If $1/x-1/y=1/z$, where $x,y,z$ are positive integers, prove that $xyz\gcd(x,y,z)$ is a perfect square I found that $(z+y)(z-x)=z^2$ but I don't know how to continue

Let the gcd of the three numbers to be $d$.
Then, $x=dx_0$ and $y=dy_0$ $z=dz_0$ where $x_0$ and $y_0$ and $z_0$ are relatively prime.
by rewritting the condition on $x,z,y$, we can get
$yz-xz=xy$
Now, substitute the variables in the last equations like this
$d^2y_0z_0-d^2x_0z_0=d^2x_0y_0$
after cancelling $d$ out, from both sides
$y_0z_0-x_0z_0=x_0y_0$
Now, according to the last equation
$x_0|y_0z_0-x_0z_0$
therefore
$x_0|y_0z_0$
but $x_0$ and $y_0$ and $z_0$ are relatively prime.so, $x_0=1$
with the same argument $y_0=1$ and $z_0=1$
now consider the multiplication again
$gcd(x,y,z)xyz=d^4x_0y_0z_0=d^4$
