<s>[INST] Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).   
Next, for any $n \in Z$, if $5$ doesn't divide $n$, then $(n^2)^2=n^4 \equiv 1\pmod{5}$ (as you can check or quote Euler's theorem), and therefore $n^2 \equiv \pm 1\pmod{5}$. Now, if $5$ doesn't divide $xy$, then $x^2 \equiv \pm 1\pmod{5}$ and $y^2 \equiv \pm 1\pmod{5}$, so
$$x^2+y^2 \equiv -2,0,2 \pmod{5}$$
Therefore if $x^2+y^2=z^2$ and $5$ doesn't divide $xy$ then $z^2=x^2+y^2 \equiv -2,0,2\pmod{5}$, so $5\mid z^2$ (why? it looks weird to me because $z^2$ can be also congruent to $2$ and $-2$) and hence $5\mid z$ (otherwise $z^2 \equiv \pm 1\pmod{5}$). It follows that $x^2+y^2=z^2$, then either $5\mid xy$ or $5\mid z$ (how does this follow? if $5$ doesn't divide $z$ then how can $5$ divide $xy$), so in any case $5\mid xyz$
Finally, we can conclude that if $x^2+y^2=z^2$ then $2\mid xyz$ and $5\mid xyz$, so $10\mid xyz$ (intuitively it looks right but I can't prove it!)
 [/INST] No problem to give a "theoretical" proof. Just for fun let me work with a table in $\mathbb F_5^2$.
By the quite known parametrization of the Pythagorean triple, $xyz$ is even so we must prove  in 
$\mathbb F_5$ that
$$xyz=mn(m^2-n^2)(m^2+n^2)=0\qquad (1)$$
If $m=n$ and if $mn=0$ it is clear so we prove $(1)$  with $m\ne n$ and $mn\ne 0$ so we need to verify $(1)$ with $12$ couples $(m,n)$
$$\begin{array}{|c|c|}\hline  (1,1) & (1,2) & (1,3) &(1, 4) \\\hline (2,1) & (2,2) & (2,3) &(2,4)\\\hline (3,1) &(3,2) & (3,3) &(3,4)\\\hline (4,1)&(4,2)&(4,3)&(4,4)\\\hline\end{array}$$ 
If $(m,n)\in \{(1,2),(1,3),(2,1),(2,4),(3,1),(3,4),(4,2),(4,3)\}$ then $m^2+n^2=0$
If $(m,n)\in\{(1,4),(2,3),(3,2),(4,1)\}$ then $m^2-n^2=0$
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