Proving: 6 Divides (n^3-n) for All Integers n

[chan8366]

Homework Statement

prove:6 divides (n^3-n) for all integers n.

Homework Equations

n^3-n=(n)(n+1)(n-1)

The Attempt at a Solution

tried to use direct proof. Then used cases that involed n=2k for some integer k and n=2k+1 for some integer k. However, i could not get it so that 6 was factored from either odd/even of n^3-n i.e. n^3-n=6m for some integer m.
Just a hint please.

 

[Mathdope]

What's the remainder on division by 3 of the 3 factors you have exhibited?

 

[HallsofIvy]

One of any two consecutive number is even. One of any three consecutive numbers is a multiple of 3.