<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] The general solution of the above Diophantine Equation is: $x = m^2 - n^2, y = 2mn, z = m^2+n^2$. Its better if you start at this point. Then your analysis yields the followings: $m^2 = \pm 1 \pmod 5, n^2 = \pm 1 \pmod 5$. Since $5 \nmid xy\implies 5 \nmid x, 5 \nmid y\implies m^2 = 1\pmod 5, n^2 = -1\pmod 5$ or $m^2 = -1\pmod 5, n^2 = 1\pmod 5$ for otherwise $x  = m^2 - n^2 = \pm1 - \pm1 = 0\pmod 5$, contradicting the assumption that $5 \nmid x$. Thus: $m^4 + 2m^2n^2 + n^4 = 1 + 2(-1) + 1 = 0 \pmod 5\implies 5 \mid (m^2+n^2)^2 = z^2$.
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