Ytukyg

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 that $gcd(x,y)=1$ and $(xy)mid(x^2+y^2)$.
Then $y^2equiv0pmod x$. If $1=ax+by$ then $byequiv1pmod x$ and so
$1equiv(by)^2=b^2y^2equiv0pmod x$. So $x=pm1$. Likewise, $y=pm1$.

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?

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 $xy=pm 1$.

Rational Algebraic Integer Approach

Suppose that
$$
frac{x^2+y^2}{xy}=frac xy+frac yxinmathbb{Z}tag1
$$
Note that if $q=frac xyinmathbb{Q}$ and $q+frac1q=ninmathbb{Z}$, then
$$
left(q-frac1qright)^2=n^2-4inmathbb{Z}tag2
$$
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, $zinmathbb{Z}$ (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
$$
begin{align}
n
&=frac{x^2+y^2}{xy}\
&=frac{u^2+v^2}{uv}\
&=frac{b^2u^2+(1-au)^2}{bu(1-au)}\
&=frac{left(a^2+b^2right)u^2-2au+1}{bu-abu^2}tag3
end{align}
$$
Then
$$
frac1u=n(b-abu)+2a-left(a^2+b^2right)uinmathbb{Z}tag4
$$
Thus, $ucdotfrac1u=1$ is an integral factorization of $1$. That is, $u=pm1$. Similarly, $v=pm1$.

Therefore, $x=pm d$ and $y=pm d$, which means that $x=pm y$.

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 $Y=pm1$