For which $n \in \mathbb{N}$ do we have $\sum\limits^n_{k=0} k^3 = (\sum\limits^n_{k=0} k)^2$?

Question

I have a question that I don't know how to solve.

For which $n \in \mathbb{N}$ do we have $$\sum\limits^n_{k=0} k^3 = (\sum\limits^n_{k=0} k)^2\quad ?$$.

Here's what I've tried:

\begin{align*}\sum\limits^n_{k=0} k^3 = (\sum\limits^n_{k=0} k)^2 & \Leftrightarrow 1+8+27+\cdots+n^3 = (1+2+3+\cdots+n)^2\\ & \Leftrightarrow \sqrt{1+8+27+\cdots+n^3} = 1+2+3+\cdots+n \end{align*}

Answer 1

It holds for all $n\in\mathbb N$. Is this sufficiently convincing:

Note. This formula has been known for 2000 years. It is due to Nicomachus of Gerasa.

Answer 2

There is a formula for the LHS (sum of consecutive cubes -- proof here) that says it's equal to $\left(\frac{n(n+1)}{2}\right)^2$, which is therefore equal to the RHS for all $n \in \mathbb{N}$