🧠:[<think>Okay, so I have this problem from the IMO 1973, problem A3. The question is: given that a and b are real numbers for which the equation x⁴ + a x³ + b x² + a x + 1 = 0 has at least one real solution, find the least possible value of a² + b². First, I need to understand the problem. We have a quartic equation with real coefficients, and it's given that there's at least one real root. We need to find the minimum value of a² + b². So, essentially, among all pairs (a, b) that make the equation have a real root, find the one where a² + b² is as small as possible. Let me start by analyzing the equation. It's a quartic polynomial: x⁴ + a x³ + b x² + a x + 1. Hmm, interestingly, the coefficients are symmetric. The coefficients of x⁴ and the constant term are both 1, and the coefficients of x³ and x are both a. This symmetry suggests that maybe the polynomial is a reciprocal polynomial. Let me check that. A reciprocal polynomial satisfies P(x) = xⁿ P(1/x), where n is the degree. For this polynomial, let's compute x⁴ P(1/x):x⁴ P(1/x) = x⁴ ( (1/x)^4 + a (1/x)^3 + b (1/x)^2 + a (1/x) + 1 ) = x⁴ (1/x⁴ + a/x³ + b/x² + a/x + 1) = 1 + a x + b x² + a x³ + x⁴ = P(x). Yes, so this is indeed a reciprocal polynomial. Reciprocal polynomials have roots that come in reciprocal pairs. That is, if r is a root, then 1/r is also a root. Since the polynomial is of even degree (4), it's possible that reciprocal roots could be real or complex. But since we need at least one real root, let's think about what that implies. If r is a real root, then 1/r is also a root. So unless r = 1 or r = -1, which are their own reciprocals, the roots come in pairs (r, 1/r). So, if there's a real root, either it's 1 or -1, or there are two real roots r and 1/r. But the problem states that there's at least one real solution, so we need to consider both cases: either the real roots are 1 or -1, or there are other real roots which are reciprocals of each other. Let me first check if 1 or -1 could be roots. Let's substitute x = 1 into the equation: P(1) = 1 + a + b + a + 1 = 2 + 2a + b. Similarly, substituting x = -1:P(-1) = 1 - a + b - a + 1 = 2 - 2a + b. So, for 1 to be a root, 2 + 2a + b = 0, and for -1 to be a root, 2 - 2a + b = 0. If either of these is true, then we have a real root. But if the real root is not 1 or -1, then there must be another pair of reciprocal roots. But maybe working directly with the reciprocal polynomial is more complicated. Another approach is to use substitution. Since the polynomial is reciprocal, we can use the substitution y = x + 1/x. This substitution is often helpful for reciprocal polynomials because it reduces the degree. Let's try that. Let me let y = x + 1/x. Then, note that y² = x² + 2 + 1/x², so x² + 1/x² = y² - 2. Similarly, x³ + 1/x³ = y³ - 3y. But our equation is x⁴ + a x³ + b x² + a x + 1 = 0. Let's divide both sides by x² to make it symmetric:x² + a x + b + a/x + 1/x² = 0.Which can be written as (x² + 1/x²) + a(x + 1/x) + b = 0.Using the substitution y = x + 1/x, then x² + 1/x² = y² - 2. Therefore, the equation becomes:(y² - 2) + a y + b = 0 ⇒ y² + a y + (b - 2) = 0.So, the original quartic equation in x is equivalent to the quadratic equation in y: y² + a y + (b - 2) = 0. But here, y = x + 1/x. However, for real x ≠ 0, y = x + 1/x can take certain values. Let's recall that for real numbers x, the expression x + 1/x has a minimum absolute value of 2. If x > 0, then by AM ≥ GM, x + 1/x ≥ 2. Similarly, if x < 0, let x = -t where t > 0, then x + 1/x = - (t + 1/t) ≤ -2. Therefore, y ∈ (-∞, -2] ∪ [2, ∞). Therefore, if the original equation has a real root x, then y = x + 1/x must satisfy y² + a y + (b - 2) = 0, and y must be in (-∞, -2] ∪ [2, ∞). So, the quadratic in y must have at least one real root in (-∞, -2] ∪ [2, ∞). So, now our problem reduces to: find real numbers a and b such that the quadratic equation y² + a y + (b - 2) = 0 has at least one real root y with |y| ≥ 2, and find the minimal value of a² + b². Therefore, instead of dealing with the quartic equation in x, we can work with the quadratic equation in y. This seems more manageable. So, now, the problem is equivalent to: minimize a² + b² subject to the quadratic equation y² + a y + (b - 2) = 0 having a root y with |y| ≥ 2. Let me denote c = b - 2. Then the quadratic equation becomes y² + a y + c = 0, and we need to minimize a² + (c + 2)², given that this quadratic has at least one root y with |y| ≥ 2. Alternatively, the original c is c = b - 2, so b = c + 2, so a² + b² = a² + (c + 2)^2. So, the problem becomes: minimize a² + (c + 2)^2, given that the quadratic y² + a y + c = 0 has at least one root y with |y| ≥ 2. So, perhaps this is a constrained optimization problem. Let me consider that. Let me denote the quadratic as Q(y) = y² + a y + c. We need Q(y) = 0 to have at least one solution y with |y| ≥ 2. To have such a solution, either the quadratic has a root ≥ 2 or a root ≤ -2. Alternatively, the maximum of the roots is ≥ 2 or the minimum of the roots is ≤ -2. But perhaps a better way is to use the fact that if a quadratic equation has a root outside the interval (-2, 2), then either Q(2) ≤ 0 or Q(-2) ≤ 0. Wait, is that true? Wait, the quadratic Q(y) is a parabola opening upwards (since the coefficient of y² is 1 > 0). Therefore, the quadratic will have a minimum at y = -a/(2*1) = -a/2. If the minimum of Q(y) occurs at y = -a/2. If the quadratic has a root ≥ 2, then since it opens upwards, Q(2) ≤ 0. Similarly, if it has a root ≤ -2, then Q(-2) ≤ 0. Therefore, the quadratic Q(y) has a root with |y| ≥ 2 if and only if Q(2) ≤ 0 or Q(-2) ≤ 0. Let me verify that. Suppose Q(2) ≤ 0. Then, since the parabola opens upwards, if Q(2) ≤ 0, there must be a root ≥ 2. Similarly, if Q(-2) ≤ 0, then there must be a root ≤ -2. Conversely, if there is a root ≥ 2, then Q(2) ≤ 0, and similarly for the other case. Therefore, the conditions are Q(2) ≤ 0 or Q(-2) ≤ 0. So, the quadratic equation y² + a y + c = 0 has a root with |y| ≥ 2 if and only if 2² + a*2 + c ≤ 0 or (-2)^2 + a*(-2) + c ≤ 0. Calculating these, Q(2) = 4 + 2a + c ≤ 0 and Q(-2) = 4 - 2a + c ≤ 0. So, our constraints are: 4 + 2a + c ≤ 0 or 4 - 2a + c ≤ 0. Therefore, the problem reduces to minimizing a² + (c + 2)^2 subject to either 4 + 2a + c ≤ 0 or 4 - 2a + c ≤ 0. So, now we have two separate inequality constraints. The minimal value of a² + (c + 2)^2 would be the minimum between the minimal values under each constraint. So, perhaps we can consider two cases:Case 1: 4 + 2a + c ≤ 0. Minimize a² + (c + 2)^2.Case 2: 4 - 2a + c ≤ 0. Minimize a² + (c + 2)^2.Then, take the smaller of the two minima. Alternatively, since the problem allows either of the constraints to be satisfied, the feasible region is the union of the regions defined by each inequality. So, the minimal value would be the minimal distance squared (since we are minimizing a² + (c + 2)^2) from the point (a, c) to the origin shifted by (0, -2). Wait, actually, a² + (c + 2)^2 is the square of the distance from (a, c) to (0, -2). So, we need the minimal distance from (0, -2) to the union of the regions 4 + 2a + c ≤ 0 and 4 - 2a + c ≤ 0. Therefore, geometrically, this is equivalent to finding the closest point to (0, -2) in the union of the two half-planes defined by 4 + 2a + c ≤ 0 and 4 - 2a + c ≤ 0. So, let's visualize this. The two lines are:First line: 2a + c = -4Second line: -2a + c = -4These are two straight lines in the a-c plane. The first line has a slope of -2, and the second has a slope of 2. They intersect where 2a + c = -4 and -2a + c = -4. Solving these two equations:Adding them: 2a + c -2a + c = -4 -4 ⇒ 2c = -8 ⇒ c = -4. Then substituting back, 2a + (-4) = -4 ⇒ 2a = 0 ⇒ a = 0. So, they intersect at (a, c) = (0, -4). So, the two lines intersect at (0, -4) and form a 'V' shape opening downward. The feasible region is the area below these two lines. But our objective is to find the point (a, c) in this feasible region that is closest to (0, -2). Because the minimal distance squared from (0, -2) to (a, c) is a² + (c + 2)^2, which is exactly the expression we need to minimize. So, geometrically, the minimal distance from the point (0, -2) to the union of the two half-planes defined by 2a + c ≤ -4 and -2a + c ≤ -4. But since the feasible region is the union of the two half-planes, the closest point would be the closest point on either of the two lines 2a + c = -4 or -2a + c = -4 to the point (0, -2). Therefore, the minimal value is the minimal distance from (0, -2) to either of the two lines. Wait, but in optimization, if we have to stay within the feasible region (the union of the half-planes), the closest point would lie on the boundary of the feasible region, which is the union of the two lines. So, to find the minimum, we can compute the distance from (0, -2) to each line and take the smaller one. But let's confirm this. The closest point to (0, -2) in the feasible region (which is the union of the two half-planes) must lie on the boundary, i.e., on one of the two lines. Because if the point (0, -2) is not in the feasible region, then the closest point is on the boundary. If it is in the feasible region, then the minimal distance is zero. But is (0, -2) in the feasible region? Let's check. For the first inequality 4 + 2a + c ≤ 0: substituting a = 0, c = -2, we get 4 + 0 + (-2) = 2 ≤ 0? No, 2 is not ≤ 0. Similarly, for the second inequality 4 - 2a + c ≤ 0: substituting a = 0, c = -2, 4 - 0 + (-2) = 2 ≤ 0? Also no. Therefore, (0, -2) is not in the feasible region, so the closest point must lie on the boundary. Therefore, the minimal distance is the minimal distance from (0, -2) to the two lines 2a + c = -4 and -2a + c = -4. So, let's compute the distance from (0, -2) to each line. The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). First line: 2a + c = -4 ⇒ 2a + c + 4 = 0. So, A = 2, B = 1, C = 4. Distance to (0, -2): |2*0 + 1*(-2) + 4| / sqrt(4 + 1) = | -2 + 4 | / sqrt(5) = |2| / sqrt(5) = 2/sqrt(5). Second line: -2a + c = -4 ⇒ -2a + c + 4 = 0. So, A = -2, B = 1, C = 4. Distance to (0, -2): | -2*0 + 1*(-2) + 4 | / sqrt(4 + 1) = | -2 + 4 | / sqrt(5) = 2/sqrt(5). Therefore, the distance to both lines is the same: 2/sqrt(5). Therefore, the minimal value of sqrt(a² + (c + 2)^2) is 2/sqrt(5), so the minimal value of a² + (c + 2)^2 is (2/sqrt(5))² = 4/5. Wait, but hold on. But c = b - 2, so the original expression is a² + b² = a² + (c + 2)^2, which we have just minimized to 4/5. Therefore, the minimal value of a² + b² is 4/5. But let's check this result. Let me verify. First, if the minimal distance squared is 4/5, then there exists a point (a, c) on one of the lines such that a² + (c + 2)^2 = 4/5. Let's find such a point. The closest point on the line 2a + c = -4 to (0, -2) can be found by moving along the direction perpendicular to the line. The line 2a + c = -4 has a normal vector (2, 1). So, the direction from (0, -2) to the line is along (2, 1). Parametrizing this: starting at (0, -2), moving along (2, 1) scaled by t. So, the point is (2t, -2 + t). This point must lie on the line 2a + c = -4. So:2*(2t) + (-2 + t) = -4 ⇒ 4t -2 + t = -4 ⇒ 5t = -2 ⇒ t = -2/5. Therefore, the closest point is (2*(-2/5), -2 + (-2/5)) = (-4/5, -12/5). Check if this is on the line 2a + c = 2*(-4/5) + (-12/5) = -8/5 -12/5 = -20/5 = -4. Yes, correct. Now, compute a² + (c + 2)^2:a = -4/5, c = -12/5. a² = (16/25), (c + 2) = (-12/5 + 10/5) = (-2/5), so squared is 4/25. Total: 16/25 + 4/25 = 20/25 = 4/5. Correct. Similarly, for the other line -2a + c = -4. Let's compute the closest point. The normal vector is (-2, 1). Moving from (0, -2) along (-2, 1) direction. Parametrize as (-2t, -2 + t). Plug into the line equation:-2*(-2t) + (-2 + t) = -4 ⇒ 4t -2 + t = -4 ⇒ 5t = -2 ⇒ t = -2/5. Thus, the point is (-2*(-2/5), -2 + (-2/5)) = (4/5, -12/5). Check if this is on the line: -2*(4/5) + (-12/5) = -8/5 -12/5 = -20/5 = -4. Correct. Compute a² + (c + 2)^2: a = 4/5, c = -12/5. a² = 16/25, (c + 2) = (-12/5 + 10/5) = (-2/5), squared is 4/25. Total again 20/25 = 4/5. Therefore, the minimal value is indeed 4/5. But wait, let's check if these points correspond to valid a and b. Remember that c = b - 2, so b = c + 2. So for the first point (a, c) = (-4/5, -12/5), b = -12/5 + 10/5 = -2/5. Therefore, (a, b) = (-4/5, -2/5). Similarly, for the other point (4/5, -12/5), b is the same: -12/5 + 10/5 = -2/5. So, (a, b) = (4/5, -2/5). Therefore, these two points give a² + b² = (16/25) + (4/25) = 20/25 = 4/5. Now, we need to verify that these points indeed correspond to the original equation having a real root. Let's take one of them, say (a, b) = (-4/5, -2/5). Then the quartic equation is:x⁴ - (4/5)x³ - (2/5)x² - (4/5)x + 1 = 0. Let's check if this has a real root. Using our substitution y = x + 1/x, then the quadratic in y is y² + a y + (b - 2) = y² - (4/5)y + (-2/5 - 2) = y² - (4/5)y - 12/5 = 0. Solving this quadratic: y = [4/5 ± sqrt( (16/25) + 240/25 )]/2 = [4/5 ± sqrt(256/25)]/2 = [4/5 ± 16/5]/2. So, y = (4/5 + 16/5)/2 = (20/5)/2 = 4/2 = 2, or y = (4/5 - 16/5)/2 = (-12/5)/2 = -6/5. Now, y must be in (-∞, -2] ∪ [2, ∞). So y = 2 is acceptable, but y = -6/5 is not (since |-6/5| = 6/5 < 2). Therefore, the quadratic in y has a root y = 2. Therefore, since y = 2, which corresponds to x + 1/x = 2. Solving for x: x² - 2x + 1 = 0 ⇒ (x - 1)^2 = 0 ⇒ x = 1. So, x = 1 is a real root. Therefore, the equation with a = -4/5 and b = -2/5 indeed has a real root at x = 1. Similarly, the other point (a, b) = (4/5, -2/5) would give a quadratic in y with root y = -2, leading to x = -1. Therefore, both of these points result in the quartic equation having real roots x = 1 or x = -1. Therefore, the minimal value of a² + b² is 4/5. But just to make sure there are no mistakes here, let me double-check. Suppose I take a = -4/5 and b = -2/5. Then the equation is x⁴ - (4/5)x³ - (2/5)x² - (4/5)x + 1. Let me plug x = 1 into this:1 - 4/5 - 2/5 - 4/5 + 1 = (1 + 1) + (-4/5 -2/5 -4/5) = 2 + (-10/5) = 2 - 2 = 0. So yes, x = 1 is a root. Similarly, x = -1 with a = 4/5, b = -2/5:1 + 4/5 - 2/5 + 4/5 +1 = (1 +1) + (4/5 +4/5) -2/5 = 2 + 8/5 -2/5 = 2 + 6/5 = 16/5 ≠ 0. Wait, that's not zero. Wait, wait, did I make a mistake here?Wait, if a = 4/5, then the equation is x⁴ + (4/5)x³ - (2/5)x² + (4/5)x + 1. Let's plug x = -1:(-1)^4 + (4/5)(-1)^3 - (2/5)(-1)^2 + (4/5)(-1) +1= 1 - 4/5 - 2/5 -4/5 +1= (1 +1) + (-4/5 -2/5 -4/5)= 2 - (10/5) = 2 - 2 = 0. Oh, right. Because even powers are positive, odd powers are negative. So yes, x = -1 is a root. Therefore, both a = -4/5 and a = 4/5 with b = -2/5 lead to roots at x = 1 and x = -1 respectively. Therefore, these are valid points. Thus, the minimal value of a² + b² is 4/5. Therefore, the answer should be boxed{dfrac{4}{5}}.Final AnswerThe least possible value of (a^2 + b^2) is boxed{dfrac{4}{5}}.</think>Given the quartic equation (x^4 + ax^3 + bx^2 + ax + 1 = 0) with real numbers (a) and (b) such that there is at least one real solution, we need to find the least possible value of (a^2 + b^2).First, we recognize that the polynomial is a reciprocal polynomial, meaning if (x) is a root, then (1/x) is also a root. This symmetry allows us to use the substitution (y = x + 1/x). Substituting (y) into the polynomial, we divide by (x^2) to get:[x^2 + ax + b + frac{a}{x} + frac{1}{x^2} = 0]This simplifies to:[left(x^2 + frac{1}{x^2}right) + aleft(x + frac{1}{x}right) + b = 0]Using (y = x + 1/x), we know (x^2 + 1/x^2 = y^2 - 2). Substituting this into the equation gives:[y^2 - 2 + ay + b = 0 implies y^2 + ay + (b - 2) = 0]For (x) to be real, (y = x + 1/x) must satisfy (|y| geq 2). Therefore, the quadratic equation (y^2 + ay + (b - 2) = 0) must have at least one root (y) such that (|y| geq 2). This translates to the conditions (4 + 2a + (b - 2) leq 0) or (4 - 2a + (b - 2) leq 0), simplifying to:[4 + 2a + b - 2 leq 0 implies 2a + b + 2 leq 0][4 - 2a + b - 2 leq 0 implies -2a + b + 2 leq 0]We need to minimize (a^2 + b^2) subject to these constraints. The minimal value is the squared distance from the point ((a, b)) to the point ((0, -2)) in the plane, considering the constraints. The closest points on the lines (2a + b + 2 = 0) and (-2a + b + 2 = 0) to ((0, -2)) are found using the distance formula.For the line (2a + b + 2 = 0), the distance from ((0, -2)) is:[frac{|2(0) + (-2) + 2|}{sqrt{2^2 + 1^2}} = frac{2}{sqrt{5}}]Similarly, for the line (-2a + b + 2 = 0), the distance is also (frac{2}{sqrt{5}}). The minimal value of (a^2 + b^2) is the square of this distance:[left(frac{2}{sqrt{5}}right)^2 = frac{4}{5}]Thus, the least possible value of (a^2 + b^2) is (boxed{dfrac{4}{5}}).,<think>Okay, so I need to find the least possible value of ( a^2 + b^2 ) given that the quartic equation ( x^4 + a x^3 + b x^2 + a x + 1 = 0 ) has at least one real solution. Hmm, let me think about how to approach this.First, the equation is a quartic (degree 4), which might be challenging, but maybe there's a way to factor it or simplify it. The equation is palindromic because the coefficients read the same forwards and backwards: 1, a, b, a, 1. Palindromic polynomials have some nice properties. For example, if ( x ) is a root, then ( frac{1}{x} ) is also a root. Let me verify that.Suppose ( x neq 0 ) is a root. Then substituting ( frac{1}{x} ) into the polynomial gives:( left(frac{1}{x}right)^4 + a left(frac{1}{x}right)^3 + b left(frac{1}{x}right)^2 + a left(frac{1}{x}right) + 1 )Multiplying through by ( x^4 ) to clear denominators:( 1 + a x + b x^2 + a x^3 + x^4 )Which is the same as the original polynomial. So yes, if ( x ) is a non-zero root, then ( frac{1}{x} ) is also a root. That's a useful symmetry. So the roots come in reciprocal pairs, unless ( x = 1 ) or ( x = -1 ), which are their own reciprocals. Maybe this can help factor the equation.But the problem states that there's at least one real solution. Let's consider that. If the quartic has a real root, then either it's 1 or -1, or there's another real root ( r ) such that ( frac{1}{r} ) is also a root. But if ( r ) is real, then ( frac{1}{r} ) is also real. So unless ( r = pm 1 ), the roots come in reciprocal pairs.But since we need only at least one real solution, maybe there's a case where one of the reciprocal pairs is real. For example, if there's a real root ( r ), then ( frac{1}{r} ) is also real, so either both are real or both are complex. Wait, no, if ( r ) is real, ( frac{1}{r} ) must be real as well, so the equation must have either two real roots (a reciprocal pair) or a real root at 1 or -1. But if there's a real root, it could be either one of these cases.Alternatively, maybe the equation can be factored into quadratics. Since it's a quartic, perhaps as the product of two quadratics. Let me try that.Suppose ( x^4 + a x^3 + b x^2 + a x + 1 = (x^2 + p x + 1)(x^2 + q x + 1) ). Let's check if this is possible.Expanding the right side:( (x^2 + p x + 1)(x^2 + q x + 1) = x^4 + (p + q) x^3 + (pq + 2) x^2 + (p + q) x + 1 )Comparing coefficients with the original equation:- Coefficient of ( x^3 ): ( p + q = a )- Coefficient of ( x ): ( p + q = a )- Coefficient of ( x^2 ): ( pq + 2 = b )So, this suggests that if we can factor the quartic into two quadratics of the form ( x^2 + p x + 1 ) and ( x^2 + q x + 1 ), then the coefficients satisfy ( a = p + q ) and ( b = pq + 2 ). That seems plausible. However, this might not account for all possible factorizations, but given the palindromic nature, this seems like a good starting point.Alternatively, perhaps substituting ( y = x + frac{1}{x} ). Since the polynomial is palindromic, such a substitution is often useful. Let me try that.Let ( x neq 0 ), and let ( y = x + frac{1}{x} ). Then ( y^2 = x^2 + 2 + frac{1}{x^2} ), so ( x^2 + frac{1}{x^2} = y^2 - 2 ). Similarly, ( x^3 + frac{1}{x^3} = y^3 - 3 y ).But let's see. The original equation is:( x^4 + a x^3 + b x^2 + a x + 1 = 0 )Divide both sides by ( x^2 ) (since ( x neq 0 )):( x^2 + a x + b + frac{a}{x} + frac{1}{x^2} = 0 )Which can be written as:( left( x^2 + frac{1}{x^2} right) + a left( x + frac{1}{x} right) + b = 0 )Using the substitution ( y = x + frac{1}{x} ), then ( x^2 + frac{1}{x^2} = y^2 - 2 ), so substituting:( (y^2 - 2) + a y + b = 0 )So:( y^2 + a y + (b - 2) = 0 )So now we have a quadratic in ( y ). Therefore, for real solutions ( x ), there must be real solutions ( y ) such that ( y = x + 1/x ), and then the quadratic in ( y ) must have real solutions. But also, for each real solution ( y ), the equation ( x + 1/x = y ) must have real solutions ( x ).Wait, but ( y = x + 1/x ). Let's consider the possible values of ( y ). For real ( x ), ( y = x + 1/x ). If ( x > 0 ), then by AM ≥ GM, ( x + 1/x geq 2 ). Similarly, if ( x < 0 ), let ( x = -t ), where ( t > 0 ), then ( y = -t - 1/t leq -2 ). Therefore, the possible values of ( y ) are ( y geq 2 ) or ( y leq -2 ).Therefore, for the quadratic equation ( y^2 + a y + (b - 2) = 0 ) to have a real solution ( y ) such that ( |y| geq 2 ).Therefore, the problem reduces to finding the minimal ( a^2 + b^2 ) such that the quadratic in ( y ) has at least one real root with ( |y| geq 2 ). Then, since the quadratic must have real roots, its discriminant must be non-negative, but also at least one of the roots must satisfy ( |y| geq 2 ).So, first, the quadratic ( y^2 + a y + (b - 2) = 0 ) must have real roots. The discriminant is ( D = a^2 - 4 times 1 times (b - 2) = a^2 - 4b + 8 ). So for real roots, we need ( a^2 - 4b + 8 geq 0 ).Additionally, at least one of the roots must satisfy ( |y| geq 2 ).So our problem becomes: minimize ( a^2 + b^2 ) subject to ( a^2 - 4b + 8 geq 0 ) and at least one root of ( y^2 + a y + (b - 2) = 0 ) has ( |y| geq 2 ).Alternatively, we can model this as an optimization problem. Let me denote ( c = b - 2 ), then the quadratic becomes ( y^2 + a y + c = 0 ), and the discriminant is ( a^2 - 4c geq 0 ). Also, we need at least one root with ( |y| geq 2 ). Then, since ( b = c + 2 ), the objective function becomes ( a^2 + (c + 2)^2 ).So, we need to minimize ( a^2 + (c + 2)^2 ) subject to ( a^2 - 4c geq 0 ) and that the quadratic ( y^2 + a y + c = 0 ) has a root with ( |y| geq 2 ).Alternatively, perhaps we can use the quadratic's roots. Let the roots be ( y_1 ) and ( y_2 ). Then:- ( y_1 + y_2 = -a )- ( y_1 y_2 = c )Given that ( |y_1| geq 2 ) or ( |y_2| geq 2 ). So, we need to have either ( |y_1| geq 2 ) or ( |y_2| geq 2 ).To ensure that at least one root is outside the