Providing divisibility condition given fraction identity If $x,y,z$ are positive integers satisfying 
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
prove that $20{\,\mid\,}xy$. 

My work:

Expanding, we find 
$$(xz)^2+(yz)^2=(xy)^2$$
 I know the Pythagorean triple formula and I tried applying that, but I couldn't find a way to get $20$. 
I've found a lot of questions on this website relating to similar questions, but none of them seem to refer to a divisibility condition.

If someone could help me find one, that'd also be greatly appreciated. 
 A: Suppose $x,y,z$ are positive integers such that
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
Equivalently, $x,y,z$ are positive integers such that
$$\qquad\qquad\; x^2y^2=z^2(x^2+y^2)\qquad(\textbf{eq})$$
Aqua has already shown that $5{\,\mid\,}(xy)$.

To show that $4{\,\mid\,}(xy)$, we can argue as follows . . .

If $x,y$ are both even, then $4{\,|\,}(xy)$, and we're done.

If $x,y$ are both odd, then $x^2+y^2$ is even, hence the RHS of $(\textbf{eq})$ is even, contradiction, since the LHS is odd.

It remains to resolve the case where exactly one of $x,y$ is even.

Without loss of generality, assume $x$ is even and $y$ is odd.

Let $2^k$ be the largest power of $2$ which divides $x$.

Since $y^2$ and $x^2+y^2$ are both odd, it follows from $(\textbf{eq})$ that $2^k$ is also the highest power of $2$ which divides $z$.

Thus we can write $x=2^kx_1$ and $z=2^kz_1$, where $x_1,z_1$ are both odd.
\begin{align*}
\text{Then}\;\;&x^2y^2=z^2(x^2+y^2)\\[4pt]
\implies\;&x_1^2y^2=z_1^2(x^2+y^2)\\[4pt]
\implies\;&x_1^2y^2\equiv z_1^2(x^2+y^2)\;(\text{mod}\; 8)\\[4pt]
\implies\;&(1)(1)\equiv (1)(x^2+1)\;(\text{mod}\; 8)\;\;\;\text{[since $x_1,z_1,y$ are odd]}\\[4pt]
\implies\;&x^2\equiv 0\;(\text{mod}\; 8)\\[4pt]
\implies\;&8{\,\mid\,}x^2\\[4pt]
\implies\;&16{\,\mid\,}x^2\\[4pt]
\implies\;&4{\,\mid\,}x\\[4pt]
\implies\;&4{\,\mid\,}(xy)\\[4pt]
\end{align*}
as required.

This completes the proof.
A: If $5\nmid xy$ then $x^2\equiv_5 \pm 1$ and $y^2\equiv_5 \pm 1$ 


*

*If $x^2\equiv_5  1$ and $y^2\equiv_5 1$ then $2z^2 \equiv 1$ so $z^2 \equiv 3$ which is not true.

*If $x^2\equiv_5  -1$ and $y^2\equiv_5 -1$ then $-2z^2 \equiv 1$ so $z^2 \equiv -3$ which is not true.

*If $x^2\equiv_5  1$ and $y^2\equiv_5 -1$ then $0 \equiv -1$ s which is not true.


So $5\mid xy$. 
Now try $4\mid xy$. It should be easier.
A: Some element of any triple is always divisible by $x\in\{3,4,5\}$ as shown here. If you follow the proof, it becomes clear. Sometimes one element is divisible my more than one of these. As a result, the product of some two of $A,B,C$ must be a $1$-or-more multiple of $20$.
