🧠:<think>First, I need to determine the value of ( k ) in the function ( y = kx ) using the given point ( A(2, 4) ).By substituting the coordinates of point ( A ) into the equation, I can set up the equation ( 4 = 2k ).Solving for ( k ), I divide both sides by 2, which gives ( k = 2 ).Now that I know ( k = 2 ), the function becomes ( y = 2x ).To find the value of ( y ) when ( x = -2 ), I substitute ( x = -2 ) into the equation ( y = 2x ).This results in ( y = 2 times (-2) = -4 ).Therefore, when ( x = -2 ), ( y = -4 ).</think>

🧠:<think>Okay, so I have this problem here about binomial expansions. It says that in the expansion of (1 + x)^n, the sum of the odd-numbered terms is P, and the sum of the even-numbered terms is Q. Then it asks what (1 - x^2)^n equals, with options A through D. Hmm, let me try to figure this out step by step.First, I remember that when you expand (1 + x)^n using the binomial theorem, you get terms like C(n,0), C(n,1)x, C(n,2)x^2, and so on up to C(n,n)x^n. The coefficients are the binomial coefficients. Now, the problem mentions the sum of the odd-numbered terms and the sum of the even-numbered terms. I think "odd-numbered" refers to the terms with odd powers of x, and "even-numbered" refers to the terms with even powers of x. So, P is the sum of the coefficients of the odd-powered terms, and Q is the sum of the coefficients of the even-powered terms.Let me write that down:(1 + x)^n = Q + P*x, where Q is the sum of even-powered terms and P is the sum of odd-powered terms. Wait, actually, no, that might not be accurate. Let me think again.If I plug in x = 1 into (1 + x)^n, I get the sum of all coefficients, which is 2^n. Similarly, if I plug in x = -1, I get the alternating sum of coefficients, which is (1 - 1)^n = 0. Hmm, maybe I can use these two results to find P and Q.Let me denote S_even as the sum of coefficients of even-powered terms and S_odd as the sum of coefficients of odd-powered terms. Then, when x = 1, S_even + S_odd = 2^n. When x = -1, S_even - S_odd = 0. So, from these two equations, I can solve for S_even and S_odd.Adding the two equations: (S_even + S_odd) + (S_even - S_odd) = 2^n + 0 => 2*S_even = 2^n => S_even = 2^{n-1}.Subtracting the two equations: (S_even + S_odd) - (S_even - S_odd) = 2^n - 0 => 2*S_odd = 2^n => S_odd = 2^{n-1}.Wait, so both S_even and S_odd are equal to 2^{n-1}? That seems interesting. So, in the expansion of (1 + x)^n, the sum of the coefficients of the even-powered terms is equal to the sum of the coefficients of the odd-powered terms, each being half of the total sum, which is 2^n.But in the problem, P is the sum of the odd-numbered terms, and Q is the sum of the even-numbered terms. So, P = S_odd = 2^{n-1}, and Q = S_even = 2^{n-1}. Therefore, P = Q.But wait, that seems too straightforward. Let me check if I'm interpreting the question correctly. It says "the sum of the odd-numbered terms" and "the sum of the even-numbered terms." Maybe "numbered" refers to their position in the expansion rather than their power. For example, the first term is C(n,0), the second term is C(n,1)x, and so on. So, odd-numbered terms would be the 1st, 3rd, 5th, etc., terms, which correspond to the even-powered terms because the first term is x^0, the third term is x^2, etc. Similarly, even-numbered terms would be the 2nd, 4th, 6th, etc., terms, which correspond to the odd-powered terms.Oh, so maybe I got it backwards! If that's the case, then P is the sum of the coefficients of the even-powered terms, and Q is the sum of the coefficients of the odd-powered terms. Let me verify that.If I consider the expansion:(1 + x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n.The odd-numbered terms (1st, 3rd, 5th, etc.) are C(n,0), C(n,2)x^2, C(n,4)x^4, etc., which are the even-powered terms. So, P is the sum of these coefficients: C(n,0) + C(n,2) + C(n,4) + ... Similarly, the even-numbered terms (2nd, 4th, 6th, etc.) are C(n,1)x, C(n,3)x^3, C(n,5)x^5, etc., so Q is the sum of these coefficients: C(n,1) + C(n,3) + C(n,5) + ...Therefore, P is the sum of the even-powered coefficients, and Q is the sum of the odd-powered coefficients. Earlier, I found that both sums are equal to 2^{n-1}. So, P = Q = 2^{n-1}.But the question is about (1 - x^2)^n. Let me expand that. Using the binomial theorem:(1 - x^2)^n = C(n,0) - C(n,1)x^2 + C(n,2)x^4 - C(n,3)x^6 + ... + (-1)^n C(n,n)x^{2n}.Hmm, so it's an alternating sum of even-powered terms. Now, I need to express this in terms of P and Q.Wait, since P is the sum of the even-powered coefficients in (1 + x)^n, which is C(n,0) + C(n,2) + C(n,4) + ..., and Q is the sum of the odd-powered coefficients, which is C(n,1) + C(n,3) + C(n,5) + ... So, in (1 - x^2)^n, the coefficients are alternating: C(n,0) - C(n,1) + C(n,2) - C(n,3) + ... which is similar to the expansion of (1 - x)^n, but only for even powers.Wait, actually, (1 - x^2)^n can be written as [(1 + x)(1 - x)]^n = (1 + x)^n (1 - x)^n. So, maybe I can express (1 - x^2)^n as the product of (1 + x)^n and (1 - x)^n.Let me write that down:(1 - x^2)^n = (1 + x)^n * (1 - x)^n.Now, I can express both (1 + x)^n and (1 - x)^n in terms of P and Q.From earlier, (1 + x)^n = P + Qx, where P is the sum of even-powered coefficients and Q is the sum of odd-powered coefficients. Similarly, (1 - x)^n can be thought of as (1 + (-x))^n, so its expansion would be P - Qx, because the odd-powered terms would have negative signs.Therefore, (1 - x)^n = P - Qx.So, multiplying (1 + x)^n and (1 - x)^n:(1 + x)^n * (1 - x)^n = (P + Qx)(P - Qx) = P^2 - (Qx)^2 = P^2 - Q^2 x^2.Wait, but (1 - x^2)^n is a polynomial in x^2, so all the terms are even-powered. However, in the product above, we have P^2 - Q^2 x^2, which is only up to x^2. That doesn't seem right because (1 - x^2)^n should have terms up to x^{2n}.Hmm, maybe I made a mistake in how I expressed (1 + x)^n and (1 - x)^n. Let me think again.Actually, (1 + x)^n = P + Qx, where P is the sum of even-powered coefficients and Q is the sum of odd-powered coefficients. Similarly, (1 - x)^n = P - Qx, because the odd-powered terms get a negative sign. Therefore, when we multiply them:(1 + x)^n * (1 - x)^n = (P + Qx)(P - Qx) = P^2 - (Qx)^2 = P^2 - Q^2 x^2.But wait, (1 - x^2)^n is equal to (1 + x)^n (1 - x)^n, which is indeed P^2 - Q^2 x^2. However, this seems to only account for the x^0 and x^2 terms, but (1 - x^2)^n has higher even-powered terms as well.Wait, no, actually, when we multiply (1 + x)^n and (1 - x)^n, we get a polynomial where each term is the product of terms from (1 + x)^n and (1 - x)^n. So, the x^2 term comes from multiplying the x term in (1 + x)^n with the x term in (1 - x)^n, but with a negative sign. Similarly, higher even-powered terms come from products of higher odd-powered terms. But in our case, P is the sum of even-powered coefficients, and Q is the sum of odd-powered coefficients. So, when we multiply (P + Qx)(P - Qx), we get P^2 - Q^2 x^2, but in reality, (1 - x^2)^n has more terms.Wait, maybe I need to consider that P and Q are not just scalars but polynomials themselves. Let me clarify.Actually, P is the sum of the even-powered terms, which is a polynomial in x^2, and Q is the sum of the odd-powered terms, which is a polynomial in x. So, when we write (1 + x)^n = P + Qx, where P is the sum of even-powered coefficients and Q is the sum of odd-powered coefficients, then P is a polynomial in x^2, and Q is a polynomial in x.Therefore, when we multiply (1 + x)^n and (1 - x)^n, we get:(1 + x)^n * (1 - x)^n = (P + Qx)(P - Qx) = P^2 - (Qx)^2.But since P is a polynomial in x^2, P^2 is a polynomial in x^4, and (Qx)^2 is a polynomial in x^2. However, (1 - x^2)^n is a polynomial in x^2, so all terms are even-powered. Therefore, the product P^2 - (Qx)^2 should equal (1 - x^2)^n.But wait, (1 - x^2)^n is equal to the product of (1 + x)^n and (1 - x)^n, which is indeed P^2 - Q^2 x^2. However, since P and Q are sums of coefficients, not polynomials, this might not directly translate.Wait, maybe I'm overcomplicating this. Let me think differently. Since P is the sum of the even-powered coefficients and Q is the sum of the odd-powered coefficients in (1 + x)^n, then:(1 + x)^n = P + Qx.Similarly, (1 - x)^n = P - Qx.Therefore, multiplying them:(1 + x)^n (1 - x)^n = (P + Qx)(P - Qx) = P^2 - (Qx)^2.But (1 + x)^n (1 - x)^n = (1 - x^2)^n.So, (1 - x^2)^n = P^2 - Q^2 x^2.But in the problem, we are asked to express (1 - x^2)^n in terms of P and Q, without x. So, perhaps we need to evaluate this at a specific value of x to eliminate x.Wait, but the problem doesn't specify evaluating at a particular x. It just asks what (1 - x^2)^n equals in terms of P and Q. So, maybe we need to consider that in the expansion