maggiemagnet wrote: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?

You've got the assumption about k

^{3} - k, so let's see if we can break that out somehow:

. . . . .. . . . .. . . . .By assumption, you have k

^{3} - k = 6m for some natural number m. For the whole thing to be divisible by 6, you need 3(k

^{2} + k) to be divisible by 6. Obviously it's divisible by 3.

So all that remains is to show that k

^{2} + k is divisible by 2. Consider cases. If k is even, then k = 2p for some natural p, and k

^{2} + k = (2p)

^{2} + (2p) = 4p

^{2} + 2p = 2(2p

^{2} + p). Now what if k is odd, so k = 2p + 1 for some natural p?

Have fun!