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 ...
 A: It makes little sense if either  of $x,y$ is zero. 
I will continue with $x,y \neq 0.$
If $x^2 + y^2 = kxy$ for nonzero integer $k,$ we have 
$$ x^2 - k xy + y^2 = 0 $$
We are taking $y \neq 0,$ so we may divide through by $y^2,$ define $r = \frac{x}{y},$ giving
$$  r^2 - kr + 1 = 0  $$
with integer $k$ and rational $r.$
So: what are the roots $r$ of
 $$  r^2 - kr + 1 = 0 \; ? \; $$
Can the roots actually be rational? For what values of $k$ can the roots be rational?
A: Suppose that $\gcd(x,y)=1$ and $(xy)\mid(x^2+y^2)$.
Then $y^2\equiv0\pmod x$. If $1=ax+by$ then $by\equiv1\pmod x$ and so
$1\equiv(by)^2=b^2y^2\equiv0\pmod x$. So $x=\pm1$. Likewise, $y=\pm1$.
A: 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 $xy=\pm 1$.
A: If $(x,y)=d,$ and $\dfrac xX=\dfrac yY=d$ so that $(X,Y)=1$
So, we need $XY$ to divide $X^2+Y^2$
$\implies X|(X^2+Y^2)\iff X|Y^2$ with $(X,Y)=1$ which is possible only if $X=\pm1$
Similarly $Y=\pm1$
A: Rational Algebraic Integer Approach
Suppose that
$$
\frac{x^2+y^2}{xy}=\frac xy+\frac yx\in\mathbb{Z}\tag1
$$
Note that if $q=\frac xy\in\mathbb{Q}$ and $q+\frac1q=n\in\mathbb{Z}$, then
$$
\left(q-\frac1q\right)^2=n^2-4\in\mathbb{Z}\tag2
$$
This means that $z=q-\frac1q$ is a rational solution to $z^2-(n^2-4)=0$; that is, $z$ is a rational algebraic integer. Thus, $z\in\mathbb{Z}$ (see this answer). Therefore, $(n+z)(n-z)=4$ is an integer factorization of $4$ where both factors have the same parity. That is, $n+z=n-z=\pm2$, which means $n=\pm2$ and $q-\frac1q=z=0$. Thus, $\frac{x^2}{y^2}=q^2=1$, and therefore, $x=\pm y$.

Bezout Approach
Let $d=(x,y)$ and $u=x/d$ and $v=y/d$. Then, there exist $a,b$ so that $au+bv=1$. Suppose that
$$
\begin{align}
n
&=\frac{x^2+y^2}{xy}\\
&=\frac{u^2+v^2}{uv}\\
&=\frac{b^2u^2+(1-au)^2}{bu(1-au)}\\
&=\frac{\left(a^2+b^2\right)u^2-2au+1}{bu-abu^2}\tag3
\end{align}
$$
Then
$$
\frac1u=n(b-abu)+2a-\left(a^2+b^2\right)u\in\mathbb{Z}\tag4
$$
Thus, $u\cdot\frac1u=1$ is an integral factorization of $1$. That is, $u=\pm1$. Similarly, $v=\pm1$.
Therefore, $x=\pm d$ and $y=\pm d$, which means that $x=\pm y$.
