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Prove that for any n , n^{5 }- n is divisible by 30 .

Guest Mar 9, 2019

#1**+1 **

The base case n = 0 is obviously correct, as 0^{5} - 0 = 0 is divisible by 30.

Next, we want to prove when n^{5} - n is divisible by 30, (n+1)^{5} - (n+1) is divisible by 30.

To do so, we need to prove that ((n+1)^{5} - (n+1)) - (n^{5} - n) is divisible by 30.

Making use of the fact that a^{5} - b^{5} = (a - b)(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4}),

((n+1)^{5} - (n+1)) - (n^{5} - n) = ((n+1)^{4} + n(n+1)^{3} + n^{2}(n+1)^{2} + n^{3}(n+1) + n^{4}) - 1 = 5n(n+1)(n^{2}+n+1).

If we can prove that n(n+1)(n^{2}+n+1) is divisible by 6, we are done.

First case: n = 3k for some integers k.

n(n+1)(n^{2}+n+1) = 3k(3k+1)(9k^{2 }+ 3k + 1).

As 3k is divisible by 3 and (3k+1) is divisible by 2, we proved that n^{5} - n is divisible by 30 when \(n\equiv 0 \pmod 3\)

Second case: n = 3k + 1 for some integers k.

n(n+1)(n^{2}+n+1) = (3k+1)(3k+2)(9k^{2 }+ 9k + 3) = 3(3k+1)(3k+2)(3k^{2} + 3k + 1).

As (3k+1) is divislbe by 2, we proved that n^{5} - n is divisible by 30 when \(n\equiv 1\pmod 3 \).

Last case: n = 3k - 1 for some integers k.

n(n+1)(n^{2}+n+1) = (3k)(3k-1)(9k^{2 }- 3k + 1)

When k is even, 3k is divisible by 6.

When k is odd, 3(3k-1) is divisible by 6.

Therefore, n^{5} - n is divisible by 30 when \(n\equiv 2\pmod 3 \)

Therefore, for all n, n^{5} - n is divisible by 30.

MaxWong Mar 9, 2019