# Generalisation of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$

Are there any generalisations of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ ?

For example can $\sum {k^m} = \left(\sum k\right)^n$ be valid for anything other than $m=3 , n=2$ ?

If not, is there a deeper reason for this identity to be true only for the case $m=3 , n=2$?

• – Eric Naslund Jun 27 '12 at 12:38
• Related: Proving the original identity without induction and proving it with induction. I am posting this here so that all of these questions are linked together, so that in the case someone finds only this one, they can also find the other two. – Eric Naslund Aug 30 '12 at 23:23

We can't have a relationship of the form $$\forall n\in\mathbb N^*, \sum_{k=1}^nk^a=\left(\sum_{k=1}^nk^b\right)^c$$ for $a,b,c\in\mathbb N$, except in the case $c=1$ and $a=b$ or $a=3$, $b=1$ and $c=2$. Indeed, we can write $$\sum_{k=1}^nk^a =n^{a+1}\frac 1n\sum_{k=1}^n\left(\dfrac kn\right)^a$$ hence $$\sum_{k=1}^nk^a\;\overset{\scriptsize +\infty}{\large \sim}\;n^{a+1}\int_0^1t^adt=\dfrac{n^{a+1}}{a+1}$$ and if we have the initial equality we should have $a+1 =(b+1)c$ and $a+1=(b+1)^c$. In particular, $(b+1)^{c-1}=c$. If $c>1$, then $c= (b+1)^{c-1}\geq 2^{c-1}\geq c$, and we should have $c=2$ and $b=1$, therefore $a=3$.

• Davide, I tweaked the spacing and sizes in the $\sim$ you had; the previous spacing looked too tight. Hope you don't mind. – J. M. is a poor mathematician Sep 4 '11 at 12:49
• @J.M. Thanks, it's more readable thanks to that. – Davide Giraudo Sep 4 '11 at 12:51

The Faulhaber polynomials are expressions of sums of odd powers as a polynomial of triangular numbers $T_n=\frac{n(n+1)}{2}$. Nicomachus's theorem, $\sum\limits_{k\leq n} k^3=T_n^2$, is a particular special case.

Other examples include

\begin{align*}\sum\limits_{k\leq n} k^5&=\frac{4T_n^3-T_n^2}{3}\\\sum\limits_{k\leq n} k^7&=\frac{6T_n^4-4T_n^3+T_n^2}{3}\end{align*}

Here is a curious (and related) identity which might be of interest to you. Let $D_{k} =${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\sigma_{0}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then it is relatively straightforward to prove \begin{eqnarray} \sum_{d \in D_k} \sigma_{0}^{*}(d)^{3} = \left( \sum_{d \in D_k} \sigma_{0}^{*}(d) \right)^{2} \qquad k \in \mathbb{N}. \end{eqnarray}

Note that $\sigma_{0}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$. For example, \begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray}

Not a completely rigorous answer, but you should be able to turn it into one.

By comparing the sums to their corresponding integrals $\int_0^n \mathrm{d}x x^m$, you can see that $$\sum k^m = \frac{1}{m+1} n^{m+1} + \mathcal{O}(n^m).$$ Also, $$(\sum k)^q = \frac{1}{2^{q}} n^{2q} + \mathcal{O}(n^{2q-1}).$$ By comparing leading order terms, equality can only occur if $m+1 = 2q$ and if $m+1 = 2^q,$ which implies that $q = 2$ and $m = 3.$