Show that the product of all Pythagorean Triples is divisible by 60.

Okay so I'm almost done solving this problem, but that's almost.

I've made the starting assumption that all relatively prime Pythagorean Triples are of the form 2mn,(m+n)(m-n) and m^2+n^2, but I can't prove this starting assumption.

Once this starting assumption is made, it's fairly straight forward to show that 4 and 3 will always be factors of the non-hypotenuse side, and with some modular arithmetic it's also rather straightforward to show that either the side of the form (m+n)(m-n) or the hypotenuse will be a multiple of 5.

But I can't finish the question because I have no idea how the starting assumption works. This is an assumption that is supposedly true which is why I'm using it, but how do I show that?

It's a bit late in the evening for serious thinking, but I think you can do this directly without your starting assumption.

Write down a^2 + b^2 = c^2 in the cases mod 3, mod 4 and mod 5. If you can prove that one of a,b,c always has to be divisible by 3, 4 and 5 then you've got what you want.