Is the equality $1^2+\cdots + 24^2 = 70^2$ just a coincidence?

Question

I have read a question (written in Korean. Added on September 9, 2020: the link is broken, unfortunately.) that the equality $$1^2+2^2+\cdots + 24^2 = 70^2$$ is just a coincidence or not. It is related to the integral points of the following elliptic curve (?): $$y^2 = \frac{1}{3}x^3 + \frac{1}{2}x^2+\frac{1}{6}x.$$

(To note, its determinant is $1/1296$.)

I have heard that finding the integral points of an elliptic curve is very hard in general, so my first goal is to find rational solutions to it. However, I have not learned of the number theory so I have no idea how to find rational points of that.

Is there any idea to solve this? Furthermore, is there an idea to find integral points of it? I would be appreciated for your help!

Answer 1

There is only one solution(except $1=1$). There is a proof in Mordell's book on Diophantine Equations. The problem is attributed to Lucas:

with N > 1 is when N = 24 and M = 70. This is known as the cannonball problem, since it can be visualized as the problem of taking a square arrangement of cannonballs on the ground and building a square pyramid out of them. It was not until 1918 that a proof (using elliptic functions) was found for this remarkable fact, which has relevance to the bosonic string theory in 26 dimensions.1 More recently, elementary proofs have been published.2

https://en.wikipedia.org/wiki/%C3%89douard_Lucas

Here is a discussion and proof by Bennett

Elementary proof by Anglin

The fact is critically important in the history of the Leech Lattice, the history of finite simple groups, Monstrous Moonshine and other cute things. See page 130 in Ebeling Lattices and Codes (second edition), I will try to find it in SPLAG as well. Yes page 524 in the first edition, chapter Lorentzian Forms for the Leech Lattice.

Answer 2

Not a coincidence, definitely. $70$ is a Pell number, so $2\cdot 70^2+1=99^2$, and some solutions of

$$ 1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = q^2 $$ can be derived by imposing that both $2n+1$ and $\frac{n(n+1)}{6}$ are squares: that leads to a Pell equation.