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 

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.
