Find the value of $a$ if $x^2+y^2=axy$ has positive integer solution.

Question

Find the value of $a$ if $x^2+y^2=axy$ has positive integer solution.

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My try:
Let g.c.d of $x$ and $y$ is $d$ i.e.$(x,y)=d$ and let $x=dx',y=dy'.$

Then $x'^2+y'^2=ax'y'$

I am stuck here.The answer given is $a=2$.I do not understand how.Please help.

Answer 1

Suppose that $x^2+y^2-axy=0$. Then $x\mid y^2$ and $y\mid x^2$ in positive integers. From this we see that $x$ and $y$ have the same prime divisors, possibly with different multiplicity. But then the equation implies $x=y$, hence $x^2(a-2)=0$. Since $x>0$ we obtain $a=2$.

Answer 2

$\frac{x^2+y^2}{xy}=a$. Assume $gcd(x,y)=d$ and $x=du, y=dv$ where $(u,v)=1.$ So we get $\frac{u^2+v^2}{uv}=a$. Now, $u|u^2+v^2$, so $u|v^2$ and so $u|v$. By similar argument, we have $v|u$. But $(u,v)=1$ so we must have both $u,v$ either $+1$ or $-1.[u,v$ both have the same sign because $a$ is positive$].$ In both cases, we have $a=2$.

Answer 3

Let $m=\frac{y}{x}$, the equation becomes $m+\frac{1}{m}=a$ and $m^2-ma+1=0$. In order to get rationals for $m$ discriminant must be a square; That is $a^2-4=k^2$. Only possible when $a=2$.