proof for series to the power of 5 is divisible by n+1 for even n and n congruent to 3 mod 4

Question

I really need help with this: $$\sum_{x=1}^{n} x^5 $$ is divisible by $n+1$ when $n$ is even, and when $ n \equiv 3 \pmod{4} $.
What I first did was proving by induction that $$\sum_{x=1}^{n} x^5 = \frac{n^2 (1 + n)^2 (-1 + 2n + 2n^2)}{12} $$ is right and then argued that because the term includes the factor $n+1$, it is divisible by $n+1$. However, this would imply divisibility for all n, which is not true, as it is not divisible, for example, when $n=5$. Can someone help me and tell how to proceed?

Answer 1

Note that, given a positive integer $n$, your sum is equal to $\frac{n^2(n+1)^2 (2n^2 + 2n - 1)}{12}$. Proving this using induction should be straightforward. If $n$ is even, then $4|n^2$, and if 3 does not divide $n$, then $3|(n+1)(2n^2+2n-1)$. That means $n+1$ is a factor of the sum. Similarly, one can analyze the other case.

Answer 2

Write

$$\sum_{x=1}^{n} x^5=\frac{n^2(n+1)^2\left[(2n+1)^2-3\right]}{24}$$

Then for all of the given cases

$$\frac{n^2(n+1)\left[(2n+1)^2-3\right]}{24}\in\mathbb{Z}$$

Answer 3

You want to show that $n+1$ divides $\sum\limits_{x=1}^{n} x^5 = \dfrac{n^2 (1 + n)^2 (-1 + 2n + 2n^2)}{12} .$

Equivalently, $12(n+1)$ divides ${n^2 (1 + n)^2 (-1 + 2n + 2n^2)}$.

Equivalently, $12$ divides ${n^2 (1 + n) (-1 + 2n + 2n^2)}$.

Can you see that $3$ divides $n, 1 + n,$ or $-1 + 2n + 2n^2$, if $n\equiv0,2,$ or $1\pmod3$, respectively?

Can you see that $4$ divides $n^2 (1 + n)$ when $n$ is even or $ n \equiv 3 \pmod{4} $?