For a=3, b≥1, but ab=3b>2 => b≥1. So k=3b -2 must be positive: 3b -2 >0 => b≥1. Since b≥1, k=3*1 -2=1, 3*2 -2=4, etc.But k must divide 34. The divisors of 34 are 1,2,17,34. So k=1,2,17,34.Therefore:k=1: 3b -2=1 => 3b=3 =>b=1. Then check ab=3*1=3>2, okay. Then z=(5 +34/1)/3=(5 +34)/3=39/3=13. So z=13.Check x=a+1=4, y=b+1=2, z=13. Compute arctan(1/4) + arctan(1/2) + arctan(1/13). We can compute as before, but note that this is a permutation of the previous solution (2,4,13). Since x and y are swapped here (4 and 2), but z remains 13. So (4,2,13) is a valid solution.k=2: 3b -2=2 =>3b=4 =>b=4/3. Not an integer, discard.k=17: 3b -2=17 =>3b=19 =>b=19/3. Not integer, discard.k=34: 3b -2=34 =>3b=36 =>b=12. Then z=(5 +34/34)/3=(5 +1)/3=6/3=2. So z=2. Check x=4, y=13, z=2. Compute arctan(1/4) + arctan(1/13) + arctan(1/2). This is similar to the solution (2,13,4), permuted. So (4,13,2) is a solution.Therefore, for a=3, we get solutions (4,2,13) and (4,13,2). But wait, when a=3, b=1, which gives x=4, y=2, z=13. Similarly, b=12 gives x=4, y=13, z=2. These are valid but permutations of previous solutions.Similarly, for a=4:D=ab -2=4b -2. z=1 + [2(a + b +2)]/(4b -2)=1 + [2(4 + b +2)]/(4b -2)=1 + [2(b +6)]/(4b -2)=1 + (2b +12)/(4b -2)Simplify numerator and denominator:Divide numerator and denominator by 2: 1 + (b +6)/(2b -1)So z=1 + (b +6)/(2b -1). Let's see if this can be simplified.Express (b +6)/(2b -1). Let me perform division:Express numerator as A*(2b -1) + B:b +6 = A*(2b -1) + BSolve for A and B:2A b -A + B = b +6Equate coefficients:2A=1 => A=1/2-A + B=6 => -1/2 + B=6 => B=13/2Not helpful. Alternatively, set k=2b -1, then b=(k +1)/2So numerator=( (k +1)/2 +6 )= (k +1 +12)/2=(k +13)/2Denominator=kThus, (b +6)/(2b -1)= (k +13)/(2k). Therefore, z=1 + (k +13)/(2k)=1 +1/2 +13/(2k)=3/2 +13/(2k). Therefore, z must be integer. So 3/2 +13/(2k) must be integer. Multiply by 2k:3k +13 must be divisible by 2k. Wait, 3k +13 = m*(2k), where m is integer. So 3k +13 =2mk =>13= k(2m -3). So k must divide 13. Since k=2b -1 and b≥1, k≥1 (since b=1 gives k=1). The divisors of 13 are 1 and13.Thus, k=1: 2b -1=1 =>2b=2 =>b=1. Then, z=3/2 +13/(2*1)=3/2 +13/2=16/2=8. So z=8. Check x=a+1=5, y=b+1=2, z=8. Which is (5,2,8). Which is a permutation of previous solution (2,5,8).k=13: 2b -1=13 =>2b=14 =>b=7. Then, z=3/2 +13/(2*13)=3/2 +1/2=2. So z=2. Check x=5, y=8, z=2. Which is (5,8,2), a permutation of (2,8,5).Thus, for a=4, we get solutions (5,2,8) and (5,8,2). These are permutations again.Continuing to a=5:Similarly, this pattern suggests that for higher a, we might get similar permutations. Let's check a=5.D=5b -2. z=1 + [2(5 + b +2)]/(5b -2)=1 + [2(b +7)]/(5b -2)=1 + (2b +14)/(5b -2)Let me see if this can be simplified. Maybe set k=5b -2, so b=(k +2)/5. Then:Numerator=2*( (k +2)/5 ) +14= (2k +4)/5 +14= (2k +4 +70)/5= (2k +74)/5Thus, (2b +14)/(5b -2)= (2k +74)/(5k). Therefore, z=1 + (2k +74)/(5k)=1 + 2/5 +74/(5k)=7/5 +74/(5k). Therefore, z must be integer. Multiply by 5k:7k +74 must be divisible by5k. So 7k +74= m*(5k), integer m. So 7k +74=5mk =>74= k*(5m -7). So k divides74. Since k=5b -2≥5*1 -2=3. The divisors of74 are1,2,37,74. So k=1,2,37,74. But k≥3, so possible k=37,74.k=37: 5b -2=37 =>5b=39 =>b=39/5, not integer.k=74:5b -2=74 =>5b=76 =>b=76/5, not integer. So no solutions for a=5.Thus, for a=5, no solutions.Similarly, a=6:D=6b -2. z=1 + [2(6 + b +2)]/(6b -2)=1 + [2(b +8)]/(6b -2)=1 + (2b +16)/(6b -2)=1 + (b +8)/(3b -1)Let me set k=3b -1, so b=(k +1)/3. Then:Numerator=( (k +1)/3 +8 )= (k +1 +24)/3=(k +25)/3Denominator=kThus, (b +8)/(3b -1)= (k +25)/(3k). Therefore, z=1 + (k +25)/(3k)=1 +1/3 +25/(3k)=4/3 +25/(3k). Multiply by 3k:4k +25 must be divisible by3k. So 4k +25= m*3k =>25= k*(3m -4). So k divides25. Since k=3b -1≥3*1 -1=2. The divisors of25 are1,5,25. So k=5,25.k=5:3b -1=5 =>3b=6 =>b=2. Then z=4/3 +25/(3*5)=4/3 +5/3=9/3=3. Check x=7, y=3, z=3. Compute arctan(1/7)+arctan(1/3)+arctan(1/3). Let's compute:First, arctan(1/3) + arctan(1/3)= sum as before, tan= 2/3/(1 -1/9)= 2/3 /8/9= 3/4. Then add arctan(1/7). Compute tan(sum + arctan(1/7))=(3/4 +1/7)/(1 -3/4*1/7)=(25/28)/(25/28)=1. So arctan(1)=π/4. Therefore, (7,3,3) is a solution. Which is a permutation of (3,3,7).k=25:3b -1=25 =>3b=26 =>b=26/3, not integer. So only one solution here: (7,3,3).Similarly, for a=6, we get (7,3,3).Continuing to a=7:Similarly, this might generate more permutations, but given the trend, it's likely that as a increases, the solutions are permutations of the earlier ones. Let's check for a=7.D=7b -2. z=1 + [2(7 + b +2)]/(7b -2)=1 + [2(b +9)]/(7b -2)=1 + (2b +18)/(7b -2). Set k=7b -2, then b=(k +2)/7. Then numerator=2*(k +2)/7 +18= (2k +4)/7 +18= (2k +4 +126)/7=(2k +130)/7. Therefore:(2b +18)/(7b -2)= (2k +130)/(7k). So z=1 + (2k +130)/(7k)=1 + 2/7 +130/(7k)=9/7 +130/(7k). Multiply by 7k:9k +130 must be divisible by7k. So 9k +130= m*7k =>130= k*(7m -9). Thus, k divides130. k=7b -2≥7*1 -2=5. Divisors of130 ≥5:5,10,13,26,65,130.Check each:k=5:7b -2=5 =>7b=7 =>b=1. Then z=9/7 +130/(7*5)=9/7 +26/7=35/7=5. Check x=8, y=2, z=5. Which is a permutation of (2,8,5).k=10:7b -2=10 =>7b=12 =>b=12/7, not integer.k=13:7b -2=13 =>7b=15 =>b=15/7, not integer.k=26:7b -2=26 =>7b=28 =>b=4. Then z=9/7 +130/(7*26)=9/7 +130/(182)=9/7 +65/91= (117 +65)/91=182/91=2. So z=2. Check x=8, y=5, z=2. Which is a permutation of (2,5,8).k=65:7b -2=65 =>7b=67 =>b=67/7, not integer.k=130:7b -2=130 =>7b=132 =>b=132/7, not integer.So for a=7, solutions are permutations: (8,2,5) and (8,5,2).This pattern shows that for each a, we get solutions where x and y are swapped and z adjusted accordingly, leading to permutations of the earlier solutions. Therefore, the fundamental solutions are the ones found with a=1,2,3, etc., and their permutations.Now, let's summarize all solutions found so far:From a=1 (x=2):(2,4,13), (2,5,8), (2,8,5), (2,13,4)From a=2 (x=3):(3,3,7), (3,7,3)From a=3 (x=4):(4,2,13), (4,13,2)From a=4 (x=5):(5,2,8), (5,8,2)From a=6 (x=7):(7,3,3)From a=7 (x=8):(8,2,5), (8,5,2)Wait, but also when a=5, we didn't find any new solutions, and a=6 gave (7,3,3). Similarly, a=7 gives (8,2,5) etc.Additionally, we need to check if there are solutions where a and b are both greater than certain values, but from the earlier pattern, it seems that all solutions are permutations of these. For example, is there a solution where x=3, y=4, z=something?Wait, let's check (3,4,z). Maybe compute z:Using formula z=(xy +x + y -1)/(xy -x - y -1). For x=3, y=4:Numerator=12 +3 +4 -1=18Denominator=12 -3 -4 -1=4So z=18/4=4.5. Not integer. So invalid.Similarly, x=3, y=5:Numerator=15 +3 +5 -1=22Denominator=15 -3 -5 -1=6z=22/6≈3.666, not integer.x=4, y=4:Numerator=16 +4 +4 -1=23Denominator=16 -4 -4 -1=7z=23/7≈3.285, not integer.x=3, y=6:Numerator=18 +3 +6 -1=26Denominator=18 -3 -6 -1=8z=26/8=3.25, not integer.Thus, these do not produce integer z.Therefore, it seems that the only solutions are the ones found earlier. Now, considering permutations, the variables x, y, z can be arranged in any order. Let's list all permutations of the found solutions:From (2,4,13): permutations are 6, but since 2,4,13 are distinct: 6 permutations. But wait, in our solutions, we only have (2,4,13), (2,13,4), and when a=3, (4,2,13), (4,13,2). So already covering some permutations. Similarly, (2,5,8) and permutations.Wait, the solutions we found include:For (2,4,13):(2,4,13), (2,13,4), (4,2,13), (4,13,2), (13,2,4), (13,4,2). But in our solutions above, we only have (2,4,13), (2,13,4), (4,2,13), (4,13,2). Missing (13,2,4) and (13,4,2). But z in those cases would be x or y, which are larger numbers. However, when a=1, b=3 gives x=2, y=4, z=13. To get z=13 as x or y, we need to see if there are solutions where z=13, and x and y are other numbers.But according to the earlier substitution, when we set a= x -1 and b= y -1, we found solutions for certain a and b, but if we set z=13, would that lead to different a and b?Alternatively, perhaps there's a different approach.Wait, but maybe we have found all solutions where x is the smallest variable. For example, when x=2, we found certain y and z. When x=3, we found y=3,7 etc. But permutations where x is larger might not have been considered. For example, (13,2,4): here x=13, which is larger. Do we have such solutions?To check, let's see if there are solutions where x=13. Using the formula:z=(13*y +13 + y -1)/(13*y -13 - y -1)=(14y +12)/(12y -14)Simplify numerator and denominator:Factor numerator: 2(7y +6)Denominator: 2(6y -7)So z=(7y +6)/(6y -7). For z to be integer, 6y -7 must divide 7y +6. Let’s compute:Let’s compute 7y +6 = k*(6y -7). So 7y +6 =6ky -7k => 7y -6ky = -7k -6 => y(7 -6k)= -7k -6 => y=(7k +6)/(6k -7)We need y to be a positive integer. So 6k -7 divides7k +6, and 6k -7>0 =>k≥2 (k=2:6*2 -7=5>0)Try k=2: y=(14 +6)/(12 -7)=20/5=4. Thus y=4. Then z=(7*4 +6)/(6*4 -7)=34/17=2. So z=2. Thus, triplet (13,4,2), which is indeed a permutation of (2,4,13). Similarly, k=3: y=(21 +6)/(18 -7)=27/11, not integer.k=4: y=(28 +6)/(24 -7)=34/17=2. Then z=(7*2 +6)/(12 -7)=20/5=4. So triplet (13,2,4). Which is another permutation.Therefore, these permutations are valid solutions. However, in our earlier approach by fixing a and b, we might not have captured all permutations, but they are still valid. So all permutations of the found solutions are valid, but we need to count each ordered triplet only once.Wait, but the problem asks for the number of all positive integer solutions (x, y, z). So ordered triples. Therefore, (2,4,13), (2,13,4), (4,2,13), (4,13,2), (13,2,4), (13,4,2) are all distinct solutions.Similarly, for the solution (2,5,8), permutations are 6:(2,5,8), (2,8,5), (5,2,8), (5,8,2), (8,2,5), (8,5,2)Similarly, (3,3,7) has permutations where two variables are 3 and the third is7. There are 3 permutations: (3,3,7), (3,7,3), (7,3,3).So let's count all permutations:For the solutions with all distinct x, y, z:(2,4,13) and its permutations: 6 permutations(2,5,8) and its permutations: 6 permutationsFor the solutions with two variables same:(3,3,7) and its permutations: 3 permutationsBut wait, in our found solutions, we have:From a=1:(2,4,13), (2,5,8), (2,8,5), (2,13,4)From a=2:(3,3,7), (3,7,3)From a=3:(4,2,13), (4,13,2)From a=4:(5,2,8), (5,8,2)From a=6:(7,3,3)From a=7:(8,2,5), (8,5,2)So in total:For the distinct triplets:(2,4,13), (2,13,4), (4,2,13), (4,13,2), (13,2,4), (13,4,2) → 6 permutations(2,5,8), (2,8,5), (5,2,8), (5,8,2), (8,2,5), (8,5,2) → 6 permutationsFor the triplet with two same numbers:(3,3,7), (3,7,3), (7,3,3) → 3 permutationsBut in our found solutions, we have:For distinct ones:(2,4,13), (2,13,4), (4,2,13), (4,13,2) →4(2,5,8), (2,8,5), (5,2,8), (5,8,2), (8,2,5), (8,5,2)→6Wait, no. Wait, in our found solutions, we have:From a=1: 4 solutions (2,4,13), (2,5,8), (2,8,5), (2,13,4)From a=2:2 solutions (3,3,7), (3,7,3)From a=3:2 solutions (4,2,13), (4,13,2)From a=4:2 solutions (5,2,8), (5,8,2)From a=6:1 solution (7,3,3)From a=7:2 solutions (8,2,5), (8,5,2)Total:4 +2 +2 +2 +1 +2=13 solutions.But these are not all permutations. For example, (13,2,4) is missing from the found solutions. However, in reality, when we consider x=13, y=2, z=4, it's another permutation which should be valid but wasn't generated in our initial search because we fixed a=x-1 and varied a starting from 1. Therefore, our approach missed some permutations. Therefore, perhaps we need to consider that all permutations are valid solutions and calculate accordingly.But the equation is symmetric in x, y, z. Therefore, all permutations of a solution are also solutions. Therefore, the total number of solutions would be the number of distinct ordered triples, considering permutations.But wait, the problem says "the number of all positive integer solutions (x, y, z)". So ordered triples. Therefore, each permutation counts as a separate solution.However, in our approach, by fixing a=x-1 and b=y-1, we found solutions for certain x and y, and their permutations. But some permutations require x or y to be larger, which we might not have covered. For example, (13,4,2) would require a=12, b=3, but in our earlier approach, a starts from 1 and increases, but a=12 would mean x=13. Let's check if such a solution exists.Take x=13, y=4. Then z=(13*4 +13 +4 -1)/(13*4 -13 -4 -1)=(52 +13 +4 -1)/(52 -13 -4 -1)=68/34=2. So z=2, which gives (13,4,2). Which is a permutation of (2,4,13). So this solution exists but wasn't generated in our earlier a=1, b=3 case because we fixed a=1, but here a=12 (x=13), which wasn't considered.Therefore, our initial approach missed some permutations. Hence, to get all ordered solutions, we need to consider all permutations of the fundamental solutions.Thus, the fundamental solutions (up to ordering) are:1. (2,4,13)2. (2,5,8)3. (3,3,7)Each of these has a certain number of permutations:1. (2,4,13): all distinct elements. Number of permutations: 62. (2,5,8): all distinct elements. Number of permutations:63. (3,3,7): two elements same, one different. Number of permutations:3Therefore, total solutions:6 +6 +3=15.However, in our earlier generated solutions, we have 13 solutions. Missing two permutations. Which ones? For example, (13,4,2) and (13,2,4). These were not generated in our initial approach because we didn't set a=12 or higher. But they are valid solutions. Therefore, the correct count should be 15.But wait, let's verify:For the distinct triple (2,4,13):Permutations:(2,4,13), (2,13,4), (4,2,13), (4,13,2), (13,2,4), (13,4,2) →6For (2,5,8):Similarly, 6 permutations.For (3,3,7):(3,3,7), (3,7,3), (7,3,3) →3Total:6+6+3=15.But in our earlier generated solutions, we have:From a=1 to a=7, we found 13 solutions. Missing two permutations of (2,4,13) and (2,5,8). For example, (13,4,2) and (13,2,4) weren't generated because when a=1, b=3, we get x=2, y=4, z=13. To get x=13, y=4, z=2, we need to set a=12, which wasn't covered in our initial loop. Similarly, permutations like (13,5,2) etc. But wait, no, (13,5,2) is not a solution. Wait, only permutations of the fundamental solutions.Wait, but actually, the solutions generated by the fundamental triplets via permutation will include all 15. Our initial approach found 13 because it stopped at a=7, but if we continue for higher a, we would find the remaining permutations. For example, setting a=12 (x=13), b=3 (y=4):z=1 + [2(a + b +2)]/(ab -2)=1 + [2(12 +3 +2)]/(12*3 -2)=1 + [2*17]/34=1 +34/34=2. So z=2. Thus, (13,4,2) is a valid solution, and this would be found when a=12. Similarly, for other permutations.Therefore, in reality, there are 15 ordered solutions. However, in our initial computation, we found 13, missing two. Therefore, the correct total is 15.But wait, let's confirm by checking with specific examples:For the triplet (2,4,13):All permutations are 6:1. (2,4,13)2. (2,13,4)3. (4,2,13)4. (4,13,2)5. (13,2,4)6. (13,4,2)Similarly, for (2,5,8):1. (2,5,8)2. (2,8,5)3. (5,2,8)4. (5,8,2)5. (8,2,5)6. (8,5,2)For (3,3,7):1. (3,3,7)2. (3,7,3)3. (7,3,3)Total 15.Therefore, the correct answer should be 15 ordered triples.But in our earlier generated list, we have 13 solutions. Missing two: (13,2,4) and (13,4,2). These would be generated when a=12 and a=13 respectively, but since we stopped at a=7, we didn't reach those. Therefore, the correct total is 15.But how can we be sure that there are no other fundamental solutions? For example, are there solutions with three equal numbers? Let's check.Suppose x=y=z. Then, 3*arctan(1/x)=π/4 => arctan(1/x)=π/12. So tan(π/12)=1/x. tan(π/12)=2 -√3≈0.2679. Therefore, 1/x≈0.2679 =>x≈3.732. Not integer. Therefore, no solution with x=y=z.Similarly, are there solutions with two variables equal and the third different, other than (3,3,7)? Let's check.Suppose x=y. Then, 2*arctan(1/x) + arctan(1/z)=π/4. Let's attempt to solve for x and z.Let’s let x=y. Then:2*arctan(1/x) + arctan(1/z)=π/4.Let’s use the formula for two arctans first:Let A = 2*arctan(1/x). Then tan(A)=2*(1/x)/(1 - (1/x)^2)= (2/x)/(1 -1/x^2)=2x/(x^2 -1)Then, the equation becomes:arctan(2x/(x^2 -1)) + arctan(1/z) = π/4.Using the addition formula:tan(arctan(α) + arctan(β))=(α + β)/(1 - αβ)=1.Therefore:(2x/(x^2 -1) +1/z)/(1 - (2x/(x^2 -1))(1/z))=1Multiply numerator and denominator by z(x^2 -1):[2x z + x^2 -1]/[z(x^2 -1) -2x] =1Therefore:2x z + x^2 -1 = z(x^2 -1) -2xRearrange:2xz +x² -1 -z x² + z +2x=0Group terms:-z x² +2xz + z + x² +2x -1=0Factor z:z(-x² +2x +1) +x² +2x -1=0Solve for z:z= ( -x² -2x +1 ) / (-x² +2x +1 ) = (x² +2x -1)/(x² -2x -1)For z to be positive integer, numerator and denominator must have same sign. Since x ≥2:Denominator=x² -2x -1. For x=2: 4 -4 -1=-1For x=3:9 -6 -1=2For x=4:16 -8 -1=7, etc.So denominator is negative for x=2, positive for x≥3.Numerator=x² +2x -1. For x=2:4 +4 -1=7For x=3:9 +6 -1=14So for x=2:z=(7)/(-1)=-7, invalid.For x=3:z=(14)/(2)=7. So z=7. So triplet (3,3,7), which we already have.For x=4:z=(16 +8 -1)/(16 -8 -1)=23/7≈3.285, not integer.x=5: numerator=25 +10 -1=34, denominator=25 -10 -1=14, z=34/14=17/7≈2.428, not integer.x=6: numerator=36 +12 -1=47, denominator=36 -12 -1=23, z=47/23≈2.043, not integer.x=7: numerator=49 +14 -1=62, denominator=49 -14 -1=34, z=62/34=31/17≈1.823, not integer.x=8: numerator=64 +16 -1=79, denominator=64 -16 -1=47, z=79/47≈1.68, not integer.Therefore, the only solution with x=y is x=3, z=7, and permutations. Hence, the triplet (3,3,7) and permutations are the only ones with two variables equal.Therefore, total number of solutions is indeed 15: 6 permutations for each of the two distinct triples and 3 permutations for the triple with two equals.But wait, to confirm, let's list all 15 solutions:From (2,4,13):1. (2,4,13)2. (2,13,4)3. (4,2,13)4. (4,13,2)5. (13,2,4)6. (13,4,2)From (2,5,8):7. (2,5,8)8. (2,8,5)9. (5,2,8)10. (5,8,2)11. (8,2,5)12. (8,5,2)From (3,3,7):13. (3,3,7)14. (3,7,3)15. (7,3,3)These are all 15 ordered solutions.Therefore, the number of all positive integer solutions is 15.Final AnswerThe number of positive integer solutions is boxed{15}.</think>To find the number of all positive integer solutions ((x, y, z)) to the equation (arctan frac{1}{x} + arctan frac{1}{y} + arctan frac{1}{z} = frac{pi}{4}), we start by using the tangent addition formula and simplifying the resulting expressions.1. Key Observations: - Each term (arctan frac{1}{x}) must be less than (frac{pi}{4}), implying (x, y, z geq 2). - Using the tangent addition formula, we derive that the equation can be transformed into a form involving integer solutions for (x, y, z).2. Transformation and Simplification: - By combining two arctangent terms at a time and using the addition formula, we derive the equation: [ frac{(x + y + z + xy + yz + zx - 1)}{(xyz - x - y - z - 2)} = 1 ] - This simplifies to finding integer solutions for (z = frac{xy + x + y - 1}{xy - x - y - 1}).3. Finding Solutions: - We systematically check values of (x) and (y) to find valid integer solutions for (z). - Fundamental solutions found are ((2, 4, 13)), ((2, 5, 8)), and ((3, 3, 7)).4. Permutations: - Each fundamental solution is permuted to account for all ordered triples: - ((2, 4, 13)) and its permutations: 6 solutions. - ((2, 5, 8)) and its permutations: 6 solutions. - ((3, 3, 7)) and its permutations: 3 solutions.5. Total Solutions: - Summing all permutations gives (6 + 6 + 3 = 15) solutions.Thus, the number of all positive integer solutions ((x, y, z)) is (boxed{15}).]

🔑:10. By symmetry, let x leqslant y leqslant z.Taking the tangent of both sides of the given equation, we getfrac{frac{1}{y}+frac{1}{z}}{1-frac{1}{y z}}=frac{1-frac{1}{x}}{1+frac{1}{x}}Rightarrow frac{y+z}{y z-1}=frac{x-1}{x+1}=1-frac{2}{x+1}.If x geqslant 5, then1-frac{2}{x+1} geqslant 1-frac{2}{5+1}=frac{2}{3} Rightarrow frac{y+z}{y z-1} geqslant frac{2}{3}.However, when z geqslant y geqslant x geqslant 5,frac{y+z}{y z-1} leqslant frac{5+5}{25-1}=frac{5}{12}<frac{2}{3},which is a contradiction. Therefore, x=2,3,4.When x=2,begin{array}{l}y z-1=3(y+z) Rightarrow(y-3)(z-3)=10 Rightarrow(y, z)=(4,13),(5,8) .end{array}When x=3,begin{array}{l}y z-1=2(y+z) Rightarrow(y-2)(z-2)=5 Rightarrow(y, z)=(3,7) .end{array}When x=4,begin{array}{l}3 y z-3=5(y+z) Rightarrow 3=5left(frac{1}{y}+frac{1}{z}right)+frac{3}{y z} leqslant 5left(frac{1}{4}+frac{1}{4}right)+frac{3}{16},end{array}which is a contradiction.Thus, the ordered triples (x, y, z) that satisfy z geqslant y geqslant x are(x, y, z)=(2,4,13),(2,5,8),(3,3,7) text {. }By permuting the order, we can obtain 6+6+3=15 solutions.In conclusion, the number of ordered positive integer solutions to the equation is 15.

🧠:[<think>Okay, so I need to find a tangent line to the curve y = e^x that passes through the origin. The question is asking for the coordinates of the tangent point. Hmm, let me think about how to approach this.First, I remember that the equation of a tangent line to a curve at a specific point involves the derivative of the function at that point. Since the curve here is y = e^x, I should start by finding its derivative. The derivative of e^x is e^x, right? So, the slope of the tangent line at any point (a, e^a) on the curve will be e^a. Now, the equation of the tangent line at the point (a, e^a) can be written using the point-slope form. That would be y - e^a = e^a(x - a). Let me check that. Yeah, point-slope is y - y1 = m(x - x1), where m is the slope. Here, y1 is e^a, and x1 is a, so substituting gives that equation.But the problem states that this tangent line must pass through the origin, which is (0, 0). So, if I plug x = 0 and y = 0 into the equation of the tangent line, it should satisfy the equation. Let me do that:0 - e^a = e^a(0 - a)Simplifying the left side: -e^aRight side: e^a*(-a) = -a e^aSo, the equation becomes: -e^a = -a e^aIf I multiply both sides by -1, that would give: e^a = a e^aHmm, now divide both sides by e^a (assuming e^a ≠ 0, which it never is). So, 1 = aWait, does that mean a = 1? So, the tangent point is at x = 1, and y = e^1 = e. Therefore, the coordinates of the tangent point are (1, e). Is that the answer? Let me verify.Let's write the equation of the tangent line at (1, e). The slope is e^1 = e. So, the equation is y - e = e(x - 1). Simplifying this: y = e x - e + e → y = e x. Wait, that can't be right. If I plug in x = 0 into y = e x, I get y = 0, which is the origin. So, yes, the tangent line at (1, e) is y = e x, which passes through the origin. So, that seems to check out.But wait, hold on. When I solved -e^a = -a e^a, I divided both sides by e^a, leading to 1 = a. So, a = 1. That seems straightforward, but maybe I missed something?Alternatively, could there be another solution? Let me see. Suppose we have e^a = a e^a. Subtract e^a from both sides: e^a - e^a = a e^a - e^a → 0 = e^a(a - 1). Since e^a is never zero, we have 0 = a - 1, so a = 1. So, that's the only solution. Therefore, only at a = 1 does the tangent line pass through the origin.Therefore, the tangent point is (1, e). So, the coordinates are (1, e). Let me double-check the tangent line equation again. At x = 1, y = e. The slope is e. So, the equation is y = e x + b. To find b, plug in x = 1, y = e: e = e*1 + b → b = 0. Therefore, the tangent line is y = e x, which indeed passes through (0,0). Perfect.Hmm, so that seems correct. Let me recap to make sure I didn't skip any steps. The problem is to find a point on y = e^x where the tangent line passes through the origin. The steps I took were:1. Let the point of tangency be (a, e^a).2. Compute the derivative at that point, which is e^a, giving the slope of the tangent line.3. Write the equation of the tangent line using point-slope: y - e^a = e^a(x - a).4. Since the line passes through (0,0), substitute x = 0 and y = 0 into the equation.5. Solve for a, leading to a = 1.6. Confirm by plugging back into the equation and checking that the tangent line is y = e x, which does pass through the origin.Therefore, all steps check out, and the tangent point is (1, e). I don't see any mistakes here. Maybe the answer is (1, e), which would be written as boxed{(1, e)}. Yeah, that seems right. Let me think if there's another way this could be approached, just to confirm.Alternative approach: Suppose we want a line passing through the origin that is tangent to y = e^x. The general equation of a line through the origin is y = mx. We need this line to be tangent to y = e^x. For two curves to be tangent, they must intersect at exactly one point, and at that point, their slopes are equal.So, set mx = e^x. The slope of the line is m, and the slope of the curve at the point of tangency is e^x. Therefore, m = e^x. So, we have two equations:1. mx = e^x2. m = e^xSubstituting equation 2 into equation 1: (e^x)x = e^x → x e^x = e^x. Subtract e^x from both sides: x e^x - e^x = 0 → e^x (x - 1) = 0. Since e^x ≠ 0, we have x - 1 = 0 → x = 1. Then, m = e^1 = e. So, the line is y = e x, and the point of tangency is x = 1, y = e. So, again, the coordinates are (1, e). Same result. So, this confirms the previous answer.Therefore, definitely, the answer is (1, e). So, the coordinates of the tangent point are boxed{(1, e)}.Final AnswerThe coordinates of the tangent point are boxed{(1, e)}.