integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of  $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.

If $(x,y)=k$ then $k^4(x_1-y_1)(x_1+y_1)x_1y_1= z^2$ which implies an equation (we use the same notation) $(x-y)(x+y)xy=z^2$ with $(x,y)=1$; it follows that $(x+y,x-y)=1$  ($x+y=mX$ and $x-y=mY$ give clearly a contradiction). Hence $x-y,  x+y, x, y$ are four coprime integers and each of them must be a square, $x, y$ being of different parity (if not then $2|(x-y,x+y)=1$).
We have three pythagorean triples to which we apply the well known parametrization
$(a,b,c) =(p^2-q^2, 2pq, p^2+q^2)$:
(1)………………$x^2=y^2+u^2$
(2)……………..$x^2+y^2=v^2$
(3)…………….$x^4+y^4=(uv)^2$
If $y$ were odd then $x$ would be odd in (1) and even in (2), absurde, so $y$ must be even.
Now from (3)
$x^2=t^2+s^2, y^2=2ts$ implies $ {x^4-y^4=(t^2-s^2)^2}$ , $t$ and $s$ of distinct parity.
Since $t^2+s^2=x^2$, we write $t=m^2-n^2, s=2mn$ and $x=m^2+n^2$  with $(m,n)=1$.
It follows $y^2=2(m^2-n^2)(2mn)=4(m^2-n^2)mn$
Hence $m=p^2, n=q^2$ and $y^2=4(p^2-q^2)p^2q^2$ and since y is even,$(p^2-q^2)(p^2+q^2)p^2q^2= w^2$.
Thus, from $(x^2-y^2)(x^2+y^2)x^2y^2=z^2$ we have got another equation
$(p^2-q^2)(p^2+q^2)p^2q^2= w^2$ similar to the first one with the values of $p,q$ less than the values of $x, y$. This process can be iterated indefinitely therefore, by infinite descent,  we finish the proof.  
