Show that $(2n+1)$ is a factor of the sum of even powers of the first $n$ integers

Question

We can easily show that $n$ is a factor of the sum of $p$-th powers $(p\in\mathbb N)$ of the first $n$ integers , by assuming that the sum is a general polynomial of order $p+1$, and setting $n=0$, giving a zero constant term (as the sum is the same whether counted from $0$ or from $1$). In fact, $(n+1)$ is also a factor.

However, it is interesting to note that for even values of $p$, $(2n+1)$ is a factor as well.

Is there a simple way of showing that $(2n+1)$ is a factor of the sum of even powers of the first $n$ integers, without evaluating the entire summation or equating coefficients for the entire polynomial (and, preferably, without using Faulhabner's formulas and Bernoulli numbers)?

@HagenvonEitzen Yes, it is. It was not mentioned that it isn't! But will reword to clarify. In fact, will move the $(n+1)$ to a different question, so that this question is focussed on $(2n+1)$. — Hypergeometricx, Jun 23 '17 at 06:31
Related : https://math.stackexchange.com/questions/2332151/show-that-n2-is-a-factor-of-the-sum-of-odd-powers-1-of-first-n-integer — lab bhattacharjee, Jun 23 '17 at 06:52

score 1 · Accepted Answer · answered Jun 23 '17 at 06:34

Again, it is clear that there is a polynomial $f$ of degree $p+1$ such that $$\sum_{k=1}^nk^{p}=f(n),$$ i.e., $$\tag1f(x)-f(x-1)=x^{p}$$ holds for all $x\in\Bbb N$, and necessarily for all $x\in\Bbb R$ (a nonzero polynomial can have only finitely many zeroes). Thus, $$f(0)=f(1)-1^p=0$$ and (assuming $p\ne0$) $$f(-1)=f(0)-0^p=0,$$ which implies that $x$ and $(x+1)$ are linear factors of $f(x)$.

If $p$ is even (and positive), then we can use $(1)$ to show by induction that $$\tag2f(-1-n)=-f(n)$$ for all $n\in \Bbb N$. Again, $(2)$ then necessarily holds for all $x\in\Bbb R$. In particular, $f(-\frac12)=0$, which implies that $(x+\frac12)$ or equivalently $(2x+1)$ is a linear factor of $f(x)$.

Can you adapt this method to show that $n^2$ and $(n+1)^2$ are factors for the closed form of the summation for cases where $p$ is odd, i.e. a double root? — Hypergeometricx, Jun 26 '17 at 16:28

Show that $(2n+1)$ is a factor of the sum of even powers of the first $n$ integers

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