of (1 - x^2)^n, the coefficients can be expressed in terms of P and Q.Alternatively, perhaps I should consider substituting x = 1 and x = -1 into the expressions for P and Q to find their values.Wait, earlier I found that P = Q = 2^{n-1}, but that was under the assumption that P and Q are the sums of even and odd coefficients respectively. However, if P is the sum of the even-powered coefficients and Q is the sum of the odd-powered coefficients, then P = Q = 2^{n-1}.But then, (1 - x^2)^n evaluated at x = 1 is (1 - 1)^n = 0, and at x = -1 is (1 - 1)^n = 0 as well. But that doesn't seem helpful.Wait, maybe I need to consider the generating functions. Let me think.We have:(1 + x)^n = P + Qx,(1 - x)^n = P - Qx.Multiplying them gives:(1 - x^2)^n = P^2 - Q^2 x^2.But we need to express (1 - x^2)^n in terms of P and Q without x. So, perhaps we can consider evaluating at x = 1 or x = -1.If we set x = 1:(1 - 1^2)^n = 0 = P^2 - Q^2 * 1^2 => 0 = P^2 - Q^2.But we already know that P = Q, so this makes sense: P^2 - Q^2 = 0.But that doesn't help us express (1 - x^2)^n in terms of P and Q for general x.Wait, maybe I'm approaching this wrong. Let me think about the original problem again.The problem states that in the expansion of (1 + x)^n, the sum of the odd-numbered terms is P, and the sum of the even-numbered terms is Q. Then it asks what (1 - x^2)^n equals.Wait, perhaps "odd-numbered" refers to the position in the expansion, not the power. So, the first term is C(n,0), the second term is C(n,1)x, the third term is C(n,2)x^2, etc. So, the odd-numbered terms are the 1st, 3rd, 5th, etc., which correspond to the even-powered terms (x^0, x^2, x^4, ...). Therefore, P is the sum of the coefficients of the even-powered terms, and Q is the sum of the coefficients of the odd-powered terms.So, P = C(n,0) + C(n,2) + C(n,4) + ..., and Q = C(n,1) + C(n,3) + C(n,5) + ... As I found earlier, both P and Q equal 2^{n-1}.Now, (1 - x^2)^n can be written as (1 + x)^n (1 - x)^n. From earlier, (1 + x)^n = P + Qx, and (1 - x)^n = P - Qx. Therefore, their product is:(1 + x)^n (1 - x)^n = (P + Qx)(P - Qx) = P^2 - (Qx)^2.But (1 - x^2)^n is equal to this product, so:(1 - x^2)^n = P^2 - Q^2 x^2.However, the problem asks for (1 - x^2)^n in terms of P and Q, not involving x. So, perhaps we need to consider that (1 - x^2)^n is a polynomial in x^2, and we can express it as P^2 - Q^2, but that doesn't seem right because P and Q are sums of coefficients, not polynomials.Wait, maybe I'm missing something. Let me think about the generating functions again.If I consider (1 + x)^n = P + Qx, then (1 - x)^n = P - Qx. Therefore, multiplying them gives:(1 - x^2)^n = (P + Qx)(P - Qx) = P^2 - Q^2 x^2.But (1 - x^2)^n is a polynomial in x^2, so all the terms are even-powered. Therefore, if I set x = 1, I get:(1 - 1^2)^n = 0 = P^2 - Q^2 * 1^2 => P^2 - Q^2 = 0.But we already know that P = Q, so this is consistent. However, this doesn't help us express (1 - x^2)^n in terms of P and Q for general x.Wait, maybe I need to consider that (1 - x^2)^n is equal to P^2 - Q^2, but that would ignore the x^2 term. Alternatively, perhaps the problem is asking for the sum of the coefficients of (1 - x^2)^n, which would be (1 - 1)^n = 0, but that's not helpful.Alternatively, maybe I need to consider that (1 - x^2)^n can be expressed as (P + Qx)(P - Qx) = P^2 - Q^2 x^2, but since we're asked for (1 - x^2)^n in terms of P and Q, perhaps it's simply P^2 - Q^2, treating P and Q as scalars.But earlier, I found that P = Q = 2^{n-1}, so P^2 - Q^2 = 0, which is consistent with (1 - x^2)^n evaluated at x = 1 being 0. However, this doesn't capture the entire polynomial.Wait, perhaps the problem is asking for the value of (1 - x^2)^n when x is set to a particular value, but it's not specified. Alternatively, maybe it's asking for the expression in terms of P and Q without x, which would be P^2 - Q^2, but that seems to only hold when x = 1.I'm a bit confused here. Let me try a different approach. Let's consider specific values of n to test the options.Let's take n = 1. Then:(1 + x)^1 = 1 + x. So, P is the sum of the odd-numbered terms, which are the 1st term (1) and the 3rd term (which doesn't exist), so P = 1. Q is the sum of the even-numbered terms, which is the 2nd term (x), so Q = 1.Then, (1 - x^2)^1 = 1 - x^2.Now, let's check the options:A. P * Q = 1 * 1 = 1. But (1 - x^2) is not equal to 1, so A is incorrect.B. P^2 - Q^2 = 1^2 - 1^2 = 0. But (1 - x^2) is not equal to 0, so B is incorrect.C. P + Q = 1 + 1 = 2. But (1 - x^2) is not equal to 2, so C is incorrect.D. P^2 + Q^2 = 1^2 + 1^2 = 2. Again, not equal to (1 - x^2).Hmm, none of the options seem to work for n = 1. Maybe I made a mistake in defining P and Q.Wait, for n = 1, (1 + x)^1 = 1 + x. The odd-numbered terms are the 1st term (1) and the 3rd term (which doesn't exist), so P = 1. The even-numbered terms are the 2nd term (x), so Q = 1.But (1 - x^2)^1 = 1 - x^2. So, none of the options A through D give 1 - x^2 when n = 1. Therefore, perhaps my initial assumption about P and Q is incorrect.Wait, maybe P and Q are not the sums of the coefficients but the sums of the terms themselves, including the variables. So, in (1 + x)^n, P is the sum of the odd-numbered terms (1st, 3rd, etc.), which are 1, C(n,2)x^2, C(n,4)x^4, etc., and Q is the sum of the even-numbered terms (2nd, 4th, etc.), which are C(n,1)x, C(n,3)x^3, etc.So, P = 1 + C(n,2)x^2 + C(n,4)x^4 + ..., and Q = C(n,1)x + C(n,3)x^3 + C(n,5)x^5 + ... Then, (1 + x)^n = P + Q.Similarly, (1 - x)^n = P - Q, because the odd-powered terms get a negative sign.Therefore, (1 - x^2)^n = (1 + x)^n (1 - x)^n = (P + Q)(P - Q) = P^2 - Q^2.Ah, that makes sense! So, (1 - x^2)^n is equal to P^2 - Q^2.Let me verify this with n = 1 again. For n = 1:P = 1 (only the 1st term), Q = x (only the 2nd term).Then, P^2 - Q^2 = 1^2 - x^2 = 1 - x^2, which is indeed (1 - x^2)^1. So, it works for n = 1.Let me try n = 2:(1 + x)^2 = 1 + 2x + x^2.P is the sum of the odd-numbered terms: 1 (1st term) and x^2 (3rd term). So, P = 1 + x^2.Q is the sum of the even-numbered terms: 2x (2nd term). So, Q = 2x.Then, P^2 - Q^2 = (1 + x^2)^2 - (2x)^2 = 1 + 2x^2 + x^4 - 4x^2 = 1 - 2x^2 + x^4.On the other hand, (1 - x^2)^2 = 1 - 2x^2 + x^4. So, it matches.Therefore, the correct answer is B. P^2 - Q^2.</think>

🧠:<think>First, Melanie gave away 3 plums to Sam.After giving away the plums, she has 4 plums remaining.To find out how many plums she picked initially, we add the plums she gave away to the plums she has left.So, 3 plums (given to Sam) + 4 plums (left with Melanie) equals 7 plums.Therefore, Melanie picked 7 plums initially.</think>

🧠:<think>Okay, so I have this equation to prove: there are infinitely many sets of positive integers x, y, and z such that (x + y + z)^2 + 2(x + y + z) equals 5(xy + yz + zx). Hmm, that seems a bit complicated, but let me try to break it down.First, I should probably start by understanding what the equation is saying. It relates the sum of the variables squared plus twice their sum to five times the sum of their pairwise products. Maybe if I can find some patterns or specific values that satisfy this equation, I can see a way to generate infinitely many solutions.Let me try plugging in some small positive integers for x, y, and z to see if they work. Let's start with the simplest case where all variables are equal. So, let’s set x = y = z = 1.Substituting into the equation:Left-hand side (LHS): (1 + 1 + 1)^2 + 2(1 + 1 + 1) = 3^2 + 2*3 = 9 + 6 = 15Right-hand side (RHS): 5(1*1 + 1*1 + 1*1) = 5(1 + 1 + 1) = 5*3 = 15Okay, so (1, 1, 1) is a solution. That's good. But I need infinitely many solutions, not just one. Maybe if I can find a way to generate more solutions from this one.Let me think about how to manipulate the equation. Maybe I can express one variable in terms of the others. Let's say I solve for z. The equation is:(x + y + z)^2 + 2(x + y + z) = 5(xy + yz + zx)Let me expand the left-hand side:(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yzSo, LHS becomes x^2 + y^2 + z^2 + 2xy + 2xz + 2yz + 2x + 2y + 2zRHS is 5(xy + yz + zx) = 5xy + 5yz + 5zxNow, let's set LHS equal to RHS:x^2 + y^2 + z^2 + 2xy + 2xz + 2yz + 2x + 2y + 2z = 5xy + 5yz + 5zxLet me bring all terms to one side:x^2 + y^2 + z^2 + 2xy + 2xz + 2yz + 2x + 2y + 2z - 5xy - 5yz - 5zx = 0Simplify like terms:x^2 + y^2 + z^2 + (2xy - 5xy) + (2xz - 5xz) + (2yz - 5yz) + 2x + 2y + 2z = 0Which simplifies to:x^2 + y^2 + z^2 - 3xy - 3xz - 3yz + 2x + 2y + 2z = 0Hmm, that looks a bit messy. Maybe I can rearrange terms:x^2 - 3xy - 3xz + y^2 - 3yz + z^2 + 2x + 2y + 2z = 0I wonder if I can factor this somehow. It's a quadratic in terms of x, y, and z. Maybe I can fix two variables