# Power Sum of Integers and Relationship with Sum of Squares and Sum of Cubes

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?

# Disclaimer

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

$$\sigma_m(x)=\sigma_m(x-1)+x^m$$

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

$$\sigma_m(-1)=\sigma_m(0)=0$$

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

$$\sigma_{2m}(x)=x(x+1)(2x+1)P_m(x)\\\sigma_{2m+1}(x)=x^2(x+1)^2Q_m(x)$$

• Very elegant solution! (+1) Aug 10, 2017 at 16:26
• @Simply Beautiful Art: Can you exaplain how do you get symmetry about $x=-\frac 12$, please? We have $\sigma_m(-1/2+x)=\sigma_m(-3/2+x)+(-1/2+x)^m$. Jul 9, 2018 at 9:01
• @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. Jul 12, 2018 at 12:01
• 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 $\sigma_m(-1-x)=-(-1)^m\sigma_m(x)$? Jul 12, 2018 at 15:59
• @FabioLucchini yes my bad. Use also the fact that $\sigma_m(-1)=\sigma_m(0)=0$ and you may prove it by induction. Jul 12, 2018 at 16:24