🧠:[<think>Okay, so I need to find the minimum value of λ such that the inequality 1/x + 1/y + λ/(x + y) is always greater than or equal to zero for all positive real numbers x and y. Hmm, let me start by understanding what this means. First, since x and y are positive, all the denominators here are positive, so there's no division by zero issues. The left side of the inequality is a combination of three terms: 1/x, 1/y, and λ/(x + y). The question is, what's the smallest λ that makes this sum non-negative no matter what positive x and y we pick. Let me think. If I can rewrite the inequality in terms of a single variable, maybe that would help. Sometimes, with two variables, you can set t = y/x or something like that to reduce the problem to one variable. Let's try that substitution. Let me set t = y/x, so since x and y are positive, t is positive as well. Then y = t x. Substituting into the inequality: 1/x + 1/(t x) + λ/(x + t x) >= 0. Let's factor out x from the denominators. First term: 1/x. Second term: 1/(t x) = 1/(x t). Third term: λ/(x(1 + t)) = λ/(x(1 + t)). So, factoring out 1/x, we get: (1 + 1/t + λ/(1 + t)) * (1/x) >= 0. But x is positive, so 1/x is positive. Therefore, multiplying both sides by x (which is positive and doesn't change the inequality direction), we get: 1 + 1/t + λ/(1 + t) >= 0. So the inequality reduces to 1 + 1/t + λ/(1 + t) >= 0 for all t > 0. Therefore, we need to find the minimal λ such that this holds for all t > 0. So now the problem is transformed into a single variable problem. So now, the problem is equivalent to finding the minimal λ such that for all t > 0, 1 + 1/t + λ/(1 + t) >= 0. Alternatively, rearranging terms, we can write:1 + 1/t >= -λ/(1 + t)But since 1 + 1/t is always positive (since t > 0), and (1 + t) is positive, so the right-hand side is -λ/(1 + t). So if λ is positive, then the RHS is negative, and since the LHS is positive, the inequality will automatically hold. Wait, but λ could be negative. Wait, but the problem says find the minimal real number λ such that the inequality holds. So maybe λ can be negative? But we need the minimal λ, which would be the most negative λ such that the inequality still holds. Wait, but maybe I need to check.Wait, if λ is positive, then the term λ/(x + y) is positive, so adding it to 1/x + 1/y which is positive, the entire expression is positive. But if λ is negative, then the term λ/(x + y) is negative, so we need to ensure that the sum of 1/x + 1/y is greater than or equal to -λ/(x + y). So the problem is equivalent to finding the minimal λ (which could be negative) such that 1/x + 1/y >= -λ/(x + y) for all positive x, y. So rearranged, 1/x + 1/y + λ/(x + y) >= 0.But to find the minimal λ such that this holds, meaning the smallest (most negative) λ that doesn't make the left-hand side negative. Wait, but perhaps λ needs to be positive? Let's test some values.Suppose λ is a very large negative number. Then λ/(x + y) would be a large negative term, so 1/x + 1/y + λ/(x + y) could be negative. Hence, we need to find the minimal λ (the most negative one) such that even when λ/(x + y) is subtracted, the sum remains non-negative. Wait, but the problem says "the minimum value of the real number λ such that the inequality ... always holds". If λ can be as small as possible (negative), then the inequality might not hold. So actually, we need the smallest λ (the least possible, which could be negative) such that the inequality still holds. But maybe λ is bounded below? Let me see.Wait, but the question says "find the minimum value of the real number λ such that the inequality ... always holds". So, minimum λ. If λ is allowed to be negative, but making it more negative would violate the inequality, then the minimal λ (the smallest possible λ) is the one where the inequality just holds. So we need to find the infimum of all λ such that the inequality holds for all x, y > 0. Then, that infimum is the minimal λ.Alternatively, perhaps it's phrased as the maximum lower bound for λ? Wait, maybe I need to be careful here.Alternatively, maybe it's equivalent to finding the greatest lower bound (infimum) of λ such that the inequality holds for all x, y > 0. But since we want the minimum λ, which would be the smallest λ that works, which is actually the infimum. But maybe the infimum is attainable. Let's see.But perhaps the question is looking for the smallest λ (could be negative) such that for all x, y > 0, the inequality holds. So we need to find the minimal λ such that 1/x + 1/y + λ/(x + y) >= 0 always. To find that λ, we can consider the worst case scenario where the left-hand side is minimized. So the minimal λ is such that the minimal value of 1/x + 1/y + λ/(x + y) over all x, y > 0 is zero. Therefore, to find λ, we can set up the problem as minimizing 1/x + 1/y + λ/(x + y), set it equal to zero, and find λ such that this minimal value is zero.But perhaps another approach is to use calculus. Let's consider t as before. Let me define t = y/x, so t > 0. Then, as before, the inequality reduces to 1 + 1/t + λ/(1 + t) >= 0. So we can define f(t) = 1 + 1/t + λ/(1 + t). We need f(t) >= 0 for all t > 0. So, the minimal λ is the minimal λ such that f(t) >= 0 for all t > 0. Therefore, to find the minimal λ, we need to find the maximum of - (1 + 1/t) * (1 + t)/1, because if we solve for λ:From f(t) >= 0,λ/(1 + t) >= - (1 + 1/t)Multiply both sides by (1 + t) (which is positive, so inequality direction remains):λ >= - (1 + 1/t)(1 + t)Simplify RHS:- (1 + 1/t)(1 + t) = - [ (1)(1 + t) + (1/t)(1 + t) ] = - [ 1 + t + 1/t + 1 ] = - [ 2 + t + 1/t ]Wait, but let's compute (1 + 1/t)(1 + t):(1 + 1/t)(1 + t) = 1*(1 + t) + (1/t)(1 + t) = 1 + t + 1/t + 1 = 2 + t + 1/t. Wait, that's right.Therefore, λ >= - (2 + t + 1/t). So to satisfy this inequality for all t > 0, λ must be greater than or equal to the maximum of - (2 + t + 1/t) over all t > 0. Hence, the minimal λ is equal to the maximum of - (2 + t + 1/t) over t > 0. But the maximum of - (2 + t + 1/t) is the same as - minimum of (2 + t + 1/t). Therefore, λ_min = - min_{t > 0} [2 + t + 1/t]. So we need to find the minimum value of 2 + t + 1/t for t > 0, and then take the negative of that.Let's compute min_{t>0} [2 + t + 1/t]. Let's set f(t) = 2 + t + 1/t. To find its minimum, take derivative f’(t) = 1 - 1/t². Set derivative equal to zero: 1 - 1/t² = 0 => 1 = 1/t² => t² = 1 => t = 1 (since t > 0). Then, the second derivative is f''(t) = 2/t³, which is positive at t = 1, so it's a minimum. Therefore, the minimal value of f(t) is f(1) = 2 + 1 + 1 = 4. Hence, λ_min = -4. Therefore, the minimal λ is -4. Wait, but let me check if that's correct. If λ = -4, then the inequality becomes 1/x + 1/y - 4/(x + y) >= 0. Let's see if this holds for all x, y > 0. Let me test some specific cases. For example, take x = y = 1. Then LHS = 1 + 1 - 4/(1 + 1) = 2 - 2 = 0. So equality holds here. Another test case: take x approaching 0, y = 1. Then LHS becomes 1/x + 1 - 4/(x + 1). As x approaches 0, 1/x tends to infinity, and 4/(x + 1) tends to 4. So the whole expression tends to infinity, which is greater than zero. Similarly, if x approaches infinity, y = 1. Then 1/x tends to 0, 1 stays 1, and 4/(x + 1) tends to 0. So LHS tends to 1, which is greater than zero. Another test case: let’s take t = 1, which gives equality. What if t is different? For example, t = 2. Then, f(t) = 1 + 1/2 + (-4)/(1 + 2) = 1 + 0.5 - 4/3 ≈ 1.5 - 1.333... ≈ 0.166..., which is positive. If t = 1/2, then f(t) = 1 + 2 + (-4)/(1 + 0.5) = 3 - 4/1.5 ≈ 3 - 2.666... ≈ 0.333..., which is still positive. What about t where the function f(t) = 1 + 1/t - 4/(1 + t) might be minimized? Let's find the critical points. Let's set f(t) = 1 + 1/t - 4/(1 + t). Then, take derivative: f’(t) = -1/t² + 4/(1 + t)^2. Set derivative to zero:-1/t² + 4/(1 + t)^2 = 0 => 4/(1 + t)^2 = 1/t² => 4t² = (1 + t)^2 => 4t² = 1 + 2t + t² => 3t² - 2t - 1 = 0.Solving quadratic equation: 3t² - 2t -1 = 0. Discriminant D = 4 + 12 = 16. Solutions: t = [2 ± 4]/6. So t = (2 + 4)/6 = 1, or t = (2 -4)/6 = -1/3. But t > 0, so t = 1. So the critical point is at t = 1, which we already checked gives f(t) = 0. Therefore, the minimum of f(t) is zero, achieved at t = 1, and for other t, f(t) is positive. Therefore, when λ = -4, the inequality holds for all t > 0, and it's tight at t = 1. Therefore, the minimal λ is indeed -4.But wait, let me confirm again. If I set λ = -4, then the original inequality is 1/x + 1/y - 4/(x + y) >= 0. Let me manipulate this. Multiply both sides by (x + y), which is positive, so inequality direction remains:(x + y)(1/x + 1/y) - 4 >= 0.Compute (x + y)(1/x + 1/y) = (x + y)(y + x)/(xy) = (x + y)^2 / (xy). So:(x + y)^2 / (xy) - 4 >= 0.Multiply both sides by xy (positive):(x + y)^2 - 4xy >= 0.Which simplifies to x² + 2xy + y² - 4xy = x² - 2xy + y² = (x - y)^2 >= 0.Which is always true, since squares are non-negative. And equality holds when x = y. So this shows that when λ = -4, the inequality reduces to (x - y)^2/(xy) >= 0, which is always true. Hence, the minimal λ is indeed -4.Therefore, the answer should be boxed{-4}.Final Answerboxed{-4}</think>Given that ( x > 0 ) and ( y > 0 ), we need to find the minimum value of the real number ( lambda ) such that the inequality ( frac{1}{x} + frac{1}{y} + frac{lambda}{x+y} geqslant 0 ) always holds.First, we substitute ( t = frac{y}{x} ) (where ( t > 0 )) into the inequality. This transforms the inequality into:[frac{1}{x} + frac{1}{tx} + frac{lambda}{x + tx} geq 0]Factoring out ( frac{1}{x} ), we get:[frac{1}{x} left( 1 + frac{1}{t} + frac{lambda}{1 + t} right) geq 0]Since ( x > 0 ), we can multiply both sides by ( x ) to obtain:[1 + frac{1}{t} + frac{lambda}{1 + t} geq 0]We need this inequality to hold for all ( t > 0 ). Rewriting it, we have:[frac{lambda}{1 + t} geq - left( 1 + frac{1}{t} right)]Multiplying both sides by ( 1 + t ) (which is positive), we get:[lambda geq - left( 1 + frac{1}{t} right)(1 + t)]Simplifying the right-hand side:[- left( 1 + frac{1}{t} right)(1 + t) = - left( 2 + t + frac{1}{t} right)]Thus, we need ( lambda ) to be greater than or equal to the maximum value of ( - left( 2 + t + frac{1}{t} right) ) for all ( t > 0 ). This is equivalent to finding the minimum value of ( 2 + t + frac{1}{t} ) and taking its negative.To find the minimum value of ( 2 + t + frac{1}{t} ), we take the derivative with respect to ( t ):[f(t) = 2 + t + frac{1}{t}][f'(t) = 1 - frac{1}{t^2}]Setting the derivative to zero:[1 - frac{1}{t^2} = 0 implies t^2 = 1 implies t = 1]The second derivative ( f''(t) = frac{2}{t^3} ) is positive at ( t = 1 ), confirming a minimum. Evaluating ( f(t) ) at ( t = 1 ):[f(1) = 2 + 1 + 1 = 4]Thus, the minimum value of ( 2 + t + frac{1}{t} ) is 4, and the maximum value of ( - left( 2 + t + frac{1}{t} right) ) is -4. Therefore, the minimum value of ( lambda ) is ( -4 ).Finally, verifying with the original inequality, when ( lambda = -4 ), the inequality becomes:[frac{1}{x} + frac{1}{y} - frac{4}{x+y} geq 0]This simplifies to:[frac{(x - y)^2}{xy(x + y)} geq 0]which is always true for all ( x, y > 0 ).Thus, the minimum value of ( lambda ) is (boxed{-4}).,<think>Okay, so I need to find the minimum value of λ such that the inequality 1/x + 1/y + λ/(x + y) is always greater than or equal to zero for all positive real numbers x and y. Hmm, let me start by understanding what this means.First, since x and y are both positive, all the denominators here are positive, so we don't have to worry about division by zero or negative denominators. That's good. The problem is asking for the smallest λ such that no matter what positive x and y we choose, the expression 1/x + 1/y + λ/(x + y) is non-negative. So, essentially, we need that this sum is always at least zero, and we want the minimal λ for which this is true.Let me rephrase the inequality:1/x + 1/y + λ/(x + y) ≥ 0.Since x and y are positive, all terms are real numbers. But 1/x and 1/y are positive, while λ/(x + y) could be positive or negative depending on λ. Since we want the inequality to hold always, we need to ensure that even in the worst-case scenario (i.e., when 1/x + 1/y is as small as possible and λ/(x + y) is as negative as possible), the sum is still non-negative. Therefore, we need to find the minimal λ such that the sum is non-negative for all x, y > 0.Alternatively, we can think of this as an optimization problem: minimize λ such that 1/x + 1/y + λ/(x + y) ≥ 0 for all x, y > 0. To find the minimal λ, we can consider the cases where the left-hand side is minimized, which would give us the tightest constraint on λ.So, perhaps the key is to find the minimum of the expression 1/x + 1/y + λ/(x + y) over all positive x and y, and set that minimum to zero. Then solve for λ. But how do I find the minimum of this expression?Alternatively, maybe we can use some inequality techniques, such as Cauchy-Schwarz, AM-GM, or other inequalities to bound the expression. Let's see.First, let's note that 1/x + 1/y is always positive, as x and y are positive. Therefore, the term λ/(x + y) must not be too negative to overpower the positive terms. But since λ is a real number we need to find, perhaps negative values of λ could make the third term negative. However, we need the entire expression to still be non-negative.Wait, but if λ is allowed to be negative, then λ/(x + y) could subtract from the sum. So the question is, how negative can λ be such that even when we subtract that term, the total is still non-negative. But we need the minimum λ, so maybe λ is negative? Wait, but the problem says "find the minimum value of the real number λ". So λ could be negative. But we need the minimal λ, which would be the most negative λ such that the inequality still holds. Wait, but hold on, the inequality is 1/x + 1/y + λ/(x + y) ≥ 0. If λ is very negative, then λ/(x + y) would be very negative, so the sum could become negative. Therefore, the minimum λ is the smallest (most negative) λ such that the sum is still non-negative. So we need to find the infimum of all λ such that the inequality holds. But the problem states "the minimum value of the real number λ", so maybe λ is actually bounded below, and the minimum is that lower bound.Alternatively, perhaps I need to maximize the negative term. Let me think.Let me rearrange the inequality:λ/(x + y) ≥ - (1/x + 1/y).Multiplying both sides by (x + y), which is positive, so the inequality direction remains the same:λ ≥ - (1/x + 1/y)(x + y).Therefore, λ must be greater than or equal to - (1/x + 1/y)(x + y) for all x, y > 0. Therefore, the minimal such λ is the supremum of - (1/x + 1/y)(x + y) over all x, y > 0. So if we can compute the maximum value of - (1/x + 1/y)(x + y), then λ is equal to that maximum value.But let's compute - (1/x + 1/y)(x + y). Let's compute (1/x + 1/y)(x + y):(1/x + 1/y)(x + y) = (x + y)/x + (x + y)/y = 1 + y/x + x/y + 1 = 2 + x/y + y/x.Therefore, - (1/x + 1/y)(x + y) = -2 - x/y - y/x.Wait, that seems odd. So according to this, λ must be greater than or equal to -2 - x/y - y/x for all x, y > 0. But x/y + y/x is always at least 2 by AM-GM inequality. Therefore, x/y + y/x ≥ 2, so -2 - x/y - y/x ≤ -2 - 2 = -4. Therefore, the expression -2 - x/y - y/x is always less than or equal to -4. Therefore, the supremum (least upper bound) of -2 - x/y - y/x over all x, y > 0 is -4. Therefore, λ must be greater than or equal to -4. But since we need the inequality to hold for all x, y > 0, then λ must be at least the supremum of - (1/x + 1/y)(x + y), which is -4. Therefore, the minimal λ is -4.Wait, that seems too straightforward, but let's check.Wait, let me verify the steps again. Starting from the inequality:1/x + 1/y + λ/(x + y) ≥ 0.Multiply both sides by (x + y):(1/x + 1/y)(x + y) + λ ≥ 0.Which simplifies to:( (x + y)/x + (x + y)/y ) + λ ≥ 0.Which is (1 + y/x + x/y + 1) + λ ≥ 0.Therefore, 2 + y/x + x/y + λ ≥ 0.Hence, λ ≥ -2 - y/x - x/y.Since y/x + x/y ≥ 2, the right-hand side is at most -2 - 2 = -4.Therefore, the inequality λ ≥ -4 must hold. Therefore, the minimal λ is -4.But wait, we need to check if λ = -4 indeed satisfies the original inequality for all x, y > 0.Let me check with λ = -4. Then the original inequality becomes:1/x + 1/y - 4/(x + y) ≥ 0.Is this true for all x, y > 0?Let me test with specific values. Let's take x = y = 1. Then the left-hand side is 1 + 1 - 4/(1 + 1) = 2 - 2 = 0. So equality holds here.Another test: x = y = t, where t > 0. Then the left-hand side is 2/t - 4/(2t) = 2/t - 2/t = 0. So equality holds for all x = y. That's a good sign. Maybe λ = -4 is indeed the minimal value.Let me test another case where x ≠ y. Let's take x = 1 and y approaching 0. Then 1/x + 1/y becomes 1 + 1/y, which goes to infinity as y approaches 0, and -4/(x + y) approaches -4/1 = -4. So the total expression approaches infinity - 4, which is still infinity, so it's positive. Similarly, if y approaches infinity, 1/y approaches 0, so 1/x + 0 + (-4)/(x + y) approaches 1/x - 4/(x + y). As y approaches infinity, 4/(x + y) approaches 0, so the expression approaches 1/x, which is positive. So in these edge cases, the expression is positive.What about another case: let’s take x = 2 and y = 2. Then the expression is 1/2 + 1/2 - 4/(4) = 1 - 1 = 0. Again equality holds.Another test: x = 1, y = 2. Then left-hand side is 1 + 1/2 - 4/(3) = 1.5 - 1.333... ≈ 0.166..., which is positive.If I take x = 1, y = 3. Then 1 + 1/3 - 4/4 = 1 + 0.333... - 1 = 0.333..., still positive.How about x = 2, y = 1. Then 1/2 + 1 - 4/3 ≈ 0.5 + 1 - 1.333 ≈ 0.166..., positive.What if we take x = 1, y = 1/2. Then 1 + 2 - 4/(1.5) ≈ 3 - 2.666... ≈ 0.333..., positive.Alternatively, suppose we take x approaching y. Let’s parameterize x and y such that y = kx, where k > 0. Then we can write the expression in terms of x and k.Let’s let y = kx. Then the inequality becomes:1/x + 1/(kx) + λ/(x + kx) ≥ 0.Simplify:(1 + 1/k)/x + λ/(x(1 + k)) ≥ 0.Multiply both sides by x (which is positive):(1 + 1/k) + λ/(1 + k) ≥ 0.So:( (k + 1)/k ) + λ/(k + 1) ≥ 0.Multiply both sides by k(k + 1) to eliminate denominators:(k + 1)^2 + λ k ≥ 0.So:λ k ≥ - (k + 1)^2.Thus:λ ≥ - (k + 1)^2 / k.Therefore, for the inequality to hold for all k > 0, λ must be greater than or equal to the maximum of - (k + 1)^2 / k over all k > 0.Therefore, to find the minimal λ, we need to compute the maximum value of - (k + 1)^2 / k over all k > 0, and set λ equal to that maximum.So let's compute f(k) = - (k + 1)^2 / k.We can write this as f(k) = - (k^2 + 2k + 1)/k = -k - 2 - 1/k.So f(k) = -k - 2 - 1/k.We need to find the maximum value of f(k) over k > 0.Since we are maximizing f(k) = -k - 2 - 1/k, which is equivalent to minimizing g(k) = k + 1/k + 2.Wait, because f(k) = - (k + 1/k + 2). Therefore, maximizing f(k) is equivalent to minimizing k + 1/k + 2.The minimal value of k + 1/k is 2 by AM-GM (since k > 0), so the minimal value of k + 1/k + 2 is 2 + 2 = 4. Therefore, the maximum value of f(k) is -4.Therefore, the maximum of f(k) is -4, achieved when k = 1. Therefore, the minimal λ is -4.So, this confirms the earlier result. Hence, the minimal value of λ is -4.But let me verify once more with another approach. Suppose we use calculus to find the minimum of the expression 1/x + 1/y + λ/(x + y). To find when this expression is zero, set it equal to zero and find conditions on λ.Wait, but since we need the expression to be non-negative for all x, y > 0, the minimum of the expression should be zero. Therefore, we can set up the problem as minimizing 1/x + 1/y + λ/(x + y) and setting that minimum to zero. Then solving for λ.To find the minimum, take partial derivatives with respect to x and y, set them to zero.Let’s compute the partial derivative with respect to x:d/dx [1/x + 1/y + λ/(x + y)] = -1/x² - λ/(x + y)².Similarly, the partial derivative with respect to y:d/dy [1/x + 1/y + λ/(x + y)] = -1/y² - λ/(x + y)².Set both derivatives to zero:-1/x² - λ/(x + y)² = 0,-1/y² - λ/(x + y)² = 0.Therefore,-1/x² = λ/(x + y)²,-1/y² = λ/(x + y)².Therefore, from the two equations:-1/x² = -1/y² => 1/x² = 1/y² => x = y, since x, y > 0.Therefore, at the critical point, x = y. Let’s set x = y = t.Then the expression becomes 1/t + 1/t + λ/(2t) = 2/t + λ/(2t).Set this equal to zero (since we are considering the minimum value being zero):2/t + λ/(2t) = 0.Multiply both sides by 2t:4 + λ = 0 => λ = -4.Therefore, the critical point occurs at x = y, and at that point, λ must be -4 to make the expression zero. Therefore, the minimal λ is -4.This aligns with the previous results. Therefore, the minimal λ is -4.So, through multiple approaches: algebraic manipulation using inequalities, substitution, and calculus, we arrive at the conclusion that λ must be at least -4. Therefore, the minimal value of λ is -4.Final AnswerThe minimum value of (lambda) is (boxed{-4}).</think>Given that (x > 0) and (y > 0), we need to find the minimum value of the real number (lambda) such that the inequality (frac{1}{x} + frac{1}{y} + frac{lambda}{x+y} geqslant 0) always holds.First, we rewrite the inequality by multiplying both sides by (x + y) (which is positive):[left(frac{1}{x} + frac{1}{y}right)(x + y) + lambda geq 0.]Expanding the left-hand side, we get:[left(frac{x + y}{x} + frac{x + y}{y}right) + lambda = 2 + frac{y}{x} + frac{x}{y} + lambda.]This simplifies to:[2 + frac{x}{y} + frac{y}{x} + lambda geq 0.]Using the AM-GM inequality, we know that (frac{x}{y} + frac{y}{x} geq 2). Therefore, the expression becomes:[2 + 2 + lambda geq 0 implies 4 + lambda geq 0 implies lambda geq -4.]To verify, we check if (lambda = -4) satisfies the original inequality for all (x, y > 0). Substituting (lambda = -4) into the inequality:[frac{1}{x} + frac{1}{y} - frac{4}{x + y} geq 0.]Testing with (x = y), we get:[frac{2}{x} - frac{4}{2x} = frac{2}{x} - frac{2}{x} = 0.]Testing with other values, such as (x = 1) and (y = 2), we get:[1 + frac{1}{2} - frac{4}{3} = frac{3}{2} - frac{4}{3} = frac{9}{6} - frac{8}{6} = frac{1}{6} geq 0.]Using calculus, we find the critical point by setting (x = y) and solving for (lambda):[frac{2}{t} + frac{lambda}{2t} = 0 implies 4 + lambda = 0 implies lambda = -4.]Thus, the minimum value of (lambda) is (boxed{-4}).]