</think>To find a tangent line to the curve ( y = e^x ) that passes through the origin, we start by considering a point ((a, e^a)) on the curve. The derivative of ( y = e^x ) is ( e^x ), so the slope of the tangent line at ((a, e^a)) is ( e^a ).The equation of the tangent line at this point using the point-slope form is:[y - e^a = e^a (x - a)]Since the tangent line passes through the origin ((0, 0)), we substitute ( x = 0 ) and ( y = 0 ) into the equation:[0 - e^a = e^a (0 - a)]Simplifying this, we get:[-e^a = -a e^a]Dividing both sides by ( -e^a ) (which is non-zero), we find:[1 = a]Thus, the value of ( a ) is 1. The corresponding coordinates of the tangent point are ((1, e^1)), which simplifies to ((1, e)).To verify, the equation of the tangent line at ((1, e)) with slope ( e ) is:[y - e = e(x - 1)]Simplifying this, we get:[y = ex]This line passes through the origin ((0, 0)), confirming our solution.Therefore, the coordinates of the tangent point are (boxed{(1, e)}).,<think>Okay, so I need to find the coordinates of the point where a tangent line to the curve y = e^x passes through the origin. Hmm, let's start by recalling what a tangent line is. A tangent line to a curve at a specific point is a straight line that just touches the curve at that point and has the same slope as the curve does at that point.So, first, let me denote the point of tangency on the curve y = e^x. Let's call this point (a, e^a), since for any x-coordinate a, the y-coordinate on the curve is e^a.Now, I need to find the equation of the tangent line at this point. To do that, I need the derivative of y = e^x, which gives the slope of the tangent line at any point. The derivative of e^x is still e^x, right? So, the slope of the tangent line at point (a, e^a) is e^a.Therefore, the equation of the tangent line can be written using the point-slope form. The point-slope formula is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point of tangency. Plugging in the values, the equation becomes:y - e^a = e^a(x - a)Simplifying that, we get:y = e^a(x - a) + e^aWhich can be further simplified to:y = e^a x - a e^a + e^aOr:y = e^a x + e^a(1 - a)Now, the problem states that this tangent line passes through the origin (0, 0). So, substituting x = 0 and y = 0 into the equation of the tangent line should satisfy it. Let's do that:0 = e^a * 0 + e^a(1 - a)Simplifying the right side:0 = 0 + e^a(1 - a)So:0 = e^a(1 - a)Now, e^a is always positive for any real number a, right? Because the exponential function is always positive. So, the only way this equation holds is if (1 - a) = 0. Therefore:1 - a = 0 => a = 1So, the x-coordinate of the tangent point is a = 1. Then, the y-coordinate is e^a = e^1 = e.Therefore, the coordinates of the tangent point are (1, e). Let me just check if this makes sense.If the tangent point is (1, e), then the slope of the tangent line there is e^1 = e. So, the equation of the tangent line would be:y - e = e(x - 1)Simplifying:y = e x - e + eWhich simplifies to:y = e xWait, this line passes through the origin, because when x=0, y=0. Yes, that's correct. So the tangent line at (1, e) is y = e x, which indeed passes through (0,0). Therefore, this seems to check out.But let me double-check my steps to make sure I didn't skip anything or make a mistake. Let's go through it again.1. Let the point of tangency be (a, e^a).2. The derivative at this point is e^a, so the slope is e^a.3. Equation of tangent line: y - e^a = e^a(x - a)4. Simplify to y = e^a x - a e^a + e^a => y = e^a x + e^a(1 - a)5. Passes through origin: 0 = e^a(1 - a)6. Since e^a ≠ 0, 1 - a = 0 => a =17. Therefore, the point is (1, e). Correct.Wait, but when I substituted a=1 into the tangent line equation, I got y = e x, which passes through the origin. So, yes, that works. But let me check if there are any other possible solutions. The equation was 0 = e^a(1 - a). Since e^a is never zero, the only solution is a=1. So there's only one such tangent line that passes through the origin. Therefore, the coordinates are (1, e). That seems to be the answer.I think that's solid. Let me see if there's another way to approach this problem.Alternatively, suppose I didn't remember calculus, but maybe tried to set up the equation of a line passing through the origin and tangent to y = e^x. Let's see if that's possible.Let the tangent line pass through (0,0) and (a, e^a). But wait, the tangent line is not just any line passing through the origin and the point (a, e^a); it has to be tangent at that point, which means the slope of the line has to equal the derivative at that point.So, the line passing through (0,0) and (a, e^a) has slope (e^a - 0)/(a - 0) = e^a / a. But this slope must equal the derivative at x = a, which is e^a. Therefore, setting them equal:e^a / a = e^aDivide both sides by e^a (since e^a ≠ 0):1/a = 1Therefore, a = 1. So same result. Therefore, the coordinates are (1, e). So this confirms it again.So whether using calculus or just algebra with the slope, the answer comes out the same. Therefore, the tangent point is (1, e). Therefore, the coordinates are (1, e). Just to visualize, the curve y=e^x is exponential, increasing rapidly. The tangent at (1, e) has slope e, so the line y = e x is a straight line starting from the origin and passing through (1, e). Since the slope is e (~2.718), which is quite steep, it makes sense that this line just touches the curve at that point. So yeah, that seems correct.I think I'm confident that the answer is (1, e).Final AnswerThe coordinates of the tangent point are boxed{(1, e)}.</think>To find the coordinates of the point where a tangent line to the curve ( y = e^x ) passes through the origin, we start by denoting the point of tangency as ( (a, e^a) ). The derivative of ( y = e^x ) is ( e^x ), so the slope of the tangent line at ( (a, e^a) ) is ( e^a ).Using the point-slope form of the equation of a line, the tangent line at ( (a, e^a) ) is:[y - e^a = e^a(x - a)]Simplifying this, we get:[y = e^a x - a e^a + e^a = e^a x + e^a(1 - a)]Since the tangent line passes through the origin ((0, 0)), substituting ( x = 0 ) and ( y = 0 ) into the equation gives:[0 = e^a(1 - a)]Since ( e^a ) is always positive, the equation ( 0 = e^a(1 - a) ) implies:[1 - a = 0 implies a = 1]Thus, the coordinates of the tangent point are ( (1, e^1) = (1, e) ).To verify, the tangent line at ( (1, e) ) has the equation:[y - e = e(x - 1)]Simplifying, we get:[y = ex]which indeed passes through the origin.Therefore, the coordinates of the tangent point are (boxed{(1, e)}).]

🔑:Let's assume the coordinates of the tangent point are (x_{0},e^{x_{0}}). Given y' |_{x-x_{0}} = e^{x_{0}}, we obtain the equation of the tangent line as y-e^{x_{0}}=e^{x_{0}}(x-x_{0}). Since the tangent line passes through the origin, we have 0-e^{x_{0}}=e^{x_{0}}(0-x_{0}), solving this gives x_{0}=1, thus the coordinates of the tangent point are (1,e). Therefore, the answer is: boxed{(1,e)}. To find the coordinates of the tangent point, it is only necessary to derive the equation of the tangent line. This can be done by first using the derivative to find the value of the derivative function at the tangent point, and then combining the geometric meaning of the derivative to find the slope of the tangent line. Finally, using the fact that the tangent line passes through the origin solves the problem. This question mainly examines the basic knowledge of the slope of a line, the geometric meaning of derivatives, and the use of derivatives to study the equation of the tangent line at a certain point on the curve, as well as computational problem-solving skills. It is a basic question.

🧠:[<think>Okay, let's try to solve this equation: [x] · {x} = 2005x. Hmm, first, I need to remember what [x] and {x} mean. The notation [x] is the greatest integer less than or equal to x, which is also called the floor function. And {x} is the fractional part of x, which is x minus [x]. So, any real number x can be written as [x] + {x}, right? Where [x] is an integer and {x} is between 0 (inclusive) and 1 (exclusive).So, the equation given is the product of the floor of x and the fractional part of x equals 2005 times x. Let me write that out more formally. Let’s let n = [x], which is an integer, and let f = {x} = x - n. Then the equation becomes n · f = 2005x. But since x = n + f, we can substitute that into the equation:n · f = 2005(n + f)Let me write that down: nf = 2005(n + f). Let me rearrange terms to group like terms together. Let's bring all terms to one side:nf - 2005n - 2005f = 0Hmm, maybe factor terms? Let's see:n(f - 2005) - 2005f = 0Not sure if that helps. Alternatively, maybe factor out common terms. Let me try to rearrange the original equation:nf = 2005n + 2005fDivide both sides by n? Wait, but n could be zero. Hmm, if n is zero, then let's check that case first.Case 1: n = 0If [x] = 0, then x is in [0, 1). Then {x} = x. So the equation becomes 0 · x = 2005x → 0 = 2005x. The solution is x = 0. But x must be in [0,1), so x=0 is a solution here. Let me note that.Case 2: n ≠ 0Then n is an integer not equal to zero. So, x = n + f, where n ∈ ℤ {0} and f ∈ [0,1). Let's substitute back into the equation:nf = 2005(n + f)Let me rearrange this equation:nf = 2005n + 2005fBring all terms to the left:nf - 2005n - 2005f = 0Factor terms with n and f:n(f - 2005) - 2005f = 0Hmm, maybe factor f:f(n - 2005) - 2005n = 0Wait, that might not be helpful. Let me try to isolate f.Starting from nf - 2005n - 2005f = 0Let me factor f from the first and third term:f(n - 2005) - 2005n = 0So, f(n - 2005) = 2005nThen, solving for f:f = (2005n)/(n - 2005)But remember that f is the fractional part of x, so f must satisfy 0 ≤ f < 1.Therefore, we have the equation:0 ≤ (2005n)/(n - 2005) < 1Now, since n is an integer not equal to zero, let's analyze this inequality.First, note that n ≠ 2005 because the denominator would be zero, which is undefined. So n can be any integer except 0 and 2005.So, the expression (2005n)/(n - 2005). Let's write this as:(2005n)/(n - 2005) = 2005n / (n - 2005)Let me see if I can simplify this. Maybe express it as:2005n / (n - 2005) = 2005 + (2005*2005)/(n - 2005)Wait, let's do polynomial division or decompose the fraction.Let me write:2005n / (n - 2005) = 2005*(n) / (n - 2005)Let me make a substitution. Let k = n - 2005. Then n = k + 2005. Substitute back:2005*(k + 2005)/k = 2005k/k + 2005*2005/k = 2005 + (2005²)/kTherefore, f = 2005 + (2005²)/kBut f must be less than 1 and non-negative.So,0 ≤ 2005 + (2005²)/k < 1But 2005 is a positive number (since 2005 is positive). Wait, but 2005 + something must be less than 1? That can't be possible unless the term (2005²)/k is negative enough to make the entire expression less than 1 and non-negative.Wait, so 2005 + (2005²)/k < 1 implies that (2005²)/k < 1 - 2005 = -2004Similarly, 2005 + (2005²)/k ≥ 0 implies that (2005²)/k ≥ -2005So, we have two inequalities:1) (2005²)/k ≥ -20052) (2005²)/k < -2004Let me solve these inequalities for k.First, note that k is an integer because n is an integer and k = n - 2005, so k is n shifted by 2005, hence integer. Also, k ≠ 0 because n ≠ 2005.First inequality:(2005²)/k ≥ -2005Multiply both sides by k, but we need to be careful about the sign of k.Case A: k > 0Then, multiplying both sides by k (positive, inequality direction remains):2005² ≥ -2005kBut 2005² is positive, and -2005k is negative. So this inequality is always true for k > 0. So for k > 0, first inequality holds.Second inequality:(2005²)/k < -2004Multiply both sides by k. Since k > 0, inequality remains:2005² < -2004kBut left side is positive, right side is negative. So this can never be true for k > 0. Therefore, no solutions when k > 0.Case B: k < 0Now, k is negative. Let k = -m where m is a positive integer.Then, first inequality:(2005²)/(-m) ≥ -2005Multiply both sides by -m (negative, so reverse inequality):2005² ≤ (-2005)(-m) → 2005² ≤ 2005mDivide both sides by 2005:2005 ≤ mSo, m ≥ 2005Second inequality:(2005²)/(-m) < -2004Multiply both sides by -m (positive, since m > 0, k = -m < 0). Wait, k is negative, so -m = k. Wait, m = |k|. Let me clarify.Let me re-express both inequalities with k negative.Let k = -m, m > 0 integer.First inequality:(2005²)/(-m) ≥ -2005 → -2005²/m ≥ -2005Multiply both sides by -1 (reverse inequality):2005²/m ≤ 2005Divide both sides by 2005:2005/m ≤ 1 → m ≥ 2005Second inequality:(2005²)/(-m) < -2004 → -2005²/m < -2004Multiply both sides by -1 (reverse inequality):2005²/m > 2004So,2005²/m > 2004 → m < 2005² / 2004Calculate 2005² / 2004:2005² = 2005×2005. Let me compute that.2005×2000 = 4,010,0002005×5 = 10,025So, 4,010,000 + 10,025 = 4,020,025So, 2005² = 4,020,025Then, 4,020,025 / 2004 ≈ Let's divide 4,020,025 by 2004.First, compute how many times 2004 fits into 4,020,025.2004 × 2000 = 4,008,000Subtract: 4,020,025 - 4,008,000 = 12,025Now, 2004 × 6 = 12,024So, 2004 × 2006 = 4,008,000 + 12,024 = 4,020,024So, 4,020,025 - 4,020,024 = 1Therefore, 4,020,025 / 2004 = 2006 + 1/2004 ≈ 2006.000499...So, m must be less than approximately 2006.0005. But since m is a positive integer, m ≤ 2006.But from the first inequality, m ≥ 2005.Therefore, m can be 2005 or 2006.Thus, k = -m, so k = -2005 or k = -2006Therefore, k ∈ {-2006, -2005}Thus, original k = n - 2005, so n = k + 2005Therefore, if k = -2005, then n = -2005 + 2005 = 0. But n ≠ 0 in Case 2.Wait, but in Case 2, we considered n ≠ 0. However, if k = -2005, then n = 0, which is excluded. Therefore, k = -2005 is invalid in this case. Hence, only k = -2006 is valid?Wait, let me check.Original k is n - 2005.If k = -2005 → n = -2005 + 2005 = 0. But in Case 2, n ≠ 0. Therefore, k = -2005 would lead to n = 0, which is considered in Case 1. So, in Case 2, m must be ≥ 2005, but k = -m, so n = 2005 - m.Wait, perhaps m ranges?Wait, n = k + 2005. If k = -m, then n = -m + 2005. So, m = 2005 - n.Wait, this is getting a bit confusing.Wait, earlier, we found that m ≥ 2005 and m ≤ 2006.So m can be 2005 or 2006.For m=2005:k = -2005, so n = 2005 + k = 2005 - 2005 = 0, which is excluded in Case 2.For m=2006:k = -2006, so n = 2005 + (-2006) = -1So, n = -1Therefore, only m=2006 leads to a valid n in Case 2 (n ≠ 0). So, n = -1.Therefore, let's check n = -1.So, n = -1, and f = (2005n)/(n - 2005) = (2005*(-1))/(-1 -2005) = (-2005)/(-2006) = 2005/2006 ≈ 0.9995...Which is less than 1, so acceptable. Also, f must be ≥ 0, which it is.Therefore, f = 2005/2006Thus, x = n + f = -1 + 2005/2006 = (-2006/2006) + 2005/2006 = (-1)/2006 ≈ -0.000498...Wait, but x must satisfy that [x] = n = -1. Let's verify.x = -1 + 2005/2006 ≈ -0.000498... So, this is greater than -1 + 1 = 0, but wait, no:Wait, x = -1 + 2005/2006 = (-2006 + 2005)/2006 = (-1)/2006 ≈ -0.000498Wait, that is between -1 and 0, right? Because -1/2006 is approximately -0.000498, which is greater than -1 and less than 0. Therefore, [x] = -1, which matches n = -1. So that's correct.Therefore, in Case 2, the solution is x = (-1)/2006.Wait, let me compute that again.x = n + f = -1 + 2005/2006Convert -1 to 2006/2006:-2006/2006 + 2005/2006 = (-2006 + 2005)/2006 = (-1)/2006Yes, so x = -1/2006But let me check this solution in the original equation.Original equation: [x] · {x} = 2005xCompute [x]: x = -1/2006 ≈ -0.000498, so the greatest integer less than or equal to x is -1.Compute {x}: fractional part is x - [x] = (-1/2006) - (-1) = (-1/2006) + 1 = (2006/2006 - 1/2006) = 2005/2006Therefore, [x] · {x} = (-1) · (2005/2006) = -2005/20062005x = 2005 · (-1/2006) = -2005/2006Therefore, both sides equal -2005/2006. So, the equation holds. Great, so x = -1/2006 is a solution.So, in Case 2, we have x = -1/2006.In Case 1, x = 0.But let's verify x = 0 in the original equation. [0] = 0, {0} = 0. So left side is 0·0 = 0. Right side is 2005·0 = 0. So 0 = 0. Therefore, x = 0 is also a solution.Therefore, are there two solutions: x = 0 and x = -1/2006?Wait, but let me check x = 0. When x = 0, [x] = 0, {x} = 0, so left side is 0*0 = 0, right side is 2005*0 = 0. So yes, valid.But earlier, when we considered Case 1 (n=0), x=0 is a solution, and in Case 2, we have x=-1/2006.Therefore, the solutions are x = 0 and x = -1/2006.But let me check if there are any other possible cases.Wait, in Case 2, we derived that n = -1 is the only possibility. Are there any other possible n?Wait, let me recap.In Case 2, we set n ≠ 0, and through substitution, found that the only possible n is -1, leading to x = -1/2006. Then in Case 1, n=0 leads to x=0.Therefore, the solutions are x=0 and x=-1/2006.But let me check if there could be other solutions.Wait, when we considered k, which was n - 2005, we found that m (which is |k|) had to be 2005 or 2006. But m=2005 led to n=0, which is considered in Case 1, and m=2006 led to n=-1.Therefore, no other integer m satisfies the inequalities. So, no other solutions.Wait, but let me check m=2007. Then k=-2007, so n=2005 -2007= -2. Then f=2005*(-2)/(-2 -2005)= (-4010)/(-2007)= 4010/2007≈1.998...But f must be less than 1, so this is invalid.Similarly, m=2004: Then k=-2004, n=2005 -2004=1. Then f=2005*1/(1 -2005)=2005/-2004≈-1.000..., which is negative, invalid.So m=2005 and m=2006 only. As discussed.Therefore, only two solutions: x=0 and x=-1/2006.But wait, x=0 is in Case 1, and x=-1/2006 is in Case 2.But let me check if x=-1/2006 is indeed in the interval [n, n+1) where n=-1. Since -1/2006 ≈-0.000498 is in [-1,0), so yes, [x]=-1.Therefore, both solutions are valid.But let me check once again the original problem. The equation is [x] · {x} = 2005x. So, for x=0, both sides are zero. For x=-1/2006, both sides equal -2005/2006. So, both are valid.Wait, but is x=-1/2006 the only solution in Case 2? Let me confirm.From the above steps, in Case 2, after solving for f, we found that the only possible n is -1, leading to x=-1/2006. So, yes, only one solution in Case 2.Therefore, the solutions are x=0 and x=-1/2006.Wait, but let me check for positive n. Suppose n is positive.Suppose n is a positive integer. Then, f = (2005n)/(n - 2005). Let's check the denominator: n - 2005.If n > 2005, then denominator is positive, so f = positive/positive = positive. But f must be less than 1. So, 2005n/(n - 2005) < 1.So, 2005n < n - 2005 → 2005n - n < -2005 → 2004n < -2005. But n is positive, so left side positive, right side negative. Impossible.If n is positive and less than 2005, denominator n -2005 is negative. So, f = (2005n)/(negative) = negative. But fractional part can't be negative. Hence, impossible. Therefore, no solutions for positive n.Therefore, only solutions are x=0 and x=-1/2006.But wait, x=0: let's check if x=0 is considered in both Cases. Wait, in Case 1, n=0, and we found x=0. In Case 2, n≠0, leading to x=-1/2006.Therefore, these are the two solutions.But the problem says "Solve the equation". So, we need to present all solutions. So, x=0 and x=-1/2006.Wait, but let me check x=0 again. If x=0, then [x]=0, {x}=0. So, 0*0=0, and 2005*0=0. So 0=0. Correct.But wait, the problem didn't specify any restrictions, so both solutions are valid. But let me check x=-1/2006. Let's compute [x] and {x}:x = -1/2006 ≈ -0.000498So, [x] is the greatest integer less than or equal to x, which is -1.{x} = x - [x] = (-1/2006) - (-1) = 1 - 1/2006 = 2005/2006Therefore, [x] * {x} = (-1) * (2005/2006) = -2005/20062005x = 2005*(-1/2006) = -2005/2006Thus, equality holds. So, both solutions are valid.Hence, the solutions are x=0 and x=-1/2006.Wait, but the problem says "Solve the equation". So, are both solutions acceptable?Wait, but let me check if there are any other solutions. For example, if n is negative but not -1.Suppose n=-2.Then, f = (2005*(-2))/(-2 -2005) = (-4010)/(-2007) ≈ 1.998..., which is greater than or equal to 1, which is invalid because {x} must be less than 1. Therefore, invalid.Similarly, n=-3:f=(2005*(-3))/(-3 -2005)= (-6015)/(-2008)=6015/2008≈2.996..., which is even larger. So, invalid.For n=-1, f=2005/2006≈0.9995, which is valid.If n=1, as before, f=2005/(1 -2005)= negative, invalid.n=2006: f=(2005*2006)/(2006 -2005)=2005*2006/1=2005*2006, which is way larger than 1, invalid.So, indeed, only n=0 and n=-1 give valid f.Therefore, the equation has two solutions: x=0 and x=-1/2006.But let me check once again for n=0. If x=0, then [x]=0 and {x}=0, so product is 0, which equals 2005*0=0. Correct.For x=-1/2006, we already checked that it works.Thus, the solutions are x=0 and x=-1/2006.But wait, in some contexts, the fractional part {x} is defined as x - floor(x). For negative numbers, floor(x) is the greatest integer less than or equal to x. For example, if x = -0.5, floor(x) = -1, and {x} = x - floor(x) = (-0.5) - (-1) = 0.5. So, fractional part is always in [0,1).Thus, in the case of x=-1/2006 ≈ -0.000498, floor(x) is -1, and {x}=x - (-1)=x +1≈0.9995, which is correct.Therefore, calculations are correct.Therefore, the equation has two solutions: x=0 and x=-1/2006.But let me check the original problem statement again. It says "Solve the equation: [x] · {x} = 2005x".Wait, perhaps there is a mistake here. Let me check if x=0 and x=-1/2006 are both valid.Yes, for x=0:Left side: 0 * 0 = 0Right side: 2005 * 0 = 0So, 0=0. Correct.For x=-1/2006:Left side: (-1) * (2005/2006) = -2005/2006Right side: 2005 * (-1/2006) = -2005/2006Thus, equality holds. Correct.Therefore, both solutions are valid. Therefore, the answer is x=0 and x=-1/2006.But wait, maybe I should present them as a set.Alternatively, is there a reason to discard x=0? Let me check. If x=0, then [x]=0, {x}=0. The equation becomes 0*0=0, which is valid. Therefore, x=0 is a valid solution.Therefore, the solutions are x=0 and x=-1/2006.However, sometimes in these equations, x=0 is considered a trivial solution, but in this case, it is indeed a valid solution.Therefore, the final answer should include both solutions. But let me check once again.Wait, but wait, when I considered Case 1 (n=0), I found x=0. In Case 2 (n ≠0), I found x=-1/2006.Therefore, these are the two solutions.But let me check if x=0 is the only solution in Case 1. For n=0, the equation is [x]{x} = 0 * x =0, and 2005x = 0. So, 0=0. So, all x in [0,1) would satisfy 0=0? Wait, hold on.Wait, hold on. If n=0, then x ∈ [0,1). The equation becomes [x] · {x} = 2005x. Since [x]=0, the left side is 0·{x}=0. The right side is 2005x. Therefore, the equation becomes 0=2005x, which implies x=0. Therefore, in Case 1, only x=0 is a solution, not the entire interval [0,1). Because 2005x must equal zero, which only occurs at x=0.So, in Case 1, x=0 is the only solution.Therefore, the two solutions are x=0 and x=-1/2006.Thus, the answer is x=0 and x=-1/2006.Therefore, boxed answers would be boxed{0} and boxed{-dfrac{1}{2006}}.But wait, the problem says "Solve the equation", so we need to present all solutions. Therefore, both solutions must be given.But let me check once again.Yes, both x=0 and x=-1/2006 satisfy the equation.Therefore, the solutions are boxed{0} and boxed{-dfrac{1}{2006}}.But the problem might expect both answers. Alternatively, maybe the problem expects only the non-trivial solution. But according to the calculation, both are valid.Alternatively, perhaps in the original problem, x=0 is considered trivial, but in the context of the problem, since [x] and {x} are defined, x=0 is a valid solution.Therefore, the answer should include both.But let me check once more.Original problem: "Solve the equation: [x] · {x} = 2005x"When x=0, [x]=0, {x}=0, so 0*0=0, 2005*0=0. Correct.When x=-1/2006, [x]=-1, {x}=2005/2006, so (-1)*(2005/2006)= -2005/2006, and 2005*(-1/2006)= -2005/2006. Correct.Therefore, both solutions are valid. Therefore, the answer is both x=0 and x=-1/2006.But in some contexts, equations with floor functions might have multiple solutions, and we need to list all of them. So, yes, both are required.Hence, the final answer is x=0 and x=-1/2006, boxed appropriately.Final AnswerThe solutions are boxed{0} and boxed{-dfrac{1}{2006}}.</think>Given the equation ([x] cdot {x} = 2005x), where ([x]) is the greatest integer less than or equal to (x) and ({x} = x - [x]) is the fractional part of (x), we start by expressing (x) as (x = n + f) where (n = [x]) is an integer and (f = {x}) is in the interval ([0, 1)). Case 1: (n = 0)If (n = 0), then (x in [0, 1)) and ({x} = x). The equation simplifies to:[0 cdot x = 2005x implies 0 = 2005x]This implies (x = 0). Since (x = 0) is within the interval ([0, 1)), it is a valid solution. Case 2: (n neq 0)For (n neq 0), substituting (x = n + f) into the equation gives:[nf = 2005(n + f)]Rearranging terms, we get:[nf - 2005n - 2005f = 0 implies f(n - 2005) = 2005n implies f = frac{2005n}{n - 2005}]Since (f) must be in ([0, 1)), we solve:[0 leq frac{2005n}{n - 2005} < 1]Analyzing the inequality, we find that (n) must be negative. Let (n = -m) where (m) is a positive integer. Solving for (m), we find (m = 2006) leading to (n = -1). Substituting back, we get:[f = frac{2005(-1)}{-1 - 2005} = frac{2005}{2006}]Thus, (x = n + f = -1 + frac{2005}{2006} = -frac{1}{2006}). Verification- For (x = 0), ([x] = 0) and ({x} = 0), so both sides of the equation are 0.- For (x = -frac{1}{2006}), ([x] = -1) and ({x} = frac{2005}{2006}), leading to both sides being (-frac{2005}{2006}). Final AnswerThe solutions are (boxed{0}) and (boxed{-dfrac{1}{2006}}).,<think>Okay, so I need to solve the equation [x] · {x} = 2005x, where [x] is the greatest integer less than or equal to x, and {x} is the fractional part of x, which is x - [x]. Alright, let me break this down.First, I know that any real number x can be written as [x] + {x}, right? So x = [x] + {x}. Let me denote [x] as n, where n is an integer, and {x} as f, where f is a fractional part between 0 (inclusive) and 1 (exclusive). So, x = n + f, with n ∈ ℤ and 0 ≤ f < 1.Given that, the original equation becomes n · f = 2005(n + f). Let me write that out:n * f = 2005(n + f)Hmm. So, substituting x with n + f. Let's rearrange this equation to solve for f or n.First, expand the right-hand side:n f = 2005n + 2005fLet me bring all terms to one side:n f - 2005n - 2005f = 0Factor terms where possible. Let's see:n(f - 2005) - 2005f = 0Hmm, not sure if that's helpful. Maybe factor f and n terms:n f - 2005n - 2005f = 0Let me factor n from the first two terms:n(f - 2005) - 2005f = 0Alternatively, factor 2005 from the last term:n(f - 2005) - 2005f = 0Wait, maybe rearrange terms:n f = 2005(n + f)Divide both sides by (n + f)? Wait, but n + f = x, which is n + f. But x could be zero? Wait, but if x is zero, then n = 0 and f = 0. Let's check if x = 0 is a solution.If x = 0, then [x] = 0, {x} = 0, so left-hand side is 0 * 0 = 0, right-hand side is 2005 * 0 = 0. So 0 is a solution. Hmm, okay. But let's see if there are other solutions.But let's proceed step by step. Let's suppose x is not zero. Then, perhaps we can divide both sides by x? Let's see.Given that x = n + f, then:n f = 2005xBut x = n + f, so:n f = 2005(n + f)Alternatively, we can write this as:n f = 2005n + 2005fLet me rearrange terms:n f - 2005n - 2005f = 0Hmm. Let's factor terms with n and terms with f:n(f - 2005) - 2005f = 0So, n(f - 2005) = 2005fThus,n = (2005f) / (f - 2005)But n is an integer. Also, since f is between 0 and 1, the denominator (f - 2005) is negative because f is less than 1, so f - 2005 is less than -2004. Therefore, the numerator is 2005f, which is non-negative because f is non-negative. So numerator is non-negative, denominator is negative, so n is non-positive.So n must be less than or equal to zero.But n is the integer part of x, which is x - f. Since f is between 0 and 1, x is in [n, n + 1). So if n is negative, then x is in [n, n + 1). But let's proceed.Let me write the expression again:n = (2005f) / (f - 2005)Let me denote this as n = (2005f) / (f - 2005)We can rewrite the denominator as -(2005 - f), so:n = - (2005f) / (2005 - f)So n is equal to negative 2005f divided by (2005 - f). Since n is an integer, then (2005f)/(2005 - f) must be an integer as well, because n is - that.Let me set k = (2005f)/(2005 - f), where k is a non-negative integer because 2005f is non-negative (f ≥ 0) and denominator 2005 - f is positive (since f < 1, 2005 - f > 2004). Therefore, k is non-negative, and k must be integer. Therefore, n = -k.So, n = -k, where k is a non-negative integer. Then, since n is the integer part of x, x = n + f = -k + f. Since f ∈ [0, 1), x is in [-k, -k + 1). But also, since x = -k + f, and f = x - [x] = x - (-k) = x + k. Therefore, f = x + k. Wait, but f is also equal to x - n = x - (-k) = x + k. So f = x + k.But f = x + k, and x = -k + f. So substituting back, f = (-k + f) + k ⇒ f = f. So that's consistent.But maybe this isn't helpful. Let's go back to the expression k = (2005f)/(2005 - f). Since k is integer, let's write this as:k = (2005f)/(2005 - f)Multiply both sides by (2005 - f):k(2005 - f) = 2005fExpand left side:2005k - k f = 2005fBring all terms to left side:2005k - k f - 2005f = 0Factor f terms:2005k - f(k + 2005) = 0Therefore,f(k + 2005) = 2005kThus,f = (2005k)/(k + 2005)But f must be in [0, 1). So, f = (2005k)/(k + 2005) < 1Therefore,(2005k)/(k + 2005) < 1Multiply both sides by (k + 2005) (which is positive because k is non-negative):2005k < k + 2005Subtract k from both sides:2004k < 2005Divide both sides by 2004:k < 2005/2004 ≈ 1.000499...But k is a non-negative integer, so k can be 0 or 1.Therefore, possible k values are 0 and 1.Let's check k = 0:If k = 0, then f = (2005*0)/(0 + 2005) = 0. So f = 0. Then n = -k = 0. Therefore, x = n + f = 0 + 0 = 0. Which is the solution we found earlier. So x = 0 is a solution.Now check k = 1:k = 1. Then f = (2005*1)/(1 + 2005) = 2005/2006 ≈ 0.999501...So f ≈ 0.9995, which is less than 1, so valid. Then n = -k = -1. Therefore, x = n + f = -1 + 2005/2006 = (-2006/2006) + 2005/2006 = (-1)/2006 ≈ -0.000498...So x ≈ -0.000498. Let's verify if this is a solution.Compute [x] · {x} and 2005x.First, [x] is the greatest integer less than or equal to x. Since x ≈ -0.000498, which is between -1 and 0. So [x] = -1.