<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 with divisibility by $2$, good.
Suppose $5\nmid xyz$, so $5$ divides none of the numbers $x$, $y$ and $z$.
Then $x^2\equiv\pm1\pmod{5}$ and the same for $y$ and $z$. However, it can't be $x^2\equiv1$ and $y^2\equiv-1$, otherwise $z^2\equiv0\pmod{5}$. Similarly it's impossible that $x^2\equiv-1$ and $y^2\equiv1$.
Suppose $x^2\equiv1$ and $y^2\equiv1$. Then $z^2\equiv1+1\pmod{5}$, which is a contradiction. Similarly $x^2\equiv-1\equiv y^2$ can be dismissed.
Hence $5\mid xyz$.

Alternatively, which is what the book seems to be doing, suppose $5\nmid xy$. Then, since $x^2\equiv\pm1$ and $y^2\equiv\pm1$, we can conclude that
\begin{align}
z^2&=x^2+y^2\equiv2\pmod5 &&\text{or} \\[4px]
z^2&=x^2+y^2\equiv-2\pmod5 &&\text{or} \\[4px]
z^2&=x^2+y^2\equiv0\pmod5
\end{align}
The first two cases are impossible, as seen above, so only the third case remains: since $5\mid z^2$, we conclude $5\mid z$.
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