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

Question

I found that $(z+y)(z-x)=z^2$ but I don't know how to continue

Answer 1

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

$gcd(x,y,z)xyz=d^4x_0y_0z_0=d^4$