Question 349548
If n=m^3-m for some integer m, then n is a multiple of 6.


My work so far:

Suppose n=m^3 - m. Thus n=m(m-1)(m+1).
This is correct. From this point I would say<ul><li>If m is an integer, then (m-1), m, and (m+1) are 3 consecutive integers</li><li>In <i>any</i> set of 3 consecutive integers, there will be a multiple of 2 and a multiple of 3.</li><li>Any number that is the product of a multiple of 2 and a multiple of 3 will be a multiple of 6.</li></ul>
This may not be particularly formal as a proof but the logic is perfectly sound.