Find all integer numbers $\displaystyle x,y$ such that
$\displaystyle (x^2+y)(x+y^2)=(x-y)^2$
Well, here's three for you for $\displaystyle x<y$, none of which are positive though: $\displaystyle (-1,2), (-1,-1), (0,1)$. I'm not finding any more by blind searching. By making the substitutions $\displaystyle y=ax, ab=1$, you can get a quadratic on $\displaystyle x$: $\displaystyle x^2+(a+b^2)x-(b^2-3b+1)=0$ as long as $\displaystyle x\ne 0$. Solve for $\displaystyle x$ using quadratic formula and replace $\displaystyle a,b$ back in terms of $\displaystyle x,y$ and you can simplify to $\displaystyle x=-\frac{1}{2}\frac{x^3+y^3}{xy^2}\pm\frac{x-y}{2xy^2}\sqrt{x^4+2x^3y+7x^2y^2+2xy^3+y^4}$. Not sure if that's any better, but you might be able to parametrize that bit under the radical to get a list of $\displaystyle x,y$ pairs that give you a perfect square. That's as far as I've gotten so far. Was there a context for this?
There are exactly 4 integer solutions to this eqaution as calculated by WolframAlpha search engine ... here's the result:
(x^2+y)*(x+y^2) ;=(x-y)^2 - Wolfram|Alpha