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
 A: Working in modulo $x$, we have $y \equiv 0 \pmod x$, i.e. $x$ divides $y$. By symmetry, $y$ divides $x$. So we must have $y = \pm x$.
$y=x$ gives $2(x+1)^2 x = 0$, which yields solutions $(0,0)$ and $(-1,-1)$.
$y=-x$ gives $-x(x+1)^2+(1-x)^2x=0$, which we factorise as $x((1-x)^2-(x+1)^2)=0$. So either $x=0$, yielding $(0,0)$ again, or $(1-x)^2 = (x+1)^2$, which again yields $x=0$. 
So $(-1,-1)$ and $(0,0)$ are indeed the only solutions.
A: As you have seen, $(0,0)$ and $(-1,-1)$ are solutions. 
Therefore, we assume that $(x,y)\neq (0,0), (-1,-1)$. 
By assumption, we have 
$$(x+1)^2y=-(y+1)^2x.\tag{1}$$
Since $x$ and $(x+1)$ is relatively prime, $x$ must divide $y$. 
Similarly, since $y$ and $(y+1)$ are relatively prime, $y$ must divide $x$. Therefore, 
we must have $x=\pm y$. But this is impossible: 
If $x=y$, it follows from $(1)$ that $(x+1)^2x=-(x+1)^2x$, which gives $1=-1$ since $x\neq 0,-1$. 
Similarly, if $x=-y$, it follows from $(1)$ that $(x+1)^2x=(-x+1)^2x$, which gives $(x+1)^2=(-x+1)^2$ which is only true when $x=0$. 
To conclude, $(0,0)$ and $(-1,-1)$ are only solutions. 
A: 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 $a\in \{-8,-6,-5,-3,-2,0\}$


*

*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.

A: Here's an interesting solution:
Define $a,b \in \mathbb{Z}$ such that $a=x+1$ and $b=y+1$, we get
$$(x+1)^2y+(y+1)^2x=0 \implies a^2(b-1)+b^2(a-1)=0$$
$$\iff ab(a+b)=a^2+b^2$$
Now, define $m,n \in \mathbb{Z}$ such that $m=a+b$ and $n=ab$, we get
$$mn=m^2-2n\iff n=\frac{m^2}{m+2}$$
$$m \equiv -2 \pmod{m+2} \implies m^2 \equiv4 \equiv0 \pmod{m+2}$$
So $k(m+2)=4 \implies m+2 \in \{1,4,-4,-1,2,-2\}$
$$\implies (m,n) \in \{(-1,1),(2,1),(-6,-9),(-3,-9),(0,0),(-4,-8)\}$$
It can be investigated noting that the equation
$$x^2-mx+n=0 \tag{1}$$
has roots $a,b$ and we shall find such integers $a,b$
We get that the only solutions that work for $(m,n)$ are $(0,0)$ and $(2,1)$ getting $a=b=0 \implies x=y=-1$ or $a=b=1 \implies x=y=0$
