# Transition to higher mathematics course, proof

1. Aug 26, 2008

### duke_nemmerle

1. The problem statement, all variables and given/known data
Show that if $$n = m^3 - m$$ for some integer $$m$$, then $$n$$ is a multiple of 6.

2. Relevant equations
The relevant information is we don't know modular arithmetic yet, and the only methods of proof we have available are direct proof, contradiction, and counterexample.

3. The attempt at a solution

I was thinking of trying to prove it by contradiction, by finding something strange happening in all of the cases where n isn't a multiple of six i.e. for n = 6k + 1 for some integer k on up through n = 6k +5 for some integer k.

For example, in the case where n was one greater than a multiple of 6 we'll have
$$m^3 - m - 6k - 1 = 0$$

I was really hoping to be able to find that the roots of the cubic violated our hypothesis that m be an integer, but I'm having a hard time doing this and I don't want to go nuts on it if it won't be right.

I don't want an outright solution, but a hint would be appreciated.

2. Aug 26, 2008

### Hitman2-2

Look at $$m^3 - m$$ and factor it. What do you notice about each of the factors?

3. Aug 26, 2008

### duke_nemmerle

Ahh, I bet that will do it for me. I had never gotten it completely factored until now, and even if I had I may not have noticed it. I'm going to walk down the hall real quick and I'm sure I'll figure out how the product of three consecutive integers should be a multiple of both 2 and 3, ie of 6 in the same way that the product of two consecutive integers must be even (a multiple of two).

Thanks