interval (-2, 2), which is equivalent to having a root with ( |y| geq 2 ).There's a theorem in quadratic equations that relates to roots lying outside an interval. Maybe we can use that.Alternatively, if both roots are within (-2, 2), then the minimum of ( a^2 + (c + 2)^2 ) would not satisfy our condition, so we need to consider the complement.But perhaps a better approach is to consider that for the quadratic equation ( y^2 + a y + c = 0 ), if there exists a root ( y ) with ( |y| geq 2 ), then either the quadratic evaluated at 2 or -2 is non-positive (since the quadratic opens upwards). That is, if ( f(2) leq 0 ) or ( f(-2) leq 0 ), where ( f(y) = y^2 + a y + c ).Wait, let's verify that. Since the quadratic ( f(y) = y^2 + a y + c ) opens upwards (coefficient of ( y^2 ) is positive), then if there exists a root with ( y geq 2 ), then ( f(2) leq 0 ). Similarly, if there's a root with ( y leq -2 ), then ( f(-2) leq 0 ).Therefore, the condition that there exists a real root with ( |y| geq 2 ) is equivalent to ( f(2) leq 0 ) or ( f(-2) leq 0 ).Therefore, we have two inequalities:1. ( 4 + 2a + c leq 0 )2. ( 4 - 2a + c leq 0 )But actually, since the quadratic can have one root in ( y geq 2 ) and another in ( y leq -2 ), but given the symmetry, maybe we can assume without loss of generality that one of them is the case.But perhaps considering both possibilities, the minimal ( a^2 + b^2 ) would be achieved when the quadratic touches the boundary, i.e., when ( f(2) = 0 ) or ( f(-2) = 0 ), since minima often occur at boundaries.Alternatively, if the quadratic has a root exactly at y = 2 or y = -2, then that would be the case when the equality holds. Let me explore that.Case 1: Suppose y = 2 is a root.Then, substituting y = 2 into the quadratic:( 4 + 2a + c = 0 implies c = -2a -4 )Then, since the quadratic is ( y^2 + a y + c = 0 ), the other root can be found via ( y_1 + y_2 = -a ), so if one root is 2, the other is ( -a - 2 ). Then, their product is ( c = 2 times (-a - 2) = -2a -4 ), which matches our earlier substitution.But since the quadratic must also have real roots, so discriminant ( a^2 - 4c geq 0 ). Let's compute discriminant:( a^2 - 4c = a^2 -4(-2a -4) = a^2 +8a +16 = (a +4)^2 geq 0 ), which is always true since squares are non-negative. So in this case, the discriminant is always non-negative, so no additional constraint here.Similarly, for Case 2: y = -2 is a root.Substituting y = -2:( 4 - 2a + c = 0 implies c = 2a -4 )Then, the other root is ( -a - (-2) = -a +2 ). Product is ( -2 times (-a +2 ) = 2a -4 = c ), which matches.Again, discriminant is ( a^2 -4c = a^2 -4(2a -4) = a^2 -8a +16 = (a -4)^2 geq 0 ), which is always non-negative.Therefore, if the quadratic has a root at y = 2 or y = -2, then c is determined in terms of a, and the discriminant is a square, hence non-negative.Therefore, our problem reduces to two cases:Case 1: Quadratic has a root at y = 2, so c = -2a -4. Then, since b = c + 2, we have b = -2a -4 + 2 = -2a -2.Therefore, in this case, we need to minimize ( a^2 + b^2 = a^2 + (-2a -2)^2 ).Similarly, Case 2: Quadratic has a root at y = -2, so c = 2a -4. Then, b = c + 2 = 2a -4 + 2 = 2a -2.So, in Case 2, we need to minimize ( a^2 + b^2 = a^2 + (2a -2)^2 ).Therefore, compute the minimal value for both cases and take the smaller one.Let me compute both.Case 1: ( a^2 + (-2a -2)^2 )Expand the second term:( (-2a -2)^2 = (2a + 2)^2 = 4a^2 + 8a + 4 )Therefore, total expression is:( a^2 + 4a^2 + 8a + 4 = 5a^2 + 8a + 4 )This is a quadratic in a. To find its minimum, take derivative or complete the square.The derivative with respect to a is 10a +8. Setting to zero: 10a +8 =0 → a = -8/10 = -4/5.Alternatively, since it's a quadratic in a, the vertex is at a = -B/(2A) where A=5, B=8. So a = -8/(2*5) = -4/5.Compute the minimal value:5*(-4/5)^2 + 8*(-4/5) +4 = 5*(16/25) - 32/5 +4 = (80/25) - (32/5) +4 = (16/5) - (32/5) + (20/5) = (16 -32 +20)/5 = 4/5 = 0.8So minimal value in Case 1 is 4/5.Case 2: ( a^2 + (2a -2)^2 )Expand the second term:( (2a -2)^2 = 4a^2 -8a +4 )Total expression:( a^2 +4a^2 -8a +4 =5a^2 -8a +4 )Similarly, find the minimum. Again, a quadratic in a. Vertex at a = B/(2A) where here A=5, B=-8. So a = 8/(2*5) = 4/5.Compute the minimal value:5*(4/5)^2 -8*(4/5) +4 = 5*(16/25) -32/5 +4 = (80/25) - (32/5) +4 = (16/5) - (32/5) + (20/5) = (16 -32 +20)/5 = 4/5 =0.8Same as Case 1. So both cases give minimal value of 4/5.Therefore, the minimal value of ( a^2 + b^2 ) is 4/5.But wait, is this correct? Let me check.Wait, but we have two different expressions for b in terms of a in the two cases, but the minimal value ends up being the same. So regardless of whether we have a root at y=2 or y=-2, the minimal a^2 +b^2 is 4/5. Therefore, the answer is 4/5.But let me verify if these values actually correspond to a quartic equation with a real root.In Case 1, when a = -4/5, b = -2a -2 = -2*(-4/5) -2 = 8/5 -10/5 = -2/5.So, a = -4/5, b = -2/5.Then the original equation is x^4 - (4/5)x^3 - (2/5)x^2 - (4/5)x +1 =0.We need to check if this equation has a real root.Alternatively, since we substituted y = x + 1/x, and set y=2 as a root, then x + 1/x =2 implies x=1 (double root). Let's check x=1:1 -4/5 -2/5 -4/5 +1 = (1+1) + (-4/5 -2/5 -4/5) = 2 - (10/5) = 2 -2 =0. So x=1 is a root. Similarly, x=1 is a real root. So yes, in this case, the quartic has x=1 as a real root.Similarly, in Case 2, a=4/5, b=2a -2= 8/5 -10/5= -2/5. Then, the equation is x^4 + (4/5)x^3 - (2/5)x^2 + (4/5)x +1 =0.Check x=-1:1 -4/5 -2/5 -4/5 +1 = 2 - (4/5 +2/5 +4/5) = 2 - (10/5) = 2 -2 =0. So x=-1 is a root. Hence, the quartic has x=-1 as a real root.Therefore, both minimal cases correspond to quartics with real roots (x=1 and x=-1). Thus, the minimal value is indeed 4/5.But wait, the problem says "at least one real solution". But in these cases, the quartic actually has a double root at x=1 or x=-1. Let me confirm.Wait, when y=2, x +1/x=2 implies x=1 (a double root), since x=1 is a solution and since the quadratic in y would have a repeated root if discriminant is zero.Wait, but in our calculation above, the discriminant in Case 1 was (a +4)^2, which when a = -4/5, gives discriminant ( (-4/5) +4 )^2 = (16/5)^2 ≠0. Wait, but if we set y=2 as a root, but then the other root is -a -2. When a=-4/5, the other root is 4/5 -2 = -6/5. Then, the quadratic equation in y would have roots at 2 and -6/5.But -6/5 is -1.2, which is within (-2, 2). But then y=-6/5 is not acceptable since |y| <2, so the equation x +1/x = -6/5 must have real solutions?Wait, if y=-6/5, then x +1/x = -6/5. Let's solve for x:x +1/x = -6/5Multiply both sides by x:x^2 + (6/5)x +1=0Discriminant: (6/5)^2 -4*1*1 = 36/25 -4 = (36 -100)/25 = -64/25 <0. Therefore, no real solutions. Therefore, even though the quadratic in y has roots at 2 and -6/5, the root at y=-6/5 doesn't correspond to real x. Therefore, only y=2 corresponds to real x (x=1). Similarly, in this case, the quartic would have x=1 as a real root (double root?), but also the quadratic factor would correspond to complex roots.Wait, but if the quartic factors into (x^2 + p x +1)(x^2 + q x +1), and if one quadratic has a real root (x=1), then the other quadratic would have roots corresponding to y=-6/5, which are complex. Therefore, the quartic would have two real roots (x=1, multiplicity 2) and two complex roots.But in the problem statement, it's only required that there's at least one real solution. So even if the other roots are complex, it's okay. Therefore, in this case, the minimal value of 4/5 is achieved when a=-4/5, b=-2/5, and the quartic has x=1 as a real root.Similarly, in Case 2, a=4/5, b=-2/5, the quartic has x=-1 as a real root. So both cases are valid and give the minimal a^2 +b^2.Therefore, the minimal value is 4/5.But let me check if there's a possibility that a lower value could be achieved if the quadratic in y has a root with |y| >2, but not exactly at 2 or -2. Maybe when the quadratic has one root inside (-2,2) and one outside, but the minimal a^2 +b^2 could be lower. Wait, but we considered the case where the quadratic touches y=2 or y=-2. Maybe that's the minimal point.Alternatively, perhaps using Lagrange multipliers to minimize ( a^2 + b^2 ) subject to the constraint that the quartic has a real root. Let me consider that approach.But since the quartic having a real root is equivalent to the quadratic in y having a real root with |y| ≥2. Therefore, our problem is equivalent to minimizing ( a^2 + b^2 ) subject to ( y^2 + a y + (b -2) =0 ) having a real solution y with |y| ≥2. So we can model this as minimizing ( a^2 + b^2 ) with ( y^2 + a y + (b -2) =0 ) and |y| ≥2.Alternatively, treat y as a variable and set up the equations. Let me consider y as a variable such that |y| ≥2, and for each such y, the equations a = -(y + (b -2)/y). Wait, no, from the quadratic equation, we have:From ( y^2 + a y + (b -2) =0 ), we can express b as ( b = -y^2 -a y +2 ).Therefore, substituting into ( a^2 + b^2 ):( a^2 + (-y^2 - a y +2)^2 )So we can express the objective function in terms of a and y, and then minimize over a and y with |y| ≥2.Let me denote ( f(a, y) = a^2 + (-y^2 -a y +2)^2 ). To find the minimum of f(a,y) subject to |y| ≥2.For a fixed y, we can find the a that minimizes f(a,y). Let's compute derivative of f with respect to a:df/da = 2a + 2(-y^2 -a y +2)(-y)Set derivative to zero:2a + 2(-y^2 -a y +2)(-y) =0Divide by 2:a + (-y^2 -a y +2)(-y) =0Simplify:a + y(y^2 + a y -2) =0Expand:a + y^3 + a y^2 -2y =0Collect terms with a:a(1 + y^2) + y^3 -2y =0Solve for a:a = (2y - y^3)/(1 + y^2)Therefore, for a given y with |y| ≥2, the optimal a is ( a = frac{2y - y^3}{1 + y^2} ).Then, substitute this back into b:From earlier, ( b = -y^2 -a y +2 )Substitute a:( b = -y^2 - left( frac{2y - y^3}{1 + y^2} right) y +2 )Simplify:( b = -y^2 - frac{2y^2 - y^4}{1 + y^2} +2 )Combine terms:First, express -y^2 as ( -y^2 times frac{1 + y^2}{1 + y^2} = frac{ -y^2 - y^4 }{1 + y^2 } )So:( b = frac{ -y^2 - y^4 }{1 + y^2 } - frac{2y^2 - y^4}{1 + y^2 } +2 )Combine the fractions:( frac{ -y^2 - y^4 -2y^2 + y^4 }{1 + y^2 } +2 = frac{ -3y^2 }{1 + y^2 } +2 )Express 2 as ( frac{2(1 + y^2)}{1 + y^2} ):( frac{ -3y^2 + 2(1 + y^2) }{1 + y^2 } = frac{ -3y^2 + 2 + 2y^2 }{1 + y^2 } = frac{ -y^2 + 2 }{1 + y^2 } )Therefore, ( b = frac{2 - y^2}{1 + y^2} )Therefore, now we can express ( a^2 + b^2 ) in terms of y:First, compute a:( a = frac{2y - y^3}{1 + y^2} = frac{ -y^3 + 2y }{1 + y^2 } )Compute a^2:( left( frac{ -y^3 + 2y }{1 + y^2 } right)^2 = frac{ y^6 -4y^4 +4y^2 }{(1 + y^2)^2 } )Compute b:( b = frac{2 - y^2}{1 + y^2} )Compute b^2:( left( frac{2 - y^2}{1 + y^2} right)^2 = frac{ y^4 -4y^2 +4 }{(1 + y^2)^2 } )Therefore, ( a^2 + b^2 = frac{ y^6 -4y^4 +4y^2 + y^4 -4y^2 +4 }{(1 + y^2)^2 } )Simplify numerator:Combine like terms:( y^6 -4y^4 +4y^2 + y^4 -4y^2 +4 = y^6 -3y^4 +0y^2 +4 )So:( a^2 + b^2 = frac{ y^6 -3y^4 +4 }{(1 + y^2)^2 } )Factor numerator if possible. Let's see.Let me check if y^6 -3y^4 +4 can be factored.Let me try substituting z = y^2, then numerator is z^3 -3z^2 +4. Let's factor this.Looking for rational roots using Rational Root Theorem. Possible roots are ±1, ±2, ±4.Test z=1: 1 -3 +4=2≠0.z=2: 8 -12 +4=0. So z=2 is a root.Therefore, divide z^3 -3z^2 +4 by (z-2):Using polynomial division or synthetic division.Using synthetic division:2 | 1 -3 0 4 2 -2 -4 1 -1 -2 0So quotient is z^2 - z -2, which factors as (z -2)(z +1). Therefore:z^3 -3z^2 +4 = (z -2)(z^2 - z -2) = (z -2)(z -2)(z +1) = (z -2)^2(z +1)Therefore, substituting back z=y^2:Numerator becomes (y^2 -2)^2(y^2 +1)Therefore, ( a^2 + b^2 = frac{ (y^2 -2)^2(y^2 +1) }{(1 + y^2)^2 } = frac{(y^2 -2)^2}{y^2 +1} )So ( a^2 + b^2 = frac{(y^2 -2)^2}{y^2 +1} )Therefore, we need to minimize ( frac{(y^2 -2)^2}{y^2 +1} ) over real numbers y with |y| ≥2.Let me set t = y^2. Then, since |y| ≥2, t ≥4.So the function becomes ( frac{(t -2)^2}{t +1} ), t ≥4.We need to minimize this function for t ≥4.Let me compute derivative with respect to t.Let ( f(t) = frac{(t -2)^2}{t +1} )Compute derivative:f’(t) = [2(t -2)(t +1) - (t -2)^2(1)] / (t +1)^2Simplify numerator:2(t -2)(t +1) - (t -2)^2Factor out (t -2):(t -2)[2(t +1) - (t -2)] = (t -2)[2t +2 -t +2] = (t -2)(t +4)Therefore, f’(t) = (t -2)(t +4)/(t +1)^2For t ≥4, the derivative is positive since (t -2) >0, (t +4) >0, denominator positive. Therefore, f(t) is increasing for t ≥4.Therefore, the minimal value occurs at t=4.Therefore, substituting t=4:f(4) = (4 -2)^2 / (4 +1 )= 4 /5 = 0.8Therefore, the minimal value of ( a^2 + b^2 ) is 4/5, achieved when t=4, i.e., y^2=4, so y=±2. Therefore, going back, when y=2 or y=-2, which corresponds to the earlier Cases 1 and 2. Hence, this confirms that the minimal value is indeed 4/5.Therefore, the answer is boxed{dfrac{4}{5}}.Final Answerboxed{dfrac{4}{5}}</think>Given the quartic equation ( x^4 + ax^3 + bx^2 + ax + 1 = 0 ) with at least one real solution, we need to find the least possible value of ( a^2 + b^2 ).1. Symmetry and Substitution: - The equation is palindromic, so if ( x ) is a root, then ( frac{1}{x} ) is also a root. - Substitute ( y = x + frac{1}{x} ). For real ( x ), ( y geq 2 ) or ( y leq -2 ).2. Quadratic Transformation: - Dividing the original equation by ( x^2 ) and substituting ( y = x + frac{1}{x} ) gives the quadratic equation ( y^2 + ay + (b - 2) = 0 ).3. Conditions for Real Roots: - The quadratic equation ( y^2 + ay + (b - 2) = 0 ) must have real roots with ( |y| geq 2 ). - The discriminant must be non-negative: ( a^2 - 4(b - 2) geq 0 ).4. Boundary Cases: - Consider ( y = 2 ) and ( y = -2 ) as roots. This leads to solving for ( a ) and ( b ) in these cases. - For ( y = 2 ), substituting into the quadratic gives ( 4 + 2a + (b - 2) = 0 ), leading to ( b = -2a - 2 ). - For ( y = -2 ), substituting gives ( 4 - 2a + (b - 2) = 0 ), leading to ( b = 2a - 2 ).5. Minimizing ( a^2 + b^2 ): - For ( y = 2 ), minimize ( a^2 + (-2a - 2)^2 ). This results in ( 5a^2 + 8a + 4 ), minimized at ( a = -frac{4}{5} ), giving ( a^2 + b^2 = frac{4}{5} ). - For ( y = -2 ), minimize ( a^2 + (2a - 2)^2 ). This results in ( 5a^2 - 8a + 4 ), minimized at ( a = frac{4}{5} ), giving ( a^2 + b^2 = frac{4}{5} ).6. Verification: - Both cases yield the minimal value ( frac{4}{5} ), confirmed by checking the quadratic transformation and ensuring the quartic equation has real roots.Thus, the least possible value of ( a^2 + b^2 ) is (boxed{dfrac{4}{5}}).]

🔑:Put y = x + 1/x and the equation becomes y 2 + ay + b - 2 = 0, which has solutions y = -a/2 ±√(a 2 + 8 - 2b)/2. We require |y| ≥ 2 for the original equation to have a real root and hence we need |a| + √(a 2 + 8 - 4b) ≥ 4. Squaring gives 2|a| - b ≥ 2. Hence a 2 + b 2 ≥ a 2 + (2 - 2|a|) 2 = 5a 2 - 8|a| + 4 = 5(|a| - 4/5) 2 + 4/5. So the least possible value of a 2 + b 2 is 4/5, achieved when a = 4/5, b = -2/5. In this case, the original equation is x 4 + 4/5 x 3 - 2/5 x 2 + 4/5 x + 1 = (x + 1) 2 (x 2 - 6/5 x + 1). s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 15th IMO 1973 © John Scholes jscholes@kalva.demon.co.uk 10 Oct 1998

🧠:[<think>Okay, so I need to find the indefinite integral of the rational function (2x³ + 4x² + 2x - 1) divided by ((x + 1)²(x² + 2x + 2)). Hmm, since it's a rational function, I think partial fractions might be the way to go here. Let me recall how partial fraction decomposition works for denominators with irreducible quadratic factors and repeated linear factors.First, the denominator here is (x + 1)² times (x² + 2x + 2). The quadratic factor x² + 2x + 2 doesn't factor further because its discriminant is (2)² - 4*1*2 = 4 - 8 = -4, which is negative. So, it's irreducible over the real numbers. So, according to partial fraction rules, the decomposition should look something like this:(2x³ + 4x² + 2x - 1)/[(x + 1)²(x² + 2x + 2)] = A/(x + 1) + B/(x + 1)² + (Cx + D)/(x² + 2x + 2)Where A, B, C, D are constants to be determined. Okay, so I need to set up this equation and solve for A, B, C, D. Let me write the numerator on the left side as the sum of the numerators on the right side multiplied out:2x³ + 4x² + 2x - 1 = A(x + 1)(x² + 2x + 2) + B(x² + 2x + 2) + (Cx + D)(x + 1)²Alright, now I need to expand each term on the right-hand side and then collect like terms to equate coefficients with the left-hand side.First, expand A(x + 1)(x² + 2x + 2). Let's do that step by step. Multiply (x + 1) with (x² + 2x + 2):(x + 1)(x² + 2x + 2) = x(x² + 2x + 2) + 1(x² + 2x + 2) = x³ + 2x² + 2x + x² + 2x + 2 = x³ + 3x² + 4x + 2So, A times that is A(x³ + 3x² + 4x + 2) = A x³ + 3A x² + 4A x + 2ANext term: B(x² + 2x + 2) = Bx² + 2Bx + 2BThird term: (Cx + D)(x + 1)². Let's expand (x + 1)² first: x² + 2x + 1. Then multiply by (Cx + D):(Cx + D)(x² + 2x + 1) = Cx(x² + 2x + 1) + D(x² + 2x + 1) = Cx³ + 2Cx² + Cx + Dx² + 2Dx + DSo combining terms: Cx³ + (2C + D)x² + (C + 2D)x + DNow, combine all terms from the right-hand side:A x³ + 3A x² + 4A x + 2A + Bx² + 2Bx + 2B + Cx³ + (2C + D)x² + (C + 2D)x + DNow, let's collect like terms:x³ terms: (A + C)x³x² terms: (3A + B + 2C + D)x²x terms: (4A + 2B + C + 2D)xconstant terms: (2A + 2B + D)So, setting this equal to the left-hand side, which is 2x³ + 4x² + 2x -1, we can equate coefficients for each power of x:For x³: A + C = 2For x²: 3A + B + 2C + D = 4For x: 4A + 2B + C + 2D = 2For constants: 2A + 2B + D = -1So now we have a system of four equations with four variables (A, B, C, D):1) A + C = 22) 3A + B + 2C + D = 43) 4A + 2B + C + 2D = 24) 2A + 2B + D = -1Okay, let's try to solve this system step by step.From equation 1: C = 2 - A. So we can substitute C in other equations.Let's substitute C = 2 - A into equations 2, 3, 4:Equation 2: 3A + B + 2*(2 - A) + D = 4Simplify:3A + B + 4 - 2A + D = 4(3A - 2A) + B + D + 4 = 4A + B + D + 4 = 4So, A + B + D = 0 --> Equation 2aEquation 3: 4A + 2B + (2 - A) + 2D = 2Simplify:4A + 2B + 2 - A + 2D = 2(4A - A) + 2B + 2D + 2 = 23A + 2B + 2D + 2 = 23A + 2B + 2D = 0 --> Equation 3aEquation 4: 2A + 2B + D = -1 --> Equation 4 remains as is.So now we have:2a) A + B + D = 03a) 3A + 2B + 2D = 04) 2A + 2B + D = -1Let me write these three equations:Equation 2a: A + B + D = 0Equation 3a: 3A + 2B + 2D = 0Equation 4: 2A + 2B + D = -1Hmm. Let's try to solve these.First, from equation 2a: D = -A - BSubstitute D = -A - B into equations 3a and 4.Equation 3a: 3A + 2B + 2*(-A - B) = 0Simplify:3A + 2B - 2A - 2B = 0(3A - 2A) + (2B - 2B) = 0A = 0Hmm, so A = 0. Then from equation 1, C = 2 - A = 2 - 0 = 2.Now, from equation 2a: D = -A - B = -0 - B = -B.And from equation 4: 2A + 2B + D = -1. Since A = 0, D = -B.Substitute A = 0, D = -B into equation 4:2*0 + 2B + (-B) = -10 + 2B - B = -1B = -1Then D = -B = -(-1) = 1.So, we have A = 0, B = -1, C = 2, D = 1.Let me check if these values satisfy all equations.Check equation 2a: A + B + D = 0 + (-1) + 1 = 0 ✔️Check equation 3a: 3*0 + 2*(-1) + 2*1 = 0 -2 + 2 = 0 ✔️Check equation 4: 2*0 + 2*(-1) + 1 = 0 -2 +1 = -1 ✔️Check original equation 1: A + C = 0 + 2 = 2 ✔️So all equations are satisfied. Therefore, the partial fractions decomposition is:(0)/(x + 1) + (-1)/(x + 1)^2 + (2x + 1)/(x² + 2x + 2)So simplifying, the integral becomes:∫ [ -1/(x + 1)^2 + (2x + 1)/(x² + 2x + 2) ] dxNow, let's split the integral into two parts:I1 = ∫ -1/(x + 1)^2 dxI2 = ∫ (2x + 1)/(x² + 2x + 2) dxLet me compute I1 first.I1: ∫ -1/(x + 1)^2 dx. Let u = x + 1, du = dx. Then integral becomes -∫ 1/u² du = -∫ u^(-2) du = - [ (-1)u^(-1) ] + C = 1/u + C = 1/(x + 1) + CNow, I2: ∫ (2x + 1)/(x² + 2x + 2) dx. Let me try to simplify this.First, notice that the denominator is x² + 2x + 2. Let me complete the square:x² + 2x + 2 = (x + 1)^2 + 1So, the denominator becomes (x + 1)^2 + 1. Maybe a substitution here.Also, the numerator is 2x + 1. Let me see if the numerator can be expressed as the derivative of the denominator plus some constant.The derivative of the denominator (x² + 2x + 2) is 2x + 2. So, the numerator 2x + 1 is (2x + 2) -1. So, let's split it:(2x + 1) = (2x + 2) -1Therefore, I2 can be written as:∫ [ (2x + 2) - 1 ] / (x² + 2x + 2) dx = ∫ (2x + 2)/(x² + 2x + 2) dx - ∫ 1/(x² + 2x + 2) dxLet's call these two integrals I2a and I2b.Compute I2a: ∫ (2x + 2)/(x² + 2x + 2) dxLet u = x² + 2x + 2, then du/dx = 2x + 2, so du = (2x + 2) dx. Therefore, I2a becomes ∫ du/u = ln|u| + C = ln|x² + 2x + 2| + CNow, compute I2b: ∫ 1/(x² + 2x + 2) dxAgain, complete the square in the denominator: x² + 2x + 2 = (x + 1)^2 + 1So, I2b becomes ∫ 1/[(x + 1)^2 + 1] dx. Let u = x + 1, then du = dx. So, integral becomes ∫ 1/(u² + 1) du = arctan(u) + C = arctan(x + 1) + CTherefore, combining I2a and I2b:I2 = ln|x² + 2x + 2| - arctan(x + 1) + CPutting it all together, the original integral is I1 + I2:[1/(x + 1) + C1] + [ln|x² + 2x + 2| - arctan(x + 1) + C2]Combining constants:1/(x + 1) + ln|x² + 2x + 2| - arctan(x + 1) + CSo that's the indefinite integral.Wait a second, let me just verify my steps to make sure I didn't make a mistake.First, the partial fractions: we had A=0, B=-1, C=2, D=1. Plugging back into the original equation:Left-hand side: 2x³ + 4x² + 2x -1Right-hand side: A(x³ + 3x² +4x +2) + B(x² +2x +2) + (Cx + D)(x² +2x +1)With A=0: first term is 0.Second term: B(x² + 2x + 2) = -1*(x² + 2x + 2) = -x² -2x -2Third term: (2x +1)(x² +2x +1) = 2x*(x² +2x +1) +1*(x² +2x +1) = 2x³ +4x² +2x +x² +2x +1 = 2x³ +5x² +4x +1Adding the second and third terms: (-x² -2x -2) + (2x³ +5x² +4x +1) = 2x³ +4x² +2x -1Which matches the left-hand side. Great, so partial fractions are correct.Then integrating each term:-1/(x +1)^2 integrated to 1/(x +1). Correct.Then (2x +1)/(x² +2x +2) broken into (2x +2 -1)/(x² +2x +2) = (2x +2)/(x² +2x +2) -1/(x² +2x +2). The first part integrates to ln|x² +2x +2|, second part to -arctan(x +1). All correct.So the final answer is 1/(x +1) + ln|x² +2x +2| - arctan(x +1) + C.But let me check if the coefficients are correct again. Wait, the integral of -1/(x +1)^2 is 1/(x +1) because the integral of u^{-2} is -u^{-1} + C. So that's correct. Yes.And the other terms as I split them. Yeah, looks good.So, combining all the results, the integral is:1/(x +1) + ln|x² +2x +2| - arctan(x +1) + CFinal AnswerThe indefinite integral is boxed{frac{1}{x + 1} + lnleft|x^2 + 2x + 2right| - arctan(x + 1) + C}.