{x} = x - [x] = (-0.000498) - (-1) = 0.999502. Wait, but earlier f was 2005/2006 ≈ 0.999501..., which is approximately the same.So [x] · {x} = (-1) * (2005/2006) ≈ -0.999501...Now, 2005x = 2005 * (-1 + 2005/2006) = 2005 * (-1/2006) ≈ -2005/2006 ≈ -0.999501...So they are equal. Therefore, x = -1 + 2005/2006 is indeed a solution.Are there any other k? Since k must be less than 2005/2004 ≈ 1.0005, so k can only be 0 or 1. So those are the only possible solutions.But wait, when k = 0, we get x = 0. When k = 1, we get x = -1 + 2005/2006. So are there only two solutions?Wait, but let me check if there could be other k. For example, if k is negative, but we had earlier that k is non-negative. Because k = (2005f)/(2005 - f). Since f is non-negative and denominator is positive, k is non-negative. So k can't be negative. So only k = 0 and k = 1.Therefore, the solutions are x = 0 and x = -1 + 2005/2006 = ( -2006 + 2005 ) / 2006 = (-1)/2006. Wait, hold on:Wait, x = n + f = -k + f. For k = 1, x = -1 + (2005/2006) = (-2006 + 2005)/2006 = (-1)/2006 ≈ -0.000498...So x = -1/2006. Wait, that's a simpler way to write it. Because 2005/2006 is 1 - 1/2006, so -1 + 1 - 1/2006 = -1/2006. Wait, that's correct.Wait, 2005/2006 = 1 - 1/2006, so x = -1 + (1 - 1/2006) = -1 + 1 - 1/2006 = -1/2006. So x = -1/2006.So the two solutions are x = 0 and x = -1/2006. Let me verify if x = -1/2006 is indeed a solution.Compute [x] and {x} for x = -1/2006.Since x is between -1 and 0, [x] = -1, and {x} = x - (-1) = x + 1 = (-1/2006) + 1 = 2005/2006.Therefore, [x] · {x} = (-1) * (2005/2006) = -2005/2006.On the other hand, 2005x = 2005 * (-1/2006) = -2005/2006.So both sides equal -2005/2006. Hence, x = -1/2006 is a valid solution.Now, check x = 0. Then [x] = 0, {x} = 0, so [x] · {x} = 0. And 2005x = 0. So both sides are 0. So x = 0 is a solution.But wait, in the problem statement, are there any restrictions on x? For example, if x is positive, negative, or zero? The problem didn't specify, so we need to consider all real numbers.But according to our analysis, the only possible solutions are x = 0 and x = -1/2006.Wait, but let me check if there are any other possibilities. For example, when k is 0, then x = 0. When k is 1, x = -1/2006. What if k is greater than 1? But we saw that k < 2005/2004 ≈ 1.0005, so k can't be 2 or higher. Therefore, only k = 0 and 1 are possible.But let me check if there's another approach. Let me suppose that x is positive. If x is positive, then [x] is a non-negative integer, and {x} is in [0, 1). Then, [x] · {x} is non-negative. However, 2005x is positive. So if x is positive, then both sides are non-negative.But x = 0 gives both sides zero. If x is positive, then 2005x is positive, and [x] · {x} is non-negative. So maybe there could be positive solutions.Wait a second, when I considered k, which is (2005f)/(2005 - f), which led me to f = (2005k)/(2005 + k). Then f < 1 gives k < 2005/2004. But this was under the assumption that n = -k. Wait, but if x is positive, then n is non-negative. So in my previous reasoning, I might have assumed that n is negative because I set n = -k. Wait, perhaps I need to revisit that.Wait, let's start over.Original equation: [x] · {x} = 2005x.Express x as n + f, with n integer and 0 ≤ f < 1. Then:n * f = 2005(n + f)Which can be rewritten as:n f = 2005n + 2005fBring all terms to left-hand side:n f - 2005n - 2005f = 0Factor:n(f - 2005) - 2005f = 0So,n = (2005f)/(f - 2005)But f - 2005 is negative because f < 1, so denominator is negative. The numerator is 2005f, which is non-negative since f ≥ 0. Therefore, n is non-positive.Therefore, n can be zero or negative.Case 1: n = 0.If n = 0, then x = 0 + f = f. Then the equation becomes 0 * f = 2005 * f ⇒ 0 = 2005f. Therefore, f = 0. Therefore, x = 0. So that's the solution x = 0.Case 2: n < 0.Then n is a negative integer. Let's let n = -k where k is a positive integer (since n < 0). Then x = -k + f, with 0 ≤ f < 1.Substitute into the original equation:n * f = 2005x ⇒ (-k) * f = 2005(-k + f)Multiply both sides by -1:k f = 2005(k - f)Expand the right side:k f = 2005k - 2005fBring all terms to the left side:k f + 2005f - 2005k = 0Factor out f:f(k + 2005) = 2005kThus,f = (2005k)/(k + 2005)Since f must be less than 1, we have:(2005k)/(k + 2005) < 1 ⇒ 2005k < k + 2005 ⇒ 2004k < 2005 ⇒ k < 2005/2004 ≈ 1.0005.Since k is a positive integer (since n = -k and n < 0), k can only be 1.Therefore, k = 1.Thus, f = (2005 * 1)/(1 + 2005) = 2005/2006.Therefore, x = -k + f = -1 + 2005/2006 = (-2006 + 2005)/2006 = -1/2006.Therefore, the solutions are x = 0 and x = -1/2006.Wait, but when we considered k as (2005f)/(2005 - f), we got the same result. So that's consistent.But let's confirm again.For x positive: Suppose x is positive, so n ≥ 0 and f ∈ [0,1). Then equation [x]{x} = 2005x. So n f = 2005(n + f). Since n and f are non-negative, the left side is non-negative, and the right side is positive unless n + f = 0, which is x = 0. So x = 0 is the only non-negative solution. For positive x, n f = 2005(n + f). Let's see if this is possible.If n ≥ 1 and f ≥ 0, then n f ≥ 0, and 2005(n + f) ≥ 2005n ≥ 2005. But n f ≤ n*1 = n (since f <1). So left side is at most n, and the right side is at least 2005n. So unless n =0, which leads to x=0, otherwise, n ≥1 would require n ≥ 2005n, which is impossible unless n=0. Therefore, the only non-negative solution is x=0.Therefore, all solutions are x=0 and x=-1/2006.But let me check if there are other negative solutions. Suppose x is negative but not an integer. Then [x] is the next lower integer. For example, if x is -0.5, [x] = -1, {x} = 0.5. So in such cases, [x] is negative, {x} is positive. Then [x] · {x} is negative. On the right-hand side, 2005x is also negative. So could there be other negative solutions?Wait, in our earlier analysis, we concluded that the only negative solution is x = -1/2006. But let me check for x in other intervals.Suppose n = -1, so x ∈ [-1, 0). Then, [x] = -1, {x} = x +1. So the equation becomes (-1)(x + 1) = 2005x. Let's solve this:- (x + 1) = 2005x=> -x -1 = 2005x=> -1 = 2006x=> x = -1/2006Which is in [-1, 0). So this is the solution we found earlier.What if n = -2? Then x ∈ [-2, -1). Then [x] = -2, {x} = x + 2. Then the equation becomes (-2)(x + 2) = 2005x.Let's solve:-2x -4 = 2005x=> -4 = 2007x=> x = -4/2007 ≈ -0.001993...But x must be in [-2, -1). However, -4/2007 ≈ -0.001993 is between -0.002 and -0.001, which is in [-1, 0), not in [-2, -1). Therefore, no solution in this interval.Similarly, for n = -k where k ≥2, let's see:Suppose n = -k, where k is a positive integer. Then x ∈ [-k, -k +1). Then {x} = x +k. The equation becomes (-k)(x +k) = 2005x.So:- k x - k² = 2005xBring terms with x to one side:- k x - 2005x = k²Factor x:x(-k -2005) = k²Thus,x = k² / (-k -2005) = -k²/(k + 2005)But x must be in [-k, -k +1). Let's check:x = -k²/(k + 2005)We need:- k ≤ x < -k +1So,- k ≤ -k²/(k + 2005) < -k +1Multiply all parts by (k + 2005), which is positive, so inequality signs don't change:- k(k + 2005) ≤ -k² < (-k +1)(k + 2005)First inequality:Left side: -k(k + 2005)Right side: -k²So,- k² -2005k ≤ -k²Add k² to both sides:-2005k ≤ 0Which is true since k >0, so -2005k <0. Therefore, the first inequality holds.Second inequality:- k² < (-k +1)(k + 2005)Expand the right-hand side:(-k)(k) + (-k)(2005) +1*(k) +1*2005= -k² -2005k +k +2005= -k² -2004k +2005So the inequality becomes:- k² < -k² -2004k +2005Add k² to both sides:0 < -2004k +2005Which implies:2004k <2005=> k <2005/2004 ≈1.0005Since k is a positive integer, k=1 is the only solution. Therefore, only when k=1, x is in [-1,0). For k ≥2, x would be in [-k, -k +1), but x computed as -k²/(k +2005) is not in that interval. Thus, only k=1 gives a valid solution in its interval.Therefore, the only solutions are x=0 and x=-1/2006.Hence, the final solutions are x=0 and x=-1/2006.Final AnswerThe solutions are (boxed{0}) and (boxed{-dfrac{1}{2006}}).</think>Given the equation ([x] cdot {x} = 2005x), where ([x]) is the greatest integer less than or equal to (x) and ({x} = x - [x]) is the fractional part of (x), we start by expressing (x) as (n + f), where (n = [x]) is an integer and (f = {x}) is a fractional part such that (0 leq f < 1).Substituting (x = n + f) into the equation, we get:[n cdot f = 2005(n + f)]Rearranging terms, we have:[n f - 2005n - 2005f = 0]Factoring out (n) and (f), we get:[n(f - 2005) - 2005f = 0]Solving for (n), we find:[n = frac{2005f}{f - 2005}]Since (f - 2005) is negative (as (0 leq f < 1)), (n) must be non-positive. Let (n = -k) where (k) is a non-negative integer. Substituting (n = -k) into the equation, we get:[-k = frac{2005f}{f - 2005}]Rewriting this, we have:[k = frac{2005f}{2005 - f}]For (k) to be an integer, (f) must be such that (frac{2005f}{2005 - f}) is an integer. Let (k) be this integer. Then:[f = frac{2005k}{2005 + k}]Since (0 leq f < 1), we have:[0 leq frac{2005k}{2005 + k} < 1]Solving for (k), we find:[2005k < 2005 + k implies 2004k < 2005 implies k < frac{2005}{2004} approx 1.0005]Thus, (k) can only be 0 or 1.- For (k = 0), we get (f = 0) and (n = 0), leading to (x = 0).- For (k = 1), we get (f = frac{2005}{2006}) and (n = -1), leading to (x = -1 + frac{2005}{2006} = -frac{1}{2006}).Thus, the solutions to the equation are:[boxed{0} quad text{and} quad boxed{-dfrac{1}{2006}}],<think>Okay, so I need to solve the equation [x] · {x} = 2005x. Hmm, let's break this down. The problem involves the floor function [x], which is the greatest integer less than or equal to x, and the fractional part {x}, which is x minus [x]. So, {x} = x - [x]. That means {x} is always between 0 (inclusive) and 1 (exclusive). First, let me recall that any real number x can be written as [x] + {x}, right? So, x = n + f, where n is an integer (since [x] is an integer) and f is the fractional part, 0 ≤ f < 1. Maybe substituting this into the equation will help. Let me try that.Let x = n + f, where n = [x] and f = {x}. Then the original equation becomes:n · f = 2005(n + f)Expanding the right side: n · f = 2005n + 2005fNow, let me rearrange terms to get all terms on one side:n · f - 2005n - 2005f = 0Factor terms where possible. Let's see, n and f are in two terms. Maybe factor n from the first two terms:n(f - 2005) - 2005f = 0Hmm, but that doesn't seem immediately helpful. Alternatively, group terms with f:f(n - 2005) - 2005n = 0Wait, let me check that again. Starting from n · f - 2005n - 2005f = 0If I factor f from the first and third term: f(n - 2005) - 2005n = 0Yes, that's correct. So, f(n - 2005) = 2005nThen, solving for f:f = (2005n)/(n - 2005)But remember that f is the fractional part, so 0 ≤ f < 1. Therefore, we must have 0 ≤ (2005n)/(n - 2005) < 1.Additionally, since n is the floor of x, which is an integer. So n must be an integer, and x = n + f, so n ≤ x < n + 1. But f is non-negative, so x is at least n.So, the key equation here is f = (2005n)/(n - 2005), and we have constraints on f: 0 ≤ f < 1. Let's write that down:0 ≤ (2005n)/(n - 2005) < 1Now, we need to find all integers n such that this inequality holds, and then compute x = n + f for each such n.But we need to make sure that the denominator n - 2005 is not zero. So n ≠ 2005. Also, let's consider the sign of the denominator and numerator to make sure f is non-negative (since f is a fractional part, it must be ≥ 0). Therefore:(2005n)/(n - 2005) ≥ 0So, the fraction must be non-negative. Let's analyze when this is true. The fraction is positive if numerator and denominator have the same sign. So, two cases:1. Both numerator and denominator positive:2005n > 0 and n - 2005 > 0Which implies:n > 0 and n > 2005Since 2005 is positive, n > 2005.2. Both numerator and denominator negative:2005n < 0 and n - 2005 < 0Which implies:n < 0 and n < 2005But since 2005n < 0, n must be negative. So n < 0.So, the fraction is non-negative if either n > 2005 or n < 0.But remember that f must be less than 1, so let's consider the two cases separately.Case 1: n > 2005In this case, denominator n - 2005 is positive, numerator 2005n is positive. So f = (2005n)/(n - 2005). We need 0 ≤ f < 1, which is already satisfied for f >=0, but we need f < 1.So:(2005n)/(n - 2005) < 1Multiply both sides by (n - 2005), which is positive here, so inequality sign remains the same:2005n < n - 2005Subtract n from both sides:2005n - n < -20052004n < -2005Divide both sides by 2004 (positive, so inequality sign stays):n < -2005/2004But n is an integer greater than 2005 in this case. However, -2005/2004 is approximately -1.0005, so n < -1.0005. But in this case, n > 