🔑:Since x > 0 and y > 0, the original inequality can be rewritten as - lambda leqslant (frac{1}{x} + frac{1}{y})(x+y) = 2 + frac{y}{x} + frac{x}{y}.Note that 2 + frac{y}{x} + frac{x}{y} geqslant 2 + 2sqrt{frac{y}{x} cdot frac{x}{y}} = 4, with equality holding if and only if x = y.Therefore, the minimum value of (frac{1}{x} + frac{1}{y})(x+y) is 4, which gives us - lambda leqslant 4, or lambda geqslant -4.Consequently, the minimum value of lambda is boxed{-4}.

🧠:[<think>Alright, so I have this sequence here: 1,1,3,1,3,5,1,3,5,7,1,3,5,7,9,1,3,5,7,9,11,… and I need to find the 2020th term. Let me try to figure out the pattern here.First, let me write down the sequence with positions to see if I can spot a pattern:1 (1st term),1 (2nd),3 (3rd),1 (4th),3 (5th),5 (6th),1 (7th),3 (8th),5 (9th),7 (10th),1 (11th),3 (12th),5 (13th),7 (14th),9 (15th),1 (16th),3 (17th),5 (18th),7 (19th),9 (20th),11 (21st),...Looking at this, it seems like the sequence is made up of smaller subsequences. Each subsequence starts with 1 and increases by 2 each time. The first subsequence is just [1], the next is [1,3], then [1,3,5], then [1,3,5,7], and so on. Let me check:First group: 1 term [1]Second group: 2 terms [1,3]Third group: 3 terms [1,3,5]Fourth group: 4 terms [1,3,5,7]Fifth group: 5 terms [1,3,5,7,9]Sixth group: 6 terms [1,3,5,7,9,11]...So each new group starts with 1 and has one more term than the previous group. Each term within a group increases by 2. So the nth group has n terms, starting at 1 and increasing by 2 each term. Therefore, the kth term in the mth group is 1 + 2(k-1). Wait, let me confirm:In the second group (m=2), the terms are 1 (k=1) and 3 (k=2). For k=2, 1 + 2*(2-1) = 1 + 2*1 = 3. Correct. Similarly, in the third group (m=3), the third term is 5, which is 1 + 2*(3-1) = 5. Yep, that works.So each group m has m terms, starting at 1 and each subsequent term increases by 2. So the terms in group m are 1, 3, 5, ..., up to 1 + 2*(m-1). Wait, but that would mean that the last term of group m is 1 + 2*(m - 1) = 2m -1. For example, group 1 has last term 1 (2*1 -1 =1), group 2 has last term 3 (2*2 -1=3), group 3 has last term 5, etc. Correct.Now, to find the 2020th term in the overall sequence, I need to figure out which group the 2020th term is in and then its position within that group.First, let's understand how the terms are grouped. The total number of terms up to and including group m is the sum of the sizes of the first m groups. Since the first group has 1 term, the second 2 terms, ..., the mth group has m terms. So the total number of terms is 1 + 2 + 3 + ... + m = m(m + 1)/2.Therefore, to find the group m such that the sum 1 + 2 + ... + (m -1) < 2020 ≤ 1 + 2 + ... + m, we need to solve for m in m(m +1)/2 ≥ 2020.Let me solve the equation m(m +1)/2 = 2020.Multiply both sides by 2: m^2 + m = 4040.Then, m^2 + m - 4040 = 0.Using the quadratic formula, m = [-1 ± sqrt(1 + 4*4040)] / 2.Calculate sqrt(1 + 16160) = sqrt(16161). Let me see, 127^2 = 16129, 128^2 = 16384. So sqrt(16161) is between 127 and 128. Let's check 127^2 = 16129, so 16161 -16129 =32. So sqrt(16161) is 127 + 32/(2*127) + ... approximately 127.13. But let's check if 127*127 =16129, 127*127. So 127*127=16129. Then 127.2^2 = (127 +0.2)^2 =127^2 +2*127*0.2 +0.2^2=16129 +50.8 +0.04=16179.84. Hmm, that's more than 16161. Wait, 127.1^2= (127 +0.1)^2=16129 +25.4 +0.01=16154.41. Still less than 16161. 127.1^2=16154.41. 127.1 + delta)^2=16161. Let's compute delta:(127.1 + delta)^2 = 127.1^2 + 2*127.1*delta + delta^2 ≈16154.41 +254.2*delta. Set equal to 16161. Then 254.2*delta ≈16161 -16154.41=6.59. So delta≈6.59/254.2≈0.0259. So sqrt(16161)≈127.1 +0.0259≈127.1259. So approx 127.126.Therefore, m≈(-1 +127.126)/2≈126.126/2≈63.063. So m≈63.06. Since m must be an integer, the value of m where the cumulative sum first exceeds 2020 is m=64.Wait, let's check m=63: sum is 63*64/2=2016. Then m=63 gives sum up to group 63 as 2016 terms. Then group 64 would have 64 terms, so term 2017 is the first term of group 64, term 2017+63=2080 is the last term of group 64. But our target is term 2020, which is within group 64, since 2016 <2020≤2016+64=2080.So group 64 contains the 2020th term. Now, to find the position within group 64: term 2020 is the (2020 -2016)=4th term in group 64.Now, the terms in group m=64 are 1,3,5,..., each increasing by 2. The first term is 1, the second 3, third 5, fourth 7, etc. So the kth term in group m is 1 + 2*(k-1).Since the 2020th term is the 4th term in group 64, k=4. Therefore, the term is 1 + 2*(4-1)=1 +6=7.Wait, so the answer is 7? Let me verify this with a smaller example.For instance, the 6th term: According to the sequence, the 6th term is 5. Let's see:Sum up groups: group1 (1 term), group2 (2 terms, total 3), group3 (3 terms, total 6). So term 6 is the last term of group3, which is 5. That's correct. So according to the formula, group3, term3: 1 +2*(3-1)=5. Correct.Another check: term 10 is 7. Let's see:Sum up group1 (1), group2 (2, total 3), group3 (3, total 6), group4 (4 terms, total 10). So term10 is the 4th term of group4. 1 +2*(4-1)=7. Correct.Term 15 is 9. Sum up to group5: 1+2+3+4+5=15. So term15 is the 5th term in group5:1 +2*(5-1)=1 +8=9. Correct.Similarly, term21 is 11, which is the 6th term in group6:1 +2*(6-1)=11. Correct.So the formula seems to hold. Therefore, term2020 is in group64, position4, which is 1 +2*(4-1)=7.But let me verify once again:Total terms up to group63:63*64/2=2016. So term2017 is first term of group64:1. Term2018:3. Term2019:5. Term2020:7. Yes, that's correct. Therefore, the 2020th term is 7.Wait, but let me check with another approach to confirm. Let's model the sequence as blocks. Each block n has n terms:1,3,...,2n-1. So the first block has 1 term (1), second block 2 terms (1,3), third block 3 terms (1,3,5), etc. So the position of the last term in block m is m(m+1)/2. So to find which block the 2020th term is in, we need m such that m(m+1)/2 >=2020. As before, m≈63.06, so m=64. The starting index of block64 is 63*64/2 +1=2016 +1=2017. Therefore, term2017 is 1, term2018 is3, term2019 is5, term2020 is7. So yes, 7 is the answer.Therefore, I think the 2020th term is 7.Final AnswerThe 2020th term of the sequence is boxed{7}.</think>The given sequence is composed of subsequences where each subsequence starts with 1 and increases by 2. The nth subsequence has n terms. To find the 2020th term, we need to determine which subsequence it belongs to and its position within that subsequence.1. Identify the subsequence (group) containing the 2020th term: - The total number of terms up to the nth subsequence is given by the sum of the first n natural numbers: ( frac{n(n+1)}{2} ). - Solve ( frac{n(n+1)}{2} geq 2020 ). Using the quadratic formula, we find ( n approx 63.06 ). Thus, ( n = 64 ) (since 63.06 is not an integer, we round up to the next integer).2. Calculate the starting position of the 64th subsequence: - The total number of terms up to the 63rd subsequence is ( frac{63 times 64}{2} = 2016 ). - The 64th subsequence starts at the 2017th term.3. Determine the position within the 64th subsequence: - The 2020th term is the ( 2020 - 2016 = 4 )th term in the 64th subsequence.4. Find the value of the 4th term in the 64th subsequence: - Each term in the subsequence is given by ( 1 + 2(k-1) ) where ( k ) is the position within the subsequence. - For ( k = 4 ), the term is ( 1 + 2(4-1) = 1 + 6 = 7 ).Thus, the 2020th term of the sequence is (boxed{7}).,<think>Okay, so I need to find the 2020th term of this sequence: 1,1,3,1,3,5,1,3,5,7,1,3,5,7,9,1,3,5,7,9,11,… Let me try to understand the pattern here.Looking at the sequence, it starts with 1. Then the next two terms are 1,3. Then the next three terms are 1,3,5. Then four terms: 1,3,5,7. Then five terms: 1,3,5,7,9. Then six terms: 1,3,5,7,9,11. Hmm, so each "block" of numbers seems to be increasing by 2, starting at 1, and each block has one more term than the previous block. Let me check:First block: [1] (length 1)Second block: [1,3] (length 2)Third block: [1,3,5] (length 3)Fourth block: [1,3,5,7] (length 4)Fifth block: [1,3,5,7,9] (length 5)Sixth block: [1,3,5,7,9,11] (length 6)And so on.So each block n (starting from n=1) has length n and consists of the odd numbers starting from 1 up to 2n-1. Wait, let me verify:For block n=1: 2*1 -1 =1, which is correct.Block n=2: 2*2 -1=3, and the terms are 1,3. Correct.Similarly, n=3: 2*3 -1=5, terms are 1,3,5. Yes, that matches. So each block n starts at 1 and goes up to 2n-1 with a common difference of 2.Therefore, the sequence is formed by concatenating blocks where the nth block has n terms: 1,3,5,...,2n-1.So, to find the 2020th term, I need to determine which block the 2020th term is in and then find its position within that block.First, let's figure out how many terms are there before the kth block. The first block has 1 term, the second has 2 terms, ..., the (k-1)th block has (k-1) terms. So the total number of terms before the kth block is the sum from i=1 to i=k-1 of i, which is (k-1)*k/2.We need to find the largest k such that the total number of terms up to the kth block is less than 2020. In other words, solve for k in:(k)(k+1)/2 < 2020 ≤ (k+1)(k+2)/2Wait, hold on. Wait, actually, the sum from 1 to m is m(m+1)/2. Wait, no. Let me clarify.Each block n (starting at n=1) has n terms. So the total number of terms up to and including block n is 1 + 2 + 3 + ... + n = n(n+1)/2. Therefore, to find which block the 2020th term is in, we need to find the smallest n such that n(n+1)/2 ≥ 2020. Wait, but actually, the total terms up to block n is n(n+1)/2. Therefore, if we solve n(n+1)/2 ≥ 2020, the smallest n satisfying this will be the block where the 2020th term is located.Wait, but let's test with an example. Let's say we want to find the 5th term. The total terms up to block 2 is 1+2=3, up to block 3 is 6. So 5th term is in block 3. Which is correct, since the first three blocks have 1,2,3 terms: positions 1, 2-3, 4-6. So 5th term is in block 3. So yes, the formula is correct.Therefore, for the 2020th term, we need to solve n(n+1)/2 ≥ 2020. Let's solve for n.Multiply both sides by 2: n(n+1) ≥ 4040.Approximate n^2 ≈ 4040. So n ≈ sqrt(4040). sqrt(4000) = 63.24, so sqrt(4040) is slightly higher. Let's compute 63^2 = 3969, 64^2=4096. So between 63 and 64. Let's compute 63*64=4032. Which is 63*64=4032. Then 63*64/2=2016. So the total number of terms up to block 63 is 63*64/2=2016. Then block 64 has 64 terms. So the 2017th term is the first term of block 64, the 2018th is the second term, ..., up to 2016 +64=2080th term.But we need the 2020th term, which is 2020 -2016=4th term in block 64.Wait, but let me confirm. The total terms up to block 63 is 2016. Therefore, the next block, block 64, will have terms from 2017 to 2016+64=2080. So term 2017: first term of block 64, term 2018: second term, ..., term 2020: fourth term.Each block starts with 1 and increases by 2 each term. So the first term is 1, second term is 3, third term is 5, fourth term is 7, etc., up to the nth term of block n: 2n-1.Therefore, in block 64, the fourth term would be 1 + 2*(4-1)=1+6=7? Wait, no, wait. Wait, each term in the block is 1, 3, 5, 7, ..., 2k-1 for the kth term in the block.Wait, in the block, the first term is 1 (k=1: 2*1 -1=1), second term is 3 (k=2:2*2 -1=3), third term is 5, etc. So the mth term in block n is 2m -1.Therefore, the fourth term in block 64 is 2*4 -1=8 -1=7. So the 2020th term is 7.But let me check this step by step to make sure I didn't make a mistake.First, confirm that the total number of terms up to block n is n(n+1)/2. For block 1: 1(2)/2=1, correct. Block 2: 2(3)/2=3, which is 1+2=3, correct. Block 3: 3(4)/2=6, which is 1+2+3=6, correct. So that formula holds.So solving n(n+1)/2 ≥ 2020. Let's compute n(n+1) ≥ 4040.Let me compute 63*64=4032, which is less than 4040. Then 64*65=4160, which is greater. Therefore, n=63 gives 4032/2=2016 terms, so the 2016th term is the last term of block 63. Then the 2017th term is the first term of block 64. Therefore, term number 2020 is 2020 -2016=4th term in block 64. Then in block 64, the mth term is 2m -1. So the fourth term is 2*4 -1=7. Therefore, the answer is 7. Wait, but let me check the sequence given.Looking at the sequence:1 (block1)1,3 (block2)1,3,5 (block3)1,3,5,7 (block4)1,3,5,7,9 (block5)1,3,5,7,9,11 (block6)So each block n corresponds to n terms, starting at 1, increasing by 2. So in block4, the terms are 1,3,5,7. So the 4th term is 7. Similarly, block5 has 1,3,5,7,9, so the 5th term is9, etc. So in general, in block k, the mth term is 2m -1. So in block64, the 4th term would be 2*4 -1=7. So yes, 7. Therefore, the 2020th term is 7.But let me check if my calculation of the position is correct. If up to block63, there are 2016 terms. Then block64 starts at term2017. So term2017:1, term2018:3, term2019:5, term2020:7. So yes, term2020 is the fourth term in block64, which is7. Therefore, the answer is7.Wait, but let me confirm with a smaller example. Let's say we want the 6th term. Total terms up to block3 is6, so 6th term is block3's third term, which is5. Correct, as per the sequence given. Similarly, term7 is first term of block4:1. Term10 is fourth term of block4:7. Yes, correct. So the formula holds.Therefore, I think the answer is7. Let me check once more.Compute n(n+1)/2 <2020. Find n=63: 63*64/2=2016. Then term2020 is term4 in block64:7. Yes.Wait, but wait, the first term of block64 is1, second3, third5, fourth7. So term2017:1,2018:3,2019:5,2020:7. Yes. So that's correct.Therefore, the 2020th term is7. So I think that's the answer.Final Answerboxed{7}</think>To find the 2020th term of the sequence (1, 1, 3, 1, 3, 5, 1, 3, 5, 7, 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 11, ldots), we observe that the sequence is formed by concatenating blocks of odd numbers. Each block (n) (starting from (n=1)) contains (n) terms: (1, 3, 5, ldots, 2n-1).To determine which block the 2020th term is in, we need to find the smallest (k) such that the total number of terms up to block (k) is at least 2020. The total number of terms up to block (k) is given by the sum of the first (k) natural numbers, which is (frac{k(k+1)}{2}).We solve for (k) in the inequality:[frac{k(k+1)}{2} geq 2020]Multiplying both sides by 2:[k(k+1) geq 4040]Approximating (k) by solving (k^2 approx 4040), we find (k approx 63.5). Testing (k = 63):[frac{63 times 64}{2} = 2016]This is less than 2020. Testing (k = 64):[frac{64 times 65}{2} = 2080]This is greater than 2020. Therefore, the 2020th term is in block 64.The number of terms before block 64 is 2016, so the 2020th term is the (2020 - 2016 = 4)th term in block 64. Each term in block (n) is given by (2m - 1) where (m) is the position within the block. Thus, the 4th term in block 64 is:[2 times 4 - 1 = 7]Therefore, the 2020th term of the sequence is (boxed{7}).]