</think>To find the indefinite integral:int frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}left(x^{2}+2 x+2right)} d xwe use partial fraction decomposition. The integrand can be decomposed as:frac{2 x^{3} + 4 x^{2} + 2 x - 1}{(x + 1)^{2}(x^{2} + 2 x + 2)} = frac{A}{x + 1} + frac{B}{(x + 1)^{2}} + frac{Cx + D}{x^{2} + 2 x + 2}Solving for the constants (A), (B), (C), and (D), we find (A = 0), (B = -1), (C = 2), and (D = 1). Thus, the decomposition is:frac{-1}{(x + 1)^{2}} + frac{2x + 1}{x^{2} + 2x + 2}We split the integral into two parts:1. (int frac{-1}{(x + 1)^{2}} , dx)2. (int frac{2x + 1}{x^{2} + 2x + 2} , dx)For the first integral, substituting (u = x + 1) gives:int frac{-1}{(x + 1)^{2}} , dx = frac{1}{x + 1} + CFor the second integral, we split the numerator and complete the square in the denominator:int frac{2x + 1}{x^{2} + 2x + 2} , dx = int frac{2x + 2}{x^{2} + 2x + 2} , dx - int frac{1}{x^{2} + 2x + 2} , dxThe first part integrates to (ln|x^2 + 2x + 2|). For the second part, substituting (u = x + 1) gives:int frac{1}{(x + 1)^2 + 1} , dx = arctan(x + 1) + CCombining all results, the indefinite integral is:boxed{frac{1}{x + 1} + lnleft|x^2 + 2x + 2right| - arctan(x + 1) + C},<think>Okay, so I need to find the indefinite integral of this rational function: (2x³ + 4x² + 2x - 1) divided by (x + 1)² times (x² + 2x + 2). Hmm, this looks like a job for partial fractions. Let me remember how partial fraction decomposition works for denominators with repeated linear factors and irreducible quadratic factors.First, the denominator here is (x + 1) squared multiplied by (x² + 2x + 2). Let me check if x² + 2x + 2 factors. The discriminant is (2)^2 - 4*1*2 = 4 - 8 = -4, which is negative, so it doesn't factor over the reals. So, that means the partial fraction decomposition should be of the form:(A)/(x + 1) + (B)/(x + 1)² + (Cx + D)/(x² + 2x + 2)Right? Since (x + 1)² is a repeated linear factor, we have two terms for it: A over (x + 1) and B over (x + 1) squared. Then, for the irreducible quadratic factor (x² + 2x + 2), we need a linear term in the numerator, so Cx + D.So, we can write:(2x³ + 4x² + 2x - 1)/[(x + 1)²(x² + 2x + 2)] = A/(x + 1) + B/(x + 1)² + (Cx + D)/(x² + 2x + 2)Now, to find A, B, C, D, I need to combine the right-hand side over the common denominator and equate the numerators.So, let's do that step by step. Multiply both sides by (x + 1)²(x² + 2x + 2) to eliminate denominators:2x³ + 4x² + 2x - 1 = A(x + 1)(x² + 2x + 2) + B(x² + 2x + 2) + (Cx + D)(x + 1)²Now, expand each term on the right-hand side.First term: A(x + 1)(x² + 2x + 2)Let me expand (x + 1)(x² + 2x + 2). Multiply x by each term: x³ + 2x² + 2x. Then multiply 1 by each term: x² + 2x + 2. Adding those together: x³ + 2x² + 2x + x² + 2x + 2 = x³ + 3x² + 4x + 2. So that term becomes A(x³ + 3x² + 4x + 2)Second term: B(x² + 2x + 2). That's just Bx² + 2Bx + 2B.Third term: (Cx + D)(x + 1)². Let's expand (x + 1)² first: x² + 2x + 1. Then multiply by (Cx + D):Cx*(x² + 2x + 1) + D*(x² + 2x + 1) = Cx³ + 2Cx² + Cx + Dx² + 2Dx + D.Combine like terms: Cx³ + (2C + D)x² + (C + 2D)x + D.So, putting all terms together:Right-hand side:A(x³ + 3x² + 4x + 2) + Bx² + 2Bx + 2B + Cx³ + (2C + D)x² + (C + 2D)x + DNow, combine like terms by degree:x³ terms: A + Cx² terms: 3A + B + 2C + Dx terms: 4A + 2B + C + 2Dconstant terms: 2A + 2B + DSo, equating the coefficients to the left-hand side numerator: 2x³ + 4x² + 2x -1.Therefore, we have the following system of equations:For x³: A + C = 2 ...(1)For x²: 3A + B + 2C + D = 4 ...(2)For x: 4A + 2B + C + 2D = 2 ...(3)For constants: 2A + 2B + D = -1 ...(4)Now, we have four equations with four variables: A, B, C, D.Let me try to solve this system step by step.From equation (1): C = 2 - A ...(1a)Let's substitute C = 2 - A into equations (2), (3), (4).Equation (2): 3A + B + 2*(2 - A) + D = 4Compute 2*(2 - A) = 4 - 2ASo, 3A + B + 4 - 2A + D = 4Combine like terms: (3A - 2A) + B + D + 4 = 4Which simplifies to: A + B + D + 4 = 4Therefore: A + B + D = 0 ...(2a)Equation (3): 4A + 2B + (2 - A) + 2D = 2Substitute C = 2 - A into equation (3):4A + 2B + 2 - A + 2D = 2Combine like terms: (4A - A) + 2B + 2D + 2 = 2Which is: 3A + 2B + 2D + 2 = 2Subtract 2 from both sides: 3A + 2B + 2D = 0 ...(3a)Equation (4): 2A + 2B + D = -1 ...(4a)So now we have three equations:(2a): A + B + D = 0(3a): 3A + 2B + 2D = 0(4a): 2A + 2B + D = -1Let me write these equations:Equation (2a): A + B + D = 0Equation (3a): 3A + 2B + 2D = 0Equation (4a): 2A + 2B + D = -1Let me try to solve these. Let's use equations (2a) and (4a) first.From (2a): D = -A - BSubstitute D = -A - B into equation (4a):2A + 2B + (-A - B) = -1Simplify: 2A + 2B - A - B = -1Which becomes: (2A - A) + (2B - B) = -1Thus: A + B = -1 ...(5)Similarly, substitute D = -A - B into equation (3a):3A + 2B + 2*(-A - B) = 0Compute 2*(-A - B) = -2A - 2BSo, equation becomes: 3A + 2B - 2A - 2B = 0Simplify: (3A - 2A) + (2B - 2B) = 0 => A = 0So, A = 0. Then from equation (5): 0 + B = -1 => B = -1Then D = -A - B = -0 - (-1) = 1From equation (1a): C = 2 - A = 2 - 0 = 2So, A = 0, B = -1, C = 2, D = 1Let me check these values in equation (2a):A + B + D = 0 + (-1) + 1 = 0, which checks out.In equation (3a):3A + 2B + 2D = 0 + (-2) + 2 = 0, correct.In equation (4a):2A + 2B + D = 0 + (-2) + 1 = -1, correct.So, the coefficients are A=0, B=-1, C=2, D=1.Therefore, the partial fractions decomposition is:0/(x + 1) + (-1)/(x + 1)² + (2x + 1)/(x² + 2x + 2)Simplify that: the first term is zero, so it's:-1/(x + 1)² + (2x + 1)/(x² + 2x + 2)So, the integral becomes:∫ [ -1/(x + 1)² + (2x + 1)/(x² + 2x + 2) ] dxSplit the integral into two parts:= -∫1/(x + 1)² dx + ∫(2x + 1)/(x² + 2x + 2) dxLet's compute each integral separately.First integral: -∫1/(x + 1)² dxThat's straightforward. The integral of 1/(x + 1)^2 dx is ∫(x + 1)^(-2) dx = (-1)(x + 1)^(-1) + C = -1/(x + 1) + CSo, with the negative sign, it becomes 1/(x + 1) + CSecond integral: ∫(2x + 1)/(x² + 2x + 2) dxHmm, this seems like we can split the numerator to match the derivative of the denominator. Let me check:Let u = x² + 2x + 2, then du/dx = 2x + 2. Hmm, the numerator is 2x + 1. Close, but not exactly du. Let's write 2x +1 as (2x + 2) -1. Then:(2x +1) = (2x + 2) -1Therefore, the integral becomes:∫[ (2x + 2) -1 ] / (x² + 2x + 2) dx = ∫(2x + 2)/(x² + 2x + 2) dx - ∫1/(x² + 2x + 2) dxThe first part: ∫(2x + 2)/u du where u = x² + 2x + 2. That's ∫ du/u = ln|u| + C. So that part is ln|x² + 2x + 2| + CThe second part: -∫1/(x² + 2x + 2) dxComplete the square in the denominator:x² + 2x + 2 = (x + 1)^2 + 1So, integral becomes -∫1/[(x + 1)^2 + 1] dxThat's a standard arctangent integral: -arctan(x + 1) + CPutting it all together, the second integral is:ln|x² + 2x + 2| - arctan(x + 1) + CTherefore, combining both integrals:First integral: 1/(x + 1)Second integral: ln|x² + 2x + 2| - arctan(x + 1)So, combining all terms, the indefinite integral is:1/(x + 1) + ln|x² + 2x + 2| - arctan(x + 1) + CLet me check my work for any possible errors.Starting with partial fractions: when I set up the equations, A=0, B=-1, C=2, D=1. Let me verify by plugging back into the original equation.Original numerator: 2x³ +4x² +2x -1From the partial fractions expression: A(x+1)(x² + 2x +2) + B(x² +2x +2) + (Cx + D)(x +1)^2Substituting A=0, B=-1, C=2, D=1:0*(...) + (-1)*(x² +2x +2) + (2x +1)*(x +1)^2Compute each term:First term: 0Second term: -x² -2x -2Third term: (2x +1)*(x² + 2x +1). Let's multiply this out:2x*(x² +2x +1) +1*(x² +2x +1) = 2x³ +4x² +2x +x² +2x +1 = 2x³ +5x² +4x +1So, adding the second and third terms:(-x² -2x -2) + (2x³ +5x² +4x +1) = 2x³ + ( -x² +5x² ) + (-2x +4x ) + (-2 +1 )= 2x³ +4x² +2x -1Which matches the original numerator. So, the partial fractions are correct.Then integrating each part:First integral: -∫1/(x +1)^2 dx = 1/(x +1) + C. Correct.Second integral: split into (2x +2 -1)/(x² +2x +2). Then split into two integrals. The first part becomes ln|x² +2x +2|, since derivative of denominator is numerator. The second part becomes -∫1/((x +1)^2 +1) dx = -arctan(x +1). Correct.Therefore, combining all terms, the integral is:1/(x +1) + ln|x² +2x +2| - arctan(x +1) + CThat seems right. Let me just check if the algebra was done correctly when splitting the numerator. The split (2x +1) = (2x +2) -1 is correct, yes.So, I think that's the correct result.Final AnswerThe indefinite integral is boxed{frac{1}{x + 1} + lnleft|x^2 + 2x + 2right| - arctan(x + 1) + C}.</think>To find the indefinite integral:int frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}left(x^{2}+2 x+2right)} d xwe use partial fraction decomposition. The decomposition is of the form:frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}(x^{2}+2 x+2)} = frac{A}{x+1} + frac{B}{(x+1)^2} + frac{Cx + D}{x^2 + 2x + 2}By equating the coefficients after expanding and combining terms, we find the constants (A = 0), (B = -1), (C = 2), and (D = 1). This gives the decomposition:-frac{1}{(x+1)^2} + frac{2x + 1}{x^2 + 2x + 2}We then split the integral into two parts:1. The first integral is straightforward: -int frac{1}{(x+1)^2} , dx = frac{1}{x+1} + C 2. The second integral is split further: int frac{2x + 1}{x^2 + 2x + 2} , dx = int frac{2x + 2 - 1}{x^2 + 2x + 2} , dx This splits into: - (int frac{2x + 2}{x^2 + 2x + 2} , dx = ln|x^2 + 2x + 2|) - (-int frac{1}{(x+1)^2 + 1} , dx = -arctan(x + 1))Combining all parts, the indefinite integral is:boxed{frac{1}{x + 1} + lnleft|x^2 + 2x + 2right| - arctan(x + 1) + C}]

🔑:## Solutionint frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}left(x^{2}+2 x+2right)} d x=We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:begin{aligned}& frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}left(x^{2}+2 x+2right)}=frac{A_{1}}{x+1}+frac{A_{2}}{(x+1)^{2}}+frac{B x+C}{x^{2}+2 x+2}= & =frac{A_{1}(x+1)left(x^{2}+2 x+2right)+A_{2}left(x^{2}+2 x+2right)+(B x+C)(x+1)^{2}}{(x+1)^{2}left(x^{2}+2 x+2right)}= & =frac{A_{1}left(x^{3}+3 x^{2}+4 x+2right)+A_{2}left(x^{2}+2 x+2right)+Bleft(x^{3}+2 x^{2}+xright)+Cleft(x^{2}+2 x+1right)}{(x+1)^{2}left(x^{2}+2 x+2right)}= & =frac{left(A_{1}+Bright) x^{3}+left(3 A_{1}+A_{2}+2 B+Cright) x^{2}+left(4 A_{1}+2 A_{2}+B+2 Cright) x+left(2 A_{1}+2 A_{2}+Cright)}{(x+1)^{2}left(x^{2}+2 x+2right)} & left{begin{array}{l}A_{1}+B=2 3 A_{1}+A_{2}+2 B+C=4 4 A_{1}+2 A_{2}+B+2 C=2 2 A_{1}+2 A_{2}+C=-1end{array}right.end{aligned}Subtract the first equation multiplied by 2 from the second equation:left{begin{array}{l}A_{1}+B=2 A_{1}+A_{2}+C=0 4 A_{1}+2 A_{2}+B+2 C=2 2 A_{1}+2 A_{2}+C=-1end{array}right.Subtract the second equation multiplied by 2 from the fourth equation:left{begin{array}{l}A_{1}+B=2 A_{1}+A_{2}+C=0 4 A_{1}+2 A_{2}+B+2 C=2 -C=-1end{array}right.left{begin{array}{l}A_{1}+B=2 A_{1}+A_{2}=-1 4 A_{1}+2 A_{2}+B=0 C=1end{array}right.Subtract the second equation multiplied by 2 from the third equation:left{begin{array}{l}A_{1}+B=2 A_{1}+A_{2}=-1 2 A_{1}+B=2 C=1end{array}right.Subtract the first equation from the third equation:begin{aligned}& left{begin{array}{l}A_{1}+B=2 A_{1}+A_{2}=-1 A_{1}=0 C=1end{array}right. & left{begin{array}{l}B=2 A_{2}=-1 A_{1}=0 C=1end{array}right. & frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}left(x^{2}+2 x+2right)}=-frac{1}{(x+1)^{2}}+frac{2 x+1}{x^{2}+2 x+2}end{aligned}Thenbegin{aligned}& int frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}left(x^{2}+2 x+2right)} d x=intleft(-frac{1}{(x+1)^{2}}+frac{2 x+1}{x^{2}+2 x+2}right) d x= & =-int frac{1}{(x+1)^{2}} d x+int frac{2 x+2-1}{x^{2}+2 x+2} d x= & =frac{1}{x+1}+int frac{2 x+2}{x^{2}+2 x+2} d x-int frac{1}{x^{2}+2 x+2} d x= & =frac{1}{x+1}+int frac{dleft(x^{2}+2 x+2right)}{x^{2}+2 x+2}-int frac{1}{(x+1)^{2}+1} d x= & =frac{1}{x+1}+ln left(x^{2}+2 x+2right)-operatorname{arctg}(x+1)+Cend{aligned}Source — «http://pluspi.org/wiki/index.php/������������_�� %D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3 % mathrm{D} 1 % 80 % mathrm{D} 0 % mathrm{~B} 0 % mathrm{D} 0 % mathrm{BB} % mathrm{D} 1 % 8 mathrm{~B}+7-4>Categories: Kuznetsov's Problem Book Integrals Problem 7 | IntegralsUkrainian Banner Network- Last modified: 21:20, 12 April 2009.- Content is available under CC-BY-SA 3.0.Created by GeeTeatoo## Problem Kuznetsov Integrals 7-5## Material from PlusPi

🧠:[<think>Okay, so I have this sequence problem here. The sequence starts with 32, 8, then there are two missing terms, and finally x. The rule is that each term after the second is the average of the two terms immediately before it. I need to find the value of x from the given options. Let me try to figure this out step by step.First, let me write down what I know. The sequence is: 32 (first term), 8 (second term), then two missing terms which I'll call term3 and term4, and then x as the fifth term. The rule says each term after the second is the average of the two previous terms. So term3 should be the average of term1 and term2, term4 is the average of term2 and term3, and x (term5) is the average of term3 and term4. Let me check if I'm interpreting the problem correctly.Yes, the problem states that each term after the second is the average of the two terms immediately before it. So starting from the third term onwards, each term is the average of the two preceding terms. That makes sense. So term3 = (term1 + term2)/2, term4 = (term2 + term3)/2, term5 = (term3 + term4)/2. And term5 is x. So I need to compute term3, term4, and then x.Let me calculate term3 first. Term1 is 32, term2 is 8. So term3 is (32 + 8)/2. Let me compute that. 32 plus 8 is 40. Divided by 2 is 20. So term3 is 20. Okay, that seems straightforward.Now term4 should be the average of term2 and term3. Term2 is 8, term3 is 20. So term4 is (8 + 20)/2. Let me add those: 8 + 20 = 28. Divided by 2 is 14. So term4 is 14.Then x, which is term5, is the average of term3 and term4. Term3 is 20, term4 is 14. So x is (20 + 14)/2. Let me compute that. 20 + 14 is 34. Divided by 2 is 17. So x is 17. Hmm, 17 is option A. But let me double-check to make sure I didn't make any mistakes in my calculations.Let me go through each step again. Term1 is 32, term2 is 8. Term3 is (32 + 8)/2. 32 + 8 is 40, divided by 2 is 20. That's correct. Term4 is (8 + 20)/2. 8 + 20 is 28, divided by 2 is 14. That's also correct. Term5 is (20 + 14)/2. 20 + 14 is 34, divided by 2 is 17. So x is 17, which is option A.Wait a second, but let me check the options again. The options are (A) 17, (B) 20, (C) 44, (D) 24, (E) 14. So 17 is an option. But let me just verify once again. Maybe I miscalculated term5? Let's see: term3 is 20, term4 is 14. So (20 + 14)/2 is indeed 34/2 = 17. Yep, that seems right.Alternatively, maybe I should check if there's another way to interpret the problem. For example, does each term after the second include the fifth term? Yes, the problem says each term after the second is the average of the two terms immediately before it. So term3, term4, term5 all follow that rule. So term3 is average of term1 and term2, term4 is average of term2 and term3, term5 is average of term3 and term4. So that's exactly what I did.Alternatively, maybe the problem is written as "each term after the second is the average of the two terms immediately before it," but perhaps "immediately before it" refers to the two terms just before that term, not the entire sequence. For example, term3 is the average of term1 and term2, term4 is the average of term2 and term3, term5 is the average of term3 and term4. That seems to be correct.Alternatively, could "the two terms immediately before it" be misinterpreted as the two terms before the entire sequence? But that doesn't make sense. Each term after the second is computed based on the two terms immediately before that term, not before the sequence. So for term3, the two terms before term3 are term1 and term2. For term4, the two terms before term4 are term2 and term3. For term5, the two terms before term5 are term3 and term4. So that's exactly how I calculated.Therefore, I think my answer is correct. x is 17, which is option A. But let me check again if the problem is about arithmetic sequence or something else, but the problem specifically says each term after the second is the average of the two terms immediately before it. So that's a recursive sequence.Alternatively, maybe there's a different approach. Let me try writing all terms out:Term1: 32Term2: 8Term3: (32 + 8)/2 = 40/2 = 20Term4: (8 + 20)/2 = 28/2 = 14Term5 (x): (20 + 14)/2 = 34/2 = 17Yes, that's exactly the same as before. So x is 17. Therefore, the correct answer is option A.But wait, let me check if maybe the problem expects the sequence to have four terms instead of five? The problem says: "In the sequence 32,8, -, -, x", each term after the second is the average... So the sequence is given as first two terms, then two missing terms, then x. So that's five terms total. So term1:32, term2:8, term3: -, term4: -, term5: x. So x is term5. So my calculation is correct.Alternatively, maybe I misread the number of terms. Let me check again. The problem says: "the sequence 32,8, -, -, x". Each dash is a missing term, so it's five terms: term1=32, term2=8, term3=?, term4=?, term5=x. Each term after the second (i.e., terms 3,4,5) is the average of the two terms immediately before it. Wait, term3 is after the second term, so term3 is average of term1 and term2. Then term4 is average of term2 and term3. Then term5 is average of term3 and term4. So that's how it goes. So x is term5, which is 17.But the options include 20, which is term3. Maybe someone could miscalculate and take term3 as x? But in the sequence, x is the fifth term. The sequence is written as 32,8, -, -, x. So the two dashes are term3 and term4, and x is term5. Therefore, x must be term5, which is 17. So answer is A.Wait, but let me check if the problem says "each term after the second is the average of the two terms immediately before it". So terms 3,4,5 are after the second term, so each of them is the average of the two terms immediately before it. So term3: average of term1 and term2. Term4: average of term2 and term3. Term5: average of term3 and term4. So yes, term5 is 17. Therefore, the answer is 17, option A.Alternatively, if someone thought that the two terms immediately before the entire sequence, but that's not the case. The terms immediately before each term. So I think my answer is correct.Final Answerboxed{A}</think>The sequence starts with 32, 8, followed by two missing terms, and then ( x ). Each term after the second is the average of the two terms immediately before it. We need to find the value of ( x ).1. Term 3: The average of the first two terms (32 and 8). [ text{Term 3} = frac{32 + 8}{2} = frac{40}{2} = 20 ]2. Term 4: The average of the second and third terms (8 and 20). [ text{Term 4} = frac{8 + 20}{2} = frac{28}{2} = 14 ]3. Term 5 ( ( x ) ): The average of the third and fourth terms (20 and 14). [ x = frac{20 + 14}{2} = frac{34}{2} = 17 ]Thus, the value of ( x ) is (boxed{A}).,<think>Okay, let's see. I need to find the value of x in the sequence 32, 8, _, _, x. The rule is that each term after the second is the average of the two terms immediately before it. Hmm, so starting from the third term, each term is the average of the previous two. Let me make sure I understand that correctly.So the first term is 32, the second term is 8. Then the third term should be the average of 32 and 8. Let me calculate that. The average of two numbers is just the sum divided by 2, right? So (32 + 8)/2. That's 40/2 = 20. Wait, but the problem says "each term after the second is the average of the two terms immediately before it." So the third term is average of the first and second. Then the fourth term is the average of the second and third. Then the fifth term, which is x, is the average of the third and fourth. Let me check that again.Let me lay out the terms step by step.Term 1: 32Term 2: 8Term 3: (Term1 + Term2)/2 = (32 + 8)/2 = 40/2 = 20Term 4: (Term2 + Term3)/2 = (8 + 20)/2 = 28/2 = 14Term 5 (x): (Term3 + Term4)/2 = (20 + 14)/2 = 34/2 = 17Wait, so x is 17? Let me check the answer choices. Option A is 17. Hmm. But let me make sure I didn't make a mistake.Starting with 32, 8. Then third term is 20. Then fourth term is 14. Then fifth term is (20 + 14)/2 = 17. That seems correct. So x is 17, which is option A. But let me double-check each calculation again.Term3: (32 + 8)/2 = 40/2 = 20. Correct.Term4: (8 + 20)/2 = 28/2 = 14. Correct.Term5: (20 + 14)/2 = 34/2 = 17. Correct. So yeah, the answer should be 17, which is option A. But hold on, the problem says "each term after the second is the average of the two terms immediately before it." So starting from term3 onwards, each term is average of the two before. So term3 is average of term1 and term2. Term4 is average of term2 and term3. Term5 is average of term3 and term4. Yes, that's exactly what I did. So the fifth term is 17. So the answer should be A.But wait, let me check the options again. The options are (A) 17, (B) 20, (C) 44, (D)24, (E)14. So 17 is an option, which matches my calculation. Hmm. But why does the problem have two blanks before x? The sequence is 32, 8, _, _, x. So the third term is 20, fourth is 14, and fifth is 17. So x is 17. Therefore, the correct answer is A. Hmm.Wait a second, maybe I made a mistake in the order? Let me check again. Maybe the problem is written as "each term after the second is the average of the two terms immediately before it". So term3 is average of term1 and term2. Term4 is average of term2 and term3. Term5 is average of term3 and term4. So the sequence is 32, 8, 20, 14, 17. So x is 17, which is answer A. So why is the answer not 20? Because term3 is 20, but x is term5. So maybe someone might get confused between term3 and term5, but according to the problem, x is the last term, so term5.Therefore, I think the answer is A, 17. Let me check once more to ensure no miscalculations.32, 8. Third term: (32 + 8)/2 = 20. Fourth term: (8 + 20)/2 = 14. Fifth term: (20 + 14)/2 = 17. Yep, that's right. So x is 17. So answer A.Final Answerboxed{A}</think>Given the sequence (32, 8, longrightarrow, longrightarrow, x), where each term after the second is the average of the two terms immediately before it, we need to find the value of (x).1. The first term is (32).2. The second term is (8).3. The third term is the average of the first and second terms: [ frac{32 + 8}{2} = frac{40}{2} = 20 ]4. The fourth term is the average of the second and third terms: [ frac{8 + 20}{2} = frac{28}{2} = 14 ]5. The fifth term (x) is the average of the third and fourth terms: [ frac{20 + 14}{2} = frac{34}{2} = 17 ]Thus, the value of (x) is (boxed{A}).]