2005, which contradicts n < -1.0005. Therefore, no solutions in this case.Case 2: n < 0Here, numerator 2005n is negative (since n is negative) and denominator n - 2005 is negative (since n < 0 < 2005, so n - 2005 < -2005 < 0). So the fraction is positive, which is okay. Now, we have:0 ≤ (2005n)/(n - 2005) < 1First, let's compute (2005n)/(n - 2005). Since both numerator and denominator are negative, the fraction is positive, so it's okay for f.We need to find integers n < 0 such that (2005n)/(n - 2005) < 1.Let's solve the inequality:(2005n)/(n - 2005) < 1Multiply both sides by (n - 2005). But here, n < 0, so n - 2005 < 0. Therefore, multiplying both sides by a negative number reverses the inequality sign:2005n > 1*(n - 2005)So:2005n > n - 2005Subtract n from both sides:2004n > -2005Divide both sides by 2004 (positive, so inequality remains):n > -2005/2004Compute -2005/2004 ≈ -1.000498753. So n > approximately -1.0005. But n is an integer less than 0. Therefore, possible integers n are n = -1, -2, -3, ... but since n > -1.0005, the only integer n satisfying this is n = -1.Therefore, in Case 2, the only possible integer is n = -1.Let's check n = -1.Compute f = (2005*(-1))/(-1 - 2005) = (-2005)/(-2006) = 2005/2006 ≈ 0.9995...Which is less than 1, so that's okay. Then x = n + f = -1 + 2005/2006. Let's compute that:-1 + 2005/2006 = (-2006/2006) + 2005/2006 = (-2006 + 2005)/2006 = (-1)/2006 = -1/2006 ≈ -0.000498...Wait, but x = n + f, which is -1 + 2005/2006. That's correct. So x = -1/2006.But let's check if this x satisfies the original equation.Compute [x] and {x}. Since x = -1/2006 ≈ -0.000498, the floor of x is -1, because -1 ≤ -0.000498 < 0. Therefore, [x] = -1. The fractional part {x} = x - [x] = (-1/2006) - (-1) = (-1/2006) + 1 = (2005/2006).Therefore, [x] · {x} = (-1) * (2005/2006) = -2005/2006.On the other hand, 2005x = 2005*(-1/2006) = -2005/2006.So indeed, [x] · {x} = 2005x. Therefore, x = -1/2006 is a solution.But wait, let's check if there are other solutions. In Case 2, we found only n = -1. But let's confirm that there are no other n. When n < 0, the inequality (2005n)/(n - 2005) < 1 led us to n > -2005/2004 ≈ -1.0005. So n must be greater than approximately -1.0005, but n is an integer less than 0, so only n = -1 is possible.Therefore, the only solution is x = -1/2006.But let's check if there are any other possible cases. Wait, we considered n > 2005 and n < 0. What about n = 0? If n = 0, then x = 0 + f = f, where 0 ≤ f < 1. Let's plug n = 0 into the equation [x] · {x} = 2005x. Then, [x] = 0, {x} = x. So the equation becomes 0 · x = 2005x, which simplifies to 0 = 2005x. Therefore, x must be 0. But x = 0 is n = 0 and f = 0. Let's check: [0] · {0} = 0 · 0 = 0, and 2005·0 = 0. So 0 = 0. So x = 0 is also a solution.Wait a minute! I didn't consider n = 0. In the earlier analysis, we split into cases n > 2005 and n < 0. But n can also be between 0 and 2005 inclusive. Wait, but when n is between 0 and 2005, then the denominator n - 2005 is negative (if n < 2005) or zero (if n = 2005). But n = 2005 is excluded because denominator becomes zero, so division by zero. If n is between 0 and 2004, then denominator is negative. Then numerator 2005n is positive (since n is non-negative). Therefore, the fraction (2005n)/(n - 2005) would be negative, but f is supposed to be non-negative. Therefore, f would be negative, which is impossible. So in these cases, no solutions.However, n = 0 is a special case. Let's see. If n = 0, then the equation f = (2005*0)/(0 - 2005) = 0 / (-2005) = 0. So f = 0. Therefore, x = 0 + 0 = 0. Then, check if [0] · {0} = 2005*0. Yes, [0] = 0, {0} = 0, so 0*0 = 0 = 2005*0. So x = 0 is a valid solution.Wait, but in the previous analysis, when n = 0, the expression (2005n)/(n - 2005) is 0 / (-2005) = 0, which is valid, and f = 0. So f is 0, which is allowed because f must be in [0,1). So x = 0 is a valid solution.Therefore, we have two solutions: x = 0 and x = -1/2006.Wait, but in my earlier analysis for n < 0, I considered n < 0 and got x = -1/2006. Then, n = 0 gives x = 0. But let's check if there's more.Wait, let's check for n = 0. If n = 0, then f = 0. So x = 0. So that's a solution. What about n = 1? Let's test n = 1. Then f would be (2005*1)/(1 - 2005) = 2005/(-2004) ≈ -1.0005, which is negative. But f must be non-negative. Therefore, no solution here. Similarly, for n = 1, 2, ..., 2004, denominator n - 2005 is negative, numerator is positive, so f is negative, which is invalid. For n = 2005, division by zero. For n > 2005, as before, we saw that leads to no solution. For n = 0, x = 0. For n = -1, x = -1/2006. For n < -1, say n = -2, let's check.If n = -2, then f = (2005*(-2))/(-2 - 2005) = (-4010)/(-2007) ≈ 1.999..., which is greater than or equal to 1. But f must be less than 1, so this is invalid. Therefore, n = -2 is invalid. Similarly, n = -3 would give f = (2005*(-3))/( -3 -2005) = (-6015)/(-2008) ≈ 2.996, which is way over 1. So all n < -1 would result in f ≥1, which is invalid. Therefore, only n = -1 and n = 0 are possible.Wait, but when n = 0, f = 0. So x = 0 + 0 = 0. So x = 0 is a solution. Therefore, two solutions: x = 0 and x = -1/2006.But in the original problem statement, the equation is [x] · {x} = 2005x. Let's check x = 0:Left side: [0] · {0} = 0 · 0 = 0Right side: 2005 · 0 = 0So 0 = 0, which is valid. So x = 0 is indeed a solution.Similarly, x = -1/2006:Left side: [-1/2006] · {-1/2006} = (-1) · (2005/2006) = -2005/2006Right side: 2005 · (-1/2006) = -2005/2006So, both sides equal, which is valid.So, there are two solutions: x = 0 and x = -1/2006.Wait, but in the earlier analysis for Case 2 (n < 0), only n = -1 gives f < 1, and n = 0 gives f = 0, which is allowed. So total solutions are two. But let me check if there's another possible n where f is 0. For example, if f = 0, then {x} = 0, so x is integer. Then, [x] · {x} = [x] · 0 = 0. So 0 = 2005x, which implies x = 0. Therefore, the only integer solution is x = 0. So if x is an integer, only x = 0 works. So that's covered.Therefore, the solutions are x = 0 and x = -1/2006.Wait, but wait. When n = 0, x = 0. When n = -1, x = -1/2006. Are there any other solutions where f = 0? If f = 0, then x = n, integer. Then equation becomes [x] · {x} = 2005x. But {x} = 0, so left side is 0. Therefore, 0 = 2005x. Therefore, x = 0. So x = 0 is the only integer solution.So, in total, two solutions: 0 and -1/2006.But let me check in the problem statement. The problem says "Solve the equation: [x] · {x} = 2005x". The problem didn't specify any constraints on x, so x can be any real number. Therefore, both solutions should be valid.But let me verify once more.For x = 0: [0] = 0, {0} = 0. So 0 * 0 = 0 = 2005*0. Correct.For x = -1/2006: [-1/2006] = -1, because -1 ≤ -1/2006 < 0. Then, {-1/2006} = x - [x] = -1/2006 - (-1) = -1/2006 + 1 = 2005/2006. Then, [x] · {x} = (-1) * (2005/2006) = -2005/2006. 2005x = 2005*(-1/2006) = -2005/2006. So equality holds.Thus, both solutions are valid.But wait, in my initial analysis, when I considered n = 0, I saw that f = 0, which is allowed. So x = 0 is a solution. Then, in n < 0 case, only n = -1 gives a valid f. So total two solutions.But let me check for n = -1. When n = -1, f = 2005/2006, which is less than 1, so x = -1 + 2005/2006 = -1/2006. So that's correct.So the solutions are x = 0 and x = -1/2006. But wait, in the problem statement, is there any restriction on x? It doesn't say, for example, x is not zero or something. So both solutions are valid.Wait, but let me check for n = 0 again. If n = 0, then x = 0. Then, [x] = 0, {x} = 0. So equation holds. So 0 is a solution.Therefore, the equation has two solutions: x = 0 and x = -1/2006.But let me check the problem statement again. It says "Solve the equation: [x] · {x} = 2005x". So they might expect both solutions.Wait, but maybe I made a mistake here. Let me check again.Wait, when n = 0, x = 0, and 0 = 2005*0, which is correct.But when n = -1, x = -1/2006, which is approximately -0.000498, and indeed, [-1/2006] = -1, { -1/2006 } = 1 - 1/2006 = 2005/2006. Then, [x] * {x} = -1 * (2005/2006) = -2005/2006. 2005x = 2005*(-1/2006) = -2005/2006. So that works.So, both solutions are correct.But let me check if there's another possible n that I missed. For example, n = 2005, but denominator would be zero, which is undefined. So n = 2005 is invalid.Wait, but what about n = 2006? Let's check n = 2006.But in the case of n > 2005, we saw that:f = (2005*2006)/(2006 - 2005) = (2005*2006)/1 = 2005*2006. But 2005*2006 is a very large number, way more than 1, so f >=1, which is invalid. So n = 2006 is invalid. Similarly, any n >2005 gives f >=1, invalid.So indeed, only two solutions: x = 0 and x = -1/2006.Wait, but the problem says "Solve the equation", and equations can have multiple solutions. So I need to present both solutions.Wait, but let me check once more if n = 0 is the only solution in n >=0. For n >=0, except n =0, when n is positive, f = (2005n)/(n -2005). But if n is positive and less than 2005, denominator is negative, numerator is positive, so f is negative, which is invalid. If n = 2005, division by zero. If n >2005, we saw no solution. So only n =0 gives f =0, which is valid. So x =0 is a solution.Therefore, the solutions are x =0 and x= -1/2006.But let me check if x =0 is acceptable. Since {0} =0, and [0] =0. So 0*0 =0, which is equal to 2005*0 =0. So yes.Thus, both solutions are valid.But wait, the problem says "Solve the equation", so perhaps the answer is both x =0 and x= -1/2006. But I need to confirm.But wait, maybe x=0 is considered a trivial solution? Let me check.But the problem didn't specify any constraints, so x=0 is a valid solution. Therefore, both x=0 and x=-1/2006 satisfy the equation.Therefore, the solutions are x=0 and x= -1/2006.But let me check once again for x=0:Left-hand side: [0] * {0} =0*0=0Right-hand side:2005*0=0So 0=0. Valid.For x=-1/2006:Left-hand side: [-1/2006] * {-1/2006} = (-1) * (2005/2006)= -2005/2006Right-hand side:2005*(-1/2006)= -2005/2006Thus, both sides equal. Valid.Therefore, these are both valid solutions.But wait, maybe I made a mistake in assuming that both are solutions, but actually when n=0, x=0, but also, if x=0, then {x}=0 and [x]=0. So in the equation [x] * {x} =0, and 2005x=0. So equality holds.Therefore, x=0 is a solution, and x=-1/2006 is another solution. So the equation has two solutions.But the problem might expect both solutions. So the answer is x=0 and x=-1/2006.Wait, but the problem says "Solve the equation", so maybe I need to present both solutions.But let me check again.Wait, but in the original equation, if x is positive, then [x] is non-negative, {x} is in [0,1), so left-hand side [x] * {x} is non-negative. The right-hand side 2005x is positive. So for positive x, both sides are non-negative. Let's see if there are any positive solutions.Wait, suppose x is positive. Then, [x] = n ≥0, {x} = f ∈ [0,1). Then, the equation is n*f = 2005(n + f).Let me rearrange this as:n*f -2005n -2005f =0n(f -2005) -2005f =0Which can be rewritten as:f(n -2005) =2005nTherefore, f=2005n/(n -2005)But x is positive, so n ≥0, and f ∈ [0,1). Let's see:If n=0, then f=0, as before, leading to x=0.If n≥1, then denominator n -2005 is negative (since n <2005 for n≥1 up to n=2004). So f=2005n/(n -2005) would be negative, which is invalid. If n=2005, division by zero. If n>2005, denominator positive, numerator positive, so f=2005n/(n -2005). Then, we require f <1, so 2005n <n -2005 =>2004n < -2005, which is impossible since n>2005. Therefore, no positive solutions except x=0.But x=0 is neither positive nor negative. Wait, x=0 is non-negative.Therefore, except x=0, there's no positive solution. So x=0 is the only non-negative solution.The other solution is negative: x=-1/2006.Thus, the equation has two real solutions: x=0 and x=-1/2006.Therefore, the answer should include both solutions.But let me check once again if there are other solutions.Wait, suppose x is negative but not an integer. For example, x=-0.5. Then, [x]=-1, {x}=0.5. Then, [x] * {x}= -1*0.5= -0.5. 2005x=2005*(-0.5)= -1002.5. These are not equal. So x=-0.5 is not a solution. So the only negative solution is x=-1/2006.Therefore, final answer: x=0 and x= -1/2006.But the problem might expect the answer in boxed form. Since there are two solutions, maybe box both?But let me check the problem statement again. It says "Solve the equation". If multiple solutions, then present both.Therefore, the solutions are x=0 and x= -1/2006.But in some cases, maybe x=0 is trivial and x=-1/2006 is non-trivial. Let me confirm.If x=0: yes, trivial.But in the equation, substituting x=0 works. So both are correct.Hence, the solutions are x=0 and x=-1/2006.Therefore, the answer is boxed{0} and boxed{-dfrac{1}{2006}}.But depending on the problem's expectation, sometimes multiple solutions are presented in separate boxes or combined. Since the problem says "put your final answer within boxed{}", and there are two solutions, I should present both boxed.But let me check if the problem allows multiple solutions. Yes, as it's a equation, it can have multiple solutions. So both x=0 and x=-1/2006 are solutions.Alternatively, the problem might expect the non-trivial solution, but since x=0 is valid, it should be included.Thus, the final answers are boxed{0} and boxed{-dfrac{1}{2006}}.Final AnswerThe solutions are boxed{0} and boxed{-dfrac{1}{2006}}.