🔑:Answer: 7Solution: We have that for each n in mathbb{N}, the (1+2+cdots+n) th term is 2 n-1. The first n+1 odd positive integers are then listed. Observe that the largest triangular number less than or equal to 2020 is frac{63(64)}{2}=2016. Therefore, the 2020 th term is 7 .

🧠:[<think>Okay, so I need to solve these two problems about sets A, B, and C. Let's take them one at a time. Starting with problem (1): Find A ∩ (complement_U B). First, I need to understand what sets A and B are.Set A is defined as {x | |x - 4| ≤ 2}. The absolute value inequality |x - 4| ≤ 2 can be rewritten as -2 ≤ x - 4 ≤ 2. Adding 4 to all parts of the inequality gives 2 ≤ x ≤ 6. So, set A is the closed interval [2, 6]. That seems straightforward.Now, set B is {x | (5 - x)/(x + 1) > 0}. To solve this inequality, I need to find where the rational expression (5 - x)/(x + 1) is positive. A rational expression is positive when the numerator and denominator have the same sign. First, let's find the critical points where the expression is zero or undefined. The numerator 5 - x equals zero when x = 5. The denominator x + 1 equals zero when x = -1. These critical points divide the real number line into three intervals: (-∞, -1), (-1, 5), and (5, ∞). Now, let's test each interval:1. Interval (-∞, -1): Let's pick x = -2. Plugging into (5 - (-2))/( -2 + 1) = (7)/(-1) = -7, which is negative. So this interval doesn't satisfy the inequality.2. Interval (-1, 5): Let's pick x = 0. (5 - 0)/(0 + 1) = 5/1 = 5, which is positive. So this interval satisfies the inequality.3. Interval (5, ∞): Let's pick x = 6. (5 - 6)/(6 + 1) = (-1)/7 = -1/7, which is negative. So this interval doesn't satisfy the inequality.Next, we check the critical points. At x = -1, the expression is undefined (denominator zero), so x = -1 is excluded. At x = 5, the expression is zero, which doesn't satisfy the strict inequality (5 - x)/(x + 1) > 0. So x = 5 is also excluded.Therefore, set B is the open interval (-1, 5). But wait, let me double-check that. The inequality was (5 - x)/(x + 1) > 0. If x is between -1 and 5, then 5 - x is positive (since x is less than 5) and x + 1 is positive (since x is greater than -1). So both numerator and denominator are positive, making the expression positive. If x < -1, both numerator and denominator are negative (since x < -1 implies x +1 < 0, and x <5 would mean 5 - x is positive, but wait x < -1: 5 - x would be 5 minus a negative number, so 5 + |x|, which is definitely positive. Wait, that contradicts my earlier test. Wait, hold on. Let me re-examine.If x < -1, let's take x = -2. Then numerator is 5 - (-2) = 7, which is positive. Denominator is -2 +1 = -1, which is negative. So positive divided by negative is negative. Therefore, (5 - x)/(x +1) is negative in (-∞, -1). Then in (-1,5), as x=0 gives 5/1=5 positive. And in (5, ∞), x=6 gives (5 -6)/7 = -1/7, which is negative. So my initial analysis was correct. So B is (-1,5).Therefore, set B is (-1,5). Then, the complement of B in the universal set U = ℝ would be all real numbers not in (-1,5). That is, complement_U B = (-∞, -1] ∪ [5, ∞).Now, we need to find A ∩ (complement_U B). Since A is [2,6] and complement_U B is (-∞, -1] ∪ [5, ∞). The intersection of [2,6] with (-∞, -1] ∪ [5, ∞) is [5,6]. Because [2,6] overlaps with [5, ∞) starting at 5, and there's no overlap with (-∞, -1]. So A ∩ complement_U B is [5,6].Wait, let's verify that. A is from 2 to 6, inclusive. complement of B is everything less than or equal to -1 or greater than or equal to 5. So overlapping parts would be the part of A that is greater than or equal to 5. So from 5 to 6, including both endpoints. Therefore, [5,6]. That seems correct.So problem (1) answer is [5,6]. Now moving on to problem (2): If the set C = {x | x < a}, and A ∩ C ≠ ∅, find the range of the real number a.First, set C is all real numbers less than a. We need A ∩ C ≠ ∅. Since A is [2,6], the intersection of [2,6] with (-∞, a) is non-empty. So that means there must be some x such that x is in both A and C. In other words, there exists an x where 2 ≤ x ≤ 6 and x < a. So for such an x to exist, the upper limit of C, which is a, must be greater than the lower limit of A, which is 2. Because if a is greater than 2, then there is overlap between [2,6] and (-∞, a). For example, if a is 3, then the intersection is [2,3). If a is 10, the intersection is [2,6]. But if a is less than or equal to 2, say a=1, then the intersection is [2,6] ∩ (-∞,1) which is empty. Therefore, for the intersection to be non-empty, a must be greater than 2.Wait, but hold on. Let me formalize this. The intersection A ∩ C is [2,6] ∩ (-∞, a). This will be non-empty if and only if there exists some x such that 2 ≤ x ≤ 6 and x < a. The smallest x in A is 2, so if a is greater than 2, then there are elements in A (starting from 2 up to the minimum of 6 and a) that are also in C. If a ≤ 2, then there's no x in A that is less than a. Hence, the condition A ∩ C ≠ ∅ requires that a > 2.Therefore, the range of a is (2, ∞). But wait, let me check with a = 2. If a = 2, then C is (-∞, 2). The intersection A ∩ C would be [2,6] ∩ (-∞,2) which is {2} ∩ (-∞,2) which is empty. So indeed, a must be greater than 2. So the range is a > 2.Therefore, the answer is a ∈ (2, ∞).Wait, but maybe the problem expects the range written in interval notation? Then it's (2, ∞). Yes, that's correct.So summarizing:1. A ∩ complement_U B is [5,6].2. The range of a is (2, ∞).But let me double-check problem (1) again. Set A is [2,6], complement of B is (-∞, -1] ∪ [5, ∞). Intersection between [2,6] and (-∞, -1] is empty. Intersection between [2,6] and [5, ∞) is [5,6]. So yes, [5,6] is correct. And for problem (2), since C is all real numbers less than a, and A starts at 2, to have an overlap, a must be greater than 2. If a is 2, C is up to but not including 2, so no overlap. If a is greater than 2, even by a little bit, then there is overlap between A and C. For example, if a is 2.1, then the intersection is [2, 2.1). Hence, the range of a is all real numbers greater than 2.Therefore, I think the answers are correct.Final Answer(1) boxed{[5, 6]}(2) boxed{(2, infty)}</think>Given the sets ( A = {x mid |x - 4| leq 2} ) and ( B = left{x mid frac{5 - x}{x + 1} > 0right} ) with the universal set ( U = mathbb{R} ):1. Finding ( A cap (complement_U B) ): - Set ( A ) is determined by solving ( |x - 4| leq 2 ), which gives the interval ([2, 6]). - Set ( B ) is determined by solving ( frac{5 - x}{x + 1} > 0 ). The critical points are ( x = -1 ) and ( x = 5 ). Testing intervals around these points shows that ( B ) is the interval ((-1, 5)). - The complement of ( B ) in ( mathbb{R} ) is ( complement_U B = (-infty, -1] cup [5, infty) ). - The intersection ( A cap (complement_U B) ) is the overlap between ([2, 6]) and ((-infty, -1] cup [5, infty)), which is ([5, 6]).2. Finding the range of ( a ) such that ( A cap C neq emptyset ) where ( C = {x mid x < a } ): - Set ( C ) is the interval ((-∞, a)). - For ( A cap C ) to be non-empty, there must be some ( x ) such that ( 2 leq x leq 6 ) and ( x < a ). This requires ( a ) to be greater than 2, as any ( a leq 2 ) would result in no overlap with ( A ).Thus, the final answers are:(1) (boxed{[5, 6]})(2) (boxed{(2, infty)}),<think>Okay, let me try to solve these two problems step by step. First, let's tackle part (1): finding ( A cap (complement_U B) ). Starting with set ( A ), it's defined as ( {x mid |x - 4| leq 2} ). The absolute value inequality ( |x - 4| leq 2 ) means that the distance between ( x ) and 4 is at most 2. To solve this, I can rewrite the inequality as ( -2 leq x - 4 leq 2 ). Adding 4 to all parts gives ( 2 leq x leq 6 ). So, set ( A ) is the closed interval [2, 6].Now, moving on to set ( B ), which is given by ( left{x mid frac{5 - x}{x + 1} > 0 right} ). This is a rational inequality. To solve it, I need to determine where the expression ( frac{5 - x}{x + 1} ) is positive. First, let's find the critical points where the numerator or denominator is zero. The numerator ( 5 - x = 0 ) when ( x = 5 ), and the denominator ( x + 1 = 0 ) when ( x = -1 ). These critical points divide the real number line into three intervals: ( (-infty, -1) ), ( (-1, 5) ), and ( (5, infty) ). Next, I need to test each interval to see where the inequality holds. 1. Interval ( (-infty, -1) ): Let's pick ( x = -2 ). Plugging into the expression: ( frac{5 - (-2)}{-2 + 1} = frac{7}{-1} = -7 ), which is negative. So, this interval doesn't satisfy the inequality.2. Interval ( (-1, 5) ): Let's choose ( x = 0 ). The expression becomes ( frac{5 - 0}{0 + 1} = 5/1 = 5 ), which is positive. So, this interval works.3. Interval ( (5, infty) ): Let's take ( x = 6 ). Then, ( frac{5 - 6}{6 + 1} = frac{-1}{7} ), which is negative. So, this interval doesn't satisfy the inequality.Additionally, we have to check the critical points. At ( x = -1 ), the denominator is zero, so the expression is undefined; hence, ( x = -1 ) is excluded. At ( x = 5 ), the numerator is zero, making the expression zero, which doesn't satisfy the strict inequality ( > 0 ). Therefore, set ( B ) is the open interval ( (-1, 5) ).The complement of ( B ) in ( mathbb{R} ) (universal set) is ( complement_U B = (-infty, -1] cup [5, infty) ).Now, we need to find ( A cap (complement_U B) ). Since ( A = [2, 6] ) and ( complement_U B = (-infty, -1] cup [5, infty) ), the intersection will be the overlap between [2, 6] and the two intervals in the complement of B. Looking at ( [2, 6] ) and ( (-infty, -1] cup [5, infty) ), the overlap occurs where ( [2, 6] ) meets ( [5, infty) ). Specifically, the intersection is ( [5, 6] ). The other part of the complement, ( (-infty, -1] ), doesn't overlap with ( [2, 6] ) at all. Therefore, ( A cap (complement_U B) = [5, 6] ).Okay, that seems straightforward. Now, part (2): If the set ( C = {x mid x < a} ), and ( A cap C neq emptyset ), find the range of the real number ( a ).First, let's recall that set ( A ) is [2, 6]. Set ( C ) is all real numbers less than ( a ). The intersection ( A cap C ) would be all numbers that are both in [2, 6] and less than ( a ). For this intersection to be non-empty, there must be some elements common to both sets.So, ( A cap C = [2, 6] cap (-infty, a) = [2, a) ) if ( a > 2 ). If ( a leq 2 ), then the intersection would be empty. Wait, but actually, if ( a ) is greater than 2, then the intersection starts at 2 and goes up to ( a ), but since ( A ) is up to 6, if ( a ) is greater than 6, then the intersection would be [2, 6). If ( a ) is between 2 and 6, the intersection is [2, a). If ( a ) is less than or equal to 2, the intersection is empty. So, for the intersection to be non-empty, ( a ) must be greater than 2. Because even if ( a ) is just slightly bigger than 2, like 2.1, then the intersection [2, 2.1) would include 2 up to 2.1, which is non-empty. Wait, but if ( a = 3 ), then [2, 3) is non-empty. Similarly, if ( a = 10 ), the intersection is [2, 6), which is also non-empty. So, the key is that ( a ) needs to be greater than the minimum of set A. Since set A starts at 2, as long as ( a > 2 ), there will be overlap between C and A. If ( a leq 2 ), then C is everything less than a, which is less than or equal to 2, but A starts at 2, so the intersection is either just the point 2 (if a = 2) or empty. Wait, but set C is defined as ( x < a ), not ( x leq a ). So, if ( a = 2 ), then C is ( (-infty, 2) ), and A is [2,6], so the intersection is empty. Therefore, the intersection is non-empty only when ( a > 2 ). Wait, let me check. Suppose ( a = 2 ), then ( C = (-infty, 2) ), and ( A = [2, 6] ). Their intersection is at x=2, but since C does not include 2, and A includes 2. So the intersection is empty. If ( a ) is greater than 2, say 3, then C is up to 3, not including 3, and A includes from 2 to 6, so the intersection is [2,3), which is non-empty. Similarly, if ( a = 6 ), then C is up to 6, so the intersection is [2,6), which is non-empty. If ( a = 10 ), the intersection is [2,10) intersected with A, which is [2,6], so [2,6). Still non-empty. If ( a ) approaches infinity, the intersection is [2,6]. So, as long as ( a > 2 ), the intersection is non-empty. If ( a leq 2 ), then the intersection is empty. Therefore, the range of real number ( a ) is ( a > 2 ). So in interval notation, ( (2, infty) ).Wait, but let me think again. Suppose ( a = 5 ), then ( C = (-infty,5) ), intersection with A is [2,5), which is non-empty. Similarly, if ( a ) is 2.1, the intersection is [2,2.1), which is non-empty. So yes, any ( a > 2 ) ensures that the intersection is non-empty. Therefore, the range of ( a ) is all real numbers greater than 2. But let me double-check. If ( a ) is exactly 2, then C is all numbers less than 2, which doesn't include 2, so the intersection with A (which starts at 2) is empty. If ( a ) is just a tiny bit more than 2, say 2 + ε where ε is a small positive number, then C includes numbers up to but not including 2 + ε, and A includes numbers from 2 onwards, so the intersection would be [2, 2 + ε), which is non-empty. Therefore, the minimal value for ( a ) is just above 2. Hence, the range is all real numbers ( a ) such that ( a > 2 ).Therefore, the answer for part (2) is ( a > 2 ), or in interval notation, ( (2, infty) ).So, summarizing my answers:1. ( A cap (complement_U B) = [5, 6] )2. The range of ( a ) is ( a > 2 )Let me quickly check if there's any mistake here.For part (1), set B was found to be (-1,5), so its complement is (-infty, -1] union [5, infty). Intersecting with A [2,6], so overlapping part is [5,6]. That seems correct.For part (2), since A is [2,6], and C is all numbers less than a. For them to have an intersection, a must be greater than 2. If a is 2 or less, then C is entirely left of A, hence no overlap. So that seems right.I think these are the correct answers.Final Answer(1) boxed{[5, 6]}(2) boxed{(2, infty)}</think>Given the sets ( A = {x mid |x - 4| leq 2} ) and ( B = left{x mid frac{5 - x}{x + 1} > 0 right} ) with the universal set ( U = mathbb{R} ):1. Finding ( A cap (complement_U B) ): - Set ( A ) is determined by solving the inequality ( |x - 4| leq 2 ). This gives the interval ([2, 6]). - Set ( B ) is determined by solving the inequality ( frac{5 - x}{x + 1} > 0 ). The critical points are ( x = -1 ) and ( x = 5 ). Testing intervals around these points shows the solution is ( (-1, 5) ). - The complement of ( B ) in ( mathbb{R} ) is ( complement_U B = (-infty, -1] cup [5, infty) ). - The intersection ( A cap (complement_U B) ) is the overlap between ([2, 6]) and ( (-infty, -1] cup [5, infty) ), which results in ([5, 6]). [ boxed{[5, 6]} ]2. Finding the range of ( a ) such that ( A cap C neq emptyset ) where ( C = {x mid x < a } ): - Set ( A ) is the interval ([2, 6]). - For ( A cap C ) to be non-empty, there must be some overlap between ([2, 6]) and ( (-infty, a) ). This requires ( a ) to be greater than the smallest element of ( A ), which is 2. If ( a leq 2 ), the intersection is empty. Therefore, the range of ( a ) is all real numbers greater than 2. [ boxed{(2, infty)} ]]

🔑:From the given conditions, we have |x-4| leq 2, which leads to -2 leq x-4 leq 2. Solving this, we get 2 leq x leq 6. For frac{5-x}{x+1} > 0, we have (x-5)(x+1) 2. Thus, the range of the real number a is (2, +infty). boxed{(2, +infty)}

🧠:[<think>Okay, so I need to figure out how many sets A there are such that the set {1, 2, 3} is a subset of A, and A is itself a subset of {1, 2, 3, 4, 5}. Hmm, let me break this down.First, let's recall what a subset means. If A is a subset of {1, 2, 3, 4, 5}, that means every element in A must be one of those numbers from 1 to 5. Also, {1, 2, 3} being a subset of A means that A has to contain all the elements 1, 2, and 3. So A must have 1, 2, 3, and can optionally include 4 and/or 5. Wait, so the problem is asking for all possible sets A that definitely include 1, 2, 3, and might include 4 or 5 or neither or both. So the elements 4 and 5 are optional. Therefore, the number of such sets depends on how many choices we have for including or excluding 4 and 5. Since each of the elements 4 and 5 can either be in the set A or not, that's two choices for each element. So for each of these two elements, there's a binary decision: include or exclude. Therefore, the number of possible combinations is 2 multiplied by 2, right? That is, 2^2 = 4. So there are 4 different possibilities.Wait, let me make sure. Let's list them out to verify. The required elements are 1, 2, 3. Then for the optional elements:1. Exclude both 4 and 5: A = {1, 2, 3}2. Include 4, exclude 5: A = {1, 2, 3, 4}3. Exclude 4, include 5: A = {1, 2, 3, 5}4. Include both 4 and 5: A = {1, 2, 3, 4, 5}So that's four different sets. Therefore, the number of such sets A is 4, which is 2^2. That makes sense because the original set {1, 2, 3, 4, 5} has 5 elements, but we're required to include the first three, so only the last two elements (4 and 5) are optional.So the formula here is that if you have a set S with n elements, and you want the number of subsets that include a specific subset T with k elements, then the number of such subsets is 2^(n - k). In this case, S is {1, 2, 3, 4, 5}, so n = 5, and T is {1, 2, 3}, so k = 3. Therefore, the number of subsets A is 2^(5 - 3) = 2^2 = 4. That matches our earlier count. So this seems right.Wait, but just to make sure I'm not missing any other elements. The problem says A is a subset of {1, 2, 3, 4, 5}, so there are no other elements besides these. So the only elements we can consider are 1 through 5. Since 1, 2, 3 must be included, and 4 and 5 can be included or excluded. Therefore, yes, each of the two elements (4 and 5) gives two options, so 2*2=4. Alternatively, another way to think about it is that once we fix the required elements {1, 2, 3}, the remaining elements that can vary are {4, 5}. The number of subsets of {4, 5} is 2^2 = 4. Each subset corresponds to whether to include 4 and 5 or not. Therefore, each such subset of {4, 5} when combined with {1, 2, 3} will give a valid A that satisfies the condition. Therefore, the total number of sets A is equal to the number of subsets of {4, 5}, which is 4. So, the answer is 4. But wait, just to be absolutely thorough, let me check again. If the original set is {1, 2, 3, 4, 5}, then all subsets A that contain {1, 2, 3} can be formed by taking {1, 2, 3} and adding any combination of the remaining elements, which are 4 and 5. So, the possible additions are:- Add nothing: {1, 2, 3}- Add 4: {1, 2, 3, 4}- Add 5: {1, 2, 3, 5}- Add both 4 and 5: {1, 2, 3, 4, 5}Yes, that's four sets. Each corresponds to a different subset of {4, 5}. Therefore, the answer is indeed 4. Alternatively, using the formula for the number of subsets between two sets: If we have two sets T and S where T is a subset of S, then the number of subsets A such that T ⊆ A ⊆ S is equal to 2^(|S| - |T|). Here, |S| is 5 and |T| is 3, so 2^(5 - 3) = 4. That's another way to arrive at the same answer. I think this is solid. So the answer should be 4. Therefore, the number of sets A is 4. Final AnswerThe number of sets ( A ) is boxed{4}.</think>To determine the number of sets ( A ) that satisfy ({1, 2, 3} subseteq A subseteq {1, 2, 3, 4, 5}), we need to consider the elements that can be included or excluded from the set ( A ). The set ( A ) must include the elements 1, 2, and 3. The elements 4 and 5 can be either included or excluded. Each of these two elements (4 and 5) has two choices: either being in the set or not. Therefore, the number of possible combinations is ( 2^2 = 4 ).Listing the possible sets:1. ( A = {1, 2, 3} )2. ( A = {1, 2, 3, 4} )3. ( A = {1, 2, 3, 5} )4. ( A = {1, 2, 3, 4, 5} )Each combination corresponds to a subset of ({4, 5}), and there are ( 2^2 = 4 ) such subsets. Thus, the number of sets ( A ) is (boxed{4}).,<think>Okay, so I need to find the number of sets A that satisfy {1, 2, 3} is a subset of A, which in turn is a subset of {1, 2, 3, 4, 5}. Hmm, let me think. First, let me recall what a subset means. If {1, 2, 3} is a subset of A, that means every element in {1, 2, 3} must also be in A. So A has to contain 1, 2, and 3. Then, A is also a subset of {1, 2, 3, 4, 5}, so A can't have any elements outside of 1, 2, 3, 4, 5. So, the set A must include 1, 2, 3, and it can optionally include 4 and 5. That means the elements 4 and 5 are the only ones that A might or might not have. Since each of these two elements can be either included or excluded, the number of possible sets A should be 2^2 = 4. Wait, let me check that again.The original big set is {1, 2, 3, 4, 5}. If we know that A must contain {1, 2, 3}, then the only flexibility we have is with elements 4 and 5. For each of these two elements, we can decide whether to include them in A or not. Since each decision is independent, the total number of combinations is 2 * 2 = 4. So the possible sets are:1. {1, 2, 3} (includes neither 4 nor 5)2. {1, 2, 3, 4} (includes 4)3. {1, 2, 3, 5} (includes 5)4. {1, 2, 3, 4, 5} (includes both 4 and 5)So that's four sets. Therefore, the answer should be 4. But wait, let me verify this with another approach.Another way to think about it is the number of subsets of {4, 5} because those are the elements that can vary. The power set of {4, 5} has 2^2 = 4 elements. Each subset of {4, 5} combined with {1, 2, 3} will give a unique set A that satisfies the given conditions. For example, the empty subset of {4, 5} combined with {1, 2, 3} is {1, 2, 3}, and the subset {4} combined with {1, 2, 3} is {1, 2, 3, 4}, and so on. Yes, this seems correct. Alternatively, if the original set had n elements and we had to include k elements, then the number of subsets would be 2^(n - k). In this case, n is 5 (elements 1 to 5) and k is 3 (elements 1, 2, 3). Therefore, the number of subsets is 2^(5 - 3) = 2^2 = 4. So regardless of the method I use, I get 4. Let me just make sure I didn't miss any elements. The original problem says {1, 2, 3} is a subset of A, so A must have all three elements. Then A can have any combination of 4 and 5. Since there are two elements that can be included or excluded, it's 2^2. I think that's solid. Maybe another example would help. Suppose instead the problem was {1} ⊆ A ⊆ {1, 2}. Then the number of sets A would be 2^(2 - 1) = 2. Indeed, the sets would be {1} and {1, 2}. That works. So applying the same logic here, 2^(5 - 3) = 4. Yes, I think the answer is 4.Final AnswerThe number of sets ( A ) is boxed{4}.</think>To find the number of sets ( A ) that satisfy ({1, 2, 3} subseteq A subseteq {1, 2, 3, 4, 5}), we start by noting that ( A ) must include the elements 1, 2, and 3. The elements 4 and 5 can be either included or excluded from ( A ).The problem reduces to finding the number of subsets of the set ({4, 5}), since each subset of ({4, 5}) can be combined with ({1, 2, 3}) to form a valid set ( A ). The number of subsets of a set with ( n ) elements is ( 2^n ). Here, the set ({4, 5}) has 2 elements, so the number of subsets is ( 2^2 = 4 ).Thus, the number of sets ( A ) is (boxed{4}).]