🔑:The third term in the sequence is, by definition, the average of the first two terms, namely 32 and 8 , or frac{1}{2}(32+8)=frac{1}{2}(40)=20.The fourth term is the average of the second and third terms, or frac{1}{2}(8+20)=14.The fifth term is the average of the third and fourth terms, or frac{1}{2}(20+14)=17.Therefore, x=17.ANSWER: (A)

🧠:[<think>Okay, let's see. I need to find all integer pairs (a, b) where both a and b are at least 1, and they satisfy the equation a^{b²} = b^{a}. Hmm, interesting. So, a raised to the power of b squared equals b raised to the power of a. Alright, let's start by plugging in some small integers to see if I can find any obvious solutions.First, let's try a = 1. If a is 1, then the left side becomes 1^{b²} which is always 1. The right side would be b^{1} which is b. So, 1 = b. Therefore, the only possible b when a is 1 is b = 1. So (1, 1) is a solution. Let's check that: 1^{1²} = 1^1 = 1, and 1^1 = 1. Yep, that works. So (1,1) is a solution.Now, let's try a = 2. Then the left side is 2^{b²} and the right side is b². So we need 2^{b²} = b². Let's try different values of b.For b=1: 2^{1} = 2, and 1² =1. Not equal.For b=2: 2^{4} = 16, and 2² =4. Not equal.b=3: 2^{9}=512, 3²=9. Not equal.b=4: 2^{16}=65536, 4²=16. Not equal. Clearly, as b increases, the left side grows much faster than the right. So maybe no solution for a=2.Wait, unless there's a larger b where 2^{b²} equals b². But that seems impossible because exponential growth vs polynomial. Probably no solutions here. Let's note that a=2 might not give any solutions.Next, try a=3. Then left side is 3^{b²}, right side is b³. Let's check b=1: 3^1=3 vs 1^3=1. No. b=2: 3^4=81 vs 2^3=8. Not equal. b=3: 3^9=19683 vs 3^3=27. Nope. Similarly, left side grows way faster. So maybe no solutions for a=3.Moving on to a=4. Left side is 4^{b²} = (2²)^{b²}=2^{2b²}, right side is b^4. Let's check for b=2: 4^{4}=256 vs 2^4=16. Not equal. b=4: 4^{16} vs 4^4=256. Not equal. Again, left side is way bigger. How about b=16? 4^{256} is an astronomically large number, whereas 16^4=65536. Not even close. So a=4 doesn't seem to work.Wait, maybe a and b are equal? Let's check that. Let a = b. Then the equation becomes a^{a²} = a^{a}, which simplifies to a^{a²} = a^{a}. Therefore, unless a=1, which we already found, or the exponents are equal. So for a ≠1, we would need a² = a. So a² =a => a=0 or a=1. But a must be at least 1, so only a=1. So the only solution when a=b is (1,1). So no other solutions where a equals b.Hmm, maybe different approach. Let's take logarithms on both sides. Taking natural log:ln(a^{b²}) = ln(b^{a}) => b² ln a = a ln b.So we have b² / a = ln b / ln a. Let me write that as (b² / a) = (ln b) / (ln a). Hmm, maybe cross-multiplied: b² ln a = a ln b.Alternatively, rearranged: (ln a)/a = (ln b)/b². So we can think of the function f(x) = (ln x)/x and g(x) = (ln x)/x². Maybe comparing these functions for integer values.Wait, so f(a) = g(b). Let me analyze these functions. For f(x) = ln x / x, the derivative is (1 - ln x)/x², so it increases up to x=e and then decreases. Similarly, g(x) = ln x / x². Derivative is (1/x²)*(1/x) - (2 ln x)/x³ = (1 - 2 ln x)/x³. So critical point when 1 - 2 ln x =0 => ln x = 1/2 => x= sqrt(e). So g(x) increases up to x=sqrt(e) and then decreases.Therefore, both f(x) and g(x) have maximums. So maybe there are limited solutions where these functions cross each other. But since a and b are integers greater or equal to 1, perhaps we can find the possible pairs.Alternatively, let's consider possible small values of a and b. Since when a or b increases, the functions might not cross again. Let's check small numbers.We already found (1,1). Let's check (2, 2). Wait, that's 2^{4}=16 vs 2^2=4. Not equal.Wait, maybe a=16 and b=2. Let's check: 16^{4} = 65536 vs 2^{16} = 65536. Oh! Wait, that's equal. So 16^4 = 2^16. Because 16 is 2^4, so 16^4 = (2^4)^4 = 2^{16}, and 2^{16} is indeed 65536. So (16, 2) is a solution. Similarly, what about (2, 16)? Let's check: 2^{256} vs 16^{2} =256. 2^{256} is way bigger. So no. So (16,2) is a solution.Similarly, is there another pair like this? Let's think. Maybe a=27, b=3. Let's check: 27^{9} vs 3^{27}. 27 is 3^3, so 27^9 = (3^3)^9 = 3^{27}, which equals 3^{27}. So 27^9 =3^27. Therefore, (27,3) is a solution. Similarly, (3,27) would be 3^{729} vs 27^3=19683. Not equal. So (27,3) is a solution.So here, we have solutions where a is a perfect power of b. Wait, let's see. For example, take a = b^k for some integer k. Then substitute into the equation:Left side: (b^k)^{b²} = b^{k b²}Right side: b^{a} = b^{b^k}So for these to be equal, we need k b² = b^k. So k b² = b^k. Let's solve for k and b. Let's divide both sides by b² (assuming b ≠0, which it isn't here):k = b^{k -2}So k = b^{k -2}. So for integers k >=1 and b >=1.Looking for integer solutions where k = b^{k -2}Hmm, let's try k=1: 1 = b^{-1}. No solution since b is at least 1.k=2: 2 = b^{0}=1. Not equal.k=3: 3 = b^{1} => b=3. So when k=3, b=3. Then a = b^k =3^3=27. Which is the solution we found earlier: (27,3).k=4: 4 = b^{2} => b=2. Then a=2^4=16. Which gives the solution (16,2).k=5: 5 =b^{3} => b needs to be a cube root of 5, which is not integer.Similarly, k=6:6 =b^{4} => b=6^(1/4). Not integer. So no solutions for k>=5.Therefore, the only solutions from this family are when k=3, b=3 and k=4, b=2, leading to a=27 and a=16.Therefore, (16,2) and (27,3) are solutions.We already found (1,1). Are there any other solutions?Let's check for a=4, b=2. 4^{4} =256 vs 2^{4}=16. Not equal. a=9, b=3: 9^{9} vs 3^{9}. 9^9 = (3^2)^9=3^{18}, which is bigger than 3^9. Not equal. So no.Wait, maybe (256, 16)? Let's check: 256^{16²} =256^{256} vs 16^{256}. But 256 is 16^2, so 256^{256} = (16^2)^{256}=16^{512}, while 16^{256} is 16^{256}. Not equal. So no.Alternatively, let's check (65536, 256), but that's getting too big. Probably not.Alternatively, perhaps there are other solutions outside of these power chains. Let's check some small numbers.Check a=16, b=2: already found. a=27, b=3: found. a=1, b=1: found.What about a=4, b=2: 4^{4}=256 vs 2^{4}=16. No.a=2, b=4: 2^{16}=65536 vs 4^2=16. Not equal.How about a=5, b= any? Let's check b=5: 5^{25} vs 5^5. 5^25 is way bigger. b=2: 5^4=625 vs 2^5=32. Not equal. b=3: 5^9 vs 3^5=243. 5^9 is 1953125, not equal. So no.a=16, b=4: 16^{16} vs 4^{16}= (2^2)^16=2^32. But 16^16= (2^4)^16=2^64. Not equal.a=8, b=2: 8^4=4096 vs 2^8=256. Not equal. a=8, b=3: 8^9=134217728 vs 3^8=6561. Not equal.a=9, b=3: 9^9=387,420,489 vs 3^9=19,683. Not equal.Wait, perhaps a=16, b=2 and a=27, b=3 are the only non-trivial ones besides (1,1). Let's see if there are any others.What about a=256, b=4: Let's check. 256^{16} vs 4^{256}. 256 is 2^8, so 256^{16}=(2^8)^16=2^{128}. 4^{256}=(2^2)^256=2^{512}. Not equal. So 2^{128} vs 2^{512}. Not equal. So no.How about a=81, b=3: 81^{9} vs 3^{81}. 81 is 3^4, so 81^9=3^{36}, while 3^{81} is much larger. Not equal.Alternatively, maybe a=4, b=16: 4^{256} vs 16^4=256. Not equal.So seems like the only solutions are (1,1), (16,2), and (27,3). Let's check if there are any more.Check a=2, b=16: 2^{256} vs 16^2=256. Not equal.a=3, b=27: 3^{729} vs 27^3=19683. Not equal.What about a=5, b= any? Let's think. For a=5, we need 5^{b²}=b^5. Let's check possible b.b=5: 5^{25} vs 5^5. Left is way bigger.b=2: 5^4=625 vs 32. No.b=3: 5^9 vs 243. Nope.b=1: 5^1=5 vs 1^5=1. No.So no solutions for a=5.Similarly, a=6: 6^{b²}=b^6. Check b=2:6^4=1296 vs 64. No. b=3:6^9 vs 729. Left is way bigger.b=6:6^{36} vs 6^6. Left way bigger. So no.How about a=4, b=2: already checked. Not equal.Wait, maybe another approach. Let's suppose that a and b are powers of the same prime. Let’s say a = p^m, b = p^n for some prime p and integers m, n. Then the equation becomes (p^m)^{(p^n)^2} = (p^n)^{p^m}. Simplify exponents:Left: p^{m p^{2n}}, Right: p^{n p^m}Therefore, for equality, the exponents must be equal: m p^{2n} = n p^m.So m p^{2n} = n p^m.Hmm, this seems similar to the equation we had before. Let me set p=2, then see if there are integers m, n such that m * 2^{2n} =n * 2^m.Wait, for example, take m=4, n=1: 4 *2^{2}=4*4=16, and n*2^m=1*16=16. So 16=16. So this gives a solution: a=2^4=16, b=2^1=2. Which is the solution we found earlier.Similarly, take p=3, m=3, n=1: m *3^{2n}=3*3^{2}=3*9=27, n*3^m=1*3^3=27. So 27=27. Therefore, a=3^3=27, b=3^1=3. Which is another solution we found.So this method gives us the two non-trivial solutions. Are there any other primes?Take p=2, m=1, n=1: 1*2^{2}=4, 1*2^1=2. Not equal.p=2, m=2, n=1: 2*2^{2}=8 vs 1*2^2=4. Not equal.p=2, m=3, n=1:3*2^{2}=12 vs 1*8=8. Not equal.p=2, m=5, n=1:5*2^{2}=20 vs1*32=32. Not equal.p=2, m=4, n=2:4*2^{4}=4*16=64 vs 2*2^4=2*16=32. Not equal.Alternatively, other combinations for p=2. If n=2, m= ?Equation: m*2^{4} =2*2^m =>16m =2^{m+1} => 8m=2^m. So 8m=2^m. Check integers m:m=1:8=2^1=2 no.m=2:16=4 no.m=3:24=8 no.m=4:32=16 no.m=5:40=32 no.m=6:48=64 no.m=0: invalid.So no solutions here.For p=3, m=3, n=1 as before.p=3, m=4, n=1:4*3^{2}=4*9=36 vs1*3^4=81. Not equal.p=3, m=2, n=1:2*9=18 vs1*9=9. Not equal.p=5, m=5, n=1:5*5^{2}=125 vs1*5^5=3125. Not equal.So seems like only for p=2 and p=3 with specific exponents do we get solutions.Therefore, the only solutions of this form are (16,2) and (27,3). Along with (1,1).Are there any other solutions not of this form? Let's check a=2, b=4. Wait, 2^{16}=65536 vs4^2=16. Not equal. How about a=4, b=2. 4^4=256 vs2^4=16. Not equal. So no.Wait, what about a=16, b=4. 16^{16} vs4^{16}. 16 is 2^4, so 16^{16}=2^{64}, and 4^{16}=2^{32}. Not equal.Alternatively, maybe a=256, b=16. 256^{256} vs16^{256}. 256 is 2^8, so 256^{256}=2^{2048}, and16^{256}=2^{1024}. Not equal.Hmm. Let's consider if there are any solutions where a and b are not powers of the same prime.For example, take a=8, b=2. Then 8^{4}=4096 vs2^{8}=256. Not equal.a=8, b=4:8^{16} vs4^8=65536. 8^16 is larger.a=9, b=3:9^9 vs3^9. 9^9 is larger.Wait, maybe a=16, b=4: no, already checked.How about a=64, b=4:64^{16}= (2^6)^{16}=2^{96} vs4^{64}= (2^2)^{64}=2^{128}. Not equal.Alternatively, a= sqrt(b^a). Wait, not sure.Wait, maybe a different approach. Let's suppose that a and b satisfy a^{b²}=b^{a}. Let's consider writing this as a^{b²}/b^{a}=1. Let's take logarithms as before: b² ln a -a ln b=0. So b² ln a =a ln b.Let me rearrange this as (ln a)/a = (ln b)/b². So the function f(x) = (ln x)/x must equal f(y) = (ln y)/y². Let's analyze the functions.We know that f(x) = ln x /x has a maximum at x=e, and then decreases. Similarly, g(x) = ln x /x² has a maximum at x=sqrt(e) and then decreases. So perhaps there are points where ln a /a = ln b /b².Given that a and b are integers >=1, let's see possible pairs where this equality holds.We already have (1,1): ln1/1=0, ln1/1²=0. So 0=0.(16,2): ln16 /16 ≈2.7726/16≈0.1733. ln2 /4≈0.6931/4≈0.1733. So equal.(27,3): ln27 /27≈3.2958/27≈0.122. ln3 /9≈1.0986/9≈0.122. Equal.Are there any other integer pairs where ln a /a ≈ ln b /b²?Let's check for a=256, b=4. ln256 /256≈5.545/256≈0.02166. ln4 /16≈1.386/16≈0.0866. Not equal.a=256, b=16: ln256 /256≈5.545/256≈0.02166. ln16 /256²=ln16/(65536). ln16≈2.7726, so 2.7726/65536≈0.0000423. Not equal.How about a=4, b= sqrt(2). But b must be integer. So no.Wait, check a= 2, b= 2. Then ln2/2≈0.3466, ln2/4≈0.1733. Not equal.a=3, b= ?ln3 /3≈0.3662, ln b /b². Let's find b where ln b /b²≈0.3662. For b=1:0, b=2:0.6931/4≈0.173, b=3:1.0986/9≈0.122, which is lower. So no.a=4: ln4 /4≈1.386/4≈0.3465. So need ln b /b²≈0.3465. For b=1:0, b=2:≈0.173, b=3:≈0.122. Not enough.a=5: ln5 /5≈1.609/5≈0.3219. Still higher than what b=2 gives. So no.a=6:ln6/6≈1.7918/6≈0.2986. Still higher. a=7:1.9459/7≈0.277. a=8:2.079/8≈0.2599. a=9:2.1972/9≈0.244. a=10:2.3026/10≈0.2303. Still higher than ln2 /4≈0.1733.So even for a=10, we need b such that ln b /b²≈0.2303. Let's check b=2:0.1733, b=3:0.122. Between b=1 and 2? But b must be integer. So no.Similarly, for a=16, we have the value≈0.1733 which matches b=2. For a=27≈0.122 which matches b=3.If we go higher a, say a=81, ln81 /81≈4.3944/81≈0.0543. Which would correspond to ln b /b²≈0.0543. Let's check possible b. For b=4: ln4 /16≈1.386/16≈0.0866. b=5:ln5 /25≈1.609/25≈0.0644. Close but not equal. b=6:ln6/36≈1.7918/36≈0.0498. So between b=5 and b=6. Not integer.a=256: as before≈0.02166. Looking for ln b /b²≈0.02166. Let's check b=5:≈0.0644, b=6:≈0.0498, b=7:≈1.9459/49≈0.0397, b=8:2.079/64≈0.0325, b=9:2.1972/81≈0.0271, b=10:2.3026/100≈0.0230. Close to a=256:0.02166. So b=10 gives≈0.0230. Close but not equal. So no integer solution here.Therefore, it seems that the only solutions where (ln a)/a = (ln b)/b² are the ones we already found: (1,1), (16,2), (27,3). Let's confirm once more.Wait, a=0 is not allowed. How about a=1, b= anything: we already saw only (1,1) works.Any other possible a and b?What about a=65536, b= 256. Let's see: a=65536, which is 2^16. Then a^{b²} = (2^16)^{(256)^2} =2^{16*65536}=2^{1048576}. On the other side, b^a=256^{65536}= (2^8)^{65536}=2^{524288}. Not equal.Similarly, a=2^{2^n}, b=2^{2^{n-1}}} recursively? Maybe not.Alternatively, let's consider the equation again: a^{b²} = b^a. If we take both sides to the power of 1/(a b²)), we get a^{1/a} = b^{1/b²}. So the function f(x) = x^{1/x} and g(x)=x^{1/x²}. Let's analyze these.f(x) =x^{1/x}, which is known to have a maximum at x=e. Similarly, g(x)=x^{1/x²}. Take natural log: (ln x)/x². As x increases, this tends to 0. So maybe there are some intersections for integer x.But since we already found the solutions for (1,1), (16,2), (27,3), and others don't seem to intersect, maybe these are the only solutions.Wait, let's check (1,1): 1^{1}=1^{1}=1. Correct.(16,2):16^{4}=65536, 2^{16}=65536. Equal.(27,3):27^{9}=27^9, 3^{27}=3^{27}. Since 27=3^3, 27^9=(3^3)^9=3^{27}. So equal.So these are valid.Are there any more?Let's check a=256, b=4. 256^{16}= (2^8)^16=2^{128}, 4^{256}= (2^2)^256=2^{512}. Not equal.Another thought: suppose that a and b are related such that a = b^k and b = a^m for some integers k and m. Then substituting, a = (a^m)^k =a^{m k}, so m k=1. Therefore, m=1, k=1. Hence a =b. Which gives the solution (1,1). But we already have other solutions where a ≠b, so this approach doesn't capture those.Alternatively, perhaps we can consider the equation in terms of exponents.Let me write the equation again: a^{b²} =b^{a}. Let's express a as b^k, similar to before. Then, (b^k)^{b²}=b^{a} =>b^{k b²}=b^{a} =>k b²=a.So, if we set a= k b², then the equation becomes equality. Wait, but then we need to have a= k b² and also the original equation a^{b²}=b^{a}. Wait, substituting a=k b² into the equation:(k b²)^{b²}=b^{k b²}.Left side: (k b²)^{b²}=k^{b²} * b^{2 b²}Right side: b^{k b²}Therefore, equate:k^{b²} * b^{2 b²}=b^{k b²}Divide both sides by b^{2 b²}:k^{b²}=b^{(k -2) b²}Take both sides to the power of 1/b²:k= b^{k -2}So we have k= b^{k -2}. Which is the same equation as before. So this brings us back to the previous analysis where k= b^{k -2} for integers k and b. As before, the solutions are when k=3, b=3 and k=4, b=2. Hence, a= k b²=3*9=27 and a=4*4=16. So the solutions (27,3) and (16,2). Along with (1,1).Therefore, these are the only solutions.Hence, the integer pairs are (1,1), (16,2), and (27,3).Final AnswerThe integer pairs are boxed{(1, 1)}, boxed{(16, 2)}, and boxed{(27, 3)}.</think>To find all integer pairs ((a, b)) where (a geqslant 1), (b geqslant 1), and satisfy the equation (a^{b^2} = b^a), we start by checking small values and then use logarithms and function analysis.1. Checking small values: - For (a = 1), we get (1^{b^2} = b^1), which simplifies to (1 = b). Thus, ((1, 1)) is a solution. - For (a = 16) and (b = 2), we check (16^{2^2} = 16^4 = 65536) and (2^{16} = 65536), which are equal. Thus, ((16, 2)) is a solution. - For (a = 27) and (b = 3), we check (27^{3^2} = 27^9 = 3^{27}) and (3^{27} = 3^{27}), which are equal. Thus, ((27, 3)) is a solution.2. Logarithmic transformation: - Taking natural logarithms of both sides, we get (b^2 ln a = a ln b), which simplifies to (frac{ln a}{a} = frac{ln b}{b^2}). - Analyzing the functions (f(x) = frac{ln x}{x}) and (g(x) = frac{ln x}{x^2}), we find that they intersect at specific points corresponding to the solutions found.3. General form analysis: - Considering (a = b^k) and solving for (k) and (b), we find solutions when (k = b^{k - 2}). This leads to the solutions (k = 3, b = 3) (giving (a = 27)) and (k = 4, b = 2) (giving (a = 16)).4. Verification: - Checking other potential values and pairs confirms that no additional solutions exist beyond the ones found.Thus, the integer pairs ((a, b)) that satisfy the equation are (boxed{(1, 1)}), (boxed{(16, 2)}), and (boxed{(27, 3)}).,<think>Okay, so I need to find all integer pairs (a, b) where both a and b are at least 1, and they satisfy the equation a^{b²} = b^{a}. Hmm, this looks a bit tricky, but let me start by plugging in some small numbers to see if I can spot a pattern or find some solutions.First, let me try a = 1. If a is 1, then the left side becomes 1^{b²} which is always 1. The right side is b^{1} which is just b. So 1 = b. Since b has to be at least 1, the only possibility here is b = 1. So (1, 1) is a solution. That seems straightforward.Now let's try a = 2. Then the left side is 2^{b²}, and the right side is b². Wait, no, the right side is b^a, which is b². Wait, no, if a is 2, then b^a is b²? Wait, hold on. The equation is a^{b²} = b^{a}. So if a is 2, it's 2^{b²} = b². Let me write that down: 2^{b²} = b². Hmm, so for b >=1, let's check possible b:b =1: 2^{1} =2 vs 1²=1. Not equal.b=2: 2^{4}=16 vs 2²=4. Not equal.b=3: 2^{9}=512 vs 3²=9. Not equal.b=4: 2^{16}=65536 vs 4²=16. Not equal. Clearly, as b increases, the left side grows exponentially while the right side is only quadratic. So no solutions here when a=2.Wait, but maybe I made a mistake here. Let me check again. If a=2, then the equation is 2^{b²} = b². Let's see for b=2, 2^{4}=16 vs 2²=4. Not equal. For b=1, 2 vs 1. Nope. So no solutions for a=2.How about a=3? Then the equation becomes 3^{b²} = b³. Let's try b=1: 3^1=3 vs 1³=1. No. b=2: 3^4=81 vs 2³=8. No. b=3: 3^9=19683 vs 3³=27. No. Similarly, the left side grows much faster. Maybe b=0? Wait, but b has to be at least 1, so no.a=4. So 4^{b²} = b^4. Let's check b=1: 4 vs 1. No. b=2: 4^4=256 vs 2^4=16. No. b=3: 4^9=262144 vs 3^4=81. No. b=4: 4^16 vs 4^4=256. Again, left side is way bigger. So no solution here.Hmm, maybe a and b need to be the same? Let me try a = b. If a = b, then the equation becomes a^{a²} = a^{a}, which simplifies to a^{a²} = a^{a}. So if a ≠ 1, then this would require a² = a, which implies a(a -1)=0. But a is at least 1, so the only solution is a=1. Wait, but a=1 gives 1^{1}=1^{1}, which is true. So (1,1) is a solution, which we already have. So no other solutions where a = b except (1,1). So that approach doesn't help.Maybe I need to consider cases where a and b are different. Let's think about possible small numbers where exponents can balance out.For example, the equation a^{b²} = b^{a}. Taking natural logarithms on both sides, we get b² * ln a = a * ln b. So (b² / a) = (ln b / ln a). Hmm, this might be helpful. Let's rearrange it: (ln a)/a = (ln b)/b². So perhaps we can think of the function f(x) = (ln x)/x and g(x) = (ln x)/x². Then, we need to find integers a, b such that f(a) = g(b).But maybe this is too abstract. Let me try specific numbers. Let's check if there are any other solutions besides (1,1). Let's try a=16 and b=2. Then, left side is 16^{4} = 65536, right side is 2^{16} = 65536. Oh, that works! So (16, 2) is a solution. Wait, how did that happen? Let me verify:16^(2^2) = 16^4 = 65536. 2^16 = 65536. Yes, they are equal. So (16, 2) is a solution. Interesting. So there's another solution here.Similarly, maybe (2, 16)? Let's check. a=2, b=16. Then left side is 2^{16²} = 2^{256}, which is a huge number. Right side is 16^2 = 256. Not equal. So that's not a solution. So the pair (16, 2) works, but swapping them doesn't. So order matters here.Are there other such pairs? Let me think. Let's try a=27 and b=3. Then left side is 27^9, which is (3^3)^9 = 3^27. Right side is 3^27. So they are equal. Therefore, (27, 3) is a solution. Similarly, (256, 4): 256^(4^2)=256^16=(2^8)^16=2^128. Right side is 4^256=(2^2)^256=2^512. Not equal. Wait, no. So maybe (n^n, n)? Let me check.Suppose a = n^n and b = n. Then left side is (n^n)^{n²} = n^{n * n²} = n^{n³}. Right side is n^{n^n}. So unless n³ = n^n, which would require n³ = n^n. So n^n = n³. Taking log base n: n = 3. So when n=3, 3^3 = 27, which is a solution. So (27, 3) is a solution. Similarly, when n=2: 2²=4, but 2^4=16. So left side is 16, right side is 16. Wait, no, when n=2, a=4, b=2. Then left side is 4^{4}=256, right side is 2^4=16. Not equal. Wait, but earlier I found that (16, 2) is a solution. Wait, maybe I need a different approach here.Wait, for (16, 2): a=16, b=2. Then a=2^4, b=2. Then, a^{b²}= (2^4)^{4}=2^{16}=65536. b^{a}=2^{16}=65536. So that works. Similarly, (27,3): a=3^3, b=3. Then a^{b²}=27^9=3^{27}, and b^{a}=3^{27}. So same thing. So in these cases, a = b^{k} and maybe there's a relation between k and b.Wait, let's suppose that a = b^k for some integer k. Then, substituting into the equation, we have (b^k)^{b²} = b^{a}. Left side is b^{k b²}, right side is b^{a} = b^{b^k}. Therefore, exponents must be equal: k b² = b^k. So k b² = b^k. Let's solve for k. Divide both sides by b² (assuming b ≠0, which it is since b >=1): k = b^{k -2}. So we have k = b^{k -2}. So now, k and b are positive integers with b >=1, k >=1 (since a = b^k and a >=1).So the equation is k = b^{k -2}. Let me write this as k = b^{k-2}. So we need integer solutions (b, k) where b >=1, k >=1, and k = b^{k -2}.Let me fix k and find b, or fix b and find k.Let's start with small k:k=1: 1 = b^{-1}, but b^{-1} is 1/b. Since b >=1, 1/b <=1. So 1 = 1/b implies b=1. So (b, k)=(1,1). Then a=1^1=1. So (a,b)=(1,1). Which is the solution we already have.k=2: 2 = b^{0}=1. So 2=1? No solution.k=3: 3 = b^{1} => b=3. So (b, k)=(3,3). Then a=3^3=27. So (a,b)=(27,3). Which is the solution we found earlier.k=4: 4 = b^{2} => b=2 (since 2^2=4). So (b, k)=(2,4). Then a=2^4=16. So (a,b)=(16,2). Which is another solution we found.k=5: 5 = b^{3}. So b^3=5. But 5 is not a perfect cube, so no integer solution.k=6: 6 = b^{4}. Then b would be the 4th root of 6, which is not integer.Similarly, for k=7,8,... it's going to require b being roots of higher numbers, which likely aren't integers.Alternatively, fix b and find k.For b=1: k =1^{k -2}=1. So k=1. Then a=1^1=1. Which is the solution (1,1).For b=2: k=2^{k -2}. Let's solve k=2^{k-2} for integer k.k=1: 1=2^{-1}=0.5 Nok=2: 2=2^{0}=1 Nok=3:3=2^{1}=2 Nok=4:4=2^{2}=4 Yes. So k=4. Then a=2^4=16. So (16,2).k=5:5=2^{3}=8 Nok=6:6=2^{4}=16 No. So only k=4 here.For b=3: k=3^{k-2}. Let's check k:k=1:1=3^{-1}≈0.333 Nok=2:2=3^{0}=1 Nok=3:3=3^{1}=3 Yes. So k=3. Then a=3^3=27. So (27,3).k=4:4=3^{2}=9 Nok=5:5=3^{3}=27 No. So only k=3 here.For b=4: k=4^{k -2}k=1:1=4^{-1}=0.25 Nok=2:2=4^{0}=1 Nok=3:3=4^{1}=4 Nok=4:4=4^{2}=16 Nok=5:5=4^{3}=64 No. No solutions here.For b=5: k=5^{k -2}Similarly, k=1: 0.2, nope. k=2:1, nope. k=3:5^1=5≠3. k=4:5²=25≠4. Etc. No solutions.So in general, for b >=4, the equation k= b^{k-2} will not have solutions because b^{k-2} grows much faster than k. For example, for b=4, even starting from k=3, we get 4^{1}=4, but k=3≠4. For k=4, 4^{2}=16≠4. So no solutions here.Therefore, the only solutions from this parametrization are (a,b)=(1,1), (16,2), and (27,3). Are there any other solutions outside of this parametrization where a is not a power of b? Let me check.Suppose a and b are not powers of each other. Let's try some examples. For instance, (a,b)=(4,2). Then 4^{4}=256 vs 2^{4}=16. Not equal. (a,b)=(9,3): 9^{9}=387,420,489 vs 3^{9}=19,683. Not equal. (a,b)=(8,2): 8^{4}=4096 vs 2^{8}=256. Not equal. Hmm.What about (a,b)=(2,4)? 2^{16}=65536 vs 4^{2}=16. Not equal. How about (a,b)=(3,2): 3^{4}=81 vs 2^{3}=8. No. (a,b)=(5, whatever). Let's see if there's a possible pair.Alternatively, maybe (a,b)=(16,16). Then left side is 16^{256}, right side is 16^{16}. Not equal. (a,b)=(256,16): 256^{256} vs 16^{256}. These are both 2^{something}, but 256=2^8, so 256^{256}=2^{8*256}=2^{2048}. 16=2^4, so 16^{256}=2^{4*256}=2^{1024}. Not equal.Alternatively, maybe (a,b)=(4, 16). 4^{256} vs 16^4=65536. Not equal.Wait, perhaps there's a solution where a and b are different and not in the form we discussed earlier. Let me try (a,b)=(64,4). Let's compute: a^{b²}=64^{16}= (2^6)^{16}=2^{96}. b^a=4^{64}= (2^2)^64=2^{128}. Not equal. So no.Alternatively, (a,b)=(81,3): 81^9= (3^4)^9=3^{36}. 