#### Question

For any positive integer n , prove that n^{3} − n divisible by 6.

#### Solution

We have n^{3} − n = n(n^{2} − 1) = (n − 1) (n) (n + 1)

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q + 5.

If n = 6q, then

(n − 1)(n)(n + 1) = (6q − 1)(6q)(6q + 1)

= 6[(6q − 1)(q)(6q + 1)]

= 6m, which is divisible by 6

If n = 6q + 1, then

(n − 1)(n + 1) = (6q)(6q + 1)(6q + 2)

= 6[(q)(6q + 1)(6q + 2)]

= 6m, which is divisible by 6

If n = 6q + 2, then

(n − 1)(n)(n + 1) = (6q + 1)(6q + 2)(6q + 3)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m, which is divisible by 6

If n = 6q + 3, then

(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)

= 6[(3q + 1)(2q + 1)(6q + 4)]

= 6m, which is divisible by 6

If n = 6q + 4, then

(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)

= 6[(2q + 1)(3q + 2)(6q + 5)]

= 6m, which is divisible by 6

If n = 6q + 5, then

(n − 1)(n)(n + 1) = (6q + 4)(6q + 5)(6q + 6)

= 6[(6q + 4)(6q + 5)(q + 1)]

= 6m, which is divisible by 6

Hence, for any positive integer n, n^{3} – n is divisible by 6.