🔑:Solution: The sets A that satisfy {1, 2, 3} subseteq A subseteq {1, 2, 3, 4, 5} are: {1, 2, 3}; {1, 2, 3, 4}; {1, 2, 3, 5}; {1, 2, 3, 4, 5}. There are a total of 4 sets. Therefore, the answer is: boxed{4}. By using the given conditions, we can list the sets A. This question tests the method of finding the number of subsets of a set and computational skills.

🧠:[<think>Okay, let me try to figure out this problem. So, we have an odd function f(x) with a period of 5 defined on all real numbers. It also says that f(1) = 8. The question is asking for f(2010) - f(2009), and the options are 6, 7, 8, or 9. Hmm, alright, let's break this down step by step.First, let's recall what an odd function is. An odd function satisfies the property that f(-x) = -f(x) for all x in its domain. So, if I know the value of the function at some point x, I can find it at -x by just taking the negative. That might come in handy here.Next, the function has a period of 5. That means f(x + 5) = f(x) for all real numbers x. So, every 5 units, the function repeats its values. That should help in simplifying f(2010) and f(2009) by subtracting multiples of 5 until we get within a single period.So, the problem is asking for f(2010) - f(2009). Let me start by reducing these arguments modulo 5 to find equivalent values within one period. Since the period is 5, f(2010) is the same as f(2010 mod 5), and similarly for f(2009).Let's compute 2010 divided by 5. 5 times 402 is 2010, so 2010 divided by 5 is exactly 402 with no remainder. Therefore, 2010 mod 5 is 0. That means f(2010) = f(0).Now, for 2009 divided by 5: 5 times 401 is 2005, so 2009 - 2005 = 4. Therefore, 2009 mod 5 is 4. So, f(2009) = f(4).Therefore, the expression f(2010) - f(2009) simplifies to f(0) - f(4).Now, we need to find f(0) and f(4). Let's see what we know. We know that f is an odd function. An odd function has the property that f(0) = 0, because f(-0) = -f(0), but -0 is 0, so f(0) = -f(0), which implies f(0) = 0. So, that's straightforward. So f(0) is 0.So now we have f(2010) - f(2009) = 0 - f(4) = -f(4). Therefore, we just need to find f(4), and then take its negative.How can we find f(4)? Well, we know the function is periodic with period 5, so f(4) is the same as f(4 - 5) because subtracting the period. So f(4) = f(-1). But since the function is odd, f(-1) = -f(1). And we know f(1) = 8. Therefore, f(-1) = -8, so f(4) = -8.Therefore, substituting back into our expression, -f(4) = -(-8) = 8.Wait, so f(2010) - f(2009) equals 8? Let me check that again to make sure I didn't make a mistake.First, 2010 divided by 5 is 402, remainder 0. So f(2010) = f(0) = 0. Then 2009 divided by 5 is 401*5 = 2005, remainder 4. So f(2009) = f(4). Then f(4) = f(-1) because 4 = -1 + 5 (since the period is 5). Then since it's odd, f(-1) = -f(1) = -8. Therefore, f(4) = -8. Then f(2010) - f(2009) = 0 - (-8) = 8. So the answer is 8, which is option C.But let me double-check everything. Sometimes when dealing with periods and odd functions, it's easy to get mixed up with the signs.So, starting over:1. The function has period 5, so f(x + 5n) = f(x) for any integer n.2. 2010 divided by 5 is 402 with remainder 0, so f(2010) = f(0).3. 2009 divided by 5 is 401*5 + 4, so remainder 4, so f(2009) = f(4).4. Since the function is odd, f(0) = 0.5. Now, f(4): since 4 is in the period from 0 to 5. Let's see if we can relate 4 to a value where we know the function. Since the function is odd, if we can express 4 as -1 + 5, which is 4. So f(4) = f(-1 + 5) = f(-1). But because the function is periodic with period 5, f(-1) is the same as f(-1 + 5) = f(4). Wait, that's redundant. Let me think.Alternatively, since the function is odd, f(-x) = -f(x). So if we can write 4 as -1 mod 5? Wait, 4 is congruent to -1 modulo 5 because 4 + 1 = 5. So 4 ≡ -1 mod 5. Therefore, f(4) = f(-1). But since the function is odd, f(-1) = -f(1). And f(1) is given as 8, so f(-1) = -8. Therefore, f(4) = -8. Therefore, f(2009) = -8.So then f(2010) - f(2009) is 0 - (-8) = 8. So the answer is 8, which is C.But let me check if there's another way to see this. Maybe by expressing 4 as a different number within the period. For example, 4 is in the period [0,5). Since the function is odd, we can relate f(4) to f(-4). But since the function is periodic, f(-4) = f(-4 + 5) = f(1). Wait, so f(-4) = f(1). But f(-4) is also equal to -f(4) because it's odd. Therefore, f(-4) = -f(4) = f(1). Therefore, -f(4) = 8, so f(4) = -8. So that's another way to see the same thing. So this confirms that f(4) = -8. Therefore, the difference is 0 - (-8) = 8. So yes, answer C.I think that makes sense. So the answer should be 8. Let me check if I considered all possibilities. The function is odd, so symmetric about the origin, but periodic with period 5. So for any x, f(x + 5) = f(x), and f(-x) = -f(x). So combining these properties, we can shift the argument by multiples of 5 to bring it within a known interval, and then use the oddness to relate it to a known value. Since we know f(1) = 8, we can use that to find f(-1) = -8, and then using periodicity, f(4) = f(-1) = -8. Then the difference is 0 - (-8) = 8. That seems correct.Therefore, the answer is C) 8.Final Answerboxed{C}</think>Given that ( f(x) ) is an odd function with a period of 5 and ( f(1) = 8 ), we need to find ( f(2010) - f(2009) ).1. Periodicity: Since ( f(x) ) has a period of 5, we can reduce the arguments modulo 5: - ( 2010 mod 5 = 0 ), so ( f(2010) = f(0) ). - ( 2009 mod 5 = 4 ), so ( f(2009) = f(4) ).2. Odd Function Property: - For ( f(0) ), since ( f ) is odd, ( f(0) = 0 ). - For ( f(4) ), using periodicity, ( f(4) = f(4 - 5) = f(-1) ). By the odd function property, ( f(-1) = -f(1) ). Given ( f(1) = 8 ), we get ( f(-1) = -8 ). Therefore, ( f(4) = -8 ).3. Calculate the Difference: - ( f(2010) - f(2009) = f(0) - f(4) = 0 - (-8) = 8 ).Thus, the answer is (boxed{C}).,<think>Okay, let's try to figure out this problem step by step. The question is about an odd function ( f(x) ) with a period of 5 defined on all real numbers. We know that ( f(1) = 8 ), and we need to find ( f(2010) - f(2009) ). The options are A through D, with numbers 6 to 9. Alright, let's start by recalling what it means for a function to be odd and periodic.First, an odd function satisfies the property ( f(-x) = -f(x) ) for all ( x ). So, if I know the value of the function at a positive point, I can find it at the corresponding negative point by just changing the sign. For example, ( f(-1) = -f(1) = -8 ).Next, the function has a period of 5. That means ( f(x + 5) = f(x) ) for all ( x ). So the function repeats every 5 units. This is key because we can use this periodicity to reduce the arguments 2010 and 2009 to something within one period, which might be easier to handle.So, let's start by simplifying ( f(2010) ) and ( f(2009) ). Since the period is 5, we can subtract multiples of 5 from these numbers until we get a value within the interval [0,5), right? Let's compute 2010 divided by 5 to find how many periods are in 2010.2010 divided by 5 is 402 with no remainder, because 402*5=2010. So, 2010 is exactly 402 periods away from 0. Therefore, ( f(2010) = f(2010 - 5*402) = f(0) ).Wait, but hold on. If the function has period 5, then subtracting 5*402 from 2010 gives 0. So ( f(2010) = f(0) ). But what is ( f(0) )? Since the function is odd, ( f(-0) = -f(0) ), but -0 is 0, so ( f(0) = -f(0) ). The only number that is equal to its own negative is 0. Therefore, ( f(0) = 0 ). So ( f(2010) = 0 ).Now let's do the same for ( f(2009) ). 2009 divided by 5: 5*401 = 2005, so 2009 - 2005 = 4. Therefore, ( f(2009) = f(4) ). So now we need to find ( f(4) ).But we only know ( f(1) = 8 ). How can we relate ( f(4) ) to ( f(1) )? Since the function is odd and periodic, maybe we can find a relation using these properties.Let me think. Since the function is periodic with period 5, ( f(4) = f(4 - 5) = f(-1) ). But the function is odd, so ( f(-1) = -f(1) = -8 ). Therefore, ( f(4) = -8 ). Thus, ( f(2009) = -8 ).Therefore, ( f(2010) - f(2009) = 0 - (-8) = 8 ). Hmm, 8 is option C. But wait, let me double-check my steps because sometimes with periodic functions and odd functions, there might be something I missed.First, checking the periodicity: yes, period 5 implies that values repeat every 5 units. So subtracting 5*402 from 2010 gives 0, and 5*401 from 2009 gives 4. So that's correct.Then, ( f(4) = f(4 - 5) = f(-1) ). That uses the periodicity to shift the argument by one period, which is valid. Then, since the function is odd, ( f(-1) = -f(1) = -8 ). That seems right. Therefore, ( f(4) = -8 ). Then, ( f(2010) = 0 ), so the difference is 0 - (-8) = 8. That would be answer C. But let me check if there's another way to look at this.Alternatively, maybe using other properties. Let's see. If the function is odd and periodic, perhaps there are symmetries over the period. For example, let's consider that since the function is periodic with period 5, then ( f(x + 5) = f(x) ). So, if we can express 2010 and 2009 in terms of their distance from multiples of 5.Wait, 2010 is a multiple of 5, as we saw, so it's 5*402. Therefore, ( f(2010) = f(0) = 0 ). Then 2009 is 5*401 + 4, so ( f(2009) = f(4) ). Then, ( f(4) = f(-1) ) because 4 is equivalent to -1 modulo 5 (since 4 -5 = -1). Then, since it's odd, ( f(-1) = -f(1) = -8 ). So yes, that still gives us 8. So, 0 - (-8) = 8. So answer C.Wait, but let me check if there's another property or maybe a mistake in assuming something about the function's behavior. For example, sometimes when functions are periodic and odd, they have certain symmetries over the period. Let me think.If a function is odd, then it's symmetric about the origin. If it's also periodic with period 5, then it's symmetric about the origin and repeats every 5 units. So, for example, ( f(5 - x) = f(-x) = -f(x) ). Wait, is that correct?Wait, let's see. Let's take ( f(5 - x) ). Since the function has period 5, ( f(5 - x) = f(-x) ). But since it's odd, ( f(-x) = -f(x) ). Therefore, ( f(5 - x) = -f(x) ). So this gives a relation between ( f(5 - x) ) and ( f(x) ). So, for example, if x=1, then ( f(5 - 1) = f(4) = -f(1) = -8 ). Which is exactly what we found earlier. So that's consistent.Alternatively, if x=4, then ( f(5 - 4) = f(1) = -f(4) ), which gives ( f(1) = -f(4) ), so ( f(4) = -f(1) = -8 ). So that's the same result. So that seems to check out.Therefore, the answer is 8, which is option C.But wait, let me check another example. For instance, take x=2. Then ( f(5 - 2) = f(3) = -f(2) ). So ( f(3) = -f(2) ). But since we don't know f(2) or f(3), maybe we can't get more info. However, in our problem, we only need f(4) and f(0), which we already determined.Therefore, I think the answer is 8. So, option C.Final Answerboxed{C}</think>Given that ( f(x) ) is an odd function with a period of 5, we need to find ( f(2010) - f(2009) ) given ( f(1) = 8 ).1. Periodicity: Since ( f(x) ) has a period of 5, we can reduce the arguments modulo 5. - For ( f(2010) ): ( 2010 div 5 = 402 ) with remainder 0. Thus, ( f(2010) = f(0) ). - For ( f(2009) ): ( 2009 div 5 = 401 ) with remainder 4. Thus, ( f(2009) = f(4) ).2. Odd Function Property: - Since ( f(x) ) is odd, ( f(-x) = -f(x) ). Therefore, ( f(0) = 0 ) because ( f(0) = -f(0) ).3. Reducing ( f(4) ) using Periodicity: - ( f(4) = f(4 - 5) = f(-1) ). - Using the odd function property, ( f(-1) = -f(1) = -8 ).4. Calculating the Difference: - ( f(2010) = f(0) = 0 ) - ( f(2009) = f(4) = -8 ) - Therefore, ( f(2010) - f(2009) = 0 - (-8) = 8 )The final answer is (boxed{C}).]

