<s>[INST] Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$ 
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.

The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
 [/INST] Let $p$ be a prime, write $x=p^au, y=p^bv$ where $u,v$ are relatively prime with $p$. $xy=p^{a+b}uv$. Suppose $a<b$ If $xy$ divides $x^2+y^2=p^{2a}u^2+p^{2b}v^2$, we have ${{p^{2a}u^2+p^{2b}v^2}\over{p^{a+b}}}$ is an integer. This implies that $p^{a+b}$ divides $p^{2a}u^2$. This is impossible. We deduce that $a\geq b$ and similarly$b\geq a$. This implies that $x=y$ or $x=-y$.
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