Prove that in any pythagorean triple, one of the side lengths is divisible by 5

The trick is to realize what your choices are.

if m or n or both are divisible by 5, then 2mn will be divisible by 5

I don't know if you know about x mod (5), but in that case there can only be 4 possibilities for m and n

m mod (5) = + / - 2 or m mod (5) = + / - 1

In the first case m^2 = 4 mod (5) and in the second case m^2 = 1 mod (5) N has the same considerations.

If m and n have the same modular results (both are [say] 2) then m^2 - n^2 which be 0

If they are different, then (m^2 + n^2) mod 5 = 0