</think>Given the equation ([x] cdot {x} = 2005x), where ([x]) is the greatest integer less than or equal to (x) and ({x} = x - [x]), we start by expressing (x) as (n + f) where (n = [x]) is an integer and (f = {x}) is the fractional part such that (0 leq f < 1).Substituting (x = n + f) into the equation, we get:[n cdot f = 2005(n + f)]Rearranging terms, we have:[n cdot f - 2005n - 2005f = 0]Factoring out terms, we get:[f(n - 2005) = 2005n]Solving for (f), we find:[f = frac{2005n}{n - 2005}]Since (f) must be a fractional part ((0 leq f < 1)), we analyze the constraints on (n):1. Case (n > 2005): - The denominator (n - 2005) is positive, but solving the inequality (frac{2005n}{n - 2005} < 1) leads to a contradiction since (n) would need to be negative. Thus, no solutions here.2. Case (n < 0): - Both the numerator (2005n) and denominator (n - 2005) are negative, making (f) positive. Solving (frac{2005n}{n - 2005} < 1) gives (n > -frac{2005}{2004} approx -1.0005). The only integer (n) here is (n = -1). Substituting (n = -1), we get (f = frac{2005 cdot (-1)}{-1 - 2005} = frac{2005}{2006}). Thus, (x = -1 + frac{2005}{2006} = -frac{1}{2006}).3. Case (n = 0): - Here, (f = frac{2005 cdot 0}{0 - 2005} = 0). Thus, (x = 0 + 0 = 0).Checking both solutions:- For (x = 0), ([0] cdot {0} = 0 cdot 0 = 0) and (2005 cdot 0 = 0), which holds true.- For (x = -frac{1}{2006}), ([- frac{1}{2006}] = -1) and ({ - frac{1}{2006} } = frac{2005}{2006}). Thus, (-1 cdot frac{2005}{2006} = -frac{2005}{2006}) and (2005 cdot -frac{1}{2006} = -frac{2005}{2006}), which also holds true.Thus, the solutions are:[boxed{0} text{ and } boxed{-dfrac{1}{2006}}],<think>Okay, so I need to solve the equation [x] · {x} = 2005x, where [x] is the greatest integer less than or equal to x, and {x} is the fractional part of x, right? Let me recall that {x} = x - [x]. So, maybe I can express everything in terms of [x] and {x}?Let me denote [x] as n, where n is an integer. Then, {x} would be x - n, which is between 0 (inclusive) and 1 (exclusive). So, x can be written as n + f, where f = {x} and 0 ≤ f < 1.Substituting x = n + f into the original equation: [x] · {x} = 2005x. That becomes n · f = 2005(n + f). Let me write that out:n · f = 2005(n + f)Hmm, so n and f are related through this equation. Let me rearrange it to see if I can solve for f in terms of n or vice versa.Starting with nf = 2005n + 2005f.Let me bring all terms to one side: nf - 2005n - 2005f = 0.Maybe factor terms? Let's see:n(f - 2005) - 2005f = 0Hmm, not sure if that helps. Alternatively, maybe factor f and n terms:nf - 2005f = 2005nf(n - 2005) = 2005nThen, solving for f:f = (2005n)/(n - 2005)Okay, so f is expressed in terms of n. But we know that f must satisfy 0 ≤ f < 1. So, substituting f into that inequality:0 ≤ (2005n)/(n - 2005) < 1So, we have two inequalities:1. (2005n)/(n - 2005) ≥ 02. (2005n)/(n - 2005) < 1Let me analyze the first inequality: (2005n)/(n - 2005) ≥ 0The sign of the fraction depends on the numerator and denominator. Since 2005 is positive, the sign of the numerator is the same as the sign of n. The denominator is (n - 2005). So, the fraction is non-negative when both numerator and denominator have the same sign.Case 1: Both numerator and denominator positive.That would mean n > 0 and n - 2005 > 0 ⇒ n > 2005.Case 2: Both numerator and denominator negative.That would mean n < 0 and n - 2005 < 0 ⇒ n < 2005. But if n is negative, n - 2005 is also negative, since 2005 is positive. So, this case is when n < 0.But wait, the numerator is 2005n, which is negative when n is negative. The denominator is n - 2005, which is also negative when n < 2005. So, if n is negative, then numerator is negative, denominator is negative (since n - 2005 < 0 when n < 2005, and n is negative here), so the fraction is positive.So, the fraction is positive in two cases: n > 2005 (both numerator and denominator positive) and n < 0 (both numerator and denominator negative). However, since [x] = n is an integer less than or equal to x, and x can be written as n + f where f is between 0 and 1, x is at least n. But the equation [x] · {x} = 2005x could have solutions where x is positive or negative? Let's check.But let's first process the first inequality. So, the first inequality (2005n)/(n - 2005) ≥ 0 holds when n > 2005 or n < 0.Now, the second inequality: (2005n)/(n - 2005) < 1Let me rewrite this as (2005n)/(n - 2005) - 1 < 0Compute:(2005n - (n - 2005))/(n - 2005) < 0Simplify numerator:2005n - n + 2005 = (2005 - 1)n + 2005 = 2004n + 2005Thus, the inequality is (2004n + 2005)/(n - 2005) < 0So, we have (2004n + 2005)/(n - 2005) < 0Again, the sign of the fraction depends on the numerator and denominator.Let me analyze numerator: 2004n + 2005Denominator: n - 2005So, the fraction is negative when numerator and denominator have opposite signs.Case 1: Numerator positive and denominator negative.2004n + 2005 > 0 and n - 2005 < 0From 2004n + 2005 > 0 ⇒ 2004n > -2005 ⇒ n > -2005/2004 ≈ -1.0005But n is an integer. So, n ≥ -1And denominator n - 2005 < 0 ⇒ n < 2005So, in this case, n must satisfy -1 ≤ n < 2005Case 2: Numerator negative and denominator positive.2004n + 2005 < 0 and n - 2005 > 0From 2004n + 2005 < 0 ⇒ 2004n < -2005 ⇒ n < -2005/2004 ≈ -1.0005Since n is integer, n ≤ -2Denominator n - 2005 > 0 ⇒ n > 2005But n ≤ -2 and n > 2005 can't happen at the same time. So, this case is impossible.Thus, the inequality (2004n + 2005)/(n - 2005) < 0 holds when -1 ≤ n < 2005But remember from the first inequality, we had that n > 2005 or n < 0. So, combining the two inequalities:First inequality: n > 2005 or n < 0Second inequality: -1 ≤ n < 2005Therefore, the intersection is n < 0 and -1 ≤ n < 2005, but n < 0 and -1 ≤ n < 2005 implies -1 ≤ n < 0But n is an integer, so n must be -1 or 0. But n < 0, so n must be -1.Wait, hold on. Let me clarify.First inequality: n > 2005 or n < 0Second inequality: -1 ≤ n < 2005So the overlap (intersection) is:For n > 2005, the second inequality restricts n < 2005, so no overlap there.For n < 0, the second inequality requires n ≥ -1. So, combining, we have -1 ≤ n < 0But n is an integer. So, n must be -1, since n is between -1 (inclusive) and 0 (exclusive). The only integer in that interval is n = -1.Therefore, possible n is -1.So, let's check n = -1.Then, f = (2005n)/(n - 2005) = (2005*(-1))/(-1 - 2005) = (-2005)/(-2006) = 2005/2006 ≈ 0.9975...But f must satisfy 0 ≤ f < 1. Here, 2005/2006 is approximately 0.9975, which is less than 1. So, this is valid.Therefore, x = n + f = -1 + 2005/2006 = ( -2006/2006 + 2005/2006 ) = (-1)/2006 ≈ -0.000498...Wait, but x is approximately -0.000498, but let's compute exactly:x = (-2006 + 2005)/2006 = (-1)/2006So, x = -1/2006 ≈ -0.000498Wait, but let's check if this x indeed satisfies the original equation [x] · {x} = 2005xFirst, compute [x]. Since x = -1/2006 ≈ -0.000498, the greatest integer less than or equal to x is -1, because -1 ≤ -0.000498 < 0. So, [x] = -1Then, {x} = x - [x] = (-1/2006) - (-1) = (-1/2006) + 1 = (2005/2006)So, [x] · {x} = (-1) · (2005/2006) = -2005/2006On the other hand, 2005x = 2005 · (-1/2006) = -2005/2006So, both sides equal -2005/2006. Hence, the equation holds.Therefore, x = -1/2006 is a solution.But wait, is this the only solution? Let me check.Earlier, we concluded that n must be -1. So, that leads to one solution. Let me confirm if there are other possibilities.Wait, when I considered the first inequality, (2005n)/(n - 2005) ≥ 0, it's possible when n > 2005 or n < 0. Then the second inequality reduces it to n = -1. But maybe I missed something?Let me check the case n > 2005. Wait, the first inequality allows n > 2005, but the second inequality requires n < 2005. So, there's no solution in n > 2005. For n < 0, the second inequality requires n ≥ -1. Since n is integer less than 0 and greater than or equal to -1, the only possible integer is n = -1. So, only one solution.Hence, the equation has only one solution at x = -1/2006.But let me check if n = 0 is allowed? Wait, in the first inequality, if n = 0, then f = (2005*0)/(0 - 2005) = 0 / (-2005) = 0. Then x = 0 + 0 = 0. Let's check if x = 0 is a solution.Compute [0] · {0} = 0 · 0 = 0, and 2005*0 = 0. So, 0 = 0. So, x = 0 is a solution?But wait, when n = 0, f = 0. So, x = 0 + 0 = 0. But in this case, {0} = 0 - [0] = 0 - 0 = 0. So, [x] · {x} = 0 · 0 = 0, and 2005x = 0. So, yes, x = 0 satisfies the equation.But in our previous analysis, when n = 0, we need to check if n = 0 is allowed. Let's go back.From the first inequality, (2005n)/(n - 2005) ≥ 0. If n = 0, then the numerator is 0, denominator is -2005. So, 0 / (-2005) = 0. So, 0 is allowed because the inequality is ≥ 0. So, n = 0 is allowed. But in the second inequality, (2004n + 2005)/(n - 2005) < 0. If n = 0, then numerator is 2005, denominator is -2005, so the fraction is 2005 / (-2005) = -1, which is less than 0. So, the second inequality is satisfied.Wait, but in our previous analysis, the second inequality (2004n + 2005)/(n - 2005) < 0 when n is between -1 and 2005. But n = 0 is in that interval. However, from the first inequality, n > 2005 or n < 0. Wait, there's a conflict here.Hold on, the first inequality (2005n)/(n - 2005) ≥ 0. For n = 0, this is 0 / (-2005) = 0, which is ≥ 0. So, n = 0 is allowed. Then, even though n = 0 is in the second inequality's allowed interval (-1 ≤ n < 2005), but in the first inequality, n = 0 is not in the regions n > 2005 or n < 0. Wait, but n = 0 is allowed here because the first inequality is satisfied (equals zero). So, maybe the initial split into n > 2005 or n < 0 was when the fraction is positive, but when the fraction is zero, it's a separate case.Wait, actually, the first inequality was (2005n)/(n - 2005) ≥ 0. So, equality occurs when numerator is zero. The numerator is 2005n, which is zero when n = 0. So, n = 0 is a valid case here, even though it's not in n > 2005 or n < 0. So, I think I missed considering the case when the fraction is zero. So, when is (2005n)/(n - 2005) = 0? When 2005n = 0, i.e., n = 0. Then, the fraction is zero, which satisfies the first inequality. So, n = 0 is a possible case.Therefore, we have three cases to consider:1. n > 20052. n < 03. n = 0But in the second inequality, when n = 0, we saw that the second inequality (2004n + 2005)/(n - 2005) < 0 becomes (2005)/(-2005) = -1 < 0, which is true. So, n = 0 is allowed.Therefore, n can be -1 or 0.Wait, but when n = 0, f = (2005 * 0)/(0 - 2005) = 0 / (-2005) = 0. So, x = 0 + 0 = 0, which we saw earlier is a solution. So, x = 0 is another solution.Therefore, there are two solutions: x = 0 and x = -1/2006.Wait, but let me check if x = 0 is indeed a solution. Let's plug into the original equation:[x] · {x} = [0] · {0} = 0 · 0 = 02005x = 2005 · 0 = 0So, 0 = 0, which holds. Therefore, x = 0 is a valid solution.But why didn't our earlier analysis catch this? Because when n = 0, f = 0, which is allowed (0 ≤ f < 1). However, in the first analysis, we considered the fraction (2005n)/(n - 2005) ≥ 0, which allowed n = 0 as a separate case where the fraction is zero. So, in addition to n = -1, n = 0 also gives a valid solution.But let me check for n = 0:From the equation nf = 2005(n + f)If n = 0, then 0 · f = 2005(0 + f) ⇒ 0 = 2005f ⇒ f = 0.So, x = 0 + 0 = 0. So, that's valid.But in the initial steps, we had f = (2005n)/(n - 2005). If n = 0, f = 0. So, that works.Therefore, there are two solutions: x = 0 and x = -1/2006.Wait, but in the original problem statement, it's to "solve the equation", which might have multiple solutions. Let me check if there are other possible solutions.Wait, when n = 0, we get x = 0. When n = -1, we get x = -1/2006. Let me check if there are other integer values of n that could satisfy the conditions.Wait, in the first inequality, when n = 0, the fraction is zero. But if n is positive but less than 2005, then (2005n)/(n - 2005) would be negative, because numerator is positive (n > 0) and denominator is negative (n - 2005 < 0). So, the fraction would be negative, which doesn't satisfy the first inequality (≥ 0). Similarly, if n is between 0 and 2005, the fraction is negative, which doesn't satisfy the first inequality. Only n > 2005 (which conflicts with the second inequality) or n < 0 (but limited to n = -1 by the second inequality) or n = 0.Wait, n = 0 is a separate case where the fraction is zero. So, the possible solutions are n = -1 and n = 0.Thus, the solutions are x = -1/2006 and x = 0.But let me check if x = 0 is the only solution with n = 0. Since when n = 0, f must be 0, so x = 0. So, that's unique.Similarly, for n = -1, f = 2005*(-1)/(-1 - 2005) = 2005/2006, so x = -1 + 2005/2006 = -1/2006.Therefore, the equation has two solutions: x = 0 and x = -1/2006.Wait, but let me check if x = 0 is indeed a solution. As above, [0] = 0, {0} = 0. So, 0 * 0 = 0, and 2005 * 0 = 0. So, yes, it works.However, in some contexts, the fractional part {x} is defined as x - [x], which for x integer would be 0. So, that's correct.But let's check x = -1/2006. [x] = -1, {x} = x - (-1) = x + 1 = (-1/2006) + 1 = 2005/2006. Then, [x] * {x} = (-1) * (2005/2006) = -2005/2006. 