After watching John Baez talking about the number 24 I got interested in the question: For which $n \in \mathbb{N}$ is there a number $m \in \mathbb{N}$ s.t.
$$ \sum_{k=1}^nk^2=m^2 $$
Or to put it in plain english, when is the sum of the first $n$ square numbers again a square.
The surprising answer to this is that this is only the case for $n=0,1,24$ and you proof this (according to him) with the formula
$$ \sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}{6} $$
and the machinery of elliptic curves. My guess is that you probably want to proof that the elliptic curve $E$ given by the above equation satisfies $E(\mathbb{Q})=\mathbb{Z}_3$ (not sure here, but something along these lines).
My main question is: can somebody name a reference for a proof of this? But I would also be interested if you could shed some light on how to proceed here or how much machinery you actually need to prove this.
Edit: Since the question got closed while I was not on a computer I just wanted to point out that the other thread doesn't answer my question. The answers in the other thread answers the question: For which $n$ are there $n$ consecutive squares that sum to a square. Note that there the consecutive squares don't have to start with 1. Part of the answer is that this is always the case when $n$ is a perfect square coprime to 6 which clearly is not the case in my situation. The problem is that the answer doesn't shed light on the number $a$ in the question. In particular it is not clear when $a=0$ and when not (which would be my situation). I'm not sure why so many people mark this as duplicate when the questions are clearly different.