3^{81} is way larger. Not equal.Alternatively, (a,b)=(256,2): 256^{4}=2^{32} vs 2^{256}. No, not equal.Wait, maybe there's a solution with a=256 and b=16? Let's check: a=256, b=16. Then left side is 256^{16²}=256^{256}= (2^8)^{256}=2^{2048}. Right side is 16^{256}= (2^4)^{256}=2^{1024}. Not equal.Hmm. Let's think differently. Suppose that a and b are both powers of the same prime. Let p be a prime, and suppose a=p^m, b=p^n. Then the equation becomes (p^m)^{(p^n)^2} = (p^n)^{p^m}. Simplifying exponents:Left side: p^{m p^{2n}}. Right side: p^{n p^{m}}. Therefore, exponents must be equal: m p^{2n} = n p^{m}.So, m p^{2n} = n p^{m}. Let's divide both sides by n p^{2n} (assuming n≠0, which it isn't):(m/n) = p^{m - 2n}.Since m and n are positive integers and p is a prime, p^{m - 2n} must be a rational number. Since p is prime, this exponent must be non-negative, so m - 2n >=0. Thus, p^{m -2n} is an integer. Therefore, (m/n) must also be an integer. Let me set k = m/n, where k is a positive integer.So, k = p^{m - 2n}. But m = k n. Substitute:k = p^{k n - 2n} = p^{n(k - 2)}.Therefore, k = p^{n(k - 2)}.We need to solve for integers k >=1, n >=1, prime p, such that k = p^{n(k -2)}.This is similar to the equation we had before when we set a = b^k.Let me consider specific primes. Let's take p=2.Then, equation is k = 2^{n(k -2)}.We need to find integers k,n >=1 such that this holds.Let me try n=1:k = 2^{k -2}So k must be a power of 2? Let's check:k=1: 1=2^{-1} nok=2: 2=2^{0}=1 nok=3:3=2^{1}=2 nok=4:4=2^{2}=4 yesk=5:5=2^{3}=8 noSo for n=1, p=2, k=4 is a solution. Then m = k n=4*1=4. So a=p^m=2^4=16, b=p^n=2^1=2. So (16,2), which we already have.n=2:k =2^{2(k -2)}. Let's check possible k:k=1:1=2^{-2}=0.25 nok=2:2=2^{0}=1 nok=3:3=2^{2}=4 nok=4:4=2^{4}=16 no. So no solution here.n=3:k=2^{3(k -2)}. For k=1: 2^{-3}=0.125. No. k=2:2^{-0}=1. So 2=1? No. k=3:2^{3}=8. So 3=8? No. Similarly, no solution.n=4:k=2^{4(k -2)}. Even larger. So no.So the only solution for p=2 is n=1, k=4, leading to (16,2).Next, p=3:k =3^{n(k -2)}.n=1:k=3^{k-2}Check k=1:1=3^{-1}≈0.333 Nok=2:2=3^{0}=1 Nok=3:3=3^{1}=3 Yes. So k=3, n=1. Then m=k n=3*1=3. So a=3^3=27, b=3^1=3. So (27,3), which is a known solution.n=2:k=3^{2(k-2)}. Let's see:k=1:1=3^{-2}=1/9 Nok=2:2=3^{0}=1 Nok=3:3=3^{2}=9 Nok=4:4=3^{4}=81 No. So no solution.n=3:k=3^{3(k-2)}. Even bigger. No solutions.So p=3 gives only the solution (27,3).Next, p=5:k=5^{n(k -2)}.n=1:k=5^{k -2}Check k=1:1=5^{-1}=0.2 Nok=2:2=5^{0}=1 Nok=3:3=5^{1}=5 Nok=4:4=5^{2}=25 NoNo solutions.Similarly, higher primes will not yield solutions because the right side grows too quickly.What about p=1? But 1 is not a prime. So, in this parametrization, the only solutions come from p=2 and p=3 with specific exponents. These correspond to the solutions (16,2) and (27,3).Additionally, we had the solution (1,1). Let's see if that fits into this parametrization. If we take p=1, but 1 is not prime. Alternatively, if a=1, b=1. If we set p=1, but 1^n=1, so maybe. But since 1 is not prime, this case isn't covered. However, (1,1) is a solution because 1^{1}=1^{1}.Therefore, combining all solutions we have so far: (1,1), (16,2), (27,3). Are there any other solutions?Let me check another approach. Suppose that a and b are not powers of the same prime. For example, a=6 and b=2. Let's compute 6^{4}=1296 vs 2^{6}=64. Not equal. a=8, b=4. 8^{16}=2^{48} vs 4^{8}=2^{16}. Not equal.Alternatively, a=9, b=3. Wait, 9^9 vs 3^9. 9^9 is 3^18, which is much larger than 3^9. So no.Wait, maybe there's a solution where a and b are both 0? But the problem states a >=1 and b >=1, so 0 is excluded.Another idea: If one of a or b is 1. We already checked a=1 gives only b=1. What about b=1? Then the equation becomes a^{1}=1^{a}, so a=1. So only (1,1) again.So perhaps these are the only solutions. Let me check for a=5, b=5. 5^{25} vs 5^5. Not equal. How about a=4, b=2. 4^{4}=256 vs 2^4=16. Not equal. Wait, but 16,2 was a solution. Wait, (16,2) gives 16^{4}=2^{16}=65536. Wait, 16^4 is 65536, and 2^16 is also 65536. So that's equal. But if I take a=4, b=2, it's 4^4 vs 2^4. So 256 vs 16. Not equal. So a=16, b=2 works, but a=4, b=2 does not. So the exponent in the left side is b squared. So for (16,2), the exponent is 4, so 16^4 vs 2^16. They both equal 65536.Wait, so maybe other pairs where a^{b²} and b^{a} can be equal even if a is not a power of b? Let me see.Suppose I set a=256 and b=16. Then a^{b²}=256^{256}= (2^8)^{256}=2^{2048}. And b^{a}=16^{256}= (2^4)^{256}=2^{1024}. Not equal. So that doesn't work.Alternatively, a=65536, b=256. Then a^{b²}=65536^{65536}= (2^{16})^{65536}=2^{1048576}. And b^{a}=256^{65536}= (2^8)^{65536}=2^{524288}. Not equal. So even though each time we take a larger power, the exponents don't match.So it seems like only (1,1), (16,2), and (27,3) work. Let me check if there are any other solutions with a and b between 1 and, say, 10.Check a=4, b=2: 4^4=256 vs 2^4=16. No.a=8, b=2: 8^4=4096 vs 2^8=256. No.a=16, b=4:16^{16} vs 4^{16}= (2^2)^16=2^32. 16^16= (2^4)^16=2^64. Not equal.a=9, b=3:9^9 vs 3^9. 9^9=387,420,489 vs 3^9=19,683. Not equal.a=5, b= any: Let's try b=5. Then a^{25} =5^{a}. Maybe a=25:25^{25} vs 5^{25}=25^{25/2}? Wait, no. 5^{25}= (5^5)^5=3125^5. 25^{25}= (5^2)^25=5^{50}. So 5^{50} vs 5^{25}. Not equal.Alternatively, a=5, b= something. a^{b²}=b^a. Let's see 5^{b²}=b^5. Let's check b=5:5^{25}=5^5. No. b=1:5 vs1. No. b=2:5^4=625 vs 32. No. b=3:5^9=1953125 vs 3^5=243. No.Similarly, a=6:6^{b²}=b^6. Trying b=6:6^{36} vs6^6. No. b=2:6^4=1296 vs64. No. b=3:6^9 vs 729. 6^9=10077696. Not equal.a=7:7^{b²}=b^7. b=7:7^{49}=7^7. No. b=2:7^4=2401 vs128. No. Etc.a=10:10^{b²}=b^{10}. b=10:10^{100}=10^{10}. No. b=2:10^4=10000 vs1024. No. b=3:10^9 vs59049. No.So, up to a=10, the only solutions besides (1,1) are (16,2) and (27,3). Let me check a=256 and b=16 again. Wait, a=256, b=16. Then a^{b²}=256^{256}= (2^8)^256=2^{2048}, and b^a=16^{256}=2^{1024}. Not equal.Wait, but (16,2) works because 16 is 2^4, and 2 is the base. So 16^{4}=2^{16} which is the same as 2^{16}=2^{16}. Similarly, 27 is 3^3, so 27^9=3^{27}, and 3^27=3^{27}. So these solutions work because a is a power of b, and when you take a^{b²}, you get a power tower that equals b^a.So in the case where a = b^k, we saw that the equation reduces to k = b^{k -2}. So solving this gives us the possible (a,b) pairs. As we saw, only k=4 with b=2 and k=3 with b=3 give solutions here. And also the trivial solution (1,1).But let's confirm if there are any other solutions where a is not a power of b. Suppose there exists a solution where a and b are not powers of the same prime. For example, a=12 and b=6. Then a^{b²}=12^{36}, and b^{a}=6^{12}. Let's see which is bigger. 12^{36} vs 6^{12}. 12 is 6*2, so 12^{36}= (6*2)^36=6^{36}*2^{36}. 6^{12} is much smaller. So 12^{36} is way larger. Not equal.Alternatively, a=6, b=12. Then 6^{144} vs12^6. 6^{144} is astronomically larger. Not equal.Another idea: Maybe there are solutions where a and b are both composite but not prime powers. Let's try a=4, b=2: already checked, doesn't work. a=9, b=3: same as above.Alternatively, a=8, b=2: 8^4=4096 vs2^8=256. No. a=25, b=5:25^25 vs5^25. 25^25=5^50 vs5^25. Not equal.Wait, so in general, if a is a square or higher power of b, unless it's specifically set as in the earlier cases, it doesn't work.Alternatively, consider a= sqrt(b). Wait, but a and b have to be integers. So if a^2 = b, then the equation becomes a^{a^4} = (a^2)^{a} = a^{2a}. So we need a^{a^4} =a^{2a}, which implies a^4=2a. So a^3=2. But a must be integer, so no solution here.Alternatively, if b = a². Then the equation becomes a^{(a²)^2} = (a²)^a => a^{a^4} = a^{2a} => a^4=2a => a^3=2. No integer solution.Alternatively, suppose that a = 2 and b is some number. Then 2^{b²} = b². We checked this earlier and saw no solutions except perhaps b=2: 2^4=16 vs 2²=4. Doesn't work. Similarly, a=3:3^{b²}=b^3. Only works if b=3:3^9=19683 vs3^3=27. Not equal. Wait, no. Wait when a=27 and b=3, which works, but a=3 and b= something else? If a=3, then 3^{b²}=b³. We can check if b=3:3^9 vs3^3. No. b=9:3^{81} vs9^3=729. No.Alternatively, consider that in the known solutions, (16,2) and (27,3), the exponents in the original equation a^{b²} and b^{a} become equal when expressed as powers of the same base. For example, 16 is 2^4, so 16^{2²}= (2^4)^4=2^{16}, and 2^{16}=2^{16}. Similarly, 27 is 3^3, so 27^{3²}=3^{27}, and 3^{27}=3^{27}. So, in both cases, the equation simplifies to the same base raised to the same exponent. So maybe the only way this can happen is if a and b are related in such a way that allows the exponents to align when expressed in terms of the same prime base.Therefore, if we consider a and b such that a = p^{m} and b = p^{n}, then substituting into the equation gives (p^{m})^{p^{2n}} = p^{n p^{m}}. Which simplifies to p^{m p^{2n}} = p^{n p^{m}}. Therefore, exponents must be equal: m p^{2n} = n p^{m}. Which is the same equation as before. So solving m p^{2n} = n p^{m}. Which led us to the previous parametrization. Hence, the only solutions are the ones we found.Alternatively, if a and b are not powers of the same prime, then it's hard to see how their exponents would align. Because if they have different prime factors, the prime factors on both sides would have to match, which is unlikely unless one side is a power of the other. But even then, exponents would have to coincide.Therefore, after checking various cases and approaches, it seems that the only integer solutions with a >=1 and b >=1 are (1,1), (16,2), and (27,3). I don't see any other pairs that satisfy the equation, and the attempts to find more solutions by trial or through parametrization haven't yielded anything else.Final AnswerThe integer pairs are boxed{(1, 1)}, boxed{(16, 2)}, and boxed{(27, 3)}.</think>To find all integer pairs ((a, b)) where (a geqslant 1), (b geqslant 1), and satisfy the equation (a^{b^2} = b^a), we start by checking small values and using logarithms to analyze the equation.1. Case (a = 1): - The equation simplifies to (1^{b^2} = b^1), which gives (1 = b). Thus, ((1, 1)) is a solution.2. Case (a = b): - The equation becomes (a^{a^2} = a^a). For (a neq 1), this implies (a^2 = a), which is only true for (a = 1). Hence, no new solutions here.3. Using logarithms: - Taking natural logarithms on both sides, we get (b^2 ln a = a ln b). Rearranging gives (frac{ln a}{a} = frac{ln b}{b^2}). This suggests analyzing the functions (f(x) = frac{ln x}{x}) and (g(x) = frac{ln x}{x^2}).4. Parametrization: - Suppose (a = b^k). Substituting into the equation gives ((b^k)^{b^2} = b^{b^k}), leading to (b^{kb^2} = b^{b^k}). Equating exponents, we get (kb^2 = b^k), which simplifies to (k = b^{k-2}). Solving for (k) and (b): - For (b = 2), (k = 4) gives (a = 2^4 = 16), leading to ((16, 2)). - For (b = 3), (k = 3) gives (a = 3^3 = 27), leading to ((27, 3)).5. Verification: - Checking other possible values and parametrizations confirms no additional solutions. For example, checking pairs where (a) and (b) are not powers of each other or different primes shows no valid solutions.After thorough analysis and verification, the integer pairs ((a, b)) that satisfy the equation are:[boxed{(1, 1)}, boxed{(16, 2)}, boxed{(27, 3)}],<think>Okay, so I need to find all integer pairs (a, b) where both a and b are at least 1, and they satisfy the equation a^{b²} = b^a. Hmm, let's see. Alright, let's start by understanding what this equation means. First, both sides are exponentials, but the exponents are different: on the left side, it's b squared, and on the right side, it's a. Maybe I can take logarithms of both sides to simplify it? Let's try that. Taking natural logs, we get:ln(a^{b²}) = ln(b^a)Which simplifies to:b² * ln(a) = a * ln(b)Hmm, that might be a bit easier to handle. Let me rearrange this equation. Maybe divide both sides by a * b²? Let's see:(ln(a)/a) = (ln(b)/b²)So, that's interesting. It suggests that the function f(x) = ln(x)/x is equal to f(y)/y, where x is a and y is b²? Wait, maybe not exactly. Let me think. Actually, if I let f(x) = ln(x)/x^k for some exponent k, but maybe that's complicating it. Alternatively, perhaps we can look for pairs where a and b are related in a way that makes these ratios equal.Alternatively, maybe I can express a in terms of b or vice versa. Let's see. Let's suppose that a and b are powers of the same prime. For example, maybe a = p^m and b = p^n. Then we can substitute into the equation. But I don't know if that approach will necessarily cover all solutions. Maybe it's better to test small integers first.Let me try small values of a and b and see if they satisfy the equation.Start with a = 1. Then the equation becomes 1^{b²} = b^1. Left side is 1, right side is b. So 1 = b. Thus, b must be 1. So (1, 1) is a solution. Check: 1^{1²} = 1^1 = 1. Yep, that works.Next, a = 2. Let's see what b would satisfy 2^{b²} = b². Wait, hold on. The equation is a^{b²} = b^a. So for a=2, it's 2^{b²} = b². Let's check possible b.b=1: 2^{1} = 2 vs 1² = 1. Not equal.b=2: 2^{4} = 16 vs 2² = 4. 16 ≠ 4.b=3: 2^{9} = 512 vs 3² = 9. 512 ≠ 9.b=4: 2^{16} = 65536 vs 4² = 16. Not equal. So no solution for a=2.Wait, maybe I made a mistake here. Let me check again. For a=2, the equation is 2^{b²} = b^2. So when b=2, left side is 2^4=16, right side is 2^2=4. Not equal. For b=4, left side is 2^{16}=65536, right side is 4^2=16. Nope. So a=2 doesn't seem to have a solution.How about a=3?For a=3, equation is 3^{b²} = b^3.Check b=1: 3^1=3 vs 1^3=1. Not equal.b=2: 3^4=81 vs 8. Not equal.b=3: 3^9=19683 vs 27. 19683 vs 27. Not equal.b=4: 3^{16} is huge, 43046721 vs 64. Nope. So no solution for a=3.a=4. Then equation is 4^{b²} = b^4.Check b=1: 4 vs 1. Nope.b=2: 4^4=256 vs 16. 256 vs 16. No.b=4: 4^{16}=4294967296 vs 256. Not equal.Wait, maybe b=16? 4^{256} vs 16^4. 16^4 is 65536, and 4^{256} is way larger. So no. Hmm.a=5: 5^{b²}=b^5.Similarly, no luck with small b.Wait, maybe trying a=16. Let's check b=2. Then 16^{4}=65536 vs 2^{16}=65536. Oh! Wait a second. 16^{4} is (2^4)^4 = 2^{16}=65536. And 2^{16}=65536. So 16^{4}=2^{16}. So if a=16 and b=2, then 16^{2²}=16^4=65536 and 2^{16}=65536. So that works. So (16, 2) is a solution.Similarly, maybe (2, 16)? Let's check. a=2, b=16. Then 2^{16²}=2^{256} vs 16^2=256. 2^{256} is way larger than 256. So that doesn't work. So the solution is (16, 2). Hmm.So maybe there is a pair here where a and b are swapped with exponents? Wait, not exactly. Wait, 16 and 2. So 16 is 2^4, and 2 is 2^1. Maybe there's a pattern here. Let's check if there are other such pairs.Wait, let's think about the equation again. a^{b²} = b^a. Let me take logarithms again: ln(a^{b²}) = ln(b^a) => b² ln a = a ln b. So rearranged, (ln a)/a = (ln b)/b². So the ratio (ln x)/x for x=a equals the ratio (ln y)/y² for y=b. So maybe these functions cross at some points.Alternatively, set a = b^k for some exponent k. Let's try substituting a = b^k into the equation. Then:Left side: (b^k)^{b²} = b^{k b²}Right side: b^{a} = b^{b^k}Therefore, setting exponents equal: k b² = b^kSo k b² = b^k => k = b^{k - 2}So we have k = b^{k - 2}, where k is a positive integer (since a and b are integers greater than or equal to 1, and a = b^k, so k must be a non-negative integer, but since a >=1, k can be 0 but then a=1. Wait, but a = b^k, if k=0, then a=1 for any b. Then check if that works. Let's see. If a=1, then equation is 1^{b²}=b^1, so 1 = b. So (1,1). That's already considered. So k must be positive. So k >=1.So for a = b^k, k >=1, and then we have k = b^{k - 2}So solving for integer k and b >=1: k = b^{k - 2}Looking for integer solutions (k, b). Let's see:Case 1: k=1. Then 1 = b^{-1}. Which would require b=1. Since b >=1, b=1. So k=1, b=1. Then a = 1^1=1. So that gives the solution (1,1). Which we already have.Case 2: k=2. Then 2 = b^{0} =1. But 2≠1. So no solution here.Case 3: k=3. Then 3 = b^{1} => b=3. So then a = 3^3=27. Let's check if (27, 3) is a solution.Compute 27^{3²}=27^9 vs 3^{27}. Hmm. 27 is 3^3, so 27^9 = (3^3)^9 = 3^{27}, and 3^{27} is 3^{27}. So yes, they are equal. So (27, 3) is another solution.Case 4: k=4. Then 4 = b^{2}. So b=2 (since 2²=4). So then a=2^4=16. Which gives the solution (16,2). Which we found earlier.Case 5: k=5. Then 5 = b^{3}. So b^3=5. But 5 is not a perfect cube, so no integer solution.k=6: 6 = b^{4}. Similarly, b would need to be 6^{1/4}, which is not integer. So no solutions here.Similarly, for k>4, we have k = b^{k-2}. The right side grows exponentially with k if b>=2, but the left side is linear in k. So for b>=2 and k-2 >=1 (i.e., k>=3), then b^{k-2} grows exponentially, which will outpace k. But wait, for k=3, b=3, which works. For k=4, b=2, which works. For k=5, as above, no solution. For k=6, no solution, etc.Alternatively, maybe for b=1. If b=1, then k =1^{k-2}=1. So k=1. So that gives a=1^1=1, which is the solution (1,1). Which we already have.Alternatively, check for b=1. For b=1, original equation is a^{1}=1^{a} => a=1. So (1,1). Which is covered.So in this substitution approach, we get solutions (1,1), (16,2), (27,3). Are there other solutions outside of this substitution?Let me check a=16, b=2 and a=27, b=3. Also, (1,1). Maybe these are the only ones?Wait, let's check (2,4). So a=2, b=4. Then 2^{16} vs 4^2=16. 2^{16}=65536 vs 16. Not equal.What about (4,2)? a=4, b=2. Then 4^{4}=256 vs 2^4=16. 256≠16. So nope.Wait, but maybe other pairs. Let's see.Check (a,b)=(27,3). 27^{9} vs 3^{27}. As above, 27^9=3^{27}, so they are equal.Check (a,b)=(256,4). Wait, let's see. If I follow the pattern, next solution would be with k=5? But that didn't work. Wait, but if k=4, we had a=16, b=2. If I set k=5, need b^3=5, which is not integer. So no.Alternatively, maybe a=256, b=4. Let's check. 256^{16} vs 4^{256}. 256 is 2^8, so 256^{16}=(2^8)^16=2^{128}, and 4^{256}=(2^2)^256=2^{512}. 2^{128} vs 2^{512}. Not equal. So no.Alternatively, maybe check for b=4. Let's see if there's an a such that a^{16}=4^a.So we need a^{16} =4^a. Let's take logarithms: 16 ln a = a ln 4 => (ln a)/a = (ln 4)/16.We can compute (ln 4)/16 ≈ (1.386)/16 ≈0.0866.We need to find a such that (ln a)/a ≈0.0866.Let me compute (ln a)/a for a=16: ln16/16≈2.7726/16≈0.173. That's larger than 0.0866.a=32: ln32/32≈3.4657/32≈0.108. Still higher.a=64: ln64/64≈4.158/64≈0.065. Lower. So between 32 and 64. Let's check a=48: ln48≈3.871/48≈0.0806. Closer. a=56: ln56≈4.025/56≈0.0719. Hmm. But a must be integer. So seems like no integer a satisfies (ln a)/a ≈0.0866. So no solution here.Alternatively, maybe b=5. Let's check if there exists a such that a^{25}=5^a. Similarly, take logs: 25 ln a = a ln5 => (ln a)/a = (ln5)/25≈1.6094/25≈0.0644. Let's check a=25: ln25/25≈3.2189/25≈0.1287. Too high. a=50: ln50≈3.912/50≈0.078. Still higher. a=75: ln75≈4.317/75≈0.0575. Lower. Crosses around 0.0644. So between 50 and 75. But no integer a would make this exact. So no solution here.Alternatively, check a=5 and b= some number. a=5: 5^{b²}=b^5. For b=5: 5^{25}=5^5. 5^{25} is way bigger. For b=2: 5^4=625 vs 32. No. b=3:5^9=1953125 vs 243. Nope. So no solutions here.How about a= b? Let’s see. If a = b, then the equation becomes a^{a²} = a^a. Which implies a^{a² - a} =1. Since a >=1, the only way this is true is if a² - a =0, so a(a-1)=0. Thus, a=0 or a=1. But a>=1, so a=1. Then b=1. Which is the solution we already have.So that only gives (1,1).Another approach: Let's consider the equation a^{b²} = b^a. If we can write a and b as powers of the same prime, maybe that would help. Let's suppose that a = p^m and b = p^n for some prime p and integers m, n >=0.Then substituting into the equation:(p^m)^{ (p^n)^2 } = (p^n)^{p^m }Simplify exponents:p^{ m p^{2n} } = p^{n p^m }Since the bases are the same, set exponents equal:m p^{2n} = n p^mSo m p^{2n} = n p^mHmm, this might not necessarily lead to a solution unless specific conditions on m and n.Alternatively, let's try p=2. Then:m * 2^{2n} = n * 2^mLooking for integers m,n >=1. Let's check if we can find solutions here.For example, take the solution (16,2). Here, a=16=2^4, b=2=2^1. So m=4, n=1.Check the equation: 4 * 2^{2*1}=4*2^2=4*4=16. On the other side: 1 * 2^4=1*16=16. So 16=16. So that works.Similarly, take the solution (27,3). Here, a=27=3^3, b=3=3^1. So p=3, m=3, n=1.Check: 3 * 3^{2*1}=3*3^2=3*9=27. On the other side:1 *3^3=27. 27=27. So that works.Similarly, the solution (1,1) is 1=1^1, so p can be any prime, but m=n=0. But since a,b >=1, that's okay.So, in general, if we set m = k and n=1, then the equation becomes:k * p^{2*1} = 1 * p^{k}So k p² = p^kWhich is equivalent to k = p^{k -2}Which is the same equation we had earlier when we set a = b^k. So this seems like a similar approach. So if we set n=1, and p is a prime, then for m=k, we have k = p^{k -2}Which gives us the solutions when p=2, k=4 (2^{4-2}=4, so k=4). And when p=3, k=3 (3^{3-2}=3). For p>=5, k would have to satisfy k=5^{k-2} or higher, which likely doesn't have integer solutions.So, indeed, the only prime solutions here are p=2,3 with k=4,3 respectively, leading to a=16,27 and b=2,3.But since we are considering all integers, not necessarily primes, maybe there are composite numbers that can satisfy the equation. Let's check.Suppose, for example, a=256 and b=4. Wait, let's test that. a=256, b=4.Left side: 256^{16} = (2^8)^{16} = 2^{128}Right side: 4^{256} = (2^2)^{256} = 2^{512}Not equal. So that's not a solution.How about a=65536 and b=16? Then left side: 65536^{256} = (2^{16})^{256}=2^{4096}Right side:16^{65536}= (2^4)^{65536}=2^{262144}. Not equal.So scaling up seems to not work. What about other composite numbers?Take a=16 and b=4. Then 16^{16} vs 4^16. 16^{16}=2^{64}, 4^16=2^{32}. Not equal.a=9, b=3. 9^9 vs 3^9. 9^9=3^{18}, 3^9=3^9. Not equal.Wait, but maybe a=27 and b=3. 27^9 vs 3^27. 27 is 3^3, so 27^9=3^{27}, which equals 3^27. So that works. But that's the solution we already have.Alternatively, check a=8, b=2. 8^{4}=4096 vs 2^8=256. Not equal.How about a=81, b=3. 81^{9} vs 3^{81}. 81=3^4, so 81^9=3^{36}, which is much less than 3^{81}. Not equal.So it seems that aside from the solutions we found earlier, there are no others. But how to be sure?Let’s think about the equation again: b² ln a = a ln b.We can write this as (ln a)/a = (ln b)/b². Let’s define two functions f(x) = (ln x)/x and g(x) = (ln x)/x². Then the equation is f(a) = g(b).So we need to find integers a, b >=1 such that f(a) = g(b). Let's analyze the behavior of these functions.First, f(x) = (ln x)/x. The derivative of f(x) is (1 - ln x)/x². So f(x) increases on (0, e) and decreases on (e, ∞). The maximum at x=e is about 0.3679/x.Similarly, g(x) = (ln x)/x². The derivative is (1 - 2 ln x)/x³. So g(x) increases on (0, e^{1/2}) and decreases on (e^{1/2}, ∞). Maximum at x=sqrt(e) ≈1.6487.So f(x) is decreasing for x >=3, and g(x) is decreasing for x >=2. So for x >=3, f(x) is decreasing, and for x >=2, g(x) is decreasing.So, for a >=3, f(a) is decreasing, and for b >=2, g(b) is decreasing.Therefore, for a >=3 and b >=2, both f(a) and g(b) are decreasing. So if we can show that after certain points, f(a) < g(b) for all possible a and b, then there are no solutions beyond that.First, let's compute f(a) for a=1,2,3,...f(1)=0f(2)=ln2/2≈0.3466f(3)=ln3/3≈0.3662f(4)=ln4/4≈0.3466f(5)=ln5/5≈1.6094/5≈0.3219f(6)=ln6/6≈1.7918/6≈0.2986f(7)=ln7/7≈1.9459/7≈0.2779Continuing, it's decreasing from a=3 onwards.g(b)= (ln b)/b²:g(1)=0g(2)=ln2/4≈0.1733g(3)=ln3/9≈0.1220g(4)=ln4/16≈0.0866g(5)=ln5/25≈0.0644g(6)=ln6/36≈0.0498g(7)=ln7/49≈0.0398Etc. So g(b) is decreasing from b=2 onwards.So, to find solutions where f(a)=g(b), we can look for pairs (a,b) where these decreasing functions cross.We have the known solutions:(1,1): f(1)=0, g(1)=0.(16,2): f(16)=ln16/16≈2.7726/16≈0.1733, which is equal to g(2)=0.1733.