<s>[INST] Solve $(x+1)^2y + (y+1)^2x=0, x, y \in \mathbb{Z} $ Solve $$(x+1)^2y + (y+1)^2x=0, x, y \in \mathbb{Z} $$
I kind of know the only integer solutions are $\{0, 0\}, \{-1, -1\}$ but I don't know how to prove that
 [/INST] Rewrite it like this: $$xy(x+y)+4xy+x+y=0$$ and let $a=x+y$ so $$xy = -{a\over a+4}$$
then $a+4\mid a$ so $$a+4\mid (a+4)-a =4\implies  a+4\in \{-4,-2,-1,1,2,4\}$$
So $a\in \{-8,-6,-5,-3,-2,0\}$


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*If $a=0$ then $xy =0$ and $y=-x$ so $x=y=0$

*If $a=-2$ then $xy =1$ and $y=-x-2$ so $x=y=-1$

*If $a=-3$ then $xy =3$ and $y=-x-3$ and no solution

*If $a=-5$ then $xy =-5$ and $y=-x-5$ and no solution

*If $a=-6$ then $xy =-3$ and $y=-x-6$ and no solution

*If $a=-8$ then $xy =-2$ and $y=-x-8$ and no solution.

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