Prove by induction that, for any natural umber n, 6 divides n^3 - n.

Let n = 1. Then n^3 - n = 1 - 1 = 0, and 6(0) = 0, so 6 divides n^3 - n.

Let n = k, and assume 6 divides k^3 - k, so 6m = k^3 - k for some natural number m.

Let n = k + 1. Then (k + 1)^3 - (k + 1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^3 + 3k^2 + 2k.

What then?