and solve for the third. Let's try fixing x and y and solving for z.Let me treat this as a quadratic equation in z:z^2 + (-3x - 3y + 2)z + (x^2 + y^2 - 3xy + 2x + 2y) = 0Yes, that's a quadratic in z. Let me write it as:z^2 + (-3x - 3y + 2)z + (x^2 + y^2 - 3xy + 2x + 2y) = 0Now, for this quadratic to have integer solutions, the discriminant must be a perfect square. The discriminant D is:D = [(-3x - 3y + 2)]^2 - 4*1*(x^2 + y^2 - 3xy + 2x + 2y)Let me compute D:D = (9x^2 + 18xy + 9y^2 - 12x - 12y + 4) - 4x^2 - 4y^2 + 12xy - 8x - 8ySimplify term by term:First, expand [(-3x - 3y + 2)]^2:= ( -3x - 3y + 2 )^2= ( (-3x)^2 ) + ( (-3y)^2 ) + (2)^2 + 2*(-3x)*(-3y) + 2*(-3x)*2 + 2*(-3y)*2= 9x^2 + 9y^2 + 4 + 18xy - 12x - 12ySo, D = 9x^2 + 9y^2 + 4 + 18xy - 12x - 12y - 4x^2 - 4y^2 + 12xy - 8x - 8yNow, combine like terms:9x^2 - 4x^2 = 5x^29y^2 - 4y^2 = 5y^218xy + 12xy = 30xy-12x - 8x = -20x-12y - 8y = -20y+4 remainsSo, D = 5x^2 + 5y^2 + 30xy - 20x - 20y + 4Hmm, that's still complicated. Maybe I can factor out a 5:D = 5(x^2 + y^2 + 6xy - 4x - 4y) + 4Not sure if that helps. Maybe I can look for a pattern or see if D can be expressed as a square.Alternatively, maybe I can assume some relationship between x, y, and z to simplify the equation. For example, suppose x = y. Let's see what happens.Let me set x = y. Then, the equation becomes:(x + x + z)^2 + 2(x + x + z) = 5(x*x + x*z + x*z)Simplify:(2x + z)^2 + 2(2x + z) = 5(x^2 + 2xz)Expand the left side:(4x^2 + 4xz + z^2) + (4x + 2z) = 5x^2 + 10xzSo:4x^2 + 4xz + z^2 + 4x + 2z = 5x^2 + 10xzBring all terms to one side:4x^2 + 4xz + z^2 + 4x + 2z - 5x^2 - 10xz = 0Simplify:- x^2 - 6xz + z^2 + 4x + 2z = 0Multiply both sides by -1:x^2 + 6xz - z^2 - 4x - 2z = 0Hmm, still a bit messy. Maybe I can rearrange terms:x^2 + 6xz - 4x - z^2 - 2z = 0Let me group terms with x:x^2 + (6z - 4)x - z^2 - 2z = 0This is a quadratic in x. Let me write it as:x^2 + (6z - 4)x - (z^2 + 2z) = 0For x to be an integer, the discriminant must be a perfect square. The discriminant D is:D = (6z - 4)^2 + 4(z^2 + 2z)Compute D:= 36z^2 - 48z + 16 + 4z^2 + 8z= (36z^2 + 4z^2) + (-48z + 8z) + 16= 40z^2 - 40z + 16Factor out 8:= 8(5z^2 - 5z + 2)Hmm, 5z^2 - 5z + 2 needs to be such that 8 times it is a perfect square. Let me see if 5z^2 - 5z + 2 can be a square.Let me denote k^2 = 5z^2 - 5z + 2So, 5z^2 - 5z + 2 = k^2This is a Diophantine equation. Maybe I can find integer solutions for z and k.Let me rearrange:5z^2 - 5z + (2 - k^2) = 0Treating this as a quadratic in z:5z^2 - 5z + (2 - k^2) = 0Discriminant D_z:D_z = 25 - 20(2 - k^2) = 25 - 40 + 20k^2 = 20k^2 - 15For z to be integer, D_z must be a perfect square. Let me set D_z = m^2:20k^2 - 15 = m^2So, m^2 = 20k^2 - 15This can be rewritten as:m^2 + 15 = 20k^2Or,m^2 = 20k^2 - 15Hmm, this seems tricky. Maybe I can look for small integer values of k and see if m^2 becomes a perfect square.Let me try k = 1:m^2 = 20*1 -15 = 5 → Not a square.k = 2:m^2 = 80 -15 = 65 → Not a square.k = 3:m^2 = 180 -15 = 165 → Not a square.k = 4:m^2 = 320 -15 = 305 → Not a square.k = 5:m^2 = 500 -15 = 485 → Not a square.k = 0:m^2 = 0 -15 = -15 → Not possible.Negative k won't help since k^2 is same as positive.Hmm, maybe this approach isn't working. Maybe assuming x = y isn't the best way.Let me try another approach. Maybe I can look for solutions where one variable is much larger than the others. For example, let me fix x and y and let z be large.Wait, but z is positive integer, so maybe I can express z in terms of x and y and see if I can find a recursive way to generate solutions.Going back to the quadratic in z:z^2 + (-3x - 3y + 2)z + (x^2 + y^2 - 3xy + 2x + 2y) = 0Let me denote this as:z^2 + Bz + C = 0Where B = -3x -3y + 2 and C = x^2 + y^2 -3xy + 2x + 2yIf I have one solution z1, then the other solution z2 can be found using Vieta's formula:z1 + z2 = -B = 3x + 3y - 2z1*z2 = CSo, if I know one solution z1, then z2 = 3x + 3y - 2 - z1This suggests that if I have a solution (x, y, z1), then (x, y, z2) is also a solution where z2 = 3x + 3y - 2 - z1Now, if I start with a solution, I can generate another solution by replacing z with z2. If I can show that this process can be repeated indefinitely, then I can generate infinitely many solutions.Let's test this with our initial solution (1,1,1). So, x=1, y=1, z1=1Then, z2 = 3*1 + 3*1 -2 -1 = 3 + 3 -2 -1 = 3So, z2=3. Let's check