🔑:- 1. C.Given that f(x) is an odd function with a period of 5 on mathbf{R}, thenbegin{array}{l}f(2010)-f(2009)=f(0)-f(-1) =f(0)+f(1)=8 .end{array}

🧠:[<think>Okay, so I need to solve this problem where we have two positive integers a and b such that both 15a + 16b and 16a - 15b are perfect squares. And we need to find the smallest possible value of the smaller of these two squares. Hmm, let's start by understanding the problem step by step.First, let's note that both 15a + 16b and 16a - 15b are perfect squares. Let me denote them as:Let’s set:15a + 16b = m² ...(1)16a - 15b = n² ...(2)where m and n are positive integers, and we need to find the smallest possible value of the smaller one between m² and n².Since a and b are positive integers, let's check the equations. From equation (2), 16a - 15b must be positive because it's a square of a positive integer. So, 16a > 15b. Therefore, a must be greater than (15/16)b. Similarly, from equation (1), 15a + 16b is a square, so since a and b are positive, m² is at least 15*1 + 16*1 = 31. So m must be at least 6 (since 5²=25 is less than 31 and 6²=36 is 36). Similarly, n² must be at least 16*1 -15*1 = 1, so n can be 1 or more. But since a and b need to be positive, n can't be 1 if a and b are to be integers. Wait, maybe n can be 1? Let's see.But let's think of how to approach this. Since we have two equations with two variables, maybe we can solve for a and b in terms of m and n, and then find constraints on m and n such that a and b are positive integers.So, let me try to solve equations (1) and (2) for a and b. Let's write them again:15a + 16b = m² ...(1)16a - 15b = n² ...(2)Let me try to solve this system. Let's multiply equation (1) by 16 and equation (2) by 15, so that the coefficients of a will be 15*16 and 16*15, which are the same. That way, we can subtract the equations to eliminate a.Wait, but actually, let me try to use linear algebra. Let's set up the equations:Equation (1): 15a + 16b = m²Equation (2): 16a - 15b = n²Let me arrange them in a matrix form:[15 16][a] [m²][16 -15][b] = [n²]To solve for a and b, we can compute the inverse of the coefficient matrix, provided that the determinant is non-zero. The determinant of the matrix [15, 16; 16, -15] is (15)(-15) - (16)(16) = -225 - 256 = -481. Since the determinant is -481, which is non-zero, the inverse exists.Therefore, the solution is:[a] (1/-481) [ -15 -16 ][m²][b] = [ -16 15 ][n²]Wait, let's compute that more carefully. The inverse of a 2x2 matrix [a b; c d] is (1/det)[d -b; -c a]. So here, the inverse is (1/-481)[ -15 -16; -16 15 ].Therefore,a = ( -15 * m² - 16 * n² ) / (-481 )b = ( -16 * m² + 15 * n² ) / (-481 )Simplify these:a = (15m² + 16n²)/481b = (16m² - 15n²)/481Since a and b must be positive integers, both 15m² + 16n² and 16m² - 15n² must be divisible by 481, and the results must be positive integers.So, we have:15m² + 16n² ≡ 0 mod 48116m² - 15n² ≡ 0 mod 481Also, 16m² - 15n² must be positive, so 16m² > 15n².Therefore, the problem reduces to finding positive integers m and n such that:1) 15m² + 16n² ≡ 0 mod 4812) 16m² - 15n² ≡ 0 mod 4813) 16m² > 15n²Additionally, a = (15m² +16n²)/481 and b = (16m² -15n²)/481 must be positive integers.So, perhaps we can consider these two congruence conditions. Let me see:From 15m² + 16n² ≡ 0 mod 481 and 16m² -15n² ≡ 0 mod 481.Let me denote x = m² and y = n². Then, we have:15x + 16y ≡ 0 mod 48116x -15y ≡ 0 mod 481So, solving this system modulo 481. Let's write this as:15x +16y ≡ 0 ...(1)16x -15y ≡ 0 ...(2)Let me solve for x and y. Let's multiply equation (1) by 16 and equation (2) by 15:15*16x +16*16y ≡ 016*15x -15*15y ≡ 0Which simplifies to:240x + 256y ≡ 0240x - 225y ≡ 0Subtract the second equation from the first:(240x +256y) - (240x -225y) = 0 - 0 => 481y ≡ 0 mod 481Which is 481y ≡0 mod 481. Since 481 divides 481y, this holds for any y. Similarly, if we subtract in a different way, perhaps.Alternatively, let me solve equations (1) and (2) for x and y.From equation (1): 15x ≡ -16y mod 481From equation (2): 16x ≡ 15y mod 481Let me express x from equation (1):x ≡ (-16/15)y mod 481But fractions modulo 481 require finding the inverse of 15 mod 481. Let me compute the inverse of 15 modulo 481.We need to find an integer k such that 15k ≡ 1 mod 481.Using the extended Euclidean algorithm:481 = 32*15 + 1 (since 32*15 = 480, so 481 - 480 =1)Therefore, 1 = 481 - 32*15Hence, inverse of 15 mod 481 is -32 mod 481, which is 481 -32 = 449.Therefore, inverse of 15 mod 481 is 449.So, x ≡ -16 * 449 * y mod 481Compute -16*449:First, compute 16*449:16*400=6400, 16*49=784, so 6400+784=7184So, -16*449 = -7184 mod 481Compute 7184 divided by 481:481*14 = 67347184 - 6734 = 450So, 7184 ≡ 450 mod 481Thus, -7184 ≡ -450 mod 481 ≡ 481 -450 = 31 mod 481Therefore, x ≡ 31y mod 481So, x ≡ 31y mod 481Similarly, from equation (2): 16x ≡15y mod 481But x ≡31y mod481, so substituting:16*31y ≡15y mod481Compute 16*31 = 496So, 496y ≡15y mod481Then, 496y -15y =481y ≡0 mod481, which holds true. So, consistent.Therefore, the solution is x ≡31y mod481.Since x and y are squares, let me think. x = m², y = n². So, m² ≡31n² mod481.Therefore, (m/n)² ≡31 mod481, assuming n is invertible mod481. Since 481 is 13*37. So, 481 is composite. So, if n is coprime with 481, then we can write (m/n)² ≡31 mod481. Therefore, 31 must be a quadratic residue modulo 481. Since 481 =13*37, 31 must be a quadratic residue modulo 13 and modulo37.Let me check if 31 is a quadratic residue modulo13 and modulo37.First modulo13:31 mod13= 31-2*13=5. So, 31≡5 mod13. So, is 5 a quadratic residue modulo13?The quadratic residues modulo13 are 1,4,9,3,12,10,10,12,3,9,4,1. Wait, let me compute them properly:Squares modulo13: 0²=0, 1=1, 2=4, 3=9, 4=16≡3, 5=25≡12, 6=36≡10, 7=49≡10, 8=64≡12, 9=81≡3, 10=100≡9, 11=121≡4, 12=144≡1.So quadratic residues modulo13 are 0,1,3,4,9,10,12. 5 is not a quadratic residue modulo13. Therefore, since 31≡5 mod13, and 5 is not a quadratic residue modulo13, 31 is not a quadratic residue modulo481. Therefore, the equation m² ≡31n² mod481 has solutions only if n is divisible by both 13 and 37, i.e., n is divisible by 481. Wait, but maybe not exactly. Let's think again.Since 31 is a quadratic residue modulo481 only if it's a quadratic residue modulo13 and modulo37. But since it's not a quadratic residue modulo13, it cannot be a quadratic residue modulo481. Therefore, the congruence m² ≡31n² mod481 implies that 31n² is a quadratic residue modulo481. But since 31 is a non-residue modulo13, and 13 divides 481, this would require n² ≡0 mod13, so n≡0 mod13, and similarly mod37.Wait, perhaps. Let me think. Let's factor 481=13*37. So, by the Chinese Remainder Theorem, solving m² ≡31n² mod13 and mod37.First, modulo13:m² ≡31n² mod13 ⇒ m² ≡5n² mod13.Since 5 is not a quadratic residue mod13, as we saw earlier. Therefore, unless n ≡0 mod13, which would make 5n²≡0 mod13, then m must also be ≡0 mod13. Similarly, m ≡0 mod13. So, if n ≡0 mod13, then 5n² ≡0 mod13, so m²≡0 mod13, so m≡0 mod13. Similarly, modulo37.Let’s check modulo37:31 mod37=31. Is 31 a quadratic residue modulo37?Compute the Legendre symbol (31/37). Since both are primes, using quadratic reciprocity:(31/37) = (37/31) * (-1)^[(31-1)(37-1)/4] = (6/31) * (-1)^(30*36/4) = (6/31)*(-1)^270 = (6/31)*(-1)^270. Since 270 is even, (-1)^270=1. Now, (6/31) = (2/31)*(3/31).Compute (2/31): Since 31≡7 mod8, so (2/31)=-1.Compute (3/31): Using reciprocity, (3/31) = (31/3)*(-1)^[(3-1)(31-1)/4] = (1/3)*(-1)^(2*30/4) = (1/3)*(-1)^15 = (1/3)*(-1). Since 31≡1 mod3, (1/3)=1. Therefore, (3/31)= -1. Therefore, (6/31)= (-1)*(-1)=1. Therefore, (31/37)=1. Therefore, 31 is a quadratic residue modulo37.So, modulo37, 31 is a quadratic residue, so the equation m²≡31n² mod37 can be written as (m/n)²≡31 mod37, which is possible if m/n ≡ some root of 31 mod37. So, there are solutions for m and n modulo37, provided that n is not 0 mod37. But if n≡0 mod37, then m must also be ≡0 mod37.So, putting together:For the equation m² ≡31n² mod481, since 31 is a non-residue mod13 and a residue mod37, the congruence holds if and only if n≡0 mod13 and m≡0 mod13, and m ≡k*n mod37 for some k where k²≡31 mod37.Therefore, m and n must both be divisible by13, and m ≡k*n mod37, where k is a square root of31 mod37.So, to find solutions, we can set n=13*37*t=481*t, but that would make n very large. Wait, but maybe n must be divisible by13 but not necessarily by37. Wait, let's analyze again.Since modulo13, m² ≡5n² mod13. Since 5 is a non-residue, the only solution is n≡0 mod13 and m≡0 mod13.Similarly, modulo37, 31 is a quadratic residue, so there exist non-trivial solutions where n is not 0 mod37. Therefore, overall, n must be divisible by13, and m must be divisible by13. Let’s set n=13n', m=13m'. Then, substituting into the equation m² ≡31n² mod481:(13m')² ≡31*(13n')² mod481169m'² ≡31*169n'² mod481Divide both sides by169. Since 169 and481: 481=13*37, and 169=13², so gcd(169,481)=13. Therefore, dividing both sides by13:13m'² ≡31*13n'² mod37Wait, maybe this is getting too complicated. Let me try another approach.Since n must be divisible by13, let’s set n=13k. Similarly, m must be divisible by13, set m=13j. Then, substituting into the equation:m² ≡31n² mod481(13j)² ≡31*(13k)^2 mod481169j² ≡31*169k² mod481Divide both sides by13:13j² ≡31*13k² mod37Wait, 481/13=37. So, mod37:13j² ≡31*13k² mod37Since 13≡13 mod37 and 31≡31 mod37.Multiply both sides by the inverse of13 mod37. The inverse of13 mod37: find x such that13x≡1 mod37. 13*26=338≡338-9*37=338-333=5≡5 mod37. 13*17=221≡221-5*37=221-185=36≡-1 mod37. So, 13*17≡-1, so 13*(-17)≡1 mod37. Since -17 mod37 is20. So inverse of13 mod37 is20.Therefore, multiply both sides by20:j² ≡31*13*20*k² mod37Calculate31*13*20 mod37:First, compute31*13=403≡403-10*37=403-370=33 mod37.33*20=660≡660-17*37=660-629=31 mod37.Therefore, j²≡31*k² mod37So, (j/k)²≡31 mod37. Since31 is a quadratic residue mod37, as established earlier, there are solutions. Let's find sqrt(31) mod37.Find x such thatx²≡31 mod37. Let's test some numbers:Compute 16²=256≡256-6*37=256-222=34≡34 mod37. 17²=289≡289-7*37=289-259=30. 18²=324≡324-8*37=324-296=28. 19²=361≡361-9*37=361-333=28. Wait, same as 18? Wait, 19²=361. 361-9*37=361-333=28. 20²=400≡400-10*37=400-370=30. 21²=441≡441-11*37=441-407=34. 22²=484≡484-13*37=484-481=3. 23²=529≡529-14*37=529-518=11. 24²=576≡576-15*37=576-555=21. 25²=625≡625-16*37=625-592=33. 26²=676≡676-18*37=676-666=10. 27²=729≡729-19*37=729-703=26. 28²=784≡784-21*37=784-777=7. 29²=841≡841-22*37=841-814=27. 30²=900≡900-24*37=900-888=12. 31²=961≡961-25*37=961-925=36. 32²=1024≡1024-27*37=1024-999=25. 33²=1089≡1089-29*37=1089-1073=16. 34²=1156≡1156-31*37=1156-1147=9. 35²=1225≡1225-33*37=1225-1221=4. 36²=1296≡1296-35*37=1296-1295=1. So, none of these squares are 31 mod37. Wait, perhaps I made a mistake.Wait, perhaps it's better to use the fact that 31≡-6 mod37. So, looking for x such thatx²≡-6 mod37. Alternatively, maybe I missed some numbers.Wait, let me check 15²=225≡225-6*37=225-222=3 mod37. 14²=196≡196-5*37=196-185=11. 13²=169≡169-4*37=169-148=21. 12²=144≡144-3*37=144-111=33. 11²=121≡121-3*37=121-111=10. 10²=100≡100-2*37=100-74=26. 9²=81≡81-2*37=81-74=7. 8²=64≡64-1*37=27. 7²=49≡49-1*37=12. 6²=36≡36. 5²=25. 4²=16. 3²=9. 2²=4. 1²=1. 0²=0. Hmm, still not finding 31. Wait, maybe my approach is wrong.Wait, but we know that 31 is a quadratic residue modulo37, as established earlier. So, there must be a solution. Maybe using a different method. Let's use Tonelli-Shanks algorithm or something. But perhaps trial and error is too tedious here.Alternatively, maybe we can note that since 31≡-6 mod37, and we can look for x such thatx²≡-6 mod37. Let's try x=16: 16²=256≡256-6*37=256-222=34≡-3 mod37. Not -6. x=17: 17²=289≡289-7*37=289-259=30≡-7 mod37. x=18: 18²=324≡324-8*37=324-296=28≡-9. x=19: 19²=361≡361-9*37=361-333=28≡-9. x=20: 20²=400≡400-10*37=400-370=30≡-7. x=21: 21²=441≡441-11*37=441-407=34≡-3. x=22: 22²=484≡484-13*37=484-481=3. x=23:23²=529≡529-14*37=529-518=11. x=24:24²=576≡576-15*37=576-555=21. x=25:25²=625≡625-16*37=625-592=33≡-4. x=26:26²=676≡676-18*37=676-666=10. x=27:27²=729≡729-19*37=729-703=26≡-11. x=28:28²=784≡784-21*37=784-777=7. x=29:29²=841≡841-22*37=841-814=27≡-10. x=30:30²=900≡900-24*37=900-888=12. x=31:31²=961≡961-25*37=961-925=36≡-1. x=32:32²=1024≡1024-27*37=1024-999=25. x=33:33²=1089≡1089-29*37=1089-1073=16. x=34:34²=1156≡1156-31*37=1156-1147=9. x=35:35²=1225≡1225-33*37=1225-1221=4. x=36:36²=1296≡1296-35*37=1296-1295=1. Still no x with x²≡31 mod37. Wait, maybe I made a mistake in the quadratic reciprocity part.Earlier, we said that (31/37)=1, but according to the computations, there's no x such thatx²≡31 mod37. Hmm, that's a contradiction. Let's check again.Using quadratic reciprocity:(31/37) = (37/31)*(-1)^{(31-1)(37-1)/4} = (6/31)*(-1)^{(30)(36)/4} = (6/31)*(-1)^{270}.Compute (-1)^{270} = (-1)^{even} = 1. So, (31/37) = (6/31). Then, (6/31) = (2/31)(3/31).(2/31): Since 31 ≡ 7 mod8, so (2/31) = -1.(3/31): Since 31 ≡1 mod3, (3/31) = (1/3) =1. Wait, but quadratic reciprocity says (p/q)(q/p) = (-1)^{(p-1)(q-1)/4}. So, (3/31) = (31/3)*(-1)^{(3-1)(31-1)/4} = (1/3)*(-1)^{(2)(30)/4} = (1/3)*(-1)^{15} = (1/3)*(-1) = -1. Wait, here's a mistake. Earlier, I thought (3/31)= -1, but here's the correct computation:(3/31) = (31/3)*(-1)^{(3-1)(31-1)/4} = (1/3)*(-1)^{(2)(30)/4} = (1/3)*(-1)^{15} = (1/3)*(-1). Since 31 ≡1 mod3, (1/3)=1. Therefore, (3/31)=1*(-1)= -1. So, (6/31)= (-1)*(-1)=1. Therefore, (31/37)=1. Therefore, 31 is a quadratic residue modulo37, which contradicts our trial. So, there must be a solution. Let's try x=10:10²=100≡26 mod37. x=15:15²=225≡3 mod37. x=16:256≡34. x=17:289≡30. x=18:324≡28. x=19:361≡28. x=20:400≡30. x=21:441≡34. x=22:484≡3. x=23:529≡11. x=24:576≡21. x=25:625≡33. x=26:676≡10. x=27:729≡26. x=28:784≡7. x=29:841≡27. x=30:900≡12. x=31:961≡36. x=32:1024≡25. x=33:1089≡16. x=34:1156≡9. x=35:1225≡4. x=36:1296≡1. Hmm, none of these are 31. So, maybe my mistake was in the quadratic reciprocity step.Wait, let's compute (31/37) again.(31/37) = (37/31)*(-1)^{(31-1)(37-1)/4} = (6/31)*(-1)^{(30)(36)/4} = (6/31)*(-1)^{270}. Since 270 is even, (-1)^{270}=1. Then, (6/31) = (2/31)(3/31). (2/31)= -1 (since31≡7 mod8), (3/31)= (31 mod3=1, so (1/3)=1, and the exponent is (3-1)(31-1)/4= (2)(30)/4=15, odd, so (3/31)= -1. Therefore, (6/31)= (-1)(-1)=1. Therefore, (31/37)=1. But in reality, 31 is not a quadratic residue modulo37? That's conflicting. Where is the mistake?Wait, no. Let's actually compute 14² mod37: 14²=196≡196-5*37=196-185=11. 11≠31. 26²=676≡676-18*37=676-666=10. 10≠31. 25²=625≡625-16*37=625-592=33. 33≡31? No. 25²≡33. 31≡-6. Maybe 16²=256≡34≡-3. 17²=289≡30≡-7. 18²=324≡28≡-9. 19²=361≡28. 20²=400≡30. 21²=441≡34. 22²=484≡3. 23²=529≡11. 24²=576≡21. 25²=625≡33. 26²=676≡10. 27²=729≡26. 28²=784≡7. 29²=841≡27. 30²=900≡12. 31²=961≡36. 32²=1024≡25. 33²=1089≡16. 34²=1156≡9. 35²=1225≡4. 36²=1296≡1. So, none of the squares mod37 give 31. Therefore, contradiction. That suggests that maybe my earlier conclusion was wrong. But according to quadratic reciprocity, (31/37)=1, which should mean that 31 is a quadratic residue modulo37. However, practical computation shows no x satisfies x²≡31 mod37. This is strange.Wait, perhaps I made a mistake in the quadratic reciprocity steps. Let me check again.(31/37) = (37/31) * (-1)^{(31-1)(37-1)/4}(37 mod31)=6, so (37/31)=(6/31)(-1)^{(30)(36)/4}= (-1)^{270}=1 because 270 is even.So, (31/37)=(6/31)= (2/31)(3/31). (2/31)=-1 as before.(3/31)= ?Using quadratic reciprocity: (3/31) = (31/3)*(-1)^{(3-1)(31-1)/4} = (1/3)*(-1)^{(2)(30)/4} = (1/3)*(-1)^{15} = (1/3)*(-1) = -1*(1/3). Since 31 ≡1 mod3, (1/3)=1. Hence, (3/31)= -1. So, (6/31)= (-1)(-1)=1. Therefore, (31/37)=1. Which suggests that 31 is a quadratic residue modulo37. However, in reality, from the computations above, there is no x such thatx²≡31 mod37. So, this is a contradiction.Wait, maybe my mistake is in the calculation of (31/37). Wait, quadratic reciprocity says that for odd primes p and q, (p/q)= (q/p)*(-1)^{(p-1)(q-1)/4}. Here, p=31, q=37. So, (31/37)= (37/31)*(-1)^{(31-1)(37-1)/4} = (6/31)*(-1)^{(30)(36)/4} = (6/31)*(-1)^{270}. 270 is 30*36/4=270, yes. 270 is even, so (-1)^{270}=1. So, (31/37)= (6/31). Then, (6/31)= (2*3/31)= (2/31)(3/31)= (-1)(-1)=1. So, (31/37)=1. Therefore, 31 is a quadratic residue modulo37, but my trial shows otherwise. There must be a solution. Let me try x= 9: 9²=81≡81-2*37=7≡7 mod37. x=10:10²=100≡26. x=11:121≡10. x=12:144≡33. x=13:169≡21. x=14:196≡11. x=15:225≡3. x=16:256≡34. x=17:289≡30. x=18:324≡28. x=19:361≡28. x=20:400≡30. x=21:441≡34. x=22:484≡3. x=23:529≡11. x=24:576≡21. x=25:625≡33. x=26:676≡10. x=27:729≡26. x=28:784≡7. x=29:841≡27. x=30:900≡12. x=31:961≡36. x=32:1024≡25. x=33:1089≡16. x=34:1156≡9. x=35:1225≡4. x=36:1296≡1. x=37:1369≡0. Still no 31. Wait, maybe the mistake is in my QR computation. Let's check another way.Suppose 31 is a quadratic residue modulo37. Then, there exists some x with x²=31+37k. Let me try k=1:31+37=68. sqrt(68)=8.246, not integer. k=2:31+74=105. sqrt(105)=10.24. k=3:31+111=142. sqrt≈11.92. k=4:31+148=179. sqrt≈13.37. k=5:210+31=241. sqrt≈15.52. k=6:31+222=253. sqrt≈15.90. k=7:31+259=290. sqrt≈17.02. k=8:31+296=327. sqrt≈18.08. k=9:31+333=364. sqrt≈19.08. k=10:31+370=401. sqrt≈20.02. k=11:31+407=438. sqrt≈20.92. k=12:31+444=475. sqrt≈21.79. k=13:31+481=512. sqrt=22.627. Not integer. k=14:31+518=549. sqrt≈23.43. k=15:31+555=586. sqrt≈24.21. k=16:31+592=623. sqrt≈24.95. k=17:31+629=660. sqrt≈25.69. k=18:31+666=697. sqrt≈26.4. k=19:31+703=734. sqrt≈27.09. k=20:31+740=771. sqrt≈27.77. Hmm, none of these are perfect squares. So, perhaps quadratic reciprocity gave a wrong result? Or maybe I need to check higher k? But this seems impossible. Wait, perhaps 31 is not a quadratic residue modulo37, contradicting QR. But that can't be. QR is a theorem. Therefore, my mistake must be somewhere else.Wait, perhaps I messed up the QR formula. Let me check again. The Law of Quadratic Reciprocity states that for distinct odd primes p and q:(p/q) = (q/p) * (-1)^{(p-1)(q-1)/4}So, (31/37) = (37/31) * (-1)^{(31-1)(37-1)/4} = (37 mod31 /31) * (-1)^{(30)(36)/4} = (6/31) * (-1)^{270} = (6/31)*1 = (6/31)Then, (6/31) = (2/31)(3/31). As before, (2/31) = -1 (since 31 ≡7 mod8). (3/31):Using QR again for (3/31):(3/31) = (31/3)*(-1)^{(3-1)(31-1)/4} = (1/3)*(-1)^{(2)(30)/4} = (1/3)*(-1)^{15} = (1/3)*(-1) = -1*(1/3). Since 31 ≡1 mod3, (1/3)=1. Therefore, (3/31)= -1. Hence, (6/31)= (-1)(-1)=1. Therefore, (31/37)=1. So, according to QR, 31 is a quadratic residue mod37, but in practice, up to k=20, I don't see any x²=31+37k. Maybe it's a very large x. Let's try k= (x²-31)/37 must be integer. Let's suppose x=25: x²=625. 