(27,3): f(27)=ln27/27≈3.2958/27≈0.1220, equal to g(3)=0.1220.So these solutions correspond to points where f(a) intersects with g(b) for integer a and b.Now, let's check if there could be any more intersections.Looking at f(a) for a=4: 0.3466 (same as f(2)), but since a >=1, and we need f(a)=g(b). For a=4, f(a)=0.3466, which is higher than g(2)=0.1733. So maybe there's a b where g(b)=0.3466? Let's see.Solve (ln x)/x² =0.3466. Let's try x=1: 0. x=2: 0.1733. x=1.5: (ln1.5)/2.25≈0.4055/2.25≈0.1802. So it's decreasing from x>sqrt(e). So the maximum value of g(x) is at x= sqrt(e)≈1.6487, where g(x)= (ln1.6487)/(1.6487)^2≈0.5/(2.718)≈0.1839. So the maximum of g(x) is about 0.1839. So g(b) can never reach 0.3466. Thus, f(a) for a=2 is 0.3466, which cannot be matched by any g(b). Therefore, a=2 cannot be a solution. Similarly, a=3 has f(a)=0.3662, which is higher than the maximum of g(b). Therefore, no solution for a=3. Wait, but a=3, f(a)=0.3662, which is higher than the maximum g(b)=0.1839. So no possible b can satisfy f(3)=g(b). Therefore, the only possible solutions with a=1,16,27 and b=1,2,3. Let's check if there's another a where f(a) is equal to some g(b).For example, a=4, f(a)=0.3466. The maximum g(b) is 0.1839, so not reachable.a=5: f(a)=0.3219. Still higher than 0.1839.a=6:0.2986. Still higher.a=7:0.2779.a=8: ln8/8≈2.079/8≈0.2599.a=9: ln9/9≈2.1972/9≈0.2441.a=10: ln10/10≈2.3026/10≈0.2303.a=12: ln12/12≈2.4849/12≈0.2071.a=15: ln15/15≈2.708/15≈0.1805.a=16: ln16/16≈2.7726/16≈0.1733.So at a=15, f(a)=0.1805, which is just above g(2)=0.1733. So 0.1805 vs 0.1733. Not equal.At a=16, f(a)=0.1733 which matches g(2)=0.1733. So that's the (16,2) solution.Similarly, check a=27: f(27)=0.1220, which matches g(3)=0.1220.Check a=81: f(81)=ln81/81≈4.3944/81≈0.0543. Which would correspond to g(b)=0.0543. Let's see which b would give that.g(b)=0.0543. So ln b / b²=0.0543. Let's check b=4: ln4/16≈0.0866. b=5:0.0644. b=6:0.0498. So between b=5 and 6. No integer b here. So no solution.Similarly, check a=256: f(256)=ln256/256≈5.545/256≈0.02166. Then g(b)=0.02166. Solving ln b / b²=0.02166. Let's check b=6:0.0498. b=7:0.0398. b=8: ln8/64≈2.079/64≈0.0325. b=9: ln9/81≈2.197/81≈0.0271. b=10: ln10/100≈0.2303/100≈0.002303. Wait, that's not right. Wait, ln10/100≈2.3026/100≈0.0230. Close to 0.02166. So b=10 gives g(10)=0.0230, which is close but not equal. So a=256 would require b≈10, but not exactly. So no integer solution.Therefore, it seems that the only integer solutions are (1,1), (16,2), and (27,3). To confirm, let's check if there are any other possible a and b where f(a)=g(b).Looking at the values:We need to find a and b where (ln a)/a = (ln b)/b².From the above analysis, after a=27, f(a) continues to decrease towards zero, but g(b) also decreases towards zero. The next possible point where they might intersect is when a is such that f(a) is equal to g(b) for some b. But since both are decreasing, and f(a) for a>27 is less than 0.1220, and g(b) for b>3 is less than 0.1220, we need to see if there's a point where f(a)=g(b) for some a>27 and b>3.Suppose, for example, a=256 and b=16. Then f(256)=0.02166, and g(16)=ln16/256≈2.7726/256≈0.0108. Not equal.How about a=81 and b=9. f(81)=0.0543, g(9)=ln9/81≈2.1972/81≈0.0271. Not equal.Alternatively, a=256 and b=16: as above. Not equal.Alternatively, maybe a= 65536 and b=256. Then f(65536)=ln65536/65536≈11.0903/65536≈0.000169. g(256)=ln256/(256)^2≈5.545/65536≈0.0000846. Not equal.It seems that as a and b increase, f(a) and g(b) both approach zero, but it's unclear if they ever cross again at integer points. However, given how rapidly both functions decrease, it's highly unlikely there's another integer solution.Alternatively, check smaller a and b beyond the ones we found.Check a=4 and b=16. Wait, a=4, b=16: 4^{256} vs 16^4=65536. Not equal.Wait, but (a,b)=(16,2) and (27,3) are solutions. Maybe there's a pattern here where a = b^{k} and k satisfies k = b^{k-2}. As we saw earlier, for k=3, b=3: 3=3^{3-2}=3^1=3. For k=4, b=2:4=2^{4-2}=2^2=4. These are the only integer solutions for this equation k = b^{k-2}. For k>=5, as we saw, there are no integer solutions. So indeed, these would be the only solutions.Therefore, the integer pairs (a, b) where a,b >=1 and a^{b²}=b^a are:(1,1), (16,2), and (27,3).Final AnswerThe integer pairs are boxed{(1, 1)}, boxed{(16, 2)}, and boxed{(27, 3)}.</think>To find all integer pairs ((a, b)) where (a geqslant 1), (b geqslant 1), and satisfy the equation (a^{b^2} = b^a), we start by taking the natural logarithm of both sides:[ln(a^{b^2}) = ln(b^a)]This simplifies to:[b^2 ln(a) = a ln(b)]Rearranging, we get:[frac{ln(a)}{a} = frac{ln(b)}{b^2}]We analyze the functions (f(x) = frac{ln(x)}{x}) and (g(x) = frac{ln(x)}{x^2}) to find where their values intersect for integer arguments. Testing small values and specific cases:1. Case (a = 1): - The equation becomes (1^{b^2} = b^1), which simplifies to (1 = b). Thus, ((1, 1)) is a solution.2. Case (a = 16) and (b = 2): - Check (16^{2^2} = 16^4 = 65536) and (2^{16} = 65536). This works, so ((16, 2)) is a solution.3. Case (a = 27) and (b = 3): - Check (27^{3^2} = 27^9 = 3^{27}) and (3^{27} = 3^{27}). This works, so ((27, 3)) is a solution.We also consider the substitution (a = b^k) leading to the equation (k = b^{k - 2}). Solving this for integer values, we find:- For (k = 3) and (b = 3), we get (a = 27).- For (k = 4) and (b = 2), we get (a = 16).After verifying all possible cases and ensuring no other pairs satisfy the equation, we conclude the solutions are:[boxed{(1, 1)}, boxed{(16, 2)}, boxed{(27, 3)}],<think>Okay, so I need to find all integer pairs (a, b) where both a and b are at least 1, and they satisfy the equation a^{b²} = b^{a}. Hmm, let's start by understanding the problem. First, the equation is a^{b²} = b^{a}. Both a and b are positive integers. I need to find all such pairs. Let me try plugging in small values for a and b to see if any obvious solutions pop up.Let's start with a = 1. Then the left-hand side becomes 1^{b²} = 1. The right-hand side would be b^{1} = b. So we have 1 = b. Since b must be at least 1, the only possibility here is b = 1. So (1,1) is a solution. Let me check: 1^{1²} = 1^1 = 1, and 1^1 = 1. Yep, that works.Next, try b = 1. Then the right-hand side is 1^{a} = 1. The left-hand side is a^{1²} = a^1 = a. So we have a = 1. Again, that's the same solution (1,1). So no other solutions when either a or b is 1.Now, let's try a = 2. Then the equation becomes 2^{b²} = b². Wait, no, the right-hand side is b^{a} = b². So 2^{b²} = b². Let's see if any integer b ≥ 2 satisfies this. Let's try b = 2: left-hand side is 2^4 = 16, right-hand side is 2^2 = 4. Not equal. b = 3: left-hand side is 2^9 = 512, right-hand side is 3^2 = 9. Not equal. b = 4: left-hand side is 2^16 = 65536, right-hand side is 4^2 = 16. Definitely not equal. So no solutions when a = 2 except maybe b= something very big? But likely not. Let me check b=1, but we already saw that gives a=1.Similarly, let's check b=2. Then the equation is a^{4} = 2^{a}. So we need a^4 = 2^a. Let's try a=2: 16 = 4. Nope. a=4: 256 = 16. Nope. a=16: 16^4 = 65536, 2^16 is also 65536. Wait, 16^4 is (2^4)^4 = 2^{16}, which is equal to 2^16. So 16^4 = 2^16. Therefore, a=16, b=2 is a solution. Let me confirm: 16^{2²} = 16^4 = 65536, and 2^{16} = 65536. Yep, that works. So (16, 2) is a solution. Interesting. So a=16 and b=2 works. What about other a's for b=2? Let's see.We have a^4 = 2^a. So possible solutions where a is a power of 2? Let's check a=4: 4^4=256, 2^4=16. Not equal. a=8: 8^4=4096, 2^8=256. Not equal. a=16: 16^4=65536, 2^16=65536. So only a=16 works here. So (16,2) is a solution.Similarly, let's check b=4. Then the equation becomes a^{16} = 4^{a}. Let's see if a=4: 4^{16} vs 4^4. 4^16 is way bigger. a=2: 2^{16}=65536 vs 4^2=16. Not equal. a=16: 16^{16} vs 4^{16}= (2^2)^16=2^32. But 16^16= (2^4)^16=2^64, which is way larger. So no solution here.Alternatively, maybe a= something else? Let's see. For a^{16} =4^a, take natural logarithm: 16 ln a = a ln 4. So 16 ln a = a ln 4. Let me consider this as a function f(a) = 16 ln a - a ln 4. We can look for a where f(a)=0. Let's compute f(16): 16 ln 16 -16 ln4=16(ln16 - ln4)=16 ln(16/4)=16 ln4≈16*1.386≈22.18. Positive. f(4): 16 ln4 -4 ln4=12 ln4≈12*1.386≈16.63. Still positive. f(2):16 ln2 -2 ln4=16 ln2 -2*2 ln2=16 ln2 -4 ln2=12 ln2≈8.318. Still positive. f(1):16 ln1 -1 ln4=0 - ln4≈-1.386. So crosses zero between a=1 and a=2. But a must be integer ≥1. So no solution here. So b=4 gives no solution except if a= something non-integer. So no.Wait, but maybe other b's. Let's check b=3. Then the equation is a^{9}=3^{a}. Let's test some a's. a=3: 3^9=19683, 3^3=27. Not equal. a=9:9^9 vs 3^9. 9^9 is (3^2)^9=3^18, which is way bigger than 3^9. a=27:27^9 vs 3^27. 27^9=(3^3)^9=3^27. So 3^27=3^27. So a=27, b=3 gives a solution. Let me verify: 27^{3²}=27^9=3^{27}, and 3^{27}=3^{27}. So yes, (27,3) is a solution. Interesting. So here, a=27 and b=3.Similarly, let's check if there are more solutions like this. So when b= k, maybe a= k^k? Wait, no. Wait, for example, when b=2, we had a=16=2^4, but 16=2^4. When b=3, we have a=27=3^3. Hmm, 3^3. Wait, 2^4=16, 3^3=27. Hmm, 4 is 2 squared. Wait, 16=2^{4}, which is 2^{2^2}, and 27=3^{3}, which is 3^{3^1}. Not sure if there's a pattern here.Wait, let's see. For (16,2): 16 is 2^4, and 2 is the base. So a=2^{2^2}=2^4=16, and b=2. For (27,3): a=3^{3}=27, b=3. So maybe for b=k, a=k^{k}? Let's test that. If a=k^{k}, then the equation becomes (k^{k})^{b²}=b^{k^{k}}. But if b=k, then left side is (k^{k})^{k²}=k^{k*k²}=k^{k^3}, and the right side is k^{k^{k}}. So unless k^3 =k^{k}, which would require k^3=k^{k}, so k^{k}-k^3=0, k^3(k^{k-3}-1)=0. Since k≥1, k=1: 1^1-1^3=0, which works. k=3: 3^3 -3^3=0. So k=3 also works. For k=2: 2^2=4 vs 2^3=8. 4≠8. So only k=1 and k=3 satisfy k^{k}=k^3. Therefore, only for k=3, a=k^{k}=27, and b=k=3 works.Similarly, when k=2, a=k^{k}=4, but 4^{b²}=b^4. Let's check if that can be a solution. So a=4, then 4^{b²}=b^4. Let's check b=2: 4^4=256, 2^4=16. Nope. b=4:4^{16}=4^16 vs 4^4=256. Nope. b=16:4^{256}= (2^2)^{256}=2^{512}, and 16^4=2^{16}. Not equal. So no solution here.So only when k=3, a=k^{k} and b=k gives a solution. So maybe the solutions are (1,1), (16,2), and (27,3). Let's check if there are others.Wait, let's test b=4. Then the equation is a^{16}=4^{a}. Let me see if there's any integer a that satisfies this. Let's try a=16: 16^{16}=4^{16}. But 16=4^2, so 16^{16}=(4^2)^16=4^{32}, which is larger than 4^{16}. So no. a=4:4^{16}=4^16 vs 4^4=256. Not equal. a=2:2^{16}=65536 vs 4^2=16. Not equal. a=8:8^{16}= (2^3)^16=2^{48} vs 4^8=2^{16}. Not equal. So no solution here.What about b=5? Then the equation is a^{25}=5^{a}. Maybe a=5? 5^{25} vs 5^5. Not equal. a=25:25^{25} vs5^{25}. 25^{25}=(5^2)^25=5^{50}, which is larger than 5^{25}. a= something else? Maybe a=5^5=3125. Let's check: 3125^{25} vs 5^{3125}. 3125 is 5^5, so 3125^{25}=(5^5)^25=5^{125}, while 5^{3125} is way bigger. So not equal. So no solution here.Wait, maybe another approach. Let's take logarithms. Take natural logs of both sides: b² ln a = a ln b. Rearranged: (ln a)/a = (ln b)/b². So we have (ln a)/a equals (ln b)/b². Let me denote f(x) = (ln x)/x. Then f(a) = f(b)/b. Hmm, not sure. Alternatively, for fixed b, we can see how (ln a)/a relates to (ln b)/b². Maybe plot these functions? But since we are dealing with integers, perhaps we can find some relations.Suppose that a and b are powers of the same prime. For example, suppose a = p^m and b = p^n. Let's see if that works. Then the equation becomes (p^m)^{(p^n)^2} = (p^n)^{p^m}. Which is p^{m p^{2n}} = p^{n p^m}. Therefore, exponents must be equal: m p^{2n} =n p^m. So m/n = p^m / p^{2n} = p^{m - 2n}. So m/n = p^{m - 2n}. Let's let k = m/n, then k = p^{m - 2n}. Since m and n are integers, k must be a positive integer. So p^{m - 2n} must be integer. Let's set m - 2n = t, so t is an integer. Then k = p^t. Also, since k = m/n, we have m = k n = p^t n. Substituting back into t = m - 2n: t = p^t n - 2n = n(p^t - 2). So n(p^t - 2) = t. Since n and t are integers with n ≥1, and p is a prime. Let's try small primes.Take p=2. Then equation becomes n(2^t - 2)=t. Let's try t=1: n(2 -2)=0=1, no. t=2: n(4 -2)=2n=2 =>n=1. Then m =k n = p^t n=2^2*1=4. So m=4, n=1. Thus a=2^4=16, b=2^1=2. Which is the solution we found earlier. t=3: n(8 -2)=6n=3 =>n=0.5, not integer. t=4: n(16 -2)=14n=4 =>n=4/14=2/7, not integer. So only t=2 gives a solution here.Next, p=3. Then equation is n(3^t -2)=t. Let's try t=1: n(3 -2)=n=1. So n=1, then m=3^1 *1=3. So a=3^3=27, b=3^1=3. Which is another solution we found. t=2: n(9 -2)=7n=2 =>n=2/7, not integer. t=3:n(27 -2)=25n=3 =>n=3/25, nope. t=0: n(1 -2)= -n=0 =>n=0, invalid. So only t=1 gives solution here.p=5: n(5^t -2)=t. t=1: n(5-2)=3n=1 =>n=1/3, nope. t=2:n(25-2)=23n=2 =>n=2/23. No. t=3: n(125 -2)=123n=3 =>n=3/123=1/41. No. So no solutions here.Similarly, other primes would not give integer solutions. So perhaps only p=2 and p=3 give solutions in this setup. But maybe there are solutions where a and b are not powers of the same prime? Let's check.Suppose a and b are different primes. Let's say a=2 and b=3. Then the equation is 2^{9}=3^2 =>512=9. No. a=3 and b=2:3^{4}=2^3 =>81=8. No. a=4 and b=3:4^9=3^4 =>262144=81. Nope. a=5 and b=2:5^4=2^5 =>625=32. Nope. How about a=16 and b=3:16^9 vs 3^{16}. 16^9=2^{36}, 3^{16}= (3^8)^2=6561^2≈43 million, while 2^{36} is about 68.7 billion. Not equal.Alternatively, maybe a=27 and b=2:27^4 vs 2^{27}. 27^4=531441, 2^{27}=134217728. Not equal.Alternatively, look for other solutions where a and b are not primes. For example, a=16, b=2 (already found), a=27, b=3 (already found). Maybe a=256, b=4? Let's check:256^{16}=4^{256}. 256 is 2^8, so 256^{16}=(2^8)^16=2^{128}. 4^{256}= (2^2)^256=2^{512}. Not equal. So no.Alternatively, a=65536, b=16: a^{256}=16^{65536}. 65536 is 2^16, so 65536^{256}=2^{16*256}=2^{4096}. 16^{65536}=2^{4*65536}=2^{262144}. Not equal. So no.Alternatively, maybe non-powers? Let's see. Suppose a and b are different numbers. Let's check a=4, b=2:4^4=256 vs 2^4=16. No. a=8, b=2:8^{4}=4096 vs 2^8=256. No. a=16, b=4:16^{16} vs4^{16}= 2^{32}. 16^{16}=2^{64}, which is way bigger. No. a=9, b=3:9^9 vs3^9. 9^9=387,420,489 vs 3^9=19,683. No. a=81, b=3:81^9 vs3^81. 81 is 3^4, so 81^9=3^{36} vs3^81. No. a=256, b=2:256^4=2^{32} vs2^{256}. No.Alternatively, maybe a and b are equal. Let's check a=b. Then the equation becomes a^{a²}=a^{a}, which implies a^{a²}=a^{a}. So either a=1, which gives 1^{1}=1^{1}, which works, or if a>1, then a² =a, which would mean a² -a=0 =>a(a-1)=0. But since a≥1, the only solution here is a=1. Therefore, aside from (1,1), there are no solutions where a=b. So all other solutions must have a≠b.So so far, we have (1,1), (16,2), (27,3). Let's check if there are more.Wait, let's check a=256, b=16. a^{b²}=256^{256}= (2^8)^{256}=2^{2048}. b^a=16^{256}= (2^4)^{256}=2^{1024}. Not equal. Nope.Alternatively, what about a=3 and b=9. Then 3^{81} vs9^{3}=729. Not equal. a=9 and b=3:9^9 vs3^9. As before, not equal.Alternatively, check a=2 and b=16:2^{256} vs16^{2}=256. Not equal. a=16 and b=16:16^{256}=16^{16}. Which would require 256=16, which is false.Wait, maybe using the logarithmic approach. Let's denote x = a, y = b. Then the equation is x^{y²} = y^{x}. Take natural logs: y² ln x = x ln y. Rearranged: (ln x)/x = (ln y)/y². Let me define the function f(z) = (ln z)/z. Then for the equation to hold, we need f(x) = f(y)/y. Hmm. So f(x) must be equal to f(y)/y.Let me analyze the function f(z) = (ln z)/z. Its derivative is f’(z) = (1/z * z - ln z)/z² = (1 - ln z)/z². So f(z) is increasing when z < e and decreasing when z > e. So it has a maximum at z=e≈2.718. So f(z) increases from z=1 to z=e, then decreases afterwards.So since x and y are integers ≥1, f(z) reaches maximum at z=3 (since e≈2.718, so the closest integers are 2 and 3). Let's compute f(1)=0, f(2)=ln2/2≈0.3466, f(3)=ln3/3≈0.3662, f(4)=ln4/4≈0.3466, f(5)=ln5/5≈0.3219, etc., decreasing after z=3.Similarly, f(y)/y would be (ln y)/y². Let's compute this for some y:For y=1: (ln1)/1²=0.y=2: (ln2)/4≈0.3466/4≈0.0866y=3: (ln3)/9≈0.3662/9≈0.0407y=4: (ln4)/16≈0.3466/16≈0.02166y=5:≈0.3219/25≈0.0129So f(y)/y decreases as y increases.So we need to find x and y such that f(x) = f(y)/y.So for example, when y=2, f(y)/y≈0.0866. We need f(x)=0.0866. So find x where (ln x)/x≈0.0866. Let's check x=4: (ln4)/4≈0.3466/4≈0.0866. So exactly! So x=4 gives f(4)=0.0866≈f(2)/2. But x=4 and y=2 gives the equation 4^{4}=2^{4} => 256=16. Wait, no. Wait, hold on. Wait, x=4, y=2. Then the original equation is 4^{4}=2^{4}? 256=16. That's not true. Wait, but according to the logarithm equation, (ln4)/4≈0.3466/4≈0.0866, and (ln2)/2²=0.3466/4≈0.0866. So they are equal. But the original equation is x^{y²}=y^{x}. So 4^{4}=256 vs 2^{4}=16. Not equal. So even though (ln x)/x = (ln y)/y², the original equation doesn't hold. That's confusing. Wait, why?Wait, because when we take logarithms, we have y² ln x = x ln y. So (ln x)/x = (ln y)/y². But even if these ratios are equal, the exponents y² and x can lead to different results. So equality of the ratios is a necessary condition, but not sufficient? Wait, actually, if y² ln x = x ln y, then exponentiating both sides gives e^{y² ln x} = e^{x ln y}, which is x^{y²} = y^x. So actually, if (ln x)/x = (ln y)/y², then x^{y²}=y^x. So the ratio equality is equivalent. Therefore, if (ln x)/x = (ln y)/y², then the equation holds. So in the case of x=4, y=2, (ln4)/4=(ln2)/4, since ln4=2 ln2. So (2 ln2)/4=(ln2)/2=0.3466. Wait, but (ln y)/y²=(ln2)/4≈0.1733. Wait, wait, no. Wait, hold on:Wait, if x=4 and y=2, then (ln x)/x=(ln4)/4≈0.3466/4≈0.0866, and (ln y)/y²=(ln2)/4≈0.1733/4≈0.0433. Wait, wait, that contradicts. Wait, I must have miscalculated earlier.Wait, let's recast:If x=16 and y=2, then (ln16)/16≈2.7726/16≈0.1733, and (ln2)/4≈0.6931/4≈0.1733. So here, (ln x)/x=(ln y)/y²≈0.1733. Hence, x=16, y=2 is a solution.Similarly, x=27, y=3: (ln27)/27≈3.2958/27≈0.122, and (ln3)/9≈1.0986/9≈0.122. So equal. Therefore, x=27, y=3 is a solution.But when x=4 and y=2: (ln4)/4≈1.386/4≈0.3465, and (ln2)/4≈0.6931/4≈0.1733. Not equal. So x=4 and y=2 does not satisfy the ratio. Wait, where did I get that x=4, y=2 was a solution? It's not. So why did I think that earlier? Because when I considered a=4 and b=2, the equation is 4^4=256 vs 2^4=16. Not equal. So indeed, x=4, y=2 is not a solution. So even though for x=16, y=2, (ln x)/x=(ln y)/y²≈0.1733, so that's why 16 and 2 work.Similarly, for x=27, y=3, (ln27)/27≈(3.2958)/27≈0.122, (ln3)/9≈1.0986/9≈0.122, so equal. Therefore, 27 and 3 work.So the key is that (ln x)/x=(ln y)/y². So we can think of this as two separate equations. For given y, solve (ln x)/x=(ln y)/y². We can attempt to solve this equation for x and y integers ≥1.Alternatively, if we fix y and try to find x such that (ln x)/x = c, where c=(ln y)/y². Since the function f(z)=(ln z)/z is first increasing to a maximum at z=e, then decreasing. So for each c < f(e)=1/e, there are at most two solutions for z. But since we are dealing with integers, maybe only one solution.For example, for y=2, c=(ln2)/4≈0.1733. Then find x such that (ln x)/x=0.1733. Let's check x=3: (ln3)/3≈0.3662/3≈0.1221. x=4:≈0.3466/4≈0.0866. Wait, but 0.1733 is between x=2 and x=3? Wait, f(2)=0.3466/2≈0.1733. Wait, x=2: f(2)=0.3466/2≈0.1733. So that's exactly c. So if y=2, then (ln y)/y²=0.1733, and x=2 would give (ln x)/x=0.1733. So x=2 and y=2. But that's the same as x=y=2, which we saw doesn't work. Wait, but if x=2 and y=2, then the original equation is 2^{4}=2^{2}, which is 16=4. Not equal. Wait, but (ln2)/2≈0.3466/2≈0.1733, and (ln2)/4≈0.1733/2≈0.0866. Wait, no, (ln y)/y²=(ln2)/4≈0.1733. Wait, but when x=2, (ln x)/x=0.3466/2≈0.1733. So according to this, x=2 and y=2 should satisfy the equation. But they don't. Wait, this is confusing. Wait, let's check the original equation. If x=2 and y=2, then x^{y²}=2^{4}=16, and y^{x}=2^{2}=4. 16≠4. So even though (ln x)/x=(ln y)/y², the original equation does not hold. Why?Wait, no. Wait, if (ln x)/x=(ln y)/y², then y² lnx =x ln y, which is exactly the logarithm of the original equation. So exponentiating both sides gives x^{y²}=y^{x}. Therefore, if (ln x)/x=(ln y)/y², then the original equation must hold. But in this case, with x=2 and y=2, we have (ln2)/2=(ln2)/4? No, (ln2)/2≈0.3466, (ln2)/4≈0.1733. So they are not equal. Wait, hold on. Wait, if x=2 and y=2, then (ln x)/x=(ln2)/2≈0.3466, and (ln y)/y²=(ln2)/4≈0.1733. These are not equal. Therefore, x=2 and y=2 do not satisfy the ratio. Therefore, my mistake earlier was incorrect.Wait, in the case of y=2, (ln y)/y²=(ln2)/4≈0.1733. So to find x such that (ln x)/x=0.1733. What x gives that? Let's see. For x=3:≈0.3662/3≈0.1221. x=4:≈1.386/4≈0.3466/4≈0.0866. Wait, but 0.1733 is between x=2 and x=3. Let's check x=2: (ln2)/2≈0.3466/2≈0.1733. So x=2 gives (ln2)/2≈0.1733. Wait, but that's exactly (ln y)/y² for y=2. So (ln2)/2=(ln2)/4? No. Wait, no. (ln y)/y²=(ln2)/4≈0.1733/2≈0.0866. Wait, no. Wait, (ln y)/y² when y=2 is (ln2)/4≈0.1733. Wait, (ln2)/2≈0.3466, which is double that. So x=2 gives (lnx)/x=0.3466/2≈0.1733, which is equal to (ln y)/y² for y=2, which is (ln2)/4≈0.1733. So actually, x=2 and y=2 gives (lnx)/x=(lny)/y², but the original equation x^{y²}=y^x becomes 2^4=2^2, which is false. Wait, that can't be. There must be a miscalculation here.Wait, let's recast:If (lnx)/x=(lny)/y², then cross-multiplying: y² lnx =x lny. Which is equivalent to ln(x^{y²})=ln(y^{x}), which implies x^{y²}=y^{x}. So if (lnx)/x=(lny)/y², then the original equation holds. Therefore, if x=2 and y=2 satisfy (lnx)/x=(lny)/y², then x^{y²}=y^{x} must hold. But in reality, (ln2)/2≈0.3466, and (ln2)/4≈0.1733. So 0.3466≠0.1733. Therefore, x=2 and y=2 do not satisfy (lnx)/x=(lny)/y², hence the original equation doesn't hold.Ah, so my earlier confusion was due to miscalculating (lny)/y². For y=2, (lny)/y² is (ln2)/4≈0.1733, but (lnx)/x for x=2 is (ln2)/2≈0.3466, which is different. Therefore, x=2 and y=2 do not satisfy the ratio. Therefore, only when x and y satisfy (lnx)/x=(lny)/y² will the equation hold. So in the case of x=16 and y=2: (ln16)/16≈2.7726/16≈0.1733, which is equal to (ln2)/4≈0.1733. Therefore, x=16 and y=2 satisfy the ratio, hence the original equation. Similarly, x=27 and y=3: (ln27)/27≈3.2958/27≈0.122, and (ln3)/9≈1.0986/9≈0.122. So equal, hence original equation holds.So this suggests that to find solutions, we need to find pairs (x,y) where (lnx)/x=(lny)/y². Given that function f(z)=lnz/z is decreasing for z>e≈2.718, and since x and y are integers ≥1, we can look for pairs where f(x) = f(y)/y.Given that f(z) is decreasing for z≥3, let's check possible integer pairs.We already have:1. (1,1): trivial solution.2. (16,2): 16^{4}=2^{16}=65536.3. (27,3):27^{9}=3^{27}=7625597484987.Are there any other solutions?Let me check for y=4. Then f(y)/y²=(ln4)/16≈1.386/16≈0.0866. So we need x such that (lnx)/x≈0.0866. Let's check x=4: (ln4)/4≈0.3466/4≈0.0866. So x=4. So x=4 and y=4. But that would mean 4^{16}=4^4, which is false. Wait, but (ln4)/4≈0.3466/4≈0.0866, and (ln4)/16≈0.1733/4≈0.0433. Wait, no. Wait, wait. If y=4, then f(y)/y²=(ln4)/16≈1.386/16≈0.0866. So need x such that (lnx)/x≈0.0866. x=4: (ln4)/4≈0.3466/4≈0.0866. So x=4 and y=4 would imply (ln4)/4=(ln4)/16? No, (ln4)/4≈0.0866, (ln4)/16≈0.0866/4≈0.0217. Not equal. Wait, I'm confused again.Wait, no. Wait, if we set y=4, then f(y)/y²=(ln4)/16≈1.386/16≈0.0866. So we need x such that (lnx)/x=0.0866. Testing x=4 gives (ln4)/4≈0.3466/4≈0.0866. So x=4. So x=4 and y=4 would give the ratio equal, but check the original equation:4^{16}=4^4. Which is false. So why does the ratio hold but the equation fails? Wait, no. If (lnx)/x=(lny)/y², then x^{y²}=y^x. So if x=4 and y=4, then x^{y²}=4^{16}, and y^x=4^4. Which are not equal. But according to the ratio, they should be equal. Wait, but if x=4 and y=4, then (ln4)/4=(ln4)/4, so (lnx)/x=(lny)/y² would require (ln4)/4=(ln4)/16, which is not true. Therefore, x=4 and y=4 do not satisfy the ratio. So my mistake here is assuming that for y=4, x=4 would satisfy the ratio, but in reality, (lny)/y² for y=4 is (ln4)/16≈0.0866/4≈0.0217. Wait, no. Wait, (lny)/y²=(ln4)/(4²)= (ln4)/16≈1.386/16≈0.0866. Wait, but x=4 gives (lnx)/x=(ln4)/4≈0.3466/4≈0.0866. So (lnx)/x=(lny)/y²≈0.0866. Therefore, x=4 and y=4 do satisfy the ratio, but the original equation x^{y²}=y^x becomes4^{16}=4^4. Which is false. Therefore, there is a contradiction here. Wait, why?Wait, no. If x=4 and y=4, then (lnx)/x=0.0866, and (lny)/y²=0.0866. Therefore, according to the earlier logic, x^{y²}=y^x. But 4^{16}=4^4? No, 4^{16} is much larger. So why is there a discrepancy?Wait, because if (lnx)/x=(lny)/y², then y² lnx =x lny. Which is equivalent to ln(x^{y²})=ln(y^x), so exponentiating both sides gives x^{]

🔑:［Solution］ If a=1, then from the equation we get b=1.If b=1, then from the equation we get a=1.If a, b are both not less than 2, at this point, we lett=frac{b^{2}}{a}Then from the equation in the problem, we getdot{a}^{a t}=(a t)^{frac{a}{2}},which means squarebegin{array}{l}a^{2 t}=a t t=a^{2 t-1}end{array}Obviously, t>0. If 2 t-1 geqslant 1, thent=a^{2 t-1} geqslant(1+1)^{2 t-1} geqslant 1+(2 t-1)=2 t>tContradiction. Therefore, 2 t-1<1 and t<1. Since t=frac{b^{2}}{a}, t is a rational number. From the equation in the problem, we getbegin{array}{c}a^{frac{a}{k}}=b^{a} a=b^{k} b^{2} k=b^{k} k=b^{k-2}end{array}which means squareIf k leqslant 2, then k=b^{k-2} leqslant 1, which contradicts the previously proven k>1. Therefore, k>2. Letk=frac{p}{q}, p, q in N,(p, q)=1, p>2 qFrom equation (1), we getleft(frac{p}{q}right)^{q}=k^{q}=b^{p-2 q}Since the right side of the above equation is an integer, we have q=1, thus k is a natural number greater than 2, i.e.,k geqslant 3 text {. }If b=2, then from equation (1) we getk=2^{k-2},It is easy to see that k neq 3 at this time, thus k geqslant 4. On the other hand, from equation (1) we knowbegin{aligned}k & =(1+1)^{k-2} geqslant C_{k-2}^{0}+C_{k-2}^{1}+C_{k-2}^{2} & =1+(k-2)+frac{(k-2)(k-3)}{2} & =1+frac{(k-1)(k-2)}{2} geqslant 1+(k-1)=kend{aligned}Since the equality in the above equation holds if and only if k=4, we have k=4,a=b^{k}=2^{4}=16If b geqslant 3, thenk=b^{k-2} geqslant(1+2)^{k-2} geqslant 1+2(k-2)=2 k-3which means squarek leqslant 3 text {. }Therefore, at this time, we must have k=3, substituting into (1) we getbegin{aligned}b & =3 a & =b^{k}=3^{3}=27end{aligned}In summary, all positive integer pairs (a, b) that satisfy the equation in the problem are (a, b)=(1,1),(16, 2), (27,3).

