Integer solutions to $x^2(y-1)+y^2(x-1)=1$

Question

Find all values of $(x,y)$ where $x,y$ belongs to integers if: $$x^2(y-1)+y^2(x-1)=1$$

I m a beginner so I do need some help.

Answer 1

Hint:

Write as quadratic on $x$ and parameter $y$: $$(y-1)x^2+y^2x-(1+y^2)=0$$

If $y\ne 1$ then it discriminat must be perfect square:$$y^4+4y^3-4y^2+4y-4=z^2$$ for some $z$. Now use: $$y^4< z^2<(y+1)^4$$

for $y>1$...and then try something similary for $y<1$.

Answer 2

Hint:

Perhaps it would be easier if you write the equation like this

$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1\implies b ={a^2+1\over a+2}$$

This means, since $b$ is integer, that $$a+2\mid a^2+1$$

and now you should continue...

---

or you could write $c=a+2$ and then $$b = {(c-2)^2+1\over c} = {c^2-4c+5\over c}= c-4 +{5\over c}$$

so $c\mid 5$....