2005x = 2005*(-1/2006) = -2005/2006. So, both sides equal -2005/2006. Therefore, correct.So, both solutions are valid.But wait, how come the initial analysis didn't catch n = 0? Because when solving for f in terms of n, we had f = (2005n)/(n - 2005). For n = 0, this gives f = 0, which is valid, but in the first inequality, we had (2005n)/(n - 2005) ≥ 0, which is satisfied when n > 2005 or n < 0, but also when n = 0 (since the fraction is zero). So, the first inequality is satisfied for n ≥ 0? Wait, no. Let me re-examine the first inequality.Original first inequality: (2005n)/(n - 2005) ≥ 0So, the fraction is non-negative. So, when is this true?Case 1: Both numerator and denominator positive: 2005n > 0 and n - 2005 > 0 ⇒ n > 0 and n > 2005 ⇒ n > 2005Case 2: Both numerator and denominator negative: 2005n < 0 and n - 2005 < 0 ⇒ n < 0 and n < 2005 ⇒ n < 0Case 3: Numerator is zero: 2005n = 0 ⇒ n = 0. In this case, the fraction is 0/(0 - 2005) = 0, which is ≥ 0.Therefore, the inequality (2005n)/(n - 2005) ≥ 0 holds when n > 2005, n < 0, or n = 0.Therefore, n can be in three cases:1. n > 20052. n < 03. n = 0Then, applying the second inequality (2004n + 2005)/(n - 2005) < 0:For n > 2005: denominator is positive. The numerator 2004n + 2005 is positive (since n > 2005). So, positive/positive is positive, which is not less than 0. So, no solutions here.For n < 0: As before, the second inequality reduces to -1 ≤ n < 2005. But n < 0, so combining gives -1 ≤ n < 0. Since n is integer, n = -1.For n = 0: (2004*0 + 2005)/(0 - 2005) = 2005/(-2005) = -1 < 0, which satisfies the inequality.Therefore, the solutions are n = -1 and n = 0.Therefore, x = -1 + 2005/2006 = -1/2006 and x = 0 + 0 = 0.So, the equation has two solutions: x = 0 and x = -1/2006.But let me confirm once more. If x = 0, then [x] = 0, {x} = 0, so 0*0 = 0, and 2005*0 = 0. Correct. If x = -1/2006, then [x] = -1, {x} = 2005/2006, so (-1)*(2005/2006) = -2005/2006, and 2005*(-1/2006) = -2005/2006. Correct.Hence, the solutions are x = 0 and x = -1/2006.But wait, the problem says "Solve the equation". Depending on the problem's context, maybe x = 0 is considered a trivial solution, but perhaps I need to check if there are other solutions. Wait, let's check with another integer.Suppose n = 1. Then, f = (2005*1)/(1 - 2005) = 2005/(-2004) ≈ -1.0005..., which is negative. But f must be ≥ 0. So, invalid.n = -2. Then f = (2005*(-2))/(-2 - 2005) = (-4010)/(-2007) ≈ 2.000..., but f must be less than 1. So, 2.000... is invalid.Similarly, n = 1: f negative, invalid.n = -1: f ≈ 0.9975, valid.n = 0: f = 0, valid.So, indeed, only n = -1 and n = 0 give valid f in [0,1). Thus, only two solutions.Therefore, the solutions are x = 0 and x = -1/2006.But let me check the problem statement again. It says "Solve the equation: [x] · {x} = 2005x". So, unless there are restrictions on x (like x must be real), but I think x is real here.But I found two solutions. But maybe the problem expects both solutions, or is there a reason to discard x = 0?Wait, when x = 0, [x] = 0 and {x} = 0. Then, 0 * 0 = 0, which equals 2005x = 0. So, valid. But maybe in some contexts, the problem is looking for non-trivial solutions? But unless specified, both solutions should be considered.But let me check once again if n = 0 is allowed. From the equation f = 2005n/(n - 2005). If n = 0, then f = 0. Since f = 0 is allowed (as {x} = 0 when x is integer), then x = 0 is a valid solution.Therefore, both x = 0 and x = -1/2006 are solutions. However, let me check the problem statement again. The original problem says "Solve the equation: [x] · {x} = 2005x". The problem didn't specify any restrictions, so both solutions are valid.But wait, let me check if x = 0 is a solution. If x = 0, then [x] = 0, {x} = 0, so 0 * 0 = 0, which equals 2005 * 0 = 0. So, yes, 0 = 0. Hence, x = 0 is a valid solution.Therefore, the equation has two solutions: 0 and -1/2006.But wait, let me check x = -1/2006. Let's compute [x] and {x} again:x = -1/2006 ≈ -0.000498So, the greatest integer less than or equal to x is -1, because -1 ≤ -0.000498 < 0. Hence, [x] = -1.Then, {x} = x - [x] = (-1/2006) - (-1) = (-1/2006) + 1 = 2005/2006 ≈ 0.9975Then, [x] * {x} = (-1) * (2005/2006) = -2005/2006On the other side, 2005x = 2005 * (-1/2006) = -2005/2006. Hence, both sides equal. So, valid.Hence, both solutions are correct.But maybe the problem expects both solutions. Let me check.Alternatively, maybe I made a mistake in considering n = 0. Let me think. When n = 0, then x = 0 + 0 = 0. So, x is an integer. But in the original equation, if x is an integer, then {x} = 0, so the left-hand side [x] * {x} = n * 0 = 0. The right-hand side is 2005x = 2005 * n. Therefore, 0 = 2005n. Thus, n must be 0. Therefore, x = 0 is the only integer solution.Therefore, x = 0 is a solution. However, when x is not an integer, then {x} ≠ 0, so the equation [x] * {x} = 2005x can have other solutions, such as x = -1/2006.Therefore, the equation has two solutions: x = 0 and x = -1/2006.But let me check again the case when x is an integer. If x is an integer, {x} = 0, so the equation reduces to [x] * 0 = 2005x ⇒ 0 = 2005x ⇒ x = 0. Therefore, the only integer solution is x = 0.For non-integer x, we can write x = n + f, n integer, 0 < f < 1. Then, the equation becomes n * f = 2005(n + f). Which leads to the solution n = -1, f = 2005/2006, so x = -1 + 2005/2006 = -1/2006.Hence, two solutions: x = 0 and x = -1/2006.But wait, is x = -1/2006 a non-integer? Yes, because it's -0.000498..., which is not an integer.Hence, the equation has two solutions.But let me check if x = -1/2006 is indeed the only non-integer solution.Suppose there exists another non-integer solution. So, x = n + f, n integer, 0 < f < 1.From the equation, nf = 2005(n + f)So, nf = 2005n + 2005fRearranged, n(f - 2005) = 2005fSo, n = (2005f)/(f - 2005)But since n is integer, (2005f)/(f - 2005) must be integer.But for n to be integer, f must be a rational number such that 2005f is divisible by (f - 2005). However, in our earlier analysis, we found that the only possible n is -1, leading to f = 2005/2006.Therefore, only one non-integer solution, x = -1/2006, and one integer solution, x = 0.Therefore, the solutions are x = 0 and x = -1/2006.Hence, the answer should be both solutions.But let me check the problem statement once again. It says "Solve the equation", so if there are multiple solutions, they should be presented. However, looking back to my initial analysis, I derived n = -1 and n = 0, leading to two solutions. Therefore, the solutions are x = 0 and x = -1/2006.But to confirm, let's check if x = -1/2006 is the only non-integer solution. Suppose there exists another n and f. Let's assume n is another integer, say n = -2.Then, f = (2005*(-2))/(-2 - 2005) = (-4010)/(-2007) ≈ 2.000498..., which is greater than 1, which is invalid because f must be less than 1. Similarly, n = 1 would give f = (2005*1)/(1 - 2005) ≈ -1.0005, which is negative. So, invalid. Therefore, only n = -1 and n = 0 gives valid f in [0,1).Therefore, the equation has exactly two solutions: x = 0 and x = -1/2006.But wait, in the original problem statement, when defining {x} = x - [x], if x is negative, how does {x} behave? For example, x = -0.5, then [x] = -1, and {x} = x - [x] = (-0.5) - (-1) = 0.5. So, {x} is always in [0,1), even for negative x. So, that's correct. Therefore, x = -1/2006 is valid, as {x} = 2005/2006, which is in [0,1).Therefore, the conclusion is the equation has two solutions: x = 0 and x = -1/2006.But let me check the problem again. It says "Solve the equation: [x] · {x} = 2005x". Since the problem is likely from a competition or textbook problem, maybe there is only one solution expected? But according to my analysis, there are two.Wait, but perhaps I made a mistake in considering x = 0. Let me check once again:If x = 0, then [x] = 0, {x} = 0. Then, 0 * 0 = 0 = 2005 * 0. So, valid.But maybe the problem considers {x} to be defined only for non-integer x? But no, the definition {x} = x - [x] holds for all real numbers, including integers. For integers, {x} = 0.Therefore, x = 0 is indeed a valid solution.However, in some contexts, fractional part functions are considered only for non-integers, but according to the given definition, {x} = x - [x], which is valid for all real x, and {x} is in [0,1). Therefore, x = 0 is a valid solution.Therefore, the equation has two solutions.But wait, let me verify with an example. Suppose x = 0. Then, the equation becomes 0 * 0 = 0, which is correct. If x = -1/2006, then as we saw, both sides equal -2005/2006. So, both are valid.But why didn't I think of x = 0 initially? Because when I first set n = [x], then x = n + f, and the equation became nf = 2005(n + f). So, for x = 0, n = 0 and f = 0. So, substituting into the equation: 0 * 0 = 2005(0 + 0), which holds. Therefore, x = 0 is a solution.Therefore, the conclusion is that the equation has two solutions: x = 0 and x = -1/2006.But to ensure completeness, let me check if there are other integer solutions. Suppose x is an integer, then {x} = 0, so the equation becomes [x] * 0 = 2005x ⇒ 0 = 2005x ⇒ x = 0. Therefore, the only integer solution is x = 0.For non-integer solutions, we found x = -1/2006. Therefore, two solutions in total.Hence, the answer is x = 0 and x = -1/2006. But let me check if the problem expects multiple answers. Since the user instruction says "put your final answer within boxed{}", which usually expects a single answer. But maybe there are two answers. Let me check.Wait, in the problem statement, it says "Solve the equation", so depending on the equation, there could be multiple solutions. The instruction here says "put your final answer within boxed{}", which in many cases can accept multiple boxed answers. But sometimes, even if there are multiple solutions, the problem might expect both. Let me check my analysis again.Alternatively, perhaps x = 0 is considered trivial and maybe the problem expects the non-trivial solution. But according to the definitions given, x = 0 is a valid solution. Unless there's a miscalculation.Alternatively, let me consider the case when x is positive. Suppose x is positive. Then, [x] is non-negative, {x} is in [0,1). So, [x] · {x} is non-negative. But 2005x is positive. So, both sides are non-negative. However, for x positive:If x is positive integer, {x} = 0, so [x] · {x} = 0, but 2005x is positive, so no solution here.If x is positive non-integer, then [x] ≥ 0, {x} > 0. Then, [x] · {x} ≥ 0, and 2005x > 0. But is there a positive solution?Suppose x is positive. Let's assume x = n + f, n ≥ 0 integer, 0 < f < 1.Then, equation: n f = 2005(n + f)Rearranged: n f - 2005n - 2005 f = 0 ⇒ n(f - 2005) - 2005 f = 0 ⇒ n = (2005 f)/(f - 2005)But since n ≥ 0, the RHS must be non-negative.So, (2005 f)/(f - 2005) ≥ 0Since f > 0, numerator is positive. Therefore, denominator must be positive. So, f - 2005 > 0 ⇒ f > 2005. But f < 1, so f > 2005 is impossible. Therefore, no positive solutions.Therefore, all solutions must be non-positive. Hence, x = 0 and x = -1/2006.Therefore, the solutions are x = 0 and x = -1/2006.But the problem might expect both solutions. However, the instruction says "put your final answer within boxed{}". In some cases, multiple answers can be boxed separately. For example:The solutions are boxed{0} and boxed{-dfrac{1}{2006}}.But maybe the problem expects both. Alternatively, maybe I missed something and there's only one solution. Let me check again.Wait, when n = 0, f = 0, so x = 0. When n = -1, f = 2005/2006, x = -1/2006. No other n gives valid f. Therefore, two solutions. So, I think both are correct.But I should check once more if x = 0 is allowed. Since {x} = x - [x], if x = 0, {0} = 0 - 0 = 0. So, yes, it's allowed. Therefore, x = 0 is a valid solution.Therefore, the conclusion is that the equation has two solutions: x = 0 and x = -1/2006.But let me check for x = -1/2006:[x] = -1, {x} = 2005/2006[x] * {x} = (-1) * (2005/2006) = -2005/20062005x = 2005 * (-1/2006) = -200]

🔑:1. Given the equation ([x] cdot {x} = 2005x), we start by expressing (x) in terms of its integer and fractional parts: [ x = [x] + {x} ] where ([x]) is the greatest integer less than or equal to (x) and ({x} = x - [x]) is the fractional part of (x).2. Substitute (x = [x] + {x}) into the given equation: [ [x] cdot {x} = 2005([x] + {x}) ]3. Distribute the 2005 on the right-hand side: [ [x] cdot {x} = 2005[x] + 2005{x} ]4. Rearrange the equation to isolate ({x}): [ [x] cdot {x} - 2005{x} = 2005[x] ] [ {x}([x] - 2005) = 2005[x] ]5. Solve for ({x}): [ {x} = frac{2005[x]}{[x] - 2005} ]6. The condition (0 leq {x} < 1) must hold. We analyze this condition for different ranges of ([x]).7. Case 1: ([x] < 2005) - If ([x] < 2005), then ([x] - 2005 < 0), making the denominator negative. - For ({x}) to be non-negative, the numerator must also be non-positive: [ 2005[x] leq 0 implies [x] leq 0 ] - Let (n = -[x]), where (n geq 0). Then: [ {x} = frac{2005(-n)}{-n - 2005} = frac{-2005n}{-n - 2005} = frac{2005n}{n + 2005} ] - The condition (0 leq {x} < 1) becomes: [ 0 leq frac{2005n}{n + 2005} < 1 ] - Simplify the inequality: [ 2005n < n + 2005 implies 2004n < 2005 implies n < frac{2005}{2004} ] - Since (n) is an integer, the only possible values are (n = 0) or (n = 1): - If (n = 0), then ([x] = 0) and ({x} = 0), so (x = 0). - If (n = 1), then ([x] = -1) and: [ {x} = frac{2005 cdot 1}{1 + 2005} = frac{2005}{2006} ] [ x = [x] + {x} = -1 + frac{2005}{2006} = frac{-1}{2006} ]8. Case 2: ([x] > 2005) - If ([x] > 2005), then ([x] - 2005 > 0), making the denominator positive. - For ({x}) to be less than 1, the numerator must be less than the denominator: [ 2005[x] < [x] - 2005 implies 2004[x] < -2005 implies [x] < frac{-2005}{2004} ] - Since ([x]) is an integer, there are no integers less than (frac{-2005}{2004}) that are greater than 2005. Thus, there are no solutions in this case.9. Therefore, the only solutions are: [ x = 0 quad text{and} quad x = frac{-1}{2006} ]The final answer is (boxed{x = 0 text{ and } x = frac{-1}{2006}}).