If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried: 
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$  divides through both $xy$ and  $x^2 + y^2$ 
This then allows me to introduce another equation $1=ax+by$ for some $a, b.$
But I then get stuck ...

Suppose $xy|x^2+y^2$. Then $xy|x^2+y^2+2xy=(x+y)^2$. But if $\gcd(x,y)=1$, then also
$$
1=\gcd(x,x+y)=\gcd(y,x+y)=\gcd(xy,x+y)=\gcd(xy,(x+y)^2)
$$
from which it then follows that $xy=\pm 1$.
