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.
 A: 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.
A: As above, let $x=ha$, $y=hb$, and $z=hc$ where $h=gcd(x,y,z)$
Then $(y-x)z=xy\implies h^2(b-a)c=h^2(ab)\implies(b-a)c=ab$ with $gcd(a,b,c)=1$.
Let $p\vert b-a$ with $p$ prime.  Then $p\vert ab\implies p\vert a$ or $p\vert b$; and 
$\textbf{1)}$ if $p\vert a$, then $p\vert a$ and $p\vert b-a\implies p\vert b\implies p^2\vert ab\implies p^2\vert (b-a)c$ with $(p^2, c)=1\implies p^2\vert b-a$.
$\textbf{2)}$ Similarly, if $p\vert b$, then $p^2\vert b-a$.
[Notice that if $b-a$ has no prime divisor, then $b-a=1$.]
Thus $b-a$ is a perfect square, so it follows that 
$hxyz=h^4abc=h^4(b-a)c^2$ and $h(y-x)=h^2(b-a)$ are perfect squares.
A: From @OP, we have $c(b-a) = ab$.
Suppose $gcd(a,b) = g$ and we write $a = ga'$ and $b = gb'$ and $g \ge h$
We have $c(b'-a') = ga'b'$. We have $gcd(c, a'b') = 1$. 
So $c$ has to divide $g$. If $g > c$, then $gcd(c,g) \gt h$. Hence $c = g = h$
$\implies b' - a' = a'b' \implies a' \mid b'$, a contradiction since $gcd(a',b') = 1 \implies g = gcd(a,b) = 1$. 
We now have $c(b-a) = ab$ where $gcd(a,b) = gcd(ab, b-a) = 1$
So $c \mid ab$. If $ab = kc$ then this contradicts $gcd(ab, b-a) = 1$ for $k \gt 1 \implies k = 1$
So $c = ab$. Hence $abc$ is a square. 
Also $b-a = 1$ which is a square. 
