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Prove that for any n , n5 - n is divisible by 30 .
+9673 The base case n = 0 is obviously correct, as 05 - 0 = 0 is divisible by 30.
Next, we want to prove when n5 - 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)) - (n5 - n) is divisible by 30.
Making use of the fact that a5 - b5 = (a - b)(a4 + a3b + a2b2 + ab3 + b4),
((n+1)5 - (n+1)) - (n5 - n) = ((n+1)4 + n(n+1)3 + n2(n+1)2 + n3(n+1) + n4) - 1 = 5n(n+1)(n2+n+1).
If we can prove that n(n+1)(n2+n+1) is divisible by 6, we are done.
First case: n = 3k for some integers k.
n(n+1)(n2+n+1) = 3k(3k+1)(9k2 + 3k + 1).
As 3k is divisible by 3 and (3k+1) is divisible by 2, we proved that n5 - n is divisible by 30 when \(n\equiv 0 \pmod 3\)
Second case: n = 3k + 1 for some integers k.
n(n+1)(n2+n+1) = (3k+1)(3k+2)(9k2 + 9k + 3) = 3(3k+1)(3k+2)(3k2 + 3k + 1).
As (3k+1) is divislbe by 2, we proved that n5 - n is divisible by 30 when \(n\equiv 1\pmod 3 \).
Last case: n = 3k - 1 for some integers k.
n(n+1)(n2+n+1) = (3k)(3k-1)(9k2 - 3k + 1)
When k is even, 3k is divisible by 6.
When k is odd, 3(3k-1) is divisible by 6.
Therefore, n5 - n is divisible by 30 when \(n\equiv 2\pmod 3 \)
Therefore, for all n, n5 - n is divisible by 30.
