If $x^2 + y^2 = z^2$, then $ xyz \equiv 0\pmod{60} $ (Pythagorean triples) 
Prove that, if $x$, $y$ and $z$ are integers, and if $x^2 + y^2 = z^2$, then $ xyz \equiv 0\pmod{60} $.

Attempt:
$ xyz \equiv 0\pmod{60} \iff 60\mid xyz$. Now notice that $60 = 3 \cdot 4 \cdot 5 $ so the statement $3 \cdot 4 \cdot 5|xyz $ is equivalent to the original statement.
Now I'd have to prove separately that $3|xyz$, $4|xyz$, and $5|xyz$, and I'm not sure how to proceed. 
At the same time, it will likely be too much work to prove each of these statements separately, so any other more concise/direct way to tackle this problem is welcome. Thanks.
 A: The easiest way to do this is to note:
If $m \equiv -1,0,1 \pmod 3$ then $m^2 \equiv 0,1\pmod 3$.  If none of them $x, y$ or $z$ are divisible by $3$ then $x^2,y^2,z^2 \equiv 1 \pmod 3$ and $1+1\equiv 1\pmod 3$ is a contradiction.  So at least one of $x,y,z$ is divisible by $3$.
Similarly if $m\equiv -2,-1,0,1,2 \pmod 5$ then $m^2 \equiv \pm 1, 0\pmod 5$.  If none of $x,y,$or $z$ are divisible by $5$ then we have $\pm 1 +\pm 1 \equiv \pm 1$ which is impossible.  So at least one of $x,y,z$ is divisible by $5$.
$4$ is a problem in not being prime we have $m\equiv 2\implies m^2 \equiv 0$. And it implies it's not enough to assume none of them are divisible by $4$ but that at most one is divisible by $2$.  
So if $m\equiv -1,0,1,2\pmod 4$ then $m^2 \equiv 0,1$ and at most one is divisible by $2$  we can have $(0,1)+(0,1)\equiv (0,1)$ only by having one of $x$ or $y$, wolog $x$, is even and $z$ and the other odd.
But then we have $x^2 = z^2 -y^2 = (z-y)(z+y)$  Both $z-y$ and $z+y$ as sum/difference of odd numbers are even.  But we can further claim that one is divisble by $4$ and the other isn't.
If $y = 2n + 1$ and $z=2m+1$ and $z -y = 2(m-n)$ with $m-n$ odd/even. Then $z+y= 2(m+n+1)$ with $m+n+1 = (m-n) + 2n + 1$ even/odd.  So one is $2*odd$ and the other is $2*even = 4*something$.
So $x^2 = (z-y)(z-x)$ is divisible by $8$ and $x$ is divisible by $4$.
A: A generalisation of Euclid's formula states that $a=k(m^2-n^2), b=2kmn$ and $c = k(m^2 + n^2)$ will generate any Pythagorean triple, where $m,n,k$ are positive integers, $m > n$, $m$ and $n$ are coprime, and at least one of $m,n$ is even.
This means you now have to prove that:
$$k^3(m+n)(m-n)(2mn)(m^2+n^2) \equiv 0 \pmod {60}$$
and you can do this by proving that the expression is congruent to $0 \pmod 3$, $0 \pmod 4$, and $0 \pmod 5$ as you mentioned.
