Question

For any positive integer n, prove that n³ − n divisible by 6.

Answer 1

n³ – n = n (n² – 1) = n (n – 1) (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, n³ – n is divisible by 6.​​