60 divides xyz for every Pythagorean triple

Idea. Squares have very few residues, so each prime power can be handled by a short case check.

Modulo $3$. Squares are $0$ or $1$. If $3\nmid x$ and $3\nmid y$, then $z^2\equiv1+1=2\pmod3$, impossible. So $3\mid xy$.

Modulo $5$. Squares are $0,1,4\equiv-1$. If $5\nmid xyz$, then $x^2,y^2,z^2\in\{\pm1\}$ and $x^2+y^2\in\{-2,0,2\}$ is never $\equiv\pm1$. So $5\mid xyz$.

Divisibility by $4$. Squares modulo $4$ are $0,1$, so $x,y$ are not both odd (else $z^2\equiv2$).
- If $x,y$ are both even, then $4\mid xy$.
- Otherwise exactly one is even, say $x$, and $y,z$ are odd. Odd squares are $\equiv1\pmod8$, so $x^2=z^2-y^2\equiv0\pmod8$. Writing $x=2m$, $4m^2\equiv0\pmod8$ forces $m$ even, so $4\mid x$.

Conclusion. $3$, $4$ and $5$ are pairwise coprime and each divides $xyz$, so $60\mid xyz$.

Check. $(3,4,5)$: $60$. $(5,12,13)$: $780=13\cdot60$. $(8,15,17)$: $2040=34\cdot60$ ✓.

Lesson. Reduce to the list of squares modulo $3$, $4$ (or $8$) and $5$: most "which numbers divide..." questions about squares are closed by such a check.