The answer is $\boxed{(0, 0), (-1, -2), (-2, -1)}.$ Multiply by $xy$ on both sides and rearrange to get $$(xy + x + y)^4 = (xy + x)^4 + (xy + y)^4.$$ By Fermat's last theorem, this only holds when one or more terms on the RHS are $0.$ Simple casework finishes. $\blacksquare$ Remark. "A nice easy problem" - Milan Haiman, I think

Solution from Twitch Solves ISL: The answer is $(-1,-2)$, $(-2,-1)$ and $(0,0)$. They work; we prove that's all. Let $a = x+y$ and $b = xy$. Then this becomes \[ b(b^2-12b-12a+2) = (2a)^2 \implies 4 \cdot a^2 + 12b \cdot a - b(b^2-12b+2) = 0. \]As a quadratic in $a$, the discriminant is \[ (12b)^2 + 4 \cdot 4b(b^2-12b+2) = 16 b \cdot (b^2 - 3b + 2) = 16b(b-1)(b-2) \]but it should also be a perfect square. This only happens if $b \in \{0,1,2\}$. From here we recover the solution set by some casework.

cw357 wrote: v_Enhance wrote: Solution from Twitch Solves ISL: The answer is $(-1,-2)$, $(-2,-1)$ and $(0,0)$. They work; we prove that's all. Let $a = x+y$ and $b = xy$. Then this becomes \[ b(b^2-12b-12a+2) = (2a)^2 \implies 4 \cdot a^2 + 12b \cdot a - b(b^2-12b+2) = 0. \]As a quadratic in $a$, the discriminant is \[ (12b)^2 + 4 \cdot 4b(b^2-12b+2) = 16 b \cdot (b^2 - 3b + 2) = 16b(b-1)(b-2) \]but it should also be a perfect square. This only happens if $b \in \{0,1,2\}$. From here we recover the solution set by some casework. This solution feels indirect, laborious, and poorly thought out. The $x$ and $y$ should be canceled before taking the discriminant, thus there will be no perfect square and you can just do a Diophantine Quadratic approximation. Would you mind editing your post to reflect this? Thanks! would you mind posting your own solution?

cw357 wrote: My Solution rjiangbz wrote: Find all ordered pairs of integers $x,y$ such that $$xy(x^2y^2 - 12xy- 12x- 12y+2) = (2x + 2y)^2.$$ Proposed by Henry Jiang divide x and y out of both sides then you get $x^2y^2-12xy-12x-12y+2=4^2$ rest is trivial $\boxed{}$ can evan apologize? henlo, (x,y) = (0,0) is a solution, tanks also wao i never knew that $\frac{(2x+2y)^2}{xy} = 4^2$ so smart

cw357 wrote: My Solution rjiangbz wrote: Find all ordered pairs of integers $x,y$ such that $$xy(x^2y^2 - 12xy- 12x- 12y+2) = (2x + 2y)^2.$$ Proposed by Henry Jiang divide x and y out of both sides then you get $x^2y^2-12xy-12x-12y+2=4^2$ rest is trivial $\boxed{}$ can evan apologize? This solution feels indirect, laborious, and poorly thought out. Would you mind editing your post to reflect this? Thanks!

No, sir please enlighten me with your true intelligence and personal greatness

cw357 wrote: My Solution rjiangbz wrote: Find all ordered pairs of integers $x,y$ such that $$xy(x^2y^2 - 12xy- 12x- 12y+2) = (2x + 2y)^2.$$ Proposed by Henry Jiang divide x and y out of both sides then you get $x^2y^2-12xy-12x-12y+2=4^2$ rest is trivial $\boxed{}$ can evan apologize? how about no?

evan thx for deleting (if that was you )

Does this solution work? It sounds ridiculously bashy Case 1: If one of x, y are 0 x = 0 forces y = 0, and vice versa. Thus our only solution from this case is (0, 0). Case 2: If x, y are non-zero Dividing both sides of the equation by xy, we get that (xy)^2 - 12xy - 12x - 12y + 2 = 4(x/y + y/x) + 8 (1). Thus we have to force x/y + y/x to be an integer. Let gcd(x, y) = q, x = q*x', y = q*y', where x' and y' are relatively prime integers. Then we must have that 4(x'/y' + y'/x') is an integer. So y'|4, x'|4, and now we apply casework on y' and x'. We assume WLOG that |y'| > |x'| (note that the given equation is symmetric, so the order of x and y doesn't really matter). By the above, |x'| is forced to be 1, since |y'| and |x'| are relatively prime. Case 1: y' = x' Then y=x, and (1) becomes x^4 - 12x^2 - 24x - 10 = 0. It is easy to see by rational root theorem that this has no integral solution. Case 2: y' = -x'. Then y = -x, and (1) becomes x^4 +12x^2 + 2 = 0. It is easy to see by rational root theorem that this has no integral solution of x. Case 3: y' = 2x' Then y = 2x, and (1) becomes 4x^4 - 24x^2 - 36x -16 = 0. This produces x = -1 by rational root theorem, so our solutions are (-1, -2). Case 4: y' = -2x' Then y = -2x, and (1) becomes 4x^4 + 24x^2 +12x +4 = 0. Rational root theorem gives no integral values of x in this solution either. Case 5: y' = 4x' Then y = 4x, and there are no roots because the LHS of (1) is a multiple of 2, while RHS of (1) becomes an odd number. Case 6: y' = -4x' Then y = -4x, and there are no roots because the LHS of (1) is a multiple of 2, while RHS of (1) becomes odd. Thus by symmetry, x,y nonzero produces the solution pairs (-1, -2) and (-2, -1). Thus all ordered pairs of integers (x, y) that satisfies the given equation are (0, 0), (-1, -2) and (-2, -1).

I have a few questions about my proof. 1. Do I necessarily need to write our my polynomial expansions for my cases? 2. Is this solution even valid? Please provide me some feedback. Thanks! chenghaohu