<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] $$ x^2 + y^2 = kxy  $$
$$ x^2 - k xy + y^2 = 0. $$
Divide by $y^2,$ let $r = x/y.$
$$ r^2 - kr + 1 = 0 $$
Discriminant is
$$ k^2 - 4. $$
Thi is not a square unless $k = \pm 2$
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