I'm trying to get to grips with mathematical proofs. I think I almost have this, but am missing something.

Q) Prove that if n=m^3-m, where m is an integer, then n is a multiple of 6.

A) A proof by contradiction:

Assume the inverse of the statement, that if n=m^3-m then n is not a multiple of 6.

This implies that n = 6k +{1,2,3,4,5} where k is an integer. In other words, n has a remainder of between 1 and 5 if you try to divide it by 6.

Now I need to show that this implies a contradiction with the original statement (n=m^3-m) for every case, {1,2,3,4,5}.

Case 1: n = 6k+1

By the orignal statement, this is equal to m^3-m. But m^3-m is always even, and 6k+1 is always odd. This is therefore obviously false. The same logic applies to cases 3 and 5.

Case 2: n = 6k+2

Again, by the original statement this is equal to m^3-m. But how do I show that this is false. This time both sides are even?

This is where I get into trouble. I am a bit confused on how to prove that 6k+2=m^3-m leads to an obvious contradiction. I can see that you can disprove it by counter example: If k=1 then n=8, but m^3-m can only take values {0,0,6,24,...}. So in general, the statement is false. In fact, it is only true when the number we add on to 6k is 6, otherwise we always 'miss' the values that m^3-m can take. So I think I can see it in words, but how do I formulate it into a proper proof?