if (1,1,3) is a solution.Compute LHS: (1+1+3)^2 + 2*(1+1+3) = 5^2 + 2*5 = 25 +10=35Compute RHS: 5*(1*1 +1*3 +3*1)=5*(1+3+3)=5*7=35Yes, it works. So, (1,1,3) is a solution.Now, let's use this new solution (1,1,3) to find another solution. Let's set z1=3, x=1, y=1z2 = 3*1 +3*1 -2 -3=3+3-2-3=1Wait, that's back to z=1. So, this seems like it's oscillating between z=1 and z=3.Hmm, that's not helpful for generating new solutions. Maybe I need to vary x and y as well.Wait, in the previous step, I fixed x and y as 1 and only changed z. Maybe if I allow x and y to change, I can get more solutions.Alternatively, maybe I can fix z and vary x and y. Let me try that.Suppose I fix z=1. Then, the equation becomes:(x + y +1)^2 + 2(x + y +1) =5(xy + y*1 +1*x)Expand LHS:(x + y +1)^2 + 2(x + y +1) = x^2 + y^2 +1 + 2xy + 2x + 2y + 2x + 2y + 2Simplify:x^2 + y^2 +1 + 2xy + 4x + 4y + 2 = x^2 + y^2 + 2xy + 4x + 4y +3RHS:5(xy + y +x) =5xy +5x +5ySet equal:x^2 + y^2 + 2xy + 4x + 4y +3 =5xy +5x +5yBring all terms to one side:x^2 + y^2 + 2xy +4x +4y +3 -5xy -5x -5y=0Simplify:x^2 + y^2 -3xy -x -y +3=0Hmm, another quadratic. Maybe I can solve for y in terms of x.Let me rearrange:y^2 - (3x +1)y + (x^2 -x +3)=0This is a quadratic in y. For y to be integer, discriminant must be a perfect square.Discriminant D:D = [-(3x +1)]^2 -4*1*(x^2 -x +3) = (9x^2 +6x +1) -4x^2 +4x -12Simplify:9x^2 +6x +1 -4x^2 +4x -12 =5x^2 +10x -11So, D=5x^2 +10x -11 must be a perfect square.Let me set D=k^2:5x^2 +10x -11 =k^2This is another Diophantine equation. Maybe I can find integer solutions.Let me rearrange:5x^2 +10x -k^2 =11Hmm, not sure. Maybe try small x values.x=1:5 +10 -k^2=11 →15 -k^2=11→k^2=4→k=±2So, k=2 or -2. Let's take k=2.Then, y = [3x +1 ±k]/2 = [4 ±2]/2So, y=(4+2)/2=3 or y=(4-2)/2=1So, y=3 or y=1Thus, solutions when x=1, z=1: y=1 or y=3We already have (1,1,1). The other solution is (1,3,1). Let's check:LHS: (1+3+1)^2 +2*(1+3+1)=5^2 +2*5=25+10=35RHS:5*(1*3 +3*1 +1*1)=5*(3+3+1)=5*7=35Yes, works.Similarly, x=1, y=3, z=1 is a solution.Now, let's try x=2:5*(4) +10*(2) -k^2=20 +20 -k^2=40 -k^2=11→k^2=29→Not a square.x=3:5*9 +10*3 -k^2=45+30 -k^2=75 -k^2=11→k^2=64→k=±8So, k=8 or -8.Then, y=[3*3 +1 ±8]/2=[10 ±8]/2So, y=(10+8)/2=9 or y=(10-8)/2=1Thus, y=9 or y=1So, solutions when x=3, z=1: y=1 or y=9Check y=9:(3+9+1)^2 +2*(3+9+1)=13^2 +2*13=169+26=195RHS:5*(3*9 +9*1 +1*3)=5*(27+9+3)=5*39=195Yes, works.Similarly, y=1 is already known.So, seems like for x=3, we get y=9 as a new solution.Similarly, x=4:5*16 +10*4 -k^2=80+40 -k^2=120 -k^2=11→k^2=109→Not a square.x=5:5*25 +10*5 -k^2=125+50 -k^2=175 -k^2=11→k^2=164→Not a square.x=6:5*36 +10*6 -k^2=180+60 -k^2=240 -k^2=11→k^2=229→Not a square.x=7:5*49 +10*7 -k^2=245+70 -k^2=315 -k^2=11→k^2=304→Not a square.x=8:5*64 +10*8 -k^2=320+80 -k^2=400 -k^2=11→k^2=389→Not a square.x=9:5*81 +10*9 -k^2=405+90 -k^2=495 -k^2=11→k^2=484→k=±22So, k=22 or -22.Then, y=[3*9 +1 ±22]/2=[28 ±22]/2So, y=(28+22)/2=25 or y=(28-22)/2=3Thus, y=25 or y=3So, solutions when x=9, z=1: y=3 or y=25Check y=25:(9+25+1)^2 +2*(9+25+1)=35^2 +2*35=1225+70=1295RHS:5*(9*25 +25*1 +1*9)=5*(225+25+9)=5*259=1295Yes, works.So, it seems that for x=1,3,9,... we get solutions where y increases. Maybe this is a pattern.Wait, x=1, y=3; x=3, y=9; x=9, y=25...Wait, 1,3,9,25... That seems like squares: 1=1^2, 3 is not a square, 9=3^2, 25=5^2. Hmm, not exactly, but maybe related.Alternatively, maybe x follows a sequence where each term is 3 times the previous? 1,3,9,27,... but y is 3,9,25,... which doesn't fit.Alternatively, maybe x and y are related in some multiplicative way.Wait, when x=1, y=3; x=3, y=9; x=9, y=25.Wait, 3=3*1, 9=3*3, 25=3*9 + something? Not sure.Alternatively, maybe y = 3x when x=1,3,9,... but 3*1=3, 3*3=9, 3*9=27≠25. So, no.Alternatively, maybe y = x + 2 when x=1: y=3; x=3: y=9; x=9: y=25. Hmm, 3=1+2, 9=3+6, 25=9+16. Not a clear pattern.Alternatively, maybe y = (x^2 + something). For x=1, y=3=1+2; x=3, y=9=9+0; x=9, y=25=81-56. Not helpful.Alternatively, maybe y = (x + 2)^2 / something. For x=1, y=3=(1+2)^2 /3=9/3=3; x=3, y=9=(3+2)^2 / something=25/ something=9→something=25/9, which is not integer. Hmm.Alternatively, maybe y = 3x when x is 1,3,9,... but as before, 3*9=27≠25.Wait, maybe there's a different relationship. Let me see:From x=1, y=3; x=3, y=9; x=9, y=25.Notice that 3=3*1, 9=3*3, 25=3*9 - 2. Hmm, not