Proving Divisibility of n^3-n by 6

[Goldenwind]

Homework Statement

Prove that n^3 - n is divisible by 6, when n is a nonnegative integer.

The Attempt at a Solution

Mathematical induction:

It works for n=0
It works for n=1 (Extra step, just in case)
Check if it works for the (k+1)th step.

For it to work, it must be expressible as 6x, where x is some integer.

In other words, to prove: (k+1)^3 - k = 6x

Can someone nudge me on this? I'm either making a mistake by calling it 6x, and maybe it should be 6k or something else...

...and/or, I'm just missing the algebraic skills to change LS into RS.

 

[tiny-tim]

Take the lazy way!

Prove that n^3 - n is divisible by 6, when n is a nonnegative integer.

The attempt at a solution
Mathematical induction

No no no no no!

Far too amibitious!

Take the lazy way!

Just factorise $$n^3 - n$$, and you'll immediately see why 6 is always a factor!

Ping!