We can easily show that $n$ is a factor of the sum of $p$-th powers 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 $1$).

However, it is interesting to note that for odd $(>1)$ values of $p$, $n^2$ is a factor as well, and in fact, $n^2(n+1)^2$ are factors.

Is there a simple way of showing that $n^2$ is a factor of the sum of odd powers $(>1)$ 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)?

For instance, in the case of the fifth power, $$\sum_{r=1}^n r^5=an^6+bn^5+cn^4+dn^3+en^2+fn+g$$ As described above, $g=0$. How can we show that $f=0$ without first evaluating the other non-zero coefficients?

  • $\begingroup$ I disagree with the phrasing of your claim, because $1 + 2 + \dots + n = n(n+1)/2$ is a sum of odd powers of the first $n$ integers, and it is not divisible by $n^2$. Odd powers $\ge 3$? $\endgroup$ – Mr. Chip Jun 22 '17 at 9:04
  • $\begingroup$ Yes, odd powers $\ge 3$. Thanks. $\endgroup$ – hypergeometric Jun 22 '17 at 9:05
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    $\begingroup$ This follows from that observation, particularly from the properties of Bernoulli polynomials mentioned in that answer. $\endgroup$ – ccorn Jun 22 '17 at 9:26
  • $\begingroup$ @ccorn - Thanks for the useful references. $\endgroup$ – hypergeometric Jun 22 '17 at 17:30

Start from $$2\sum_{r=1}^nr^k=n^k+\sum_{r=1}^n(r^k+(n-r)^k).$$ Now if $k>1$ is odd, we can write $$\sum_{r=1}^n(r^k+(n-r)^k)=n^2f(n)+\sum_{r=1}^n(nkr^{k-1}),$$ where $f$ is some polynomial. Now, using the fact that the sum of the $(k-1)$th powers is a polynomial with zero constant term, we have $2\sum_{r=1}^nr^k=n^2g(n)$ for some polynomial $g$.

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    $\begingroup$ And about the same logic can be applied with $(n+1-r)$, thus giving the factor $(n+1)^2$. Nice! $\endgroup$ – ccorn Jun 22 '17 at 10:24
  • $\begingroup$ What a brilliant proof! (+1, accepted) Thanks! $\endgroup$ – hypergeometric Jun 22 '17 at 17:28
  • $\begingroup$ @ccorn - That was going to be the next question posted :) $\endgroup$ – hypergeometric Jun 22 '17 at 17:34
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    $\begingroup$ Given the solution and comments above, it can be proven that $n^2$ and $(n+1)^2$ are factors of the sum of odd $(>1)$ powers $p$ of the first $n$ integers. Hence, for $p=3$, having easily determined the constant multiplier as $\frac 14$ (by putting $n=1$), we arrive directly at the well-known closed form of $$\dfrac 14n^2(n+1)^2$$. $\endgroup$ – hypergeometric Jun 22 '17 at 17:37

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