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