KCC's Quiz AQQ300 about Divisibility by 6

We can expand n⁵ – n into factors:

n⁵ – n = n(n² + 1)(n + 1)(n - 1)  [1]

If n is even, n = 2k and we obtain: n⁵ – n = 2k(4k² + 1)(2k + 1)(2k - 1)  [2]. Therefore, n⁵ – n = M2 since [2] contains a factor of 2.

If n is odd, n = 2k + 1 and: n⁵ – n = (2k + 1)[(2k + 1).(2k+1)+1](2k + 2)(2k)   [3]. Therefore, n⁵ – n = M2 since [3] also contains a factor of 2.

The n can be M3, M3 + 1, or M3 - 1 and

If n = M3, n⁵ – n is necessarily M3 since its expansion [1] contains the factor n.

If n = M3 + 1, then n⁵ – n is necessarily M3 since its expansion [1] contains the factor (n-1) = M3.

If n = M3 - 1, then n⁵ – n is necessarily M3 since its expansion [1] contains the factor (n+1) = M3.

Therefore, for all n, n⁵ – n is divisible by 2 and 3, and therefore by 6.

We can expand n⁵ – n into factors:

n⁵ – n = n(n² + 1)(n + 1)(n - 1)  [1]

If n is even, n = 2k and we obtain: n⁵ – n = 2k(4k² + 1)(2k + 1)(2k - 1)  [2]. Therefore, n⁵ – n = M2 since [2] contains a factor of 2.

If n is odd, n = 2k + 1 and: n⁵ – n = (2k + 1)[(2k + 1).(2k+1)+1](2k + 2)(2k)   [3]. Therefore, n⁵ – n = M2 since [3] also contains a factor of 2.

The n can be M3, M3 + 1, or M3 - 1 and

If n = M3, n⁵ – n is necessarily M3 since its expansion [1] contains the factor n.

If n = M3 + 1, then n⁵ – n is necessarily M3 since its expansion [1] contains the factor (n-1) = M3.

If n = M3 - 1, then n⁵ – n is necessarily M3 since its expansion [1] contains the factor (n+1) = M3.

Therefore, for all n, n⁵ – n is divisible by 2 and 3, and therefore by 6.

Well done MikeLh , congrats!