sure.Alternatively, maybe y = x^2 + 2. For x=1, y=1+2=3; x=3, y=9+2=11≠9; so no.Alternatively, y = x + 2x=3x. For x=1, y=3; x=3, y=9; x=9, y=27≠25. Close but not exact.Wait, but when x=9, y=25, which is close to 3x=27. Maybe y=3x -2. For x=1, y=1≠3*1-2=1. Wait, 3*1-2=1, which is y=1, but we have y=3 as another solution. Hmm, conflicting.Alternatively, maybe y = (x + 2)^2 / something. For x=1, y=3=(1+2)^2 /3=9/3=3; x=3, y=9=(3+2)^2 / something=25/ something=9→something=25/9, not integer.Alternatively, maybe y = x^2 + 2x + something. For x=1, y=3=1+2+0; x=3, y=9=9+6+ something=15 + something=9→something=-6. Not helpful.Alternatively, maybe y = x^2 - 2x + something. For x=1, y=3=1-2+ something= -1 + something=3→something=4; x=3, y=9=9-6 +4=7≠9. No.Alternatively, maybe y = x^2 + 2. For x=1, y=3; x=3, y=11≠9; no.Alternatively, maybe y = 3x when x is 1,3,9,... but as before, 3*9=27≠25.Wait, maybe y = (x + 2)^2 - something. For x=1, y=3=(3)^2 -6=9-6=3; x=3, y=9=(5)^2 -16=25-16=9; x=9, y=25=(11)^2 - something=121 - something=25→something=96. Hmm, not a clear pattern.Alternatively, maybe y = (x + 1)^2 - something. For x=1, y=3=(2)^2 -1=4-1=3; x=3, y=9=(4)^2 -7=16-7=9; x=9, y=25=(10)^2 -75=100-75=25. Hmm, 1,7,75... Not a pattern.Alternatively, maybe y = x^2 + 2. For x=1, y=3; x=3, y=11≠9; no.Alternatively, maybe y = x + 2x=3x, but as before, doesn't fit for x=9.Wait, maybe I'm overcomplicating this. Let me think differently.From the quadratic in z, we saw that if we have a solution (x,y,z), then another solution is (x,y,3x+3y-2 -z). So, starting from (1,1,1), we get (1,1,3). Then, using (1,1,3), we get back to (1,1,1). Not helpful.But earlier, when we fixed z=1 and varied x and y, we found solutions like (1,3,1), (3,9,1), (9,25,1), etc. So, maybe if we fix z=1 and let x and y follow a sequence, we can generate infinitely many solutions.From the earlier examples:(1,1,1), (1,3,1), (3,9,1), (9,25,1), (25, ?,1), etc.Wait, the pattern seems like x_{n+1} = y_n, and y_{n+1} = something.From (1,1,1) to (1,3,1): x=1, y=3From (1,3,1) to (3,9,1): x=3, y=9From (3,9,1) to (9,25,1): x=9, y=25So, x_{n+1} = y_nAnd y_{n+1} = ?From 1 to 3: 3=1*3From 3 to 9:9=3*3From 9 to 25:25=9 +16Wait, 25=9 +16=9 +4^2Hmm, not sure.Alternatively, 3=1+2, 9=3+6, 25=9+16. The differences are 2,6,16,... which are 2, 6=2*3, 16=2^4. Not a clear pattern.Alternatively, 3=1*3, 9=3*3, 25=5*5. So, y_n = (2n+1)^2? For n=0, y=1=1^2; n=1, y=3= not a square; n=2, y=9=3^2; n=3, y=25=5^2. Hmm, seems like y_n = (2n+1)^2 for n starting at 0.Wait, but for n=0, y=1=1^2; n=1, y=3≠3^2=9; n=2, y=9=3^2; n=3, y=25=5^2. So, it's alternating between y being a square and not. Maybe not.Alternatively, maybe y_n = (x_n + 2)^2 / something. For x=1, y=3=(3)^2 /3=9/3=3; x=3, y=9=(5)^2 / something=25/ something=9→something=25/9; not integer.Alternatively, maybe y_n = x_n^2 + 2. For x=1, y=3=1+2; x=3, y=9=9+0; x=9, y=25=81-56. Not helpful.Alternatively, maybe y_n = x_n + 2x_n=3x_n. For x=1, y=3; x=3, y=9; x=9, y=27≠25. Close but not exact.Wait, but when x=9, y=25, which is 3*9 -2=25. So, y=3x -2.Let me check:For x=1, y=3=3*1 -0=3For x=3, y=9=3*3 -0=9For x=9, y=25=3*9 -2=25Hmm, so maybe y=3x -2 when x is 9, but not for x=1 and x=3.Alternatively, maybe y=3x when x is 1,3,9,... but for x=9, y=25≠27.Wait, maybe y=3x - something. For x=9, y=25=27-2=3*9 -2For x=3, y=9=9-0=3*3 -0For x=1, y=3=3-0=3*1 -0So, maybe y=3x -2 when x is 9, but not for smaller x.Alternatively, maybe y=3x - 2*(x-1). For x=1, y=3*1 -2*0=3; x=3, y=9 -2*2=5≠9; no.Alternatively, maybe y=3x - 2*(x-1). For x=1, y=3-0=3; x=3, y=9-4=5≠9; no.Alternatively, maybe y=3x - 2*(x-1). For x=1, y=3-0=3; x=3, y=9-4=5≠9; no.Alternatively, maybe y=3x - 2*(x-1). For x=1, y=3-0=3; x=3, y=9-4=5≠9; no.Hmm, maybe this approach isn't working. Let me think differently.From the earlier examples, when z=1, we have solutions where x and y follow a sequence: 1,3,9,25,...Wait, 1,3,9,25 are 1^2, 3^1, 3^2, 5^2. Not a clear pattern.Alternatively, maybe x_{n+1} = y_n, and y_{n+1} = 3x_{n+1} -2.From (1,1,1):x=1, y=1Next, x=1, y=3 (since y=3x -2=3*1 -2=1, but we have y=3, so maybe not.Wait, when x=1, y=3, then next x=3, y=9.Then, x=3, y=9, next x=9, y=25.So, y=3x when x=1,3,9,... but for x=9, y=25≠27.Wait, maybe y=3x -2 when x is a multiple of 3.For x=3, y=9=3*3 -0=9For x=9, y=25=3*9 -2=25For x=27, y=3*27 - something=81 - something. Let's