625-31=594. 594/37≈16.05. Not integer. x=26:676-31=645. 645/37≈17.43. x=27:729-31=698. 698/37=18.86. x=28:784-31=753. 753/37≈20.35. x=29:841-31=810. 810/37≈21.89. x=30:900-31=869. 869/37=23.48. x=31:961-31=930. 930/37≈25.13. x=32:1024-31=993. 993/37≈26.83. x=33:1089-31=1058. 1058/37≈28.59. x=34:1156-31=1125. 1125/37≈30.4. x=35:1225-31=1194. 1194/37=32.27. x=36:1296-31=1265. 1265/37≈34.18. x=37:1369-31=1338. 1338/37≈36.16. x=38:1444-31=1413. 1413/37=38.18. x=39:1521-31=1490. 1490/37≈40.27. x=40:1600-31=1569. 1569/37≈42.4. x=41:1681-31=1650. 1650/37≈44.59. x=42:1764-31=1733. 1733/37≈46.84. x=43:1849-31=1818. 1818/37≈49.13. x=44:1936-31=1905. 1905/37≈51.48. x=45:2025-31=1994. 1994/37≈53.89. x=46:2116-31=2085. 2085/37≈56.35. x=47:2209-31=2178. 2178/37=58.86. x=48:2304-31=2273. 2273/37≈61.43. x=49:2401-31=2370. 2370/37≈64.05. 2370 divided by37: 37*64=2368, so 2370=37*64 +2. Not divisible. x=50:2500-31=2469. 2469/37≈66.73. x=51:2601-31=2570. 2570/37≈69.45. x=52:2704-31=2673. 2673/37=72.24. x=53:2809-31=2778. 2778/37=75.08. x=54:2916-31=2885. 2885/37=77.97. Not integer. This is going too far. There must be a mistake here. Wait, maybe I'm missing something.Alternatively, perhaps the equation m²≡31n² mod481 implies that m and n must both be multiples of 13 and37. Because since31 is a non-residue modulo13, as we saw earlier, so n must be a multiple of13, and m must be a multiple of13. Similarly, modulo37, since31 is a quadratic residue, but even if it's a quadratic residue, the equation requires that m and n are related by m ≡k*n mod37 for some k. However, perhaps the minimal solution requires that n is a multiple of37 as well. Therefore, n=13*37*t=481*t, and similarly m=13*something. But this would make the minimal solution quite large. Alternatively, maybe we can use the fact that the equations are homogeneous. Let's see.From x ≡31y mod481, where x=m² and y=n². So, m²=31n² +481k for some integer k. Since m and n are integers, we can think of this as a Diophantine equation. Alternatively, think of m and n as multiples of certain numbers. Since the previous approach is leading me in circles, perhaps it's better to think back to the original equations.Recall that a=(15m² +16n²)/481 and b=(16m² -15n²)/481. Both a and b must be positive integers. Therefore, 15m² +16n² must be divisible by481, and16m² -15n² must be divisible by481, and both results must be positive.So, 15m² +16n² ≡0 mod481 and16m² -15n² ≡0 mod481. Let me consider these two congruence equations.Let’s denote x=m² and y=n².So, 15x +16y ≡0 mod481 ...(1)16x -15y ≡0 mod481 ...(2)From equation (1): 15x ≡ -16y mod481From equation (2):16x ≡15y mod481Let me solve equation (2) for x: x ≡ (15/16)y mod481.To compute (15/16) mod481, first find the inverse of16 mod481.Using the extended Euclidean algorithm:481 divided by16 is30 with remainder1 (since16*30=480, 481-480=1).So,1=481 -16*30.Therefore, inverse of16 mod481 is-30 mod481=481-30=451.Therefore, x ≡15*451*y mod481.Compute15*451: 15*450=6750, 15*1=15, so total6750+15=6765.Now, 6765 mod481: 481*14=6734, 6765-6734=31. So, x ≡31y mod481.Therefore, x=31y +481k for some integer k.But since x and y are squares, this gives m²=31n² +481k.We need to find integers n and k such that31n² +481k is a perfect square. Hmm, challenging.Alternatively, perhaps assume k=0? Then m²=31n². But 31 is not a perfect square, so m=n=0, which contradicts positive integers. Therefore, k must be non-zero.Alternatively, set k=1: m²=31n² +481.We need to find positive integers n and m such thatm² -31n²=481. This is a Pell-type equation, but with a positive constant term. These are harder to solve. Maybe try small values ofn and see ifm²=31n² +481 is a square.Let's compute m²=31n² +481.Start withn=1: m²=31 +481=512. 512 is not a square.n=2:31*4 +481=124+481=605. Not a square.n=3:31*9+481=279+481=760. Not a square.n=4:31*16+481=496+481=977. Not a square.n=5:31*25+481=775+481=1256. Not a square.n=6:31*36+481=1116+481=1597. Not a square.n=7:31*49+481=1519+481=2000. 2000 is not a square.n=8:31*64+481=1984+481=2465. Not a square.n=9:31*81+481=2511+481=2992. Not a square.n=10:31*100+481=3100+481=3581. Not a square.n=11:31*121+481=3751+481=4232. Not a square.n=12:31*144+481=4464+481=4945. Not a square.n=13:31*169+481=5239+481=5720. Not a square.n=14:31*196+481=6076+481=6557. Not a square.n=15:31*225+481=6975+481=7456. 7456 is not a square.n=16:31*256+481=7936+481=8417. Not a square.n=17:31*289+481=8959+481=9440. Not a square.n=18:31*324+481=10044+481=10525. 10525 is 102.5² approximately. 102²=10404, 103²=10609. Not a square.n=19:31*361+481=11191+481=11672. Not a square.n=20:31*400+481=12400+481=12881. sqrt(12881)=113.47. Not an integer.n=21:31*441+481=13671+481=14152. Not a square.n=22:31*484+481=15004+481=15485. Not a square.n=23:31*529+481=16399+481=16880. Not a square.n=24:31*576+481=17856+481=18337. sqrt(18337)=135.4. Not integer.n=25:31*625+481=19375+481=19856. sqrt(19856)=140.9. Not integer.n=26:31*676+481=20956+481=21437. Not a square.n=27:31*729+481=22599+481=23080. Not a square.n=28:31*784+481=24304+481=24785. Not a square.n=29:31*841+481=26071+481=26552. Not a square.n=30:31*900+481=27900+481=28381. sqrt(28381)=168.47. Not integer.n=31:31*961+481=29791+481=30272. Not a square.n=32:31*1024+481=31744+481=32225. Wait, 32225 is 179.5². 179²=32041, 180²=32400. Not a square.n=33:31*1089+481=33759+481=34240. Not a square.n=34:31*1156+481=35836+481=36317. Not a square.n=35:31*1225+481=37975+481=38456. Not a square.n=36:31*1296+481=40176+481=40657. sqrt(40657)=201.63. Not integer.n=37:31*1369+481=42439+481=42920. Not a square.n=38:31*1444+481=44764+481=45245. Not a square.n=39:31*1521+481=47151+481=47632. Not a square.n=40:31*1600+481=49600+481=50081. sqrt(50081)=223.79. Not integer.This is not looking promising. Maybe k=2? So m²=31n² +962. Let's check n=17:31*289=8959+962=9921. sqrt(9921)=99.6. Not integer. Alternatively, this approach might not be feasible.Alternatively, perhaps there's a better way. Since a and b must be positive integers, and both 15a +16b and16a -15b are squares, maybe we can use substitution. Let me see.From equation (1):15a +16b =m²From equation (2):16a -15b =n²Let me solve these two equations for a and b.Multiply equation (1) by16:240a +256b=16m²Multiply equation (2) by15:240a -225b=15n²Subtract the second equation from the first:(240a +256b) - (240a -225b) =16m² -15n²481b=16m² -15n²Therefore, b=(16m² -15n²)/481Similarly, add equations (1) and (2):15a +16b +16a -15b= m² +n²31a +b =m² +n²From this, a=(m² +n² -b)/31But b=(16m² -15n²)/481Therefore, substituting:a=(m² +n² - (16m² -15n²)/481)/31= ( (481m² +481n² -16m² +15n² ) /481 ) /31= ( (465m² +496n² ) /481 ) /31= (465m² +496n² ) / (481*31 )Hmm, this is getting messy. But since a must be an integer, 465m² +496n² must be divisible by481*31=14911.Alternatively, since this approach is getting too algebraic, maybe try to find some relations between m andn.Recall that from the original equations:15a +16b =m²16a -15b =n²Let me think of these as components of a vector, and perhaps consider some transformation. For instance, if we consider a and b as variables, then the transformation (a,b) ↦ (15a+16b, 16a-15b) is linear. Its determinant is 15*(-15) -16*16= -225 -256= -481, which is the same determinant as before. Since the determinant is -481, this transformation is invertible modulo any prime not dividing481, but since we are working with integers, the inverse is as we found before.Alternatively, perhaps think of this as a system requiring both linear combinations to be squares. Such problems often have solutions related to Pythagorean triples or other Diophantine equations. But in this case, the coefficients are 15 and16, which complicates things.Alternatively, let's suppose that the two squares m² and n² are part of a Pythagorean-like triple. For instance, maybe m andn are legs or hypotenuses of a right triangle. However, I don't see a direct connection.Alternatively, consider that 15a +16b and16a -15b are squares. Let me assume that m andn are not too large, and try small values.But given the previous attempts where m andn would need to be quite large, this might not be feasible.Alternatively, perhaps the minimal solution corresponds to m andn such that15a +16b and16a -15b are both squares, and a and b are multiples of481. But since a andb are divided by481 in their expressions, perhaps the minimal solution occurs when a andb are fractions over481, but since they must be integers, the numerators must be multiples of481. Therefore, 15m² +16n² and16m² -15n² must be multiples of481.Therefore, maybe the minimal m andn are found when15m² +16n²=481*p and16m² -15n²=481*q, where p and q are positive integers. Then, a=p and b=q.Therefore, the problem reduces to solving:15m² +16n²=481p ...(1)16m² -15n²=481q ...(2)With p and q positive integers, and m andn positive integers.We need to find the minimal value of the smaller of m² andn².Alternatively, since both equations are linked, perhaps adding and subtracting them:Adding (1) and (2):31m² +n²=481(p +q)Subtracting (2) from (1): -m² +31n²=481(p -q)But not sure if this helps.Alternatively, divide the equations. Let me take equation (1) and equation (2):15m² +16n²=481p16m² -15n²=481qLet me divide these two equations:(15m² +16n²)/(16m² -15n²) = p/qBut not sure.Alternatively, express p and q in terms ofm andn:p=(15m² +16n²)/481q=(16m² -15n²)/481Since p and q must be integers, so we need:15m² +16n² ≡0 mod48116m² -15n² ≡0 mod481Which is the same as before.Therefore, we need to find m andn such that both15m² +16n² and16m² -15n² are divisible by481. Then, the minimal such m andn would give the minimal squares.Perhaps try to find such m andn.Given that m² ≡31n² mod481, and both m andn must be divisible by13.Let me set n=13k, m=13j. Then:m²=169j², n²=169k²Substituting into the equation:15*169j² +16*169k² ≡0 mod481169*(15j² +16k²) ≡0 mod13*37Since169=13², this is13²*(15j² +16k²)≡0 mod13*37.Dividing both sides by13:13*(15j² +16k²)≡0 mod37So,13*(15j² +16k²)≡0 mod37Similarly, from the other equation:16m² -15n²=16*169j² -15*169k²=169*(16j² -15k²)≡0 mod481Divide by13:13*(16j² -15k²)≡0 mod37Therefore, we have two congruence equations:13*(15j² +16k²)≡0 mod3713*(16j² -15k²)≡0 mod37Dividing both by13 (since13 and37 are coprime):15j² +16k²≡0 mod3716j² -15k²≡0 mod37So now, we have a similar system but with smaller modulus:15j² +16k²≡0 mod37 ...(A)16j² -15k²≡0 mod37 ...(B)Let me solve this system.From equation (A):15j² ≡-16k² mod37From equation (B):16j² ≡15k² mod37Let me express j² from equation (A):j²≡(-16/15)k² mod37Compute -16/15 mod37. First, inverse of15 mod37 is20, as before. So, -16*20= -320≡-320+9*37=-320+333=13 mod37.Therefore, j²≡13k² mod37Similarly, from equation (B):16j²≡15k² mod37. Substitute j²≡13k²:16*13k²≡15k² mod37208k²≡15k² mod37208 mod37=208-5*37=208-185=2323k²≡15k² mod378k²≡0 mod37Since37 is prime, and8 and37 are coprime, this impliesk²≡0 mod37. Hence, k≡0 mod37. Therefore, k=37t, and hence, j²≡13*(37t)² mod37. But37≡0 mod37, soj²≡0 mod37. Therefore, j≡0 mod37.Therefore, bothj andk must be divisible by37.Therefore, set j=37u, k=37v. Then, substituting back:n=13k=13*37v=481vm=13j=13*37u=481uTherefore, m=481u andn=481v. Then, substituting into the original equations:15a +16b =m²=(481u)²=481²u²16a -15b =n²=(481v)²=481²v²Then, solving fora andb:a=(15m² +16n²)/481=(15*(481²u²)+16*(481²v²))/481=481*(15*481u² +16*481v²)/481=15*481u² +16*481v²=481*(15u² +16v²)Similarly, b=(16m² -15n²)/481=(16*481²u² -15*481²v²)/481=481*(16*481u² -15*481v²)/481=16*481u² -15*481v²=481*(16u² -15v²)Therefore, a=481*(15u² +16v²) and b=481*(16u² -15v²). Since a and b must be positive integers, we require that16u² -15v² >0, so16u² >15v².But this leads to a and b being multiples of481, which would make the squares m² andn² equal to481²u² and481²v², which are very large. Therefore, the minimal solution would have u=1 and v=0, but v=0 gives b=481*(16*1 -15*0)=481*16, which is positive, but n=481*0=0, which is not a positive integer. Therefore, v must be at least1.But if u=1 and v=1, then16u²=16,15v²=15, so16u² -15v²=1>0. Therefore, b=481*(16*1 -15*1)=481*1=481, which is positive. a=481*(15*1 +16*1)=481*31=14911.Then, m=481*1=481, n=481*1=481. Then, the squares are m²=481²=231361 andn²=481². But both are equal, so the smaller one is the same. However, this is likely not the minimal solution.But this suggests that the minimal solution is when u andv are 1, leading to a=14911, b=481, and both squares are231361. But this seems way too large. The problem asks for the smallest possible value of the smaller of the two squares, so maybe there's a smaller solution where m andn are not multiples of481.However, our earlier analysis showed that m andn must be multiples of13 and37, hence multiples of481. But this seems to lead to very large solutions, but perhaps there is no smaller solution.Wait, but maybe the initial assumption that both m andn must be multiples of13 and37 is incorrect. Let's recap.Earlier steps showed that for m²≡31n² mod481, and since31 is a non-residue mod13, the only solution is m≡n≡0 mod13. However, modulo37, since31 is a quadratic residue, we could have non-trivial solutions where n is not 0 mod37. However, the previous deduction led us to conclude that both j andk must be0 mod37, hence leading m andn to be multiples of481. But this might not necessarily be the case.Wait, let me check the step where we derived that k≡0 mod37. In the reduced equations after setting m=13j, n=13k, we ended up with:15j² +16k²≡0 mod3716j² -15k²≡0 mod37From these, we found that j²≡13k² mod37, and substituting into the second equation led to8k²≡0 mod37, hence k≡0 mod37. Therefore, k must be0 mod37, leading j also to be0 mod37. Therefore, this implies that the minimal solution has m=13*37u andn=13*37v, i.e., multiples of481. Therefore, the minimal non-trivial solution would be whenu=v=1, giving m=n=481, leading to the squares both being481²=231361. But the problem asks for the smallest possible value of the smaller of these two squares. However, if both squares are equal, then the smaller one is231361. But perhaps there exists a solution where m andn are not equal, leading to one square being smaller than the other.But according to our previous deductions, the only solutions are multiples of481, so m andn must be multiples of481. Therefore, the minimal possible value of the smaller square would be481², but maybe there's a case where one is smaller and the other is larger. However, since both m andn are multiples of481, the minimal possible square is481². However, the problem states that a andb must be positive integers, and in the case where m=n=481, we have a=481*(15+16)=481*31=14911 and b=481*(16-15)=481*1=481, which are both positive integers. However, the two squares are both481²=231361. So, the smaller of the two squares is231361, which is very large. But perhaps there is a smaller solution where the two squares are not equal.Wait, but according to our previous analysis, the only solutions are when m andn are multiples of481. If that's the case, then the minimal solution is indeed when m=n=481, leading to both squares being481². But this seems counterintuitive. Let me verify with another approach.Suppose that there exists a solution where m andn are not multiples of481. Then, m² ≡31n² mod481. As we saw earlier, since31 is a non-residue mod13, this requires that both m andn are multiples of13. Let m=13m', n=13n'. Then, the equation becomes:(13m')² ≡31*(13n')² mod481169m'² ≡31*169n'² mod481Divide both sides by169 (which is13²):m'² ≡31n'² mod37Now, since31 is a quadratic residue mod37, there exists somek such thatk²≡31 mod37. Let’s say there exists some integerl with l²≡31 mod37. Then, m' ≡l*n' mod37.Therefore, m' = l*n' +37t for some integert.Substituting back, we can write m' = l*n' +37t. Therefore, m' andn' are related via this equation. Then, substituting into m'² ≡31n'² mod37, which is satisfied by construction.Therefore, possible solutions exist where m' andn' are not multiples of37. Therefore, this suggests that there exist solutions where m=13m' andn=13n', where m' andn' are not multiples of37, leading to m andn not multiples of481. Thus, contradicting the previous conclusion that k must be0 mod37.Therefore, my earlier deduction that k must be0 mod37 was incorrect. Let me revisit the steps.We had:From the reduced equations modulo37:15j² +16k² ≡0 mod37 ...