My approach:
let $x=3 \lambda, y=4 \lambda,z=5 \lambda $ where $\lambda \in \mathbb Z$ ,then $x^2+y^2=z^2 \implies $ $(3 \lambda)^2+(4 \lambda)^2=(5 \lambda)^2$
Since $xyz=60 \lambda^3$ & $60 \lambda^3 \equiv0 \pmod{60}$ $\implies xyz \equiv0 \pmod{60}$
Hence proved!
Is it correct?
Is there any alternative method of doing this?
Since for any $\;m\in\Bbb Z\;$ we have that $\;m^2=0,\,1\pmod 4\;$, this means one of $\;x,y\;$ is even and one is odd and then $\;z\;$ is odd, or else all three are even.
Also, $\;m^2=0,1,-1\pmod5\;$ , so either all are multiples of $\;5\;$ or else exactly one is.
Also, $\;m^2=0,1\pmod 3\;$ so either all or exactly one are multiples of $\;3\;$:
$\color{red}{Case\; 1}$ -- all three $\;x,y,z\;$ are even. Then clearly $\;3,4,5\,\mid\,xyz\;$ and thus $\;60\,\mid xyz\;$
$\color{red}{Case\; 2}\;$ -- $\;x\;$ is even, $\;y,z\;$ are odd : we then have
$$x^2=(z-y)(z+y)$$
and now check that no matter what remainder modulo $\;4\;$ both $\;y,z\;$ are, the above right product is always at least a multiple of $\;8\;$ and thus in fact $\;x\;$ must be a multiple of $\;4\;$, so again $\;60\,\mid\,xyz\;$
Hint $ $ Using the parametrization of Pythagorean triples it reduces to $\,30\mid ab(b^4-a^4),\,$ which is case $\,p=5\ $ of $\,\ 6p\ |\ ab^p\! - ba^p\,$ for all $\,a,b\in \mathbb Z,\,$ prime $\:p\ne 2,3,\,$ which I proved here.
