Question

Prove that 6 divides $n^3-n$ whenever n is a non negative integer.

Short answer

6 divides $n^3-n$ whenever n is a positive integer

Step-by-step solution

Step: 1

If n=0,

0 - 0 = 0

it is true for n=0.

Step: 2

Let P(k) be true.

$k^3-k$

We need to prove that P(k+1) is true.

Step: 3

$(k+1{)}^3-(k+1)=\left(k^3-k\right)+3k(k+1)$

It is divisible by 6.

It is true for P(k+1) is true.