see, if we follow the pattern, y=3x -2 when x=9, so maybe y=3x -2 for x=9,27,...But for x=3, y=9=3*3 -0=9So, maybe y=3x -2 when x is 9,27,... and y=3x when x is 1,3,9,...Wait, but x=9 is both 3x=27 and y=25=3x -2=25. So, maybe y=3x -2 when x is a multiple of 9.But this is getting too speculative.Alternatively, maybe I can find a recursive formula for x and y.From the earlier examples:(1,1,1) → (1,3,1) → (3,9,1) → (9,25,1) → ...So, x_{n+1} = y_nAnd y_{n+1} = ?From (1,1,1) to (1,3,1): y=3=1*3From (1,3,1) to (3,9,1): y=9=3*3From (3,9,1) to (9,25,1): y=25=9 +16Wait, 25=9 +16=9 +4^2Similarly, 9=3 +6=3 +2*33=1 +2=1 +2*1Hmm, maybe y_{n+1}= y_n + (2y_n)^2 / something.Wait, not sure.Alternatively, maybe y_{n+1}= y_n + (y_n - x_n)^2.From (1,1,1): y=1, x=1, so y_{n+1}=1 + (1-1)^2=1+0=1, but we have y=3. No.Alternatively, y_{n+1}= y_n + (x_n + y_n)^2.From (1,1,1): y=1 + (1+1)^2=1+4=5≠3. No.Alternatively, y_{n+1}= y_n + (x_n)^2.From (1,1,1): y=1 +1=2≠3. No.Alternatively, y_{n+1}= y_n + 2x_n.From (1,1,1): y=1 +2*1=3. Yes.From (1,3,1): y=3 +2*1=5≠9. No.Wait, but in the next step, x=3, so y=3 +2*3=9. Yes.From (3,9,1): y=9 +2*3=15≠25. No.Wait, but in reality, y=25. So, 25=9 +16=9 + (2*3)^2=9 +36=45≠25. No.Alternatively, y_{n+1}= y_n + (x_n)^2.From (1,1,1): y=1 +1=2≠3. No.From (1,3,1): y=3 +1=4≠9. No.Alternatively, y_{n+1}= y_n + 2x_n.From (1,1,1): y=1 +2=3. Yes.From (1,3,1): y=3 +2*1=5≠9. No.But in reality, y=9=3 +6=3 +2*3=9. So, maybe y_{n+1}= y_n +2x_n when x_{n+1}= y_n.Wait, let's see:From (1,1,1):x=1, y=1Next, x=1, y=3=1 +2*1=3Then, x=3, y=9=3 +2*3=9Then, x=9, y=25=9 +2*9=27≠25. Hmm, close but not exact.Wait, 25=9 +16=9 + (2*4)^2. Not sure.Alternatively, maybe y_{n+1}= y_n + 2x_n - something.From (9,25,1): y=25=9 +2*9 -2=9+18-2=25. Yes.So, y_{n+1}= y_n +2x_n -2Let me test this:From (1,1,1):y_{n+1}=1 +2*1 -2=1+2-2=1≠3. No.From (1,3,1):y_{n+1}=3 +2*1 -2=3+2-2=3≠9. No.From (3,9,1):y_{n+1}=9 +2*3 -2=9+6-2=13≠25. No.Hmm, not matching.Alternatively, maybe y_{n+1}= y_n +2x_n + something.From (1,1,1):y=1 +2*1 +0=3From (1,3,1):y=3 +2*1 +4=9From (3,9,1):y=9 +2*3 + something=9+6 + something=15 + something=25→something=10So, 0,4,10,... which are 0,4,10,... differences of 4,6,...Not a clear pattern.Alternatively, maybe y_{n+1}= y_n + (x_n)^2.From (1,1,1):y=1 +1=2≠3From (1,3,1):y=3 +1=4≠9No.Alternatively, maybe y_{n+1}= y_n + 2x_n + (x_n)^2.From (1,1,1):y=1 +2*1 +1=4≠3No.Alternatively, maybe y_{n+1}= y_n + 2x_n + something else.This is getting too convoluted. Maybe I should accept that the sequence is x=1,3,9,25,... and y=3,9,25,... and z=1, and that this can be extended indefinitely, giving infinitely many solutions.But how can I be sure that this sequence can continue indefinitely? Maybe by induction.Assume that for some n, (x_n, y_n,1) is a solution. Then, we can find y_{n+1} such that (y_n, y_{n+1},1) is also a solution.From the earlier quadratic in y, when z=1, we have:y^2 - (3x +1)y + (x^2 -x +3)=0Given that x_n is known, y_{n+1} can be found as the other root of this quadratic, given that y_n is one root.From Vieta's formula, the sum of roots is 3x_n +1, and the product is x_n^2 -x_n +3.So, if y_n is one root, the other root y_{n+1}= (3x_n +1) - y_nAnd y_n * y_{n+1}=x_n^2 -x_n +3So, given y_n, we can find y_{n+1}=3x_n +1 - y_nBut since x_{n+1}=y_n, we can write:y_{n+1}=3x_n +1 - y_n=3x_n +1 -x_{n+1}But x_{n+1}=y_n, so:y_{n+1}=3x_n +1 -x_{n+1}But x_{n+1}=y_n, so:y_{n+1}=3x_n +1 -y_nBut from the previous step, y_{n}=3x_{n-1} +1 -y_{n-1}Wait, this is getting recursive.Alternatively, maybe we can express y_{n+1} in terms of x_n and y_n.Given that y_{n+1}=3x_n +1 - y_n, and x_{n+1}=y_n, we can write:y_{n+1}=3x_n +1 -x_{n+1}But x_{n+1}=y_n, so:y_{n+1}=3x_n +1 -y_nBut we also have from the previous step:y_n=3x_{n-1} +1 -y_{n-1}So, substituting:y_{n+1}=3x_n +1 - (3x_{n-1} +1 -y_{n-1})=3x_n +1 -3x_{n-1} -1 +y_{n-1}=3x_n -3x_{n-1} +y_{n-1}But y_{n-1}=x_{n-2}, since x_{n-1}=y_{n-2}Wait, this is getting too tangled. Maybe I should accept that the sequence can be extended indefinitely, giving infinitely many solutions.Therefore, starting from (1,1,1), we can generate (1,3,1), (3,9,1), (9,25,1), etc., each time using the quadratic to find the next y, and thus generating infinitely many solutions.Hence, there exist infinitely many sets of positive integers x, y, z satisfying the given equation.</think>