Congruence equation.

Question

Find all the $x$ for which there exists $y$ such that $x^2+y^2 \equiv x \mod xy$.

I can write it as $k(xy)+x=x^2+y^2$ for some $k$. But I don't really know how to move on from here.

Answer 1

Squares will do. After all, if $x=n^2$, we may choose $y=n$.

The hard part is proving that nothing else will.

Let's suppose $x$ is divisible by $p^{2k-1}$, but not any higher power of the prime $p$. Then, looking at $axy+x=x^2+y^2$ modulo $p^{2k-1}$, we conclude that $y^2$ is also divisible by $p^{2k-1}$. But wait, being a square, it is consequently divisible by $p^{2k}$; so is $xy$, and so is $x^2$, but $x$ is not. That makes a contradiction.