Let $\displaystyle\sigma_m=\sum_{r=1}^n r^m$.

Refer to the tabulation of the power sum of integers here.

It is interesting to note that

$$\begin{align} \color{green}{\sigma_1}\ &=\frac 12 n(n+1)\\ \color{blue}{\sigma_2}\ &=\frac 16 n(n+1)(2n+1)\\ \color{red}{\sigma_3}\ &=\frac 14 n^2(n+1)^2&&=\color{green}{\sigma_1}^2\\ \sigma_4\ &=\frac 1{30}n(n+1)(2n+1)(3n^2+3n-1)&&=\frac 15\; \color{blue}{\sigma_2} \ (3n^2+3n-1)\\ \sigma_5\ &=\frac 1{12}n^2(n+1)^2(2n^2+2n-1)&&=\frac 13\; \color{red}{\sigma_3}\ (2n^2+2n-1)\\ \sigma_6\ &=\frac 1{42}n(n+1)(2n+1)(3n^4+6n^3-3n+1)&&=\frac 17\;\color{blue}{\sigma_2}\ (3n^4+6n^3-3n+1)\\ \sigma_7\ &=\frac 1{24}n^2(n+1)^2 (\cdots)&&=\frac 16\; \color{red}{\sigma_3}\ (\cdots)\\ \sigma_8\ &=\frac 1{90}n(n+1)(2n+1)(\cdots)&&=\frac 1{15}\color{blue}{\sigma_2}\ (\cdots)\\ \sigma_9\ &=\frac 1{20}n^2(n+1)^2(n^2+n-1)(\cdots)&&=\frac 15\; \color{red}{\sigma_3}\ (n^2+n-1)(\cdots)\\ \sigma_{10}&=\frac 1{66}n(n+1)(2n+1)(n^2+n-1)(\cdots)&&=\frac 1{11}\color{blue}{\sigma_2}\ (n^2+n-1)(\cdots) \end{align}$$ i.e.

  • the sum of squares, $\sigma_2$, is a factor of sum of even powers greater than $2$, and
  • the sum of cubes, $\sigma_3$, is a factor of sum of odd powers greater than $3$.

Is there a simple explanation for this, if possible without using Faulhaber's formula and Bernoulli numbers, etc?

and also,

Why does this occur only for $\sigma_2, \sigma_3$ but not for $\sigma_4, \sigma_5$, etc?



Its kinda hand wavy if I can't use anything really advanced, but here's an intuitive look on the situation:

Let $\sigma_m(x)$ be a polynomial of $x$ such that on $x\in\mathbb N$, it agrees with your $\sigma_m$. Note this polynomial satisfies the recursive relation


which extends it to negative values.

The phenomenon of $\sigma_2$ and $\sigma_3$ appearing in $\sigma_{m>3}$ is not too surprising, since it is easy to note that


for any $m\in\mathbb N_{>0}$.

One can see from the recursive relation that $\sigma_m(x)$ is symmetric along $x=-\frac12$.

For even $m$, the symmetry is odd, so there is a root at $x=-\frac12$.

For odd $m$, the symmetry is even, so every other root reflects over. This makes $x=0$ and $x=-1$ roots with a multiplicity of $2$.

Combine these two and you can see that


  • $\begingroup$ Very elegant solution! (+1) $\endgroup$ – hypergeometric Aug 10 '17 at 16:26
  • $\begingroup$ @Simply Beautiful Art: Can you exaplain how do you get symmetry about $x=-\frac 12$, please? We have $\sigma_m(-1/2+x)=\sigma_m(-3/2+x)+(-1/2+x)^m$. $\endgroup$ – Fabio Lucchini Jul 9 '18 at 9:01
  • $\begingroup$ @FabioLucchini notice that $$\sigma_m(-x)=\sigma_m(-x+1)-(-x)^m$$ which is almost the same as the original recursive relation but for negative integers. $\endgroup$ – Simply Beautiful Art Jul 12 '18 at 12:01
  • $\begingroup$ Sorry, but I don't get it ... it would to be $\sigma_m(-x)=\sigma_m(-x+1)-(1-x)^m$. But how this proves $\sigma_m(-1-x)=-(-1)^m\sigma_m(x)$? $\endgroup$ – Fabio Lucchini Jul 12 '18 at 15:59
  • $\begingroup$ @FabioLucchini yes my bad. Use also the fact that $\sigma_m(-1)=\sigma_m(0)=0$ and you may prove it by induction. $\endgroup$ – Simply Beautiful Art Jul 12 '18 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.