How find this postive integer $a$ such $a(x^2+y^2)=x^2y^2$ always have roots Find all postive integer numbers of $a$,such this equation
$$a(x^2+y^2)=x^2y^2,xy\neq0$$ 
always have  integer roots $(x,y)$
my try: since
$$\dfrac{x^2y^2}{x^2+y^2}\in N$$
and I can't 
Thank you 
 A: Take any two positive coprime integers $u$ and $v$, and a third positive integer $t$.
Let $q$ be the square-free part of $u^2 + v^2$: thus $u^2 + v^2 = qs^2$ for some positive integer $s$.
Let $a = u^2 v^2 t^2 q$. 
Then the equation $a(x^2+y^2) = x^2y^2$ has the non-trivial solution $x = tsqu$ and $y = tsqv$ because
$a(x^2+y^2) = u^2v^2t^2q(tsq)^2 (u^2+v^2) = u^2v^2t^2(tsq)^2q^2s^2 = (tsq)^4u^2v^2 = x^2y^2.$
Conversely, if there is a non-trivial solution to this equation for some $a$, then $a$ must necessarily have the form given above. 
To see this, let $z = \gcd(x,y)$ and write $x = zu$ and $y = zv$ for some coprime positive integers $u,v$. Then $a(u^2+v^2) = z^2u^2v^2$. Since $u^2v^2$ and $u^2+v^2$ cannot share a prime factor, $a = bu^2v^2$ for some positive integer $b$. Let $q$ be the square-free part of $u^2+v^2$ and write $u^2 + v^2 = qs^2$ for some positive integer $s$. Then $b q s^2 = z^2$ forces $q$ to divide $b$. So $b = qw$ and $w (qs)^2 = z^2$; hence $w = t^2$ is a perfect square. Putting everything together gives $a = u^2v^2t^2q$ as claimed.
A: If $(x,y)=1$, then $(x^2+y^2,x^2y^2)=1$. Suppose $(x,y)=d$, then $\left(\frac xd,\frac yd\right)=1$ and therefore
$$
\left(\frac{x^2+y^2}{d^2},\frac{x^2y^2}{d^4}\right)=1\tag{1}
$$
which implies that both
$$
\left(\frac{x^2+y^2}{d^2},\frac{x^2}{d^2}\right)=1\quad\text{and}\quad\left(\frac{x^2+y^2}{d^2},\frac{y^2}{d^2}\right)=1\tag{2}
$$
Suppose that
$$
\frac{x^2y^2}{x^2+y^2}=a\in\mathbb{Z}\tag{3}
$$
then
$$
\frac{x^2\frac{y^2}{d^2}}{\frac{x^2}{d^2}+\frac{y^2}{d^2}}
=\frac{\frac{x^2}{d^2}y^2}{\frac{x^2}{d^2}+\frac{y^2}{d^2}}
=a\tag{4}
$$
Thus, $(2)$ and $(4)$ require
$$
\left.\frac{x^2}{d^2}+\frac{y^2}{d^2}\middle|\,x^2\right.\quad\text{and}\quad\left.\frac{x^2}{d^2}+\frac{y^2}{d^2}\middle|\,y^2\right.\tag{5}
$$
which in turn require
$$
\left.\frac{x^2}{d^2}+\frac{y^2}{d^2}\middle|\,d^2\right.\tag{6}
$$
Let $u=\frac xd$ and $v=\frac yd$. Then, $(u,v)=1$ and $(6)$ guarantees that
$$
u^2+v^2\mid d^2\tag{7}
$$
and $(3)$ becomes
$$
\frac{d^2}{u^2+v^2}u^2v^2=a\tag{8}
$$
So pick any $u,v$ so that $(u,v)=1$, and let $u^2+v^2=b^2c$ where $c$ is square-free, then by setting $d$ to be any multiple of $bc$, we get that $a$ can be any square multiple of $cu^2v^2$.