(A)16j² -15k² ≡0 mod37 ...(B)From (A):15j² ≡-16k² mod37From (B):16j² ≡15k² mod37Let me solve forj² in terms ofk².From (A):j²≡ (-16/15)k² mod37From (B):j²≡ (15/16)k² mod37Therefore, these two expressions forj² must be equal:(-16/15)k² ≡ (15/16)k² mod37Multiply both sides by15*16 to eliminate denominators:-16*16k² ≡15*15k² mod37-256k² ≡225k² mod37Compute-256 mod37:256 divided by37 is6*37=222, 256-222=34. So-256≡-34 mod37. Similarly,225 mod37:225-6*37=225-222=3. So:-34k²≡3k² mod37-34k² -3k²≡0 mod37-37k²≡0 mod37Which is always true, since37≡0 mod37. Therefore, the two equations are consistent, and the solution is j²≡(-16/15)k² mod37. Therefore, there are non-trivial solutions wherek is not0 mod37.Therefore, settingk=1, thenj²≡(-16/15) mod37.As before, compute -16/15 mod37. inverse of15 mod37 is20. So, -16*20= -320≡-320+9*37= -320+333=13 mod37. Therefore, j²≡13 mod37.Now, need to check if13 is a quadratic residue mod37. Compute squares mod37:0²=01²=12²=43²=94²=165²=256²=367²=49≡128²=64≡279²=81≡710²=100≡2611²=121≡1012²=144≡3313²=169≡2814²=196≡1115²=225≡316²=256≡3417²=289≡3018²=324≡2819²=361≡2820²=400≡3021²=441≡3422²=484≡323²=529≡1124²=576≡2725²=625≡1226²=676≡3627²=729≡2528²=784≡729²=841≡2630²=900≡1231²=961≡3632²=1024≡2533²=1089≡1634²=1156≡935²=1225≡436²=1296≡1So, quadratic residues mod37 are: 0,1,3,4,7,9,10,11,12,16,25,26,27,28,30,33,34,36.13 is not among them. Therefore, j²≡13 mod37 has no solution. Therefore, there are no solutions withk=1.Similarly, tryingk=2:From j²≡13*2²=13*4=52≡52-37=15 mod37. 15 is a quadratic residue? From the list, 15 is not a QR. k=3:13*9=117≡117-3*37=117-111=6≡6 mod37. 6 is not a QR. k=4:13*16=208≡208-5*37=208-185=23≡23 mod37. 23 is not a QR. k=5:13*25=325≡325-8*37=325-296=29≡29 mod37. Not a QR. k=6:13*36=468≡468-12*37=468-444=24≡24 mod37. Not a QR. k=7:13*49=637≡637-17*37=637-629=8≡8 mod37. Not a QR. k=8:13*64=832≡832-22*37=832-814=18≡18 mod37. Not a QR. k=9:13*81=1053≡1053-28*37=1053-1036=17≡17 mod37. Not a QR. k=10:13*100=1300≡1300-35*37=1300-1295=5≡5 mod37. Not a QR.Continuing this way, it seems that for any k not divisible by37, j²≡13k² mod37 does not yield a quadratic residue. Therefore, the only solution is whenk≡0 mod37, which impliesj≡0 mod37. Therefore, the minimal solution is indeed whenk=37t, j=37s, leading tom=13j=13*37s=481s andn=13k=13*37t=481t.Therefore, the minimal positive solution is whens=t=1, leading tom=n=481, a=481*(15+16)=14911, b=481*(16-15)=481, and both squares are481²=231361. Thus, the smallest possible value of the smaller square is231361.However, the problem statement says "the smallest value that the smaller of these two squares can take." But in this solution, both squares are equal, so the smaller one is481². But perhaps there is a solution where the two squares are different, with one smaller than481².Wait, but according to our analysis, any solution requires m andn to be multiples of481, hence their squares would be at least481². Therefore, the minimal possible value is481². But the problem is from a Russian competition, and the answer is likely a much smaller number. Therefore, my approach must be missing something.Let me consider the original equations again:15a +16b =m²16a -15b =n²Adding these two equations:31a +b =m² +n²Subtracting the second from the first:-a +31b =m² -n²Let me write these as:31a +b =S ...(3)-a +31b =D ...(4)Where S=m² +n² andD=m² -n².From equation (3): a=(S -b)/31Substitute into equation (4):-(S -b)/31 +31b =DMultiply both sides by31:-(S -b) +31*31b =31D-S +b +961b =31D-S +962b =31DBut S=m² +n² andD=m² -n², so:-(m² +n²) +962b =31(m² -n²)Rearrange:962b =31m² -31n² +m² +n²962b =32m² -30n²Simplify:962b =2*(16m² -15n²)But from equation (2):16m² -15n²=481q, where q is an integer (since b=(16m² -15n²)/481). Therefore:962b =2*481qBut962=2*481. Therefore:2*481b =2*481q => b=qThus, consistent.Therefore, this doesn't give new information.Alternatively, let's assume that m andn are part of a Pythagorean triple where m² +n² andm² -n² are related to a andb. But I don't see a direct relation.Alternatively, think of the problem as looking for points (m,n) such that both15a +16b and16a -15b are squares, with a andb expressed in terms ofm andn. But since a andb must be integers, this ties back to the original equations.Let me try small values of m andn and see if I can find a solution where a andb are integers.Starting with m= 100:15a +16b=1000016a -15b=n²Let me solve for a andb:From the first equation:15a=10000 -16b => a=(10000 -16b)/15From the second equation:16*(10000 -16b)/15 -15b =n²Multiply through by15 to eliminate denominator:16*(10000 -16b) -225b =15n²160000 -256b -225b=15n²160000 -481b=15n²Therefore,481b=160000 -15n²Therefore, b=(160000 -15n²)/481We need b to be a positive integer. Therefore,160000 -15n² must be divisible by481 and positive.Compute160000/481≈332.64. So n² <160000/15≈10666.67, so n<103.26. So n can be up to103.Let me compute (160000 -15n²) mod481=0.Compute160000 mod481:First, compute481*332=481*(300+32)=481*300=144300; 481*32=15392. Total144300+15392=159692. 160000 -159692=308.So,160000≡308 mod481.15n²≡308 mod481.So,15n²≡308 mod481. Solve forn²:n²≡308/15 mod481. Compute inverse of15 mod481=449.308*449= Let's compute308*400=123200;308*49=15092. Total=123200+15092=138292.138292 mod481: Divide138292 by481.481*287=481*(200+80+7)=481*200=96200;481*80=38480;481*7=3367. Total=96200+38480=134680+3367=138047. 138292-138047=245.So,n²≡245 mod481.Check if245 is a quadratic residue mod481. Since481=13*37, check if245 is a quadratic residue mod13 and mod37.245 mod13:245-18*13=245-234=11. 11 is a quadratic residue mod13? Earlier list: quadratic residues mod13 are0,1,3,4,9,10,12. 11 is not a quadratic residue mod13. Therefore,245 is not a quadratic residue mod13, so no solution.Therefore, no solution for m=100.Similarly, trying m= 80:15a +16b=640016a -15b=n²From first equation: a=(6400 -16b)/15Second equation:16*(6400 -16b)/15 -15b =n²Multiply by15:16*(6400 -16b) -225b =15n²102400 -256b -225b=15n²102400 -481b=15n²481b=102400 -15n²b=(102400 -15n²)/481102400/481≈212.89. So, n² <102400/15≈6826.67, so n<82.62. n up to82.Compute102400 mod481:481*212=481*(200+12)=481*200=96200;481*12=5772. Total=96200+5772=101972. 102400-101972=428. So,102400≡428 mod481.Thus,15n²≡428 mod481 => n²≡428/15 mod481. Inverse of15 is449.428*449= Let's compute400*449=179600;28*449=12572. Total=179600+12572=192172.192172 mod481: 481*400=192400. 192172-192400=-228≡-228+481=253 mod481.So,n²≡253 mod481. Check quadratic residues:253 mod13=253-19*13=253-247=6. 6 is not a quadratic residue mod13. Therefore, no solution.Similarly, this approach is not fruitful.Alternatively, consider that481 is the determinant, and the problem might,<think>Okay, so I have this problem here where I need to find the smallest value that the smaller of two squares can take, given that 15a + 16b and 16a - 15b are both perfect squares. The integers a and b are positive. Alright, let me try to break this down step by step.First, let me write down the given conditions:1. 15a + 16b = m² (where m is a positive integer)2. 16a - 15b = n² (where n is a positive integer)And the goal is to find the minimal possible value of the smaller square between m² and n². So whichever of m² or n² is smaller, we need to find the smallest possible value that can occur under these conditions.Hmm. Since a and b are positive integers, both expressions 15a + 16b and 16a - 15b must also be positive. That means 16a - 15b > 0, so 16a > 15b, which implies that a > (15/16)b. So a must be at least (15/16)b + 1, maybe? But since a and b are integers, a must be at least the ceiling of (15/16)b. Not sure if that's useful yet.Let me think. Maybe I can express a and b in terms of m and n. Let's treat these two equations as a system of linear equations in variables a and b.So, we have:15a + 16b = m² ...(1)16a - 15b = n² ...(2)Let me try to solve this system for a and b. Let's use the method of elimination. Multiply equation (1) by 16 and equation (2) by 15:15*16a + 16*16b = 16m² ...(1a)16*15a - 15*15b = 15n² ...(2a)Now subtract equation (2a) from equation (1a):(15*16a - 16*15a) + (16*16b + 15*15b) = 16m² - 15n²Wait, 15*16a and 16*15a cancel each other out? No, wait:Wait, equation (1a) is 15*16a + 16*16b = 16m²Equation (2a) is 16*15a - 15*15b = 15n²So, subtract (2a) from (1a):(15*16a - 16*15a) + (16*16b + 15*15b) = 16m² - 15n²But 15*16a - 16*15a is zero. So we have:0 + (16²b + 15²b) = 16m² - 15n²Wait, 16*16b + 15*15b? Wait, no, equation (1a) is 16*16b and equation (2a) is -15*15b. So when you subtract, it's 16*16b - (-15*15b) = 16²b + 15²b.Yes, right. So:(16² + 15²)b = 16m² - 15n²Compute 16² + 15²: 256 + 225 = 481. So:481b = 16m² - 15n² ...(3)Similarly, maybe we can add the equations instead. Let's see. If we do equation (1a) + equation (2a):15*16a + 16*16b + 16*15a - 15*15b = 16m² + 15n²Which simplifies to:(15*16a + 16*15a) + (16*16b - 15*15b) = 16m² + 15n²15*16a + 16*15a = 2*15*16a, and 16*16b - 15*15b = (256 - 225)b = 31bWait, no:Wait, 15*16a + 16*15a = (15*16 + 16*15)a = 2*(15*16)a = 480aWait, 15*16 is 240, so 240a + 240a = 480a. Then 16*16b - 15*15b is 256b - 225b = 31b.So:480a + 31b = 16m² + 15n² ...(4)Hmm. So equations (3) and (4) are:481b = 16m² - 15n² ...(3)480a + 31b = 16m² + 15n² ...(4)But maybe this is getting more complicated. Alternatively, maybe solving the original two equations for a and b.Original equations:15a + 16b = m² ...(1)16a - 15b = n² ...(2)Let me solve for a and b.Multiply equation (1) by 16: 240a + 256b = 16m² ...(1b)Multiply equation (2) by 15: 240a - 225b = 15n² ...(2b)Subtract (2b) from (1b):(240a + 256b) - (240a - 225b) = 16m² - 15n²So, 240a - 240a + 256b + 225b = 16m² - 15n²Therefore, 481b = 16m² - 15n² ...(5)Similarly, if we add equations (1b) and (2b):(240a + 256b) + (240a - 225b) = 16m² + 15n²480a + 31b = 16m² + 15n² ...(6)Now, from equation (5): 481b = 16m² - 15n²From equation (6): 480a + 31b = 16m² + 15n²So, perhaps we can express a and b in terms of m and n.From equation (5):b = (16m² - 15n²)/481Since b must be a positive integer, (16m² - 15n²) must be divisible by 481, and the result must be positive. Therefore:16m² - 15n² = 481k, where k is a positive integer.Similarly, from equation (6):480a = 16m² + 15n² - 31bBut since we have b in terms of m and n, we can substitute:480a = 16m² + 15n² - 31*(16m² - 15n²)/481Multiply both sides by 481 to eliminate the denominator:480a * 481 = 481*(16m² + 15n²) - 31*(16m² - 15n²)Compute the right-hand side:481*16m² + 481*15n² - 31*16m² + 31*15n²Factor terms:(481*16 - 31*16)m² + (481*15 + 31*15)n²= 16*(481 - 31)m² + 15*(481 + 31)n²Compute 481 - 31 = 450, 481 + 31 = 512So:16*450m² + 15*512n²= 7200m² + 7680n²Therefore:480a * 481 = 7200m² + 7680n²Divide both sides by 480:a * 481 = 15m² + 16n²Therefore,a = (15m² + 16n²)/481So now we have expressions for both a and b in terms of m and n:a = (15m² + 16n²)/481b = (16m² - 15n²)/481Since a and b must be positive integers, both numerators must be divisible by 481, and the results must be positive integers. So:1. 15m² + 16n² ≡ 0 mod 4812. 16m² - 15n² ≡ 0 mod 4813. 16m² - 15n² > 0So, we have two congruence conditions and one inequality. Let me note that 481 is a prime number? Wait, 481. Let's check if 481 is prime. 481 divided by 13: 13*37 is 481. Yes, 13*37=481. So 481 factors into 13 and 37. So it's a composite number. Therefore, modulo 13 and modulo 37.Therefore, the congruences must hold modulo 13 and modulo 37. So, perhaps we can break the problem into congruences modulo 13 and 37.Given that:15m² + 16n² ≡ 0 mod 13 and mod 3716m² - 15n² ≡ 0 mod 13 and mod 37Let me handle modulo 13 first.First, modulo 13:15 ≡ 2 mod 1316 ≡ 3 mod 13So:15m² + 16n² ≡ 2m² + 3n² ≡ 0 mod 1316m² - 15n² ≡ 3m² - 2n² ≡ 0 mod 13So we have:2m² + 3n² ≡ 0 mod 13 ...(A)3m² - 2n² ≡ 0 mod 13 ...(B)Let me try to solve these two congruences.From equation (A): 2m² ≡ -3n² mod 13From equation (B): 3m² ≡ 2n² mod 13Let me solve equation (B) first: 3m² ≡ 2n² mod 13. Let's multiply both sides by the inverse of 3 modulo 13. The inverse of 3 mod13 is 9 because 3*9=27≡1 mod13.So m² ≡ 2*9 n² ≡ 18n² ≡ 5n² mod13.Therefore, m² ≡5n² mod13.Then plugging into equation (A): 2m² + 3n² ≡ 2*(5n²) + 3n² = 10n² +3n²=13n²≡0 mod13. Which is always true. So equation (A) is satisfied if equation (B) is satisfied.Therefore, the only condition modulo13 is that m² ≡5n² mod13.Which implies that (m/n)² ≡5 mod13. Assuming n is not divisible by13, we can write this as (m/n)^2 ≡5 mod13. Therefore, 5 must be a quadratic residue modulo13.Quadratic residues modulo13 are 1, 4, 9, 3, 12, 10. Let me check:1²=1, 2²=4, 3²=9, 4²=16≡3, 5²=25≡12, 6²=36≡10, 7²=49≡10, 8²=64≡12, 9²=81≡3, 10²=100≡9, 11²=121≡4, 12²=144≡1.So quadratic residues modulo13 are 1, 4, 9, 3, 12, 10. 5 is not a quadratic residue modulo13. Therefore, there is no solution unless n and m are both divisible by13.Wait, but if 5 is not a quadratic residue modulo13, then m/n can't be such that (m/n)^2 ≡5 mod13, unless both m and n are divisible by13. Because if n is not divisible by13, then 5 is not a quadratic residue, so no solution. If n is divisible by13, then m must also be divisible by13. So, in order for the congruence to hold, both m and n must be divisible by13.Similarly, if m =13k and n=13l, then substituting back into m²≡5n² mod13 gives (13k)^2≡5*(13l)^2 mod13 → 0≡0 mod13, which is always true. Therefore, the only solutions modulo13 are when both m and n are divisible by13.Therefore, m=13k and n=13l for some integers k and l.Similarly, we need to check the conditions modulo37.So moving on to modulo37.15m² + 16n² ≡0 mod3716m² -15n² ≡0 mod37Again, compute coefficients modulo37:15 mod37=1516 mod37=16So:15m² +16n² ≡0 mod37 ...(C)16m² -15n² ≡0 mod37 ...(D)Again, let's solve these equations.From equation (D):16m² ≡15n² mod37Multiply both sides by inverse of 16 mod37. The inverse of 16 mod37. Let's compute it:Find x such that 16x ≡1 mod37.Trying 16*7=112≡112-3*37=112-111=1 mod37. So 16*7≡1 mod37. Therefore, inverse of16 is7.Therefore, m² ≡15*7 n² ≡105n²≡105-2*37=105-74=31n² mod37.Therefore, m²≡31n² mod37.Similarly, substitute into equation (C):15m² +16n² ≡15*(31n²)+16n²=465n² +16n²=(465+16)n²=481n²≡481 mod37.481 divided by37: 37*13=481. So 481≡0 mod37.Therefore, 481n²≡0 mod37, which is always true. So equation (C) is satisfied if equation (D) is satisfied.Thus, similar to modulo13, the condition here is m²≡31n² mod37.But 31 is a quadratic residue modulo37?Let me check if 31 is a quadratic residue mod37. To check if there exists x such that x²≡31 mod37.Compute the Legendre symbol (31/37). Since 37≡1 mod4, (31/37)=(37/31) by quadratic reciprocity. 37 mod31=6, so (6/31). Then (6/31)=(2/31)*(3/31). (2/31)= -1^(31²-1)/8= -1^(961-1)/8= -1^120=1. Wait, no: (2/p)=1 if p≡±1 mod8, else -1. 31≡7 mod8, which is -1. So (2/31)= -1. (3/31). Since 31≡1 mod4, (3/31)=(31/3)=(1/3)=1. So (6/31)=(-1)*(1)= -1. Therefore, (31/37)= -1, meaning 31 is not a quadratic residue modulo37. Therefore, similar to modulo13, the congruence m²≡31n² mod37 has no solution unless both m and n are divisible by37.Therefore, the only solutions modulo37 occur when both m and n are divisible by37.Therefore, combining both modulo13 and modulo37 conditions, we have that m and n must be divisible by both13 and37. Since13 and37 are primes, m and n must be divisible by13*37=481. So, m=481p and n=481q for some integers p and q. But then substituting back into equations (5) and (6):Wait, let's see. If m=13k and n=13l from modulo13 condition, and m=37k' and n=37l' from modulo37 condition, so m and n must be divisible by LCM(13,37)=481. Therefore, m=481p and n=481q, where p and q are positive integers. But substituting this into equations for a and b:a = (15m² +16n²)/481 = (15*(481p)^2 +16*(481q)^2)/481 = 481*(15*481p² +16*481q²)/481 = 15*481p² +16*481q². Wait, no:Wait, a = (15m² +16n²)/481. If m=481p and n=481q, then:a = (15*(481p)^2 +16*(481q)^2)/481 = 481*(15*481p² +16*481q²)/481 = 15*481p² +16*481q². Therefore, a would be 481*(15p² +16q²). Similarly, b = (16m² -15n²)/481 = (16*(481p)^2 -15*(481q)^2)/481 = 481*(16*481p² -15*481q²)/481 = 16*481p² -15*481q². Therefore, b =481*(16p² -15q²).But since a and b must be positive integers, 16p² -15q² must be positive. So 16p² >15q². Therefore, (p/q)^2 >15/16. So p/q > sqrt(15/16)=sqrt(15)/4≈0.968. So p must be at least q.Wait, but p and q are positive integers. Let's take q=1. Then p must satisfy p² >15/16, so p must be at least1. Let me check if p=1, q=1: 16*1 -15*1=1>0. So b=481*(16*1 -15*1)=481*(1)=481. Then a=481*(15*1 +16*1)=481*31=481*30 +481=14430+481=14911.Therefore, in this case, m=481p=481*1=481, n=481q=481*1=481. Therefore, both squares would be (481)^2, so the smaller one is 481². But maybe this isn't the minimal possible? Because p and q could be fractions, but they must be integers.Wait, but since m and n must be multiples of481, then the minimal possible m and n would be481, leading to squares of481². However, the problem says "Find the smallest value that the smaller of these two squares can take." So maybe there is a smaller solution where m and n are not multiples of481? But according to our earlier analysis, modulo13 and37 conditions force m and n to be multiples of481. Therefore, the minimal squares would be481², so the answer is481². But wait, the problem states "the smallest value that the smaller of these two squares can take." But perhaps there is a solution where the squares are smaller? Let me check.Wait, perhaps my reasoning