<s>[INST] Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are perfect squares Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are both perfect squares.
My attempt:
Let $x=ha,y=xb,z=xc$, then $a,b,c$ are positive integers such that $\gcd(a,b,c)=1$.
Now, suppose $\gcd(a,b)=g$, so, $a=ga',b=gb'$, where $\gcd(a',b')=\gcd(a',a'-b')=\gcd(b',a'-b')=1$. So, we have $c(b'-a')=ga'b'$, so $g\mid c$, and $g=1$.  
Now, I do not know what to do. Please help. Thank you.
 [/INST] Let us consider
$$
\frac{1}{x} - \frac{1}{y} = \frac{1}{z},
$$
and
$$
h = \textrm{gcd}(x,y,z).
$$
Say that
$$
\begin{eqnarray}
x &=& h x',\\
y &=& h y',\\
z &=& h z',\\
\end{eqnarray}
$$
then we obtain
$$
\frac{1}{x'} - \frac{1}{y'} = \frac{1}{z'},
$$
where
$$
\textrm{gcd}(x',y',z') = 1.
$$

We can write
$$
\Big(y'-x'\Big)z' = x'y'.
$$
Let us look to odd and even:
$$
\begin{array}{ccccl}
x' & y' & z'\\
\textrm{even} & \textrm{even} & \textrm{even} & \Rightarrow &
  \textrm{gcd}(x',y',z') > 1.\\
\textrm{odd} & \textrm{odd} & \textrm{odd} & \Rightarrow&
  \textrm{contradiction as $xy$ is odd but $(x-y)z$ is even}.\\
\textrm{odd} & \textrm{even} & \textrm{even}\\
\textrm{even} & \textrm{odd} & \textrm{even}
\end{array}
$$
So we write
$$
\begin{eqnarray}
y' &=& x' + 2p + 1,\\
z' &=& 2q,
\end{eqnarray}
$$
whene
$$
x'^2 + \Big(2p+1\Big)x' = \Big(2p+1\Big)2q,
$$
so
$$
x' = - \frac{2p+1}{2} + \frac{2p+1}{2} \sqrt{1 + \frac{8q}{2p+1}},
$$
meaning that
$$
q = \frac{1}{2} r \Big( r + 1 \Big) \Big( 2p + 1 \Big),
$$
so that
$$
\begin{eqnarray}
x' &=& r \Big( 2p + 1 \Big),\\
y' &=& \Big( r + 1 \Big) \Big( 2p + 1 \Big),\\
z' &=& r \Big( r + 1 \Big) \Big( 2p + 1 \Big).\\
\end{eqnarray}
$$
However $\textrm{gcd}(x',y',z') = 1$, thus $2p+1=1$, whence
$$
\begin{eqnarray}
x &=& h r,\\
y &=& h \Big( r + 1),\\
z &=& h r \Big( r + 1 \Big).\\
\end{eqnarray}
$$

It is clear that
$$
\begin{eqnarray}
hxyz &=& \left[ h^2 r \Big( r + 1 \Big) \right]^2,\\
h(y-x) &=& \Big[ h r \Big]^2,\\
\end{eqnarray}
$$
thus both $hxyz$ and $h(y-x)$ are perfect squares.
</s>