<s>[INST] 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 $m=3 , n=2$?
 [/INST] 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}
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