🧠:[<think>Okay, let me try to figure out this problem. So, we have a = 256, and b is the product of all positive divisors of a. We need to check which equation is correct among the given options. First, I need to recall how to find the product of all positive divisors of a number. I remember that for a number n, if it has prime factorization n = p₁^k₁ * p₂^k₂ * ... * pₙ^kₙ, then the number of divisors is (k₁ + 1)(k₂ + 1)...(kₙ + 1). But the product of the divisors is something else. Hmm.Wait, there's a formula for the product of the divisors. If a number n has t divisors, then the product of all the divisors is n^(t/2). Is that right? Let me check with a small example. Take n = 4. Its divisors are 1, 2, 4. The product is 1*2*4 = 8. Now, the number of divisors t is 3. So n^(t/2) would be 4^(3/2) = 8. That matches. Another example: n = 6. Divisors are 1, 2, 3, 6. Product is 1*2*3*6 = 36. Number of divisors t = 4. n^(t/2) = 6^(4/2) = 6^2 = 36. That also works. Okay, so I think the formula is correct: product of divisors is n^(t/2), where t is the number of divisors.So in this problem, a = 256, which is 2^8. So prime factorization is 2^8. Then the number of divisors t is (8 + 1) = 9. Therefore, the product of all divisors is a^(9/2). So b = a^(9/2). Then we need to see which equation is correct.Looking at the options: (A) b = a^4(B) b = a^9(C) b² = a^7(D) b² = a^9(E) b³ = a^10So if b = a^(9/2), then squaring both sides gives b² = a^9. That's option D. Let me confirm.Wait, let's compute b step by step. If a = 256, which is 2^8, then the divisors are 2^0, 2^1, 2^2, ..., up to 2^8. So the divisors are 1, 2, 4, 8, 16, 32, 64, 128, 256. The product of these is 1*2*4*8*16*32*64*128*256. Let's compute this.Alternatively, since they are all powers of 2, we can write the exponents. Each term is 2^k where k ranges from 0 to 8. So the product is 2^0 * 2^1 * 2^2 * ... * 2^8 = 2^(0+1+2+3+4+5+6+7+8). The sum of exponents is (8*9)/2 = 36. So the product is 2^36. But a is 2^8, so 2^36 = (2^8)^(36/8) = (2^8)^4.5 = a^4.5. So b = a^4.5 = a^(9/2). Therefore, squaring both sides gives b² = a^9. So indeed, option D is correct. Let me check the other options to be safe.Option A: b = a^4. But a^4 is (2^8)^4 = 2^32. But we saw that b is 2^36, which is 2^32 * 2^4 = a^4 * 16. So 16a^4 ≠ b. So A is wrong.Option B: b = a^9. That would be 2^72, but b is 2^36. So no.Option C: b² = a^7. If b² is (2^36)^2 = 2^72. a^7 is (2^8)^7 = 2^56. Not equal.Option D: b² = a^9. a^9 is (2^8)^9 = 2^72, which is equal to b². Correct.Option E: b³ = a^10. b³ is (2^36)^3 = 2^108. a^10 is (2^8)^10 = 2^80. Not equal. So E is wrong.Therefore, the correct answer is D.Final Answerboxed{D}</think>Given ( a = 256 ), which is ( 2^8 ), we need to find the product of all positive divisors of ( a ) and determine which equation is correct.First, recall that the product of all positive divisors of a number ( n ) with ( t ) divisors is ( n^{t/2} ). For ( a = 2^8 ), the number of divisors ( t ) is ( (8 + 1) = 9 ). Therefore, the product of all divisors is ( a^{9/2} ).Calculating the product of the divisors step-by-step, the divisors of ( 256 ) are ( 1, 2, 4, 8, 16, 32, 64, 128, 256 ). These can be written as ( 2^0, 2^1, 2^2, ldots, 2^8 ). The product of these divisors is:[2^0 cdot 2^1 cdot 2^2 cdot ldots cdot 2^8 = 2^{0+1+2+ldots+8} = 2^{36}]Since ( a = 2^8 ), we have ( 2^{36} = (2^8)^{9/2} = a^{9/2} ). Therefore, ( b = a^{9/2} ).Squaring both sides, we get ( b^2 = a^9 ). Thus, the correct equation is ( b^2 = a^9 ), which corresponds to option D.[boxed{D}],<think>Okay, so the problem is: Let a = 256, and b be the product of all positive divisors of a. Then we have to determine which of the given equations is correct. The options are from A to E, with various equations involving b and a. Hmm. Let me think step by step.First, I need to recall how to find the product of all positive divisors of a number. I remember that for a number n, if it has prime factorization n = p₁^k₁ * p₂^k₂ * ... * p_m^k_m, then the number of divisors is (k₁ + 1)(k₂ + 1)...(k_m + 1). But the product of the divisors? I think there's a formula for that as well.Wait, yes! The product of all divisors of n is n^(t(n)/2), where t(n) is the number of positive divisors of n. That formula makes sense because each divisor d pairs with n/d, and their product is n. So if there are t(n) divisors, you have t(n)/2 such pairs, each multiplying to n, so the total product is n^(t(n)/2). Okay, so that's the key formula here.So in this problem, a = 256. Let's first find the number of positive divisors of a. Since 256 is a power of 2, specifically 2^8. So the prime factorization is 2^8. Therefore, the number of divisors is (8 + 1) = 9. So t(a) = 9. Therefore, the product of all divisors of a is a^(9/2). So b = a^(9/2). Then, we can see what the options are.But wait, the options are equations relating b and a. Let's check each option.Option A: b = a^4. Let's compute a^(9/2) vs a^4. Since a = 256, which is 2^8. Then a^(9/2) = (2^8)^(9/2) = 2^(8*(9/2)) = 2^(36). On the other hand, a^4 = (2^8)^4 = 2^32. So 2^36 vs 2^32, these are not equal. So A is incorrect.Option B: b = a^9. Similarly, a^9 would be (2^8)^9 = 2^72. But b is 2^36, so this is way too big. So B is incorrect.Option C: b² = a^7. Let's compute b². If b = a^(9/2), then b² = (a^(9/2))² = a^9. So b squared is a^9. But option C says b² = a^7. So a^9 vs a^7. Not equal. So C is incorrect.Option D: b² = a^9. As above, since b² = a^9, this is exactly what we have. So D would be correct.Option E: b³ = a^{10}. Let's check that. b³ = (a^(9/2))^3 = a^(27/2). a^{10} = (2^8)^10 = 2^80. a^(27/2) is (2^8)^(27/2) = 2^(8 * 27/2) = 2^(108) = 2^108. Which is much larger than 2^80, so E is incorrect.Therefore, the correct answer is D.But wait, let me verify my steps again to make sure.First, a = 256 = 2^8. The number of divisors is 8 + 1 = 9. Therefore, the product of divisors is a^(9/2). So b = a^(9/2). Therefore, squaring both sides gives b² = a^9. That's exactly option D. So yes, D is correct.Alternatively, perhaps I can compute the product of all divisors directly to verify.The divisors of 256 (which is 2^8) are 2^0, 2^1, 2^2, ..., up to 2^8. So the divisors are 1, 2, 4, 8, 16, 32, 64, 128, 256. There are 9 divisors. The product would be 1 * 2 * 4 * 8 * 16 * 32 * 64 * 128 * 256.Let's compute that. Since all are powers of 2, we can write their exponents:1 = 2^02 = 2^14 = 2^28 = 2^316 = 2^432 = 2^564 = 2^6128 = 2^7256 = 2^8So the product is 2^(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8). Let's compute the exponent: sum from 0 to 8. The sum is (8)(8 + 1)/2 = 36. So the product is 2^36.Now, a is 256 = 2^8. So a^9 = (2^8)^9 = 2^72. But b is 2^36, so b squared is (2^36)^2 = 2^72, which is equal to a^9. Therefore, b² = a^9, which is option D. That matches.Alternatively, if I compute 2^36, that is the product of the divisors. Let's see, since each pair multiplies to 256. For example, 1 * 256 = 256, 2 * 128 = 256, 4 * 64 = 256, 8 * 32 = 256, 16 * 16 = 256. Wait, but 16 is the square root of 256. So there are 4 pairs multiplying to 256, and the middle one is 16. So total product would be 256^4 * 16. Let's compute that. 256^4 is (2^8)^4 = 2^32. 16 is 2^4. So total product is 2^32 * 2^4 = 2^36. Which is indeed the same as before.Then, b = 2^36, and a = 2^8. Therefore, b = (2^8)^(36/8) = a^(36/8) = a^(4.5) = a^(9/2). Therefore, squaring both sides gives b² = a^9. Yep, so D is correct.So all steps check out. Therefore, the correct answer is D.Final Answerboxed{D}</think>Given ( a = 256 ), we need to determine the correct equation involving ( b ), the product of all positive divisors of ( a ).1. Prime Factorization of ( a ): - ( a = 256 = 2^8 ).2. Number of Divisors: - The number of divisors of ( 2^8 ) is ( 8 + 1 = 9 ).3. Product of Divisors: - The product of all divisors of a number ( n ) is ( n^{t(n)/2} ), where ( t(n) ) is the number of divisors. - For ( a = 256 ), the product of its divisors is ( a^{9/2} ). Thus, ( b = a^{9/2} ).4. Verification: - The divisors of ( 256 ) are ( 1, 2, 4, 8, 16, 32, 64, 128, 256 ). - The product of these divisors is ( 1 times 2 times 4 times 8 times 16 times 32 times 64 times 128 times 256 = 2^{0+1+2+3+4+5+6+7+8} = 2^{36} ). - Since ( a = 2^8 ), we have ( a^9 = (2^8)^9 = 2^{72} ). - Squaring ( b ) gives ( b^2 = (2^{36})^2 = 2^{72} ), which is equal to ( a^9 ).Thus, the correct equation is ( b^2 = a^9 ).[boxed{D}]]

🔑:A1. Since a=2^{8}, we have b=1 cdot 2 cdot 2^{2} cdot ldots cdot 2^{8}=2^{0+1+2+ldots+8}=2^{36}. Therefore, b^{2}=2^{72}=a^{9}. The correct answer is (D).

🧠:[<think>Okay, so I need to find the length of segment BD on a sphere with radius 1. The chord AB has length 1 and is positioned at a 60-degree angle to the diameter CD. Also, AC is given as √2, and AC is less than BC. Let me try to visualize this first.A sphere with radius 1 means all points on the sphere are 1 unit from the center. The diameter CD is a straight line passing through the center of the sphere, so C and D are endpoints of this diameter. Chord AB has length 1, so points A and B are 1 unit apart. This chord is at a 60-degree angle to diameter CD. Hmm, I need to clarify: when they say the chord is positioned at a 60-degree angle to the diameter CD, does that mean the angle between AB and CD is 60 degrees? Or is it the angle between the plane containing AB and CD or something else?Wait, in 3D geometry, the angle between two lines is the angle between their direction vectors. So if AB is at 60 degrees to CD, then the angle between AB and CD is 60 degrees. But AB is a chord, CD is a diameter. Since CD is a diameter, it passes through the center of the sphere. Let me note that the center of the sphere is O. So, OC and OD are both radius 1, and CD is length 2.Given that AC = √2 and AC < BC. Points A, C, and B are all on the sphere. AC is the distance from A to C, which is √2, and BC is longer than AC. So, point B is farther from C than A is.I need to find BD, the distance from B to D. Since D is the other end of the diameter from C, BD might be related through some spherical triangle or using the law of cosines in 3D space.Let me try to model this. Let me place the sphere in a coordinate system. Let me choose coordinates so that the diameter CD is along the z-axis. Let’s say point C is at (0, 0, 1) and D is at (0, 0, -1), since the radius is 1. Then the center O is at (0, 0, 0).Now, chord AB has length 1 and is at 60 degrees to CD. The angle between AB and CD is 60 degrees. Since CD is along the z-axis, the direction vector of CD is (0, 0, -2), but normalized, it's (0, 0, -1). The direction vector of AB can be found by the coordinates of points A and B. The angle between AB and CD is 60 degrees, so the dot product between the direction vectors of AB and CD should be equal to the cosine of 60 degrees, which is 0.5.Wait, AB is a chord, so vector AB is from A to B. CD is a diameter from C to D, vector CD is (0, 0, -2). The angle between AB and CD is 60 degrees. So if we take the unit vectors in the directions of AB and CD, their dot product is cos(60°) = 0.5.So if AB is a vector of length 1 (since chord AB is length 1), then the unit vector in the direction of AB is AB/1 = AB. The direction vector of CD is (0, 0, -2), so the unit vector is (0, 0, -1). Then the dot product of AB and (0, 0, -1) should be 0.5.Wait, but AB is a vector from A to B. Let me denote coordinates of A as (x1, y1, z1) and B as (x2, y2, z2). Then vector AB is (x2 - x1, y2 - y1, z2 - z1). The length of AB is 1, so sqrt[(x2 -x1)^2 + (y2 - y1)^2 + (z2 - z1)^2] = 1.The unit vector in the direction of AB is (x2 -x1, y2 - y1, z2 - z1). The unit vector in the direction of CD is (0, 0, -1). The dot product between these two vectors is (x2 -x1)*0 + (y2 - y1)*0 + (z2 - z1)*(-1) = -(z2 - z1). This should equal cos(60°) = 0.5.Therefore, -(z2 - z1) = 0.5 => z1 - z2 = 0.5.So the difference in z-coordinates between A and B is 0.5. So z1 = z2 + 0.5.Also, since points A, B, C, D are all on the sphere of radius 1, their coordinates satisfy x^2 + y^2 + z^2 = 1.Given that AC = √2. Point C is (0, 0, 1). So the distance from A to C is sqrt[(x1 - 0)^2 + (y1 - 0)^2 + (z1 - 1)^2] = sqrt(x1^2 + y1^2 + (z1 - 1)^2) = √2.So squaring both sides: x1^2 + y1^2 + (z1 - 1)^2 = 2.But since A is on the sphere, x1^2 + y1^2 + z1^2 = 1. Therefore, substituting:1 + (z1 - 1)^2 - z1^2 = 2. Wait, let's do that step carefully.From the sphere equation: x1^2 + y1^2 = 1 - z1^2. Substitute into the AC equation:(1 - z1^2) + (z1 - 1)^2 = 2.Compute (z1 - 1)^2 = z1^2 - 2z1 + 1. So substituting:1 - z1^2 + z1^2 - 2z1 + 1 = 2.Simplify: 1 + 1 - 2z1 = 2 => 2 - 2z1 = 2 => -2z1 = 0 => z1 = 0.So the z-coordinate of point A is 0. Then, since z1 - z2 = 0.5 (from earlier), z2 = z1 - 0.5 = 0 - 0.5 = -0.5.Therefore, point A is somewhere on the equator (z=0) of the sphere, and point B is at z=-0.5.Now, since point A is on the sphere, x1^2 + y1^2 + 0^2 = 1 => x1^2 + y1^2 = 1.Similarly, point B is on the sphere, so x2^2 + y2^2 + (-0.5)^2 = 1 => x2^2 + y2^2 = 1 - 0.25 = 0.75.Also, the distance between A and B is 1:sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (-0.5 - 0)^2] = 1.Squaring both sides:(x2 - x1)^2 + (y2 - y1)^2 + 0.25 = 1.Therefore, (x2 - x1)^2 + (y2 - y1)^2 = 0.75.So we have points A (x1, y1, 0) and B (x2, y2, -0.5) such that x1^2 + y1^2 = 1, x2^2 + y2^2 = 0.75, and (x2 - x1)^2 + (y2 - y1)^2 = 0.75.Additionally, AC = √2, which we already used to find z1=0. So point A is at (x1, y1, 0), and C is at (0, 0, 1). The distance AC is sqrt[(x1)^2 + (y1)^2 + (0 - 1)^2] = sqrt(1 + 1) = sqrt(2), which checks out because x1^2 + y1^2 = 1.Now, the problem states that AC < BC. So BC > √2. Let's compute BC.Point B is (x2, y2, -0.5), point C is (0, 0, 1). So BC = sqrt[(x2 - 0)^2 + (y2 - 0)^2 + (-0.5 - 1)^2] = sqrt(x2^2 + y2^2 + (-1.5)^2) = sqrt(0.75 + 2.25) = sqrt(3) ≈ 1.732. Wait, sqrt(3) is approximately 1.732, which is less than √2 ≈ 1.414. Wait, that contradicts AC < BC. Wait, sqrt(3) is actually greater than sqrt(2), so sqrt(3) ≈ 1.732 > sqrt(2) ≈ 1.414. Therefore, BC = sqrt(3) ≈ 1.732, which is indeed greater than AC = sqrt(2). So that satisfies AC < BC. So actually, BC is sqrt(3). Wait, how did I get BC?Wait, point B is on the sphere, so x2^2 + y2^2 + (-0.5)^2 = 1, so x2^2 + y2^2 = 0.75, as before. Then BC is sqrt[(x2)^2 + (y2)^2 + (-1.5)^2] = sqrt(0.75 + 2.25) = sqrt(3). Therefore, BC is sqrt(3), which is indeed greater than AC = sqrt(2). So that condition is satisfied. So actually, regardless of where B is, as long as it's at z=-0.5, BC is sqrt(3). So maybe this is fixed? Hmm, interesting.But wait, let me check. If point B is at z=-0.5, then BC is fixed as sqrt(3). So then the condition AC < BC is automatically satisfied, since sqrt(2) < sqrt(3). So maybe the problem is just telling us that, but actually, given the setup, AC is sqrt(2), BC is sqrt(3), so the condition is automatically true. So perhaps the problem is structured such that once you fix z1=0 and z2=-0.5, BC is sqrt(3), which is greater than AC= sqrt(2). So that's given.Therefore, the key is to find BD. Point D is (0, 0, -1). So BD is the distance from B (x2, y2, -0.5) to D (0, 0, -1). So BD = sqrt[(x2)^2 + (y2)^2 + (-0.5 - (-1))^2] = sqrt[x2^2 + y2^2 + (0.5)^2] = sqrt(0.75 + 0.25) = sqrt(1) = 1. Wait, that's interesting. Because x2^2 + y2^2 = 0.75, so BD = sqrt(0.75 + 0.25) = sqrt(1) = 1. So BD is 1? But that seems too straightforward. Is that correct?Wait, let me check again. BD is the distance between B (x2, y2, -0.5) and D (0, 0, -1). So coordinates of D are (0, 0, -1). So the difference in x: x2 - 0 = x2. Difference in y: y2 - 0 = y2. Difference in z: -0.5 - (-1) = 0.5. So BD = sqrt(x2^2 + y2^2 + 0.5^2). Since x2^2 + y2^2 = 0.75, BD = sqrt(0.75 + 0.25) = sqrt(1) = 1. So BD is 1. But the problem didn't specify anything else? Wait, but in that case, BD is 1 regardless of the positions of A and B, as long as B is at z=-0.5. But does that hold?Wait, but AB is length 1 and at 60 degrees to CD, which led us to z1 - z2 = 0.5. Then AC is sqrt(2), which fixed z1=0, hence z2=-0.5. Then x1^2 + y1^2=1, x2^2 + y2^2=0.75, and (x2 -x1)^2 + (y2 - y1)^2=0.75. So we have a system of equations here. But even so, BD seems to always come out as 1. Let me verify.Suppose points A and B are arbitrary points on their respective circles (A on equator, B at z=-0.5), with AB=1. Then BD would still be 1? Let me take a specific example.Suppose point A is (1, 0, 0). Then, since AB has length 1 and is at 60 degrees to CD. Then point B is at z=-0.5, so let's find B such that distance from (1, 0, 0) to (x, y, -0.5) is 1. So:(x - 1)^2 + y^2 + (-0.5)^2 = 1=> (x - 1)^2 + y^2 + 0.25 = 1=> (x - 1)^2 + y^2 = 0.75.Also, x^2 + y^2 = 0.75 (since B is on the sphere at z=-0.5).So subtract the two equations:(x - 1)^2 + y^2 - (x^2 + y^2) = 0.75 - 0.75 => (x^2 - 2x + 1 + y^2) - x^2 - y^2 = 0 => -2x + 1 = 0 => x = 0.5.Then, from x=0.5, substitute into x^2 + y^2 = 0.75:0.25 + y^2 = 0.75 => y^2 = 0.5 => y = ±√0.5.So point B would be (0.5, ±√0.5, -0.5). Then BD is sqrt[(0.5)^2 + (±√0.5)^2 + (0.5)^2] = sqrt(0.25 + 0.5 + 0.25) = sqrt(1) = 1. So in this case, BD is indeed 1.Alternatively, take another point A. Let me choose point A as (0, 1, 0). Then, find point B such that AB=1. Then, same steps:(x - 0)^2 + (y - 1)^2 + (-0.5)^2 = 1=> x^2 + (y - 1)^2 + 0.25 = 1=> x^2 + (y - 1)^2 = 0.75.Also, x^2 + y^2 = 0.75.Subtract the equations:x^2 + (y - 1)^2 - (x^2 + y^2) = 0.75 - 0.75 => (y^2 - 2y + 1) - y^2 = 0 => -2y + 1 = 0 => y = 0.5.Then, substitute y=0.5 into x^2 + y^2 = 0.75:x^2 + 0.25 = 0.75 => x^2 = 0.5 => x=±√0.5.So point B is (√0.5, 0.5, -0.5) or (-√0.5, 0.5, -0.5). Then BD is sqrt[(√0.5)^2 + (0.5)^2 + (0.5)^2] = sqrt(0.5 + 0.25 + 0.25) = sqrt(1) = 1.Same result. So in all cases, BD=1. Therefore, regardless of the position of A and B on their respective circles, BD is always 1. Therefore, the answer is 1. So BD=1.But let me check once more. Since BD is computed as sqrt(x2^2 + y2^2 + 0.5^2). Since x2^2 + y2^2 = 0.75, then BD = sqrt(0.75 + 0.25)=sqrt(1)=1. So yes, BD is always 1. Therefore, the answer is 1.But wait, the problem didn't mention anything about the position of A and B except that AB is at 60 degrees to CD, AC=√2, and AC < BC. So according to the calculations, BD=1. That seems straightforward, but maybe I missed something? Let me check the problem again."Chord AB of a sphere with radius 1 has a length of 1 and is positioned at an angle of 60° to the diameter CD of this sphere. It is known that AC=√2 and AC<BC. Find the length of the segment BD."Yes, so the key steps were:1. Sphere radius 1, so all points are at distance 1 from the center.2. CD is a diameter, so C and D are opposite poles.3. AB is a chord of length 1, making a 60° angle with CD.4. AC=√2, and AC < BC.Through coordinate geometry, by placing CD along the z-axis, we found that A must be on the equator (z=0) and B must be at z=-0.5. Then, regardless of their positions, BD computes to 1. Therefore, the answer is 1.But just to make sure, maybe there's another configuration? Suppose the chord AB is not in the plane perpendicular to CD but at some angle. Wait, but we considered the angle between AB and CD, so regardless of the plane, the angle is defined by their direction vectors. So by using the dot product, we derived the relation between z-coordinates. That seems correct.Therefore, I think BD is indeed 1, so the answer is boxed{1}.Final Answerboxed{1}</think>Given a sphere with radius 1, chord (AB) has a length of 1 and is positioned at an angle of (60^circ) to the diameter (CD). It is known that (AC = sqrt{2}) and (AC < BC). We need to find the length of the segment (BD).1. Sphere and Coordinates: - The sphere has radius 1, so all points on the sphere are 1 unit from the center (O). - Place the diameter (CD) along the z-axis with (C) at ((0, 0, 1)) and (D) at ((0, 0, -1)).2. Chord (AB) and Angle: - The angle between (AB) and (CD) is (60^circ). The dot product of the direction vectors of (AB) and (CD) gives (cos(60^circ) = 0.5). - The z-coordinates of (A) and (B) differ by 0.5, i.e., (z_1 - z_2 = 0.5).3. Distance (AC = sqrt{2}): - Point (A) is on the sphere, so (x_1^2 + y_1^2 + z_1^2 = 1). - Distance (AC = sqrt{2}) gives (x_1^2 + y_1^2 + (z_1 - 1)^2 = 2). Solving this, we find (z_1 = 0) (equator) and (z_2 = -0.5).4. Coordinates of (B): - Point (B) is on the sphere at (z = -0.5), so (x_2^2 + y_2^2 = 0.75).5. Distance (BD): - Point (D) is ((0, 0, -1)). The distance (BD) is calculated as: [ BD = sqrt{x_2^2 + y_2^2 + left(-0.5 - (-1)right)^2} = sqrt{0.75 + 0.25} = sqrt{1} = 1 ]Thus, the length of the segment (BD) is (boxed{1}).,<think>Okay, let's try to figure out this geometry problem. So, we have a sphere with radius 1. There's a chord AB that's 1 unit long. This chord is at a 60-degree angle to the diameter CD. Also, we know that AC is √2 and AC is less than BC. We need to find the length of BD. Hmm, okay. Let's break this down step by step.First, let's visualize the sphere. Since the radius is 1, the diameter CD must be 2 units long. The chord AB has length 1, so it's a shorter chord somewhere on the sphere. The angle between AB and CD is 60 degrees. I need to figure out how these are positioned relative to each other.Given that AC is √2 and AC < BC, point A must be closer to C than to B. Wait, but AC is a distance from A to C. Since the sphere has radius 1, the maximum distance between two points is 2 (the diameter). But √2 is approximately 1.414, which is less than 2, so that's possible. But how does AC relate to the other points?Maybe it's helpful to consider coordinates. Let me set up a coordinate system. Let's place the sphere at the origin (0,0,0) with radius 1. Let's choose CD as the z-axis for simplicity. So, points C and D are at the poles of the sphere: C is (0,0,1) and D is (0,0,-1). Then, the diameter CD is along the z-axis.Now, chord AB has length 1 and is at a 60-degree angle to CD (the z-axis). The angle between two lines in 3D space can be found using the dot product. If AB is at 60 degrees to CD (the z-axis), then the direction vector of AB makes a 60-degree angle with the z-axis.Let me denote points A and B as vectors from the origin. Let’s suppose point A has coordinates (x1, y1, z1) and point B has coordinates (x2, y2, z2). Since both are on the sphere, they satisfy x1² + y1² + z1² = 1 and x2² + y2² + z2² = 1.The chord AB has length 1, so the distance between A and B is 1. The distance formula gives √[(x2 - x1)² + (y2 - y1)² + (z2 - z1)²] = 1. Squaring both sides, (x2 - x1)² + (y2 - y1)² + (z2 - z1)² = 1.The angle between AB and CD (the z-axis) is 60 degrees. The direction vector of AB is (x2 - x1, y2 - y1, z2 - z1). The direction vector of CD is (0,0,-2), but since we're talking about the angle with the diameter, which is a line, the direction vector can be taken as (0,0,1) or (0,0,-1). The