is flawed here. Let me retrace.We have equations:15a +16b =m²16a -15b =n²We derived that m and n must be divisible by481, hence m=481p, n=481q. Then substituting into the equations:15a +16b=(481p)^216a -15b=(481q)^2Then, solving for a and b gives:a=481*(15p² +16q²)b=481*(16p² -15q²)Since a and b must be positive integers, we need 16p² -15q² >0. So, 16p² >15q². Let's take the minimal p and q such that this holds. Let's try q=1, then p must be at least1. p=1: 16 -15=1>0. Then:a=481*(15 +16)=481*31=14911b=481*(16 -15)=481*1=481Then, m=481*1=481, n=481*1=481. So both squares are481². Therefore, the smaller square is481². But is there a case where one square is smaller than the other? Suppose we take p=2, q=1. Then:16*(4)=64 >15*1=15, so okay.a=481*(15*4 +16*1)=481*(60 +16)=481*76=481*70 +481*6=33670 +2886=36556b=481*(16*4 -15*1)=481*(64 -15)=481*49=481*50 -481=24050 -481=23569Then m=481*2=962, n=481*1=481. So squares are962² and481². Then the smaller square is481², same as before. If we take p=1, q=1, both squares are equal. If p=2, q=1, squares are962² and481². If p=1, q=2, then 16p² -15q²=16*1 -15*4=16-60=-44 <0, which is invalid. Therefore, q must be less than p*sqrt(16/15). For q=2, p must be at least ceiling(q*sqrt(15/16))=ceiling(2*(sqrt(15)/4))≈ceiling(2*0.968)=ceiling(1.936)=2. So p=2, q=2:16*4 -15*4=64-60=4>0. Then:a=481*(15*4 +16*4)=481*(60 +64)=481*124=481*120 +481*4=57720 +1924=59644b=481*(16*4 -15*4)=481*(64 -60)=481*4=1924Then m=481*2=962, n=481*2=962. Both squares are962². So the smaller square is still962². Therefore, regardless, the minimal smaller square is481². But wait, is there a way to get smaller squares? If m and n must be multiples of481, then the minimal square is(481)^2, so the answer would be481²=231361. But wait, the problem is from a Russian competition, maybe the answer is smaller? Maybe my reasoning is wrong?Wait, perhaps I made a mistake in assuming that m and n must be multiples of481. Let me think again.We derived that 15m² +16n² ≡0 mod481 and16m² -15n²≡0 mod481. But instead of m and n being multiples of481, maybe there are solutions where m and n are not multiples of481 but satisfy the congruences. But since 481=13*37, and we saw that modulo13 and37 require m and n to be divisible by13 and37, so m and n must be divisible by LCM(13,37)=481. Therefore, m and n must be multiples of481. Therefore, the minimal m and n are481, leading to squares of481². Therefore, the answer should be481². But let me check for a=14911 and b=481. Then:15a +16b =15*14911 +16*481=223665 +7696=231361=481²16a -15b=16*14911 -15*481=238576 -7215=231361=481². So both are equal. So both squares are481². Therefore, the smaller of the two is481².But wait, the problem says "the smaller of these two squares". In this case, both squares are equal, so the smaller one is481². But maybe there exists a solution where one square is smaller than481². Let me see. If such a solution exists, then my earlier conclusion is wrong.Alternatively, perhaps my initial assumption that m and n must be multiples of481 is incorrect. Let me verify with an example. Suppose m and n are not multiples of481, but still satisfy the congruence conditions.Wait, if I consider the congruence conditions modulo13 and37:For modulus13, we found that m and n must be multiples of13.For modulus37, similarly, m and n must be multiples of37.Therefore, unless m and n are multiples of both13 and37, i.e., multiples of481, the congruence conditions cannot be satisfied. Therefore, m and n must indeed be multiples of481. Hence, the minimal possible squares are481². So the answer should be481²=231361. But let me check the problem statement again."Find the smallest value that the smaller of these two squares can take."Wait, but in the case where both squares are equal, the smaller one is equal to both. But maybe there's a solution where one square is smaller? For example, m²=481² and n²= (something smaller). But according to our previous analysis, both m and n must be multiples of481, so n² is at least481². Therefore, the minimal possible value is481². Therefore, the answer is481², which is231361. So I think that's the answer.Wait, but I feel like this might not be the intended answer because 481 is quite a large number, and maybe the problem expects a smaller number. Let me check if there's an error in my reasoning.Wait, the problem states that a and b are positive integers. So, in our solution, a=14911 and b=481, which are positive integers. But is there a smaller solution where m and n are not multiples of481 but still satisfy the congruence conditions? Let's suppose that there's a solution where m and n are not multiples of481, but their squares are such that 15m² +16n² and16m² -15n² are multiples of481, hence allowing a and b to be integers.But from the congruence analysis, modulo13 and37 require that m and n are multiples of13 and37. Therefore, m and n must be multiples of481. Therefore, unless the problem allows a and b to be non-integers, which it doesn't, there are no smaller solutions. Therefore, the minimal possible value is indeed481².But wait, let's consider another approach. Let me set x = m and y = n. Then the equations are:15a +16b =x²16a -15b =y²We can treat these as equations in a and b. Let's solve for a and b:From the first equation: 15a =x² -16bFrom the second equation:16a = y² +15bMultiply the first equation by16:240a=16x² -256bMultiply the second equation by15:240a=15y² +225bSet equal:16x² -256b=15y² +225b16x² -15y²=256b +225b=481bTherefore, 16x² -15y²=481bFrom the first equation:15a +16b=x² => 15a =x² -16b => a=(x² -16b)/15Since a must be a positive integer, x² -16b must be divisible by15 and positive. Similarly, from 16x² -15y²=481b, we have b=(16x² -15y²)/481. Therefore, substituting back into a:a=(x² -16*(16x² -15y²)/481)/15= (x² - (256x² -240y²)/481)/15= ( (481x² -256x² +240y²)/481 ) /15= (225x² +240y²)/ (481*15)= (15x² +16y²)/481Which matches our previous result. Therefore, a=(15x² +16y²)/481 and b=(16x² -15y²)/481. Therefore, for a and b to be integers, 15x² +16y² must be divisible by481, and16x² -15y² must be divisible by481. Therefore, as we saw earlier, x and y must be multiples of481.Therefore, the minimal possible x and y are481, leading to a= (15*(481)^2 +16*(481)^2)/481=481*(15+16)=481*31=14911 and b=(16*481² -15*481²)/481=481*(16-15)=481. Therefore, the squares arex²=481² andy²=481². Therefore, the minimal smaller square is481²=231361.Wait, but before I conclude, let me check if the problem allows both squares to be equal. If yes, then the minimal smaller square is just that value. The problem says "the smaller of these two squares", but if they are equal, then the smaller one is equal to both. Therefore, 481² is the answer.But I have a feeling that the answer is smaller. Maybe there is a way where m and n are not multiples of481 but still satisfy the divisibility conditions. Let me test with m= 481 and n=481. That works, but maybe with smaller numbers?Wait, for example, let's suppose that m= 16* something and n=15* something. Let me think. Suppose that m=16k and n=15k. Then:15a +16b=(16k)^2=256k²16a -15b=(15k)^2=225k²Let me solve these equations.From the first equation:15a =256k² -16bFrom the second equation:16a =225k² +15bMultiply first equation by16:240a=4096k² -256bMultiply second equation by15:240a=3375k² +225bSet equal:4096k² -256b=3375k² +225b4096k² -3375k²=225b +256b721k²=481bThus, b=(721/481)k²=(721/481)k². But 721 and481: gcd(721,481). Let's compute gcd(481,721-481)=gcd(481,240). gcd(240,481-2*240)=gcd(240,1)=1. Therefore, 721/481 reduces to721/481. So b is not an integer unlessk is a multiple of481, which brings us back to the previous case. Therefore, this approach doesn't help.Alternatively, let me try specific small values. Suppose that the smaller square is1. Then either15a +16b=1 or16a -15b=1. But sincea andb are positive integers,15a +16b ≥15+16=31>1, so impossible. Similarly, the smaller square can't be1,4,9,16,25, etc., up to maybe some number.Suppose the smaller square is25. Let's assume that16a -15b=25. Then we have:15a +16b=m²16a -15b=25Let me solve these equations. Let's solve for a and b.From the second equation:16a=25 +15b => a=(25 +15b)/16Substitute into the first equation:15*(25 +15b)/16 +16b =m²Multiply out:(375 +225b)/16 +16b =m²Multiply all terms by16:375 +225b +256b =16m²375 +481b=16m²Thus,481b=16m² -375Since481 divides16m² -375, we need16m² ≡375 mod481Compute375 mod481=375So,16m² ≡375 mod481. Let's solve for m.Multiply both sides by inverse of16 mod481. Earlier, we saw that inverse of16 mod37 is7. Let's compute inverse of16 mod481.Since481=13*37. Compute inverse of16 modulo13 and37.mod13: inverse of16 mod13 is inverse of3 mod13=9, since3*9=27≡1 mod13.mod37: inverse of16 mod37 is7, as before.Thus, by Chinese Remainder Theorem, inverse of16 mod481 is a number x such thatx≡9 mod13 andx≡7 mod37.Find x≡7 mod37 andx≡9 mod13.Let x=37k +7. Then, 37k +7 ≡9 mod13. 37≡11 mod13, so 11k +7 ≡9 mod13 =>11k ≡2 mod13. Multiply both sides by inverse of11 mod13, which is6, since11*6=66≡1 mod13. Thus, k≡2*6=12≡12 mod13. Therefore, k=13m +12. Therefore,x=37*(13m +12)+7=481m +444 +7=481m +451. Therefore, the inverse of16 mod481 is451.Therefore, m²≡375*451 mod481Compute375*451 mod481.First, compute375 mod481=375451 mod481=451Multiply375*451:Let's compute 375*451 = (400 -25)*(450 +1) =400*450 +400*1 -25*450 -25*1=180000 +400 -11250 -25=180000 -11250=168750 +400=169150 -25=169125Now, compute169125 mod481.Divide169125 by481:481*350=481*300=144300; 481*50=24050; total144300+24050=168350. Then169125-168350=775. Then775-481=294. So169125≡294 mod481.Therefore, m²≡294 mod481So, we need to find m such thatm²≡294 mod481. Since481=13*37, check modulo13 and37.First, modulo13:294 mod13. 13*22=286, so294-286=8, so294≡8 mod13. Check if8 is a quadratic residue mod13. Quadratic residues mod13 are1,4,9,3,12,10.8 is not a quadratic residue mod13. Therefore, no solution. Therefore, m²≡294 mod481 is impossible. Therefore, there is no solution where16a -15b=25.Similarly, trying other small squares. Suppose16a -15b=16. Then similar steps:15a +16b=m²16a -15b=16Solving for a:16a=16 +15b => a=(16 +15b)/16Plug into first equation:15*(16 +15b)/16 +16b =m²Multiply by16:15*(16 +15b) +256b =16m²240 +225b +256b=16m²240 +481b=16m²Thus,481b=16m² -240So,16m² ≡240 mod48116m² ≡240 mod481Multiply both sides by inverse of16 mod481, which is451:m² ≡240*451 mod481Compute240*451 mod481.240*451=240*(450 +1)=240*450 +240=108000 +240=108240108240 divided by481:481*225=481*(200 +25)=96200 +12025=108225108240 -108225=15. So m²≡15 mod481.Check modulo13:15 mod13=2. 2 is a quadratic residue mod13? QR are1,4,9,3,12,10. No. So15 mod13=2 is not a quadratic residue, so no solution. Therefore, impossible.Alternatively, take16a -15b=49 (7²). Then:16a -15b=49Thus,16a=49 +15b => a=(49 +15b)/16Substitute into15a +16b=m²:15*(49 +15b)/16 +16b=m²Multiply by16:15*49 +15*15b +256b=16m²735 +225b +256b=16m²735 +481b=16m²Thus,481b=16m² -735So16m² ≡735 mod481Compute735 mod481=735-481=254So16m²≡254 mod481Multiply by451:m²≡254*451 mod481254*451=254*(450 +1)=254*450 +254=114300 +254=114554Divide by481:481*238=481*(200 +38)=96200 +481*38. Compute481*30=14430, 481*8=3848, so14430+3848=18278. Total=96200 +18278=114478. 114554 -114478=76. So m²≡76 mod481.Check modulo13:76 mod13=76-5*13=76-65=11. 11 is not a quadratic residue mod13. So no solution. Therefore, impossible.This pattern suggests that for small squares, the congruence conditions modulo13 and37 cannot be satisfied, because the required quadratic residues do not exist. Therefore, the minimal possible square is when both squares are481². Hence, the answer should beboxed{231361}But wait, the problem is from Russia 37-4, which might have a more elegant solution. Alternatively, maybe there is a solution with smaller numbers that I missed. Let me think differently.Suppose that m and n are such that m² =15a +16b and n²=16a -15b. Let me consider these as equations and try to eliminate variables.Let me consider m² +n²=15a +16b +16a -15b=31a +bSimilarly, m² -n²=15a +16b -16a +15b= -a +31bSo:m² +n²=31a +b ...(S1)m² -n²= -a +31b ...(S2)Let me solve this system for a and b.From (S1) and (S2):Add (S1) and (S2):2m² =30a +32b => m²=15a +16b (which is the original equation)Subtract (S2) from (S1):2n²=32a -30b =>n²=16a -15b (original equation again). So not helpful.Alternatively, let me solve for a and b:From (S1):31a +b =m² +n²From (S2):-a +31b =m² -n²Let me solve this linear system:Let me write it as:31a + b = m² + n² ...(A)-a +31b = m² - n² ...(B)Let me multiply equation (B) by31:-31a +961b =31m² -31n² ...(B1)Add equation (A) and (B1):(31a -31a) + (b +961b) = (m² +n²) + (31m² -31n²)0 +962b =32m² -30n²Thus:962b =32m² -30n²Divide by2:481b =16m² -15n²Which is the same as equation (3). Similarly, solve for a:From equation (A):31a =m² +n² -bSubstitute b=(16m² -15n²)/481:31a =m² +n² - (16m² -15n²)/481Multiply both sides by481:31*481a =481m² +481n² -16m² +15n²31*481a =(481 -16)m² + (481 +15)n²=465m² +496n²Therefore,a=(465m² +496n²)/(31*481)Simplify:465/31=15, 496/31=16Therefore,a=(15m² +16n²)/481Which matches our previous result. So, no progress.Another approach: Let me think of Pythagorean triples. Suppose that m and n are legs of a Pythagorean triple, but not sure.Alternatively, think of the two expressions 15a +16b and16a -15b as squares. Let me denote them as x² and y². So:x² =15a +16by²=16a -15bWe can write this as a system:15a +16b =x²16a -15b =y²Let me solve for a and b in terms ofx andy.Multiply first equation by16:240a +256b=16x²Multiply second equation by15:240a -225b=15y²Subtract the second from the first:(240a +256b) - (240a -225b)=16x² -15y²Which gives:481b=16x² -15y² =>b=(16x² -15y²)/481Similarly, add the two equations:480a +31b=16x² +15y²Substitute b:480a +31*(16x² -15y²)/481=16x² +15y²Multiply both sides by481:480*481a +31*(16x² -15y²)=481*(16x² +15y²)480*481a =481*(16x² +15y²) -31*(16x² -15y²)=16x²*(481 -31) +15y²*(481 +31)=16x²*450 +15y²*512Therefore:480*481a =7200x² +7680y²Divide both sides by480:481a=15x² +16y²Which gives a=(15x² +16y²)/481. So again, the same result.Thus, the only way for a and b to be integers is that15x² +16y² and16x² -15y² are both divisible by481. As we saw earlier, this necessitates x and y to be multiples of481. Therefore, the minimal solution is when x and y are481, leading to a= (15+16)*481=31*481=14911 and b=(16 -15)*481=481. Therefore, both squares are481².Hence, the minimal value of the smaller square is481²=231361. Therefore, the answer isboxed{231361}Final Answerboxed{231361}</think>Given the problem where positive integers (a) and (b) satisfy the conditions that (15a + 16b) and (16a - 15b) are both perfect squares, we need to find the smallest value that the smaller of these two squares can take.We start with the equations:1. (15a + 16b = m^2)2. (16a - 15b = n^2)We solve these equations for (a) and (b) by treating them as a system of linear equations. By multiplying and subtracting the equations, we derive:[ 481b = 16m^2 - 15n^2 ][ 480a + 31b = 16m^2 + 15n^2 ]From these, we express (a) and (b) in terms of (m) and (n):[ a = frac{15m^2 + 16n^2}{481} ][ b = frac{16m^2 - 15n^2}{481} ]For (a) and (b) to be positive integers, both numerators must be divisible by 481. This leads to the congruences:[ 15m^2 + 16n^2 equiv 0 pmod{481} ][ 16m^2 - 15n^2 equiv 0 pmod{481} ]Analyzing these congruences modulo 13 and 37 (since 481 = 13 × 37), we find that both (m) and (n) must be multiples of 481. Thus, the smallest values for (m) and (n) are 481, leading to:[ m = 481 quad text{and} quad n = 481 ]Substituting these values back, we find:[ 15a + 16b = (481)^2 ][ 16a - 15b = (481)^2 ]Thus, both squares are equal to (481^2). Therefore, the smallest value that the smaller of these two squares can take is:[ boxed{231361} ]]

🔑:Let positive integers a, b be such that 15a + 16b and 16a - 15b are both squares of positive integers. Let15a + 16b = r^2, quad 16a - 15b = s^2,where r, s are positive integers. Thus,begin{array}{c}15^2 a + 16^2 a = 15r^2 + 16s^2, 481a = 15r^2 + 16s^2. 16^2 b + 15^2 b = 16r^2 - 15s^2, 481b = 16r^2 - 15s^2.end{array}Therefore, 481 mid 15r^2 + 16s^2, 481 mid 16r^2 - 15s^2. We will prove that 481 mid r, 481 mid s. Since 481 = 13 times 37, it suffices to prove that r, s are both multiples of 13 and 37.First, we prove that r, s are both multiples of 13. We use proof by contradiction. If not, at least one of r, s is not divisible by 13, and since 13 mid 16r^2 - 15s^2, we have 13 mid r, 13 mid s. Since16r^2 equiv 15s^2 quad (bmod 13),we have16r^2 cdot s^{10} equiv 15s^{12} equiv 15 equiv 2 quad (bmod 13).Taking the 6th power on both sides, we getbegin{array}{c}(4rs^5)^{12} equiv 64 quad (bmod 13), 1 equiv -1 quad (bmod 13),end{array}which is a contradiction.Next, we prove that 37 mid r, 37 mid s. If not, then 37 nmid r, 37 nmid s. Since37 mid 15r^2 + 16s^2, quad 37 mid 16r^2 - 15s^2,we have37 mid r^2 - 31s^2,which meansbegin{array}{c}r^2 equiv 31s^2 quad (bmod 37) r^2 s^{34} equiv 31s^{36} equiv 31 quad (bmod 37).end{array}Taking the 18th power on both sides, we get(r s^{17})^{36} equiv 31^{18} quad (bmod 37).Since (r s^{17})^{36} equiv 1 (bmod 37) and 31^{18} equiv (31^2)^9 equiv ((-6)^2)^9 equiv 36^9 equiv (-1)^9 equiv -1 quad (bmod 37), we have a contradiction.Therefore, 481 mid r, 481 mid s. Thus, these two squares are both at least 481^2.On the other hand, when a = 481 times 31, b = 481, both perfect squares are 481^2. Therefore, the minimum value is 481^2 = 231361.