How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$?

Question

How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$?

I have no clue to find solution . I tried to change into quadratic equation $$y^2(x-1)+yx^2-(x^2+1)=0\\y=\frac{-y\pm\sqrt{x^4+4(x-1)(x^2+1)(-1)}}{2(x-1)}$$ but stop here ...

Please help to me to find a method of solving . Thanks in advance. Remark : by graphing that I found $$\quad{(x,y)=\\(2,1)\\(1,2)\\(2,-5)\\(-5,2)}$$

Answer 1

I recommend to do it in terms of $\,x+y\,$ and $\,xy$. Then the equation becomes

$$xy=\frac{1+(x+y)^2}{x+y+2}.$$

And we have $\,(x-y)^2=(x+y)^2-4xy\geq 0$. After determining the ranges of $\,x+y\,$ and $\,xy\,$ with these equations and inequalities, the integer solutions can be found one by one.

Answer 2

Write it like this:

$$ xy (x+y)-(x+y)^2+2xy =1$$

If we put $a=x+y$ and $b=xy$ we get $$ b ={1+a^2\over a+2}\;\;\;\implies\;\;\;a+2\mid a^2+1$$

Now since $a+2\mid a^2-4$ we have $a+2\mid 5$ so $a+2\in\{\pm 1,\pm 5\}$ and thus $a\in\{-3,-1,-7,3\}$ \begin{array}{c|c|c|c} a & b & x&y\\\hline -7 & -10 &/&/\\ -3 & -10 &-5(2)&2(-5)\\ -1 & 2 &/&/\\ 3& 2 &2(1)&1(2)\\ \end{array}