angle between AB and the z-axis is 60 degrees, so the direction vector of AB makes a 60-degree angle with the z-axis.The formula for the angle θ between two vectors u and v is cosθ = (u · v)/(|u||v|). Let's take the direction vector of AB as vector u = (x2 - x1, y2 - y1, z2 - z1) and the z-axis direction vector as v = (0,0,1). Then:cos60° = ( (z2 - z1) ) / |u|.But |u| is the length of AB, which is 1. So:0.5 = (z2 - z1)/1 → z2 - z1 = 0.5.Wait, but AB is a chord of the sphere, so the vector AB has length 1, but we're taking the difference in z-coordinates as 0.5. Hmm, is that right?Wait, but cosθ here is equal to the dot product divided by the product of the magnitudes. Since the direction vector of AB is (x2 - x1, y2 - y1, z2 - z1), which has magnitude 1 (since AB is length 1). The z-axis direction vector is (0,0,1), which is a unit vector. So, the dot product is (z2 - z1), and the angle between them is 60 degrees. Therefore:cos60° = (z2 - z1)/1 → z2 - z1 = 0.5.So, the difference in z-coordinates between points B and A is 0.5.Okay, so z2 = z1 + 0.5.Now, we also know that AC = √2, where C is (0,0,1). So, the distance from A to C is √2. Let's compute that.Distance AC = √[(x1 - 0)² + (y1 - 0)² + (z1 - 1)²] = √[x1² + y1² + (z1 - 1)²] = √2.But since A is on the sphere, x1² + y1² + z1² = 1. Therefore, substituting that into the distance formula:√[1 - z1² + (z1 - 1)²] = √2.Let me compute inside the square root:1 - z1² + (z1 - 1)^2 = 1 - z1² + z1² - 2z1 + 1 = 1 - 2z1 + 1 = 2 - 2z1.So, √(2 - 2z1) = √2.Squaring both sides: 2 - 2z1 = 2 → 2 - 2z1 = 2 → subtract 2: -2z1 = 0 → z1 = 0.Ah! So, the z-coordinate of point A is 0. Therefore, point A is somewhere on the equator of the sphere (since z=0). Then, since z2 = z1 + 0.5, point B has z-coordinate z2 = 0 + 0.5 = 0.5.So, point A is on the equator, point B is at z=0.5. Let me note that down.So coordinates of A: (x1, y1, 0), with x1² + y1² = 1.Coordinates of B: (x2, y2, 0.5), with x2² + y2² + (0.5)^2 = 1 → x2² + y2² = 1 - 0.25 = 0.75.Also, the distance between A and B is 1:√[(x2 - x1)^2 + (y2 - y1)^2 + (0.5 - 0)^2] = 1.Squaring both sides:(x2 - x1)^2 + (y2 - y1)^2 + 0.25 = 1 → (x2 - x1)^2 + (y2 - y1)^2 = 0.75.So, we have two equations:1. x1² + y1² = 1 (since A is on the sphere)2. x2² + y2² = 0.75 (since B is on the sphere)3. (x2 - x1)^2 + (y2 - y1)^2 = 0.75 (distance between A and B squared minus the z-component)Hmm, let's see. So we have three equations here. Maybe we can expand the third equation:(x2 - x1)^2 + (y2 - y1)^2 = x2² - 2x1x2 + x1² + y2² - 2y1y2 + y1² = (x1² + y1²) + (x2² + y2²) - 2(x1x2 + y1y2) = 1 + 0.75 - 2(x1x2 + y1y2) = 1.75 - 2(x1x2 + y1y2) = 0.75.Therefore:1.75 - 2(x1x2 + y1y2) = 0.75 → -2(x1x2 + y1y2) = 0.75 - 1.75 → -2(x1x2 + y1y2) = -1 → x1x2 + y1y2 = 0.5.So, the dot product of the projections of A and B onto the xy-plane is 0.5.Let me denote vector A_proj = (x1, y1) and vector B_proj = (x2, y2). Then, their dot product is 0.5. Also, |A_proj| = √(x1² + y1²) = 1 (since z1=0, so x1² + y1² =1). The magnitude of B_proj is √(x2² + y2²) = √0.75 = (√3)/2 ≈ 0.866.So, the dot product of A_proj and B_proj is 0.5. The formula for the dot product is |A_proj||B_proj|cosθ = 0.5, where θ is the angle between them.Therefore:1 * (√3/2) * cosθ = 0.5 → (√3/2) cosθ = 0.5 → cosθ = 0.5 / (√3/2) = (0.5 * 2)/√3 = 1/√3 → θ = arccos(1/√3) ≈ 54.7356 degrees.Hmm, interesting. So the projections of A and B onto the xy-plane have an angle of approximately 54.7 degrees between them. Not sure if that's immediately useful, but maybe.Now, we also know that AC < BC. Since AC is √2, and BC is the distance from B to C (0,0,1). Let's compute BC in terms of coordinates.Point B is (x2, y2, 0.5), point C is (0,0,1). So, distance BC is √[(x2 - 0)^2 + (y2 - 0)^2 + (0.5 - 1)^2] = √[x2² + y2² + (-0.5)^2] = √[0.75 + 0.25] = √1 = 1.Wait, that's interesting! BC is always 1? But the problem states AC < BC. But we calculated AC = √2 and BC = 1. But √2 ≈ 1.414, which is greater than 1. This contradicts the given condition that AC < BC. Wait, that can't be. There must be a mistake here.Wait, hold on. Let's check again. The problem says "It is known that AC = √2 and AC < BC". But according to our calculations, BC = 1, which is less than √2. But the problem states AC < BC, which would mean √2 < BC, but our calculation shows BC =1. So, that's a contradiction. That suggests that there's an error in my reasoning.Wait, let's retrace. So, point C is at (0,0,1), point A is at (x1, y1, 0), so AC is the distance between (x1, y1, 0) and (0,0,1). So:AC = √[(x1)^2 + (y1)^2 + (0 - 1)^2] = √[x1² + y1² + 1]. But x1² + y1² =1 (since A is on the sphere). Therefore, AC = √[1 + 1] = √2. So that's correct.Then, BC is the distance from B to C. Point B is (x2, y2, 0.5). So:BC = √[x2² + y2² + (0.5 - 1)^2] = √[x2² + y2² + 0.25]. But x2² + y2² = 0.75, so BC = √[0.75 + 0.25] = √1 =1. Therefore, BC =1. Therefore, AC =√2 ≈1.414 and BC=1. But the problem states AC < BC. But √2 is greater than 1, so AC > BC. This contradicts the given condition.Hmm, this suggests that there's a mistake in my setup. Maybe my coordinate system is incorrect? Or perhaps the angle between AB and CD is not as I interpreted.Wait, the problem says "Chord AB of a sphere with radius 1 has a length of 1 and is positioned at an angle of 60° to the diameter CD of this sphere." Maybe the angle between AB and CD is 60°, but CD is a diameter, so a line. The angle between a line (AB) and another line (CD) is defined as the angle between their direction vectors. So if AB is at 60° to CD, then yes, the direction vectors have a 60° angle. But according to the calculation, BC comes out as 1, which conflicts with AC < BC.Alternatively, maybe I made a wrong assumption in placing CD along the z-axis. But since CD is a diameter, it can be placed anywhere, but perhaps the position of points A and B relative to C and D is different.Wait, the problem says "It is known that AC = √2 and AC < BC". So AC is from A to C, and BC is from B to C. Since AC < BC, point A must be closer to C than point B is. But according to our previous calculation, AC = √2 ≈1.414 and BC=1, which would mean A is farther from C than B is. Therefore, our current model contradicts the given condition. Therefore, our initial assumptions must be wrong.Hmm. Let's think again. Maybe the chord AB is not in the position we thought. Perhaps CD is not the z-axis? Wait, but we placed CD as the z-axis for simplicity. If we choose a different coordinate system, maybe the calculations would differ.Alternatively, maybe the angle between AB and CD is not in 3D space but when projected onto a plane? Wait, the problem states "positioned at an angle of 60° to the diameter CD". That should be the angle in 3D space, not a projected angle. So the angle between the two lines AB and CD is 60°, which is determined by the dot product.But according to that, we arrived at BC =1 and AC =√2, which contradicts AC < BC. Therefore, there must be an error in the reasoning.Wait, maybe I miscalculated something. Let me check again.We had point A at (x1, y1, 0) because z1=0. Then, point B is at (x2, y2, 0.5). The distance from B to C is √[x2² + y2² + (0.5 -1)^2] = √[x2² + y2² + 0.25]. Since x2² + y2² =0.75, this becomes √[0.75 +0.25] = √1 =1. So BC=1.AC is distance from A to C: √[x1² + y1² + (0 -1)^2] = √[1 +1] =√2. Therefore, AC=√2 and BC=1, so AC > BC. But the problem states AC < BC. So, contradiction. Therefore, my model is incorrect.Therefore, my initial assumption about the coordinate system or the positions of points is wrong. Maybe CD is not the z-axis? Wait, but CD is a diameter, so it's a straight line passing through the center. If I place CD along the z-axis, that should be fine. Alternatively, maybe the chord AB is not in the position I thought.Wait, maybe the angle of 60 degrees is not between the direction vector of AB and CD, but between the line AB and the plane perpendicular to CD? Wait, no, the problem says "positioned at an angle of 60° to the diameter CD", which typically refers to the angle between two lines.Alternatively, maybe the angle is measured between AB and the plane containing CD? Wait, angles between lines and planes are measured differently. The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. But the problem says "at an angle of 60° to the diameter CD", so it's more likely the angle between the two lines AB and CD.Hmm. Since this leads to a contradiction, perhaps my coordinate system is not suitable, or there's another configuration.Wait, perhaps points A and B are not placed as I assumed. Maybe point C is not one of the poles? Wait, CD is a diameter, so C and D are endpoints. If I take CD as a diameter not along the z-axis, but in some other direction, maybe that would resolve the contradiction.Alternatively, maybe I need to consider that the chord AB is at 60 degrees to diameter CD, but not in the way I thought. Let me try to approach this differently.Let me consider the sphere with diameter CD. Let me denote the center of the sphere as O. Since CD is a diameter, then O is the midpoint of CD, so CO = OD =1 (since radius is1). The chord AB has length 1 and is at 60 degrees to CD. Also, AC =√2 and AC < BC.Wait, perhaps we can use some properties of spheres and chords. The distance from the center to a chord can be found using the formula: if a chord of length l is in a sphere of radius r, then the distance d from the center to the chord is d = √[r² - (l/2)²].Here, chord AB has length 1, so the distance from the center O to AB is √[1² - (1/2)^2] = √(1 - 0.25) = √0.75 = (√3)/2 ≈0.866.So, the line AB is at a distance of √3/2 from the center.But AB is positioned at an angle of 60 degrees to diameter CD. Hmm, how can a line (AB) be at an angle to another line (CD)? The angle between two lines in space is determined by the angle between their direction vectors. So, the angle between AB and CD is 60 degrees. So, the direction vectors of AB and CD form a 60-degree angle.Given that CD is a diameter, let's take CD as a vertical line for visualization. Then AB is a chord making a 60-degree angle with CD.Alternatively, perhaps we can model this with vectors. Let me consider vectors OA and OB, since A and B are points on the sphere.But AB is a chord of length 1. So, the distance between A and B is 1. So, |A - B| =1. Also, the angle between AB and CD is 60 degrees. The direction of AB is (B - A), and the direction of CD is (D - C). Since CD is a diameter, (D - C) is a vector of length 2 in the direction from C to D.But perhaps we can represent vectors with the center at O. Let’s denote vector OC as going from O to C, and OD as from O to D. Since CD is a diameter, then vector CD = D - C, which is 2*OD (if O is the center). Wait, if O is the center, then OC = -OD, since C and D are endpoints.But maybe this is complicating things. Let me try another approach.Given that AB is a chord of length 1 at 60 degrees to CD, and AC =√2 with AC < BC. Let's try to find coordinates again, but perhaps with a different orientation.Let me assume that CD is along the z-axis. Then, point C is (0,0,1), D is (0,0,-1), center O is (0,0,0). Chord AB is at 60 degrees to CD. So, the direction vector of AB makes 60 degrees with the z-axis.Let’s denote the direction vector of AB as (a, b, c). Since the angle between AB and CD (z-axis) is 60 degrees, then the dot product between (a, b, c) and (0,0,1) is |(a, b, c)| |(0,0,1)| cos60°. Let's normalize the direction vector.Wait, AB is a vector from A to B, with length 1. So, the direction vector is (B - A), and its magnitude is 1. The z-axis direction vector is (0,0,1). The angle between them is 60 degrees. Therefore:(B - A) · (0,0,1) = |B - A| |(0,0,1)| cos60° ⇒ (z2 - z1) = 1 * 1 * 0.5 ⇒ z2 - z1 = 0.5.So, as before, z2 = z1 + 0.5.Now, AC =√2. Point A is somewhere on the sphere. Let's find coordinates of A. AC is the distance from A to C(0,0,1):√[(x1)^2 + (y1)^2 + (z1 -1)^2] =√2.But since A is on the sphere, x1² + y1² + z1² =1. Therefore, substituting:√[1 - z1² + (z1 -1)^2] =√2 ⇒ as before, 2 - 2z1 =2 ⇒ z1=0.So, point A is on the equator (z=0), and point B is at z=0.5.Then, distance from B to C is:√[x2² + y2² + (0.5 -1)^2] =√[x2² + y2² +0.25]. But since B is on the sphere, x2² + y2² +0.5²=1 ⇒ x2² + y2²=0.75. Therefore, BC=√[0.75 +0.25]=√1=1.So, AC=√2 and BC=1. Therefore, AC > BC, but the problem states AC < BC. Contradiction. So, something is wrong here.Therefore, the problem must be in the initial assumption. Wait, maybe the angle is not between AB and CD, but between their projections? Or maybe the chord AB is not straight but... No, chords are straight lines.Alternatively, perhaps the problem is in 2D? But it mentions a sphere, so 3D.Wait, the problem says "Chord AB of a sphere". So, the chord is in 3D space. Hmm.Alternatively, maybe the chord AB is not a straight chord, but an arc? But no, in geometry, a chord is a straight line connecting two points on a sphere. The length of the chord is the straight-line distance.Wait, perhaps the problem is misinterpreted in terms of the angle. The angle between AB and CD might be the angle between the planes containing AB and CD, but that's not standard terminology. Usually, the angle between two lines is the angle between their direction vectors.Alternatively, maybe the angle is between AB and the diameter CD when both are projected onto some plane? But the problem doesn't specify, so I think the standard interpretation is the 3D angle between the two lines.Hmm. Given that this leads to a contradiction, perhaps the problem has a typo or is ill-posed? But the user says it's a problem to solve, so likely my approach is missing something.Wait, let's check again. If AC =√2 and AC < BC, but according to the calculation BC=1, then perhaps there's a different configuration where BC is greater than √2. How?Wait, BC is the distance from B to C. Since the sphere has radius 1, the maximum possible distance between two points is 2 (diametrically opposite). So BC can be up to 2. In our previous model, BC=1, but maybe there's another position where BC > √2.But how? Let's think.If point B is closer to D than to C, then BC would be longer. Wait, but in our previous setup, point B is at z=0.5, which is above the equator but below C. If we instead have point B at a lower z-coordinate, maybe?Wait, but according to the angle between AB and CD (60 degrees), if AB is directed downward, then z2 - z1 would be negative. Let me see.Wait, the angle between AB and CD is 60 degrees. The direction could be either upwards or downwards. The problem says "positioned at an angle of 60°", which is ambiguous. If the angle is 60 degrees regardless of direction, then the difference in z-coordinates could be positive or negative. So, perhaps z2 - z1 = ±0.5.But in our previous calculation, we assumed z2 - z1 = 0.5. What if it's -0.5?So, if z2 - z1 = -0.5, then z2 = z1 - 0.5.But we know z1=0, so z2= -0.5.Then, point B would be at z=-0.5.Then, coordinates of B: (x2, y2, -0.5). Then, the distance from B to C (0,0,1) would be √[x2² + y2² + (-0.5 -1)^2] = √[x2² + y2² + 2.25]. But since point B is on the sphere, x2² + y2² + (-0.5)^2 =1 ⇒ x2² + y2²=1 -0.25=0.75. Therefore, BC=√[0.75 +2.25]=√3≈1.732.So, AC=√2≈1.414 and BC=√3≈1.732. Therefore, AC < BC, which satisfies the given condition.Ah! So, that's the key. I assumed that the angle was such that z2 - z1 =0.5, but if instead z2 - z1 =-0.5, then point B is below the equator (z=-0.5), making BC longer. That would satisfy AC < BC.Therefore, my initial mistake was assuming the direction of the angle. The angle between AB and CD could be 60 degrees in the downward direction, making z2 - z1 = -0.5 instead of 0.5.Therefore, let's correct that. So, z2 = z1 -0.5. Since z1=0, z2=-0.5.Now, coordinates of A: (x1, y1, 0) with x1² + y1²=1.Coordinates of B: (x2, y2, -0.5) with x2² + y2² + (-0.5)^2=1 ⇒ x2² + y2²=0.75.Distance AB=1:√[(x2 -x1)^2 + (y2 - y1)^2 + (-0.5 -0)^2] =1.Squaring:(x2 -x1)^2 + (y2 - y1)^2 +0.25=1 ⇒ (x2 -x1)^2 + (y2 - y1)^2=0.75.Which gives, expanding as before:(x2² -2x1x2 +x1²) + (y2² -2y1y2 + y1²) =0.75.Substituting x1² + y1²=1 and x2² + y2²=0.75:1 + 0.75 -2(x1x2 + y1y2) =0.75 ⇒ 1.75 -2(x1x2 + y1y2)=0.75 ⇒ -2(x1x2 + y1y2)= -1 ⇒ x1x2 + y1y2=0.5.So, same as before. Therefore, the dot product of vectors A_proj and B_proj is 0.5.Thus, the angle between their projections on the xy-plane is still arccos(1/√3)≈54.7 degrees.Now, our goal is to find the length of BD. Point D is (0,0,-1). So, BD is the distance from B(x2, y2, -0.5) to D(0,0,-1):BD=√[(x2 -0)^2 + (y2 -0)^2 + (-0.5 - (-1))^2] = √[x2² + y2² + (0.5)^2] = √[0.75 +0.25]=√1=1.Wait, so BD=1? But that seems too straightforward. But let's check.Point B is at (x2, y2, -0.5), D is at (0,0,-1). The distance BD is √[x2² + y2² + (-0.5 +1)^2] = √[0.75 +0.25] = √1=1. So BD=1.But the problem asks for the length of segment BD. So, is the answer 1? But the problem seems more complicated. Maybe I made an error.Wait, but BD is a chord from B to D. Since B is at z=-0.5 and D is at z=-1. The distance is 1. However, since the sphere has radius 1, the maximum distance between two points is 2, so BD=1 is possible.But let me verify again. If BD=1, but AB=1, and the problem is presented in a way that suggests it's not obvious, perhaps BD is not 1. Wait, but according to coordinates, yes. Wait, let me check again.Point B is at (x2, y2, -0.5). Since x2² + y2²=0.75. Therefore, BD=√[x2² + y2² + (-0.5 +1)^2] = √[0.75 +0.25] = √1=1. So BD=1.But according to the problem statement, AC=√2, which we have, and AC < BC, which is satisfied since BC=√3. Then BD=1. So, the answer is 1? But let me check if there's another way this could be interpreted.Wait, but maybe BD is not the straight line through the sphere, but an arc along the surface? But the problem says "segment BD", which in geometry usually refers to the straight line (chord) unless specified otherwise. The length of the chord BD would be 1, as calculated.But the problem is presented in a way that suggests it's non-trivial. Maybe the answer is 1, but maybe I made a mistake in the coordinate system.Wait, but let's check with another approach. Since points A and B are on the sphere, and we have their coordinates, maybe we can find coordinates of A and B explicitly.From the previous equations:A is (x1, y1, 0), x1² + y1²=1.B is (x2, y2, -0.5), x2² + y2²=0.75.And x1x2 + y1y2=0.5.Let me parametrize point A. Since A is on the equator, we can write A as (1, 0, 0) without loss of generality by choosing coordinates. Wait, because we can rotate the sphere around the z-axis so that point A lies along the x-axis.Yes, since the problem doesn't specify any particular orientation besides CD as a diameter, we can choose coordinates such that point A is at (1,0,0). That would simplify calculations.Let’s try that. Let me set point A at (1,0,0). Then, coordinates of A: (1,0,0).Then, coordinates of B: (x2, y2, -0.5), with x2² + y2²=0.75.Distance AB=1:√[(x2 -1)^2 + y2^2 + (-0.5 -0)^2] =1.Squaring:(x2 -1)^2 + y2^2 +0.25=1 ⇒ (x2 -1)^2 + y2^2=0.75.But x2² + y2²=0.75. So substituting y2²=0.75 -x2² into the above equation:(x2 -1)^2 +0.75 -x2²=0.75.Expand (x2 -1)^2:x2² -2x2 +1 +0.75 -x2²=0.75.Simplify:(-2x2 +1 +0.75)=0.75 → -2x2 +1.75=0.75 → -2x2= -1 → x2=0.5.Therefore, x2=0.5. Then, y2²=0.75 -x2²=0.75 -0.25=0.5 → y2=±√0.5=±(√2)/2≈±0.707.Therefore, coordinates of B are (0.5, ±√0.5, -0.5).Now, let's compute BD. Point D is (0,0,-1).BD=√[(0.5 -0)^2 + (±√0.5 -0)^2 + (-0.5 -(-1))^2] =√[0.25 +0.5 +0.25] =√[1]=1.So, indeed BD=1.Therefore, the length of segment BD is 1.But the problem seems to present a non-trivial problem, yet the answer is 1? Let me verify once more.Alternatively, maybe the problem expects the distance along the sphere's surface (arc length), but the problem says "segment BD", which refers to the straight line (chord) distance. Since we've computed the chord length as 1, which is correct.But given that the sphere has radius 1, the chord length BD=1 corresponds to an arc length of 2*arcsin(0.5/1)=2*30°=60°, so 60 degrees. But the problem doesn't ask for arc length.Alternatively, perhaps there's a misunderstanding in the problem's initial conditions. Wait, let's re-read the problem:"Chord AB of a sphere with radius 1 has a length of 1 and is positioned at an angle of 60° to the diameter CD of this sphere. It is known that AC=√2 and AC<BC. Find the length of the segment BD."So, the key was recognizing that the angle could be in the downward direction, making BD=1. However, without considering the direction, we initially had a contradiction, but flipping the direction resolved it. Therefore, the answer is 1.But let's check if there's another possible configuration. Suppose point A is not on the equator. Wait, but we derived that z1=0 from AC=√2. So point A must be on the equator.Yes, because AC=√2 implies that z1=0. Therefore, point A is on the equator, and the only variables are the positions of B. When we considered the direction of AB, we had to set B below the equator to satisfy AC < BC. Then, BD=1.Therefore, the answer is boxed{1}.Final Answerboxed{1}</think>Given a sphere with radius 1, chord (AB) has a length of 1 and is positioned at an angle of (60^circ) to the diameter (CD). It is known that (AC = sqrt{2}) and (AC < BC). We need to find the length of the segment (BD).1. Setting up the coordinate system: - Place the sphere at the origin with radius 1. - Choose diameter (CD) along the z-axis, with (C) at ((0, 0, 1)) and (D) at ((0, 0, -1)).2. Coordinates of points (A) and (B): - Since (AC = sqrt{2}), point (A) must be on the equator of the sphere (z-coordinate 0). - The coordinates of (A) are ((x_1, y_1, 0)) with (x_1^2 + y_1^2 = 1). - The angle between chord (AB) and diameter (CD) is (60^circ), leading to the z-coordinate difference between (B) and (A) being (pm 0.5).3. Considering the angle direction: - If the angle is downward, point (B) has a z-coordinate of (-0.5). - Coordinates of (B) are ((x_2, y_2, -0.5)) with (x_2^2 + y_2^2 = 0.75).4. Verifying distances: - Distance (BC) is calculated as (sqrt{x_2^2 + y_2^2 + (-0.5 - 1)^2} = sqrt{0.75 + 2.25} = sqrt{3} approx 1.732), satisfying (AC < BC). - Distance (BD) is calculated as (sqrt{x_2^2 + y_2^2 + (-0.5 - (-1))^2} = sqrt{0.75 + 0.25} = sqrt{1} = 1).5. Conclusion: - The length of segment (BD) is (boxed{1}).]

🔑:4.19. Let O be the center of the sphere. Take a point E such that overrightarrow{C E}=overrightarrow{A B}. Since angle O C E=60^{circ} and C E=1=O C, then O E=1. Point O is equidistant from all vertices of the parallelogram A B E C, so A B E C is a rectangle and the projection O_{1} of point O onto the plane of this rectangle coincides with its center, i.e., with the midpoint of segment B C. Segment O O_{1} is the midline of triangle C B D, therefore B D=20 O_{1}=2 sqrt{O C^{2}-B C^{2} / 4} Rightarrow =2 sqrt{1-left(A B^{2}+A C^{2}right) / 4}=1.