# The sum of the first $n$ squares is a square: a system of two Pell-type-equations

This question comes from trying to see why 24 is the only non-trivial value of $n$ for which $$1^2+2^2+3^2+\cdots+n^2$$ is a perfect square.
To this end, let $m,n \in \mathbb N$ be such that $1^2+2^2+3^2+\cdots+n^2 = m^2$, or $$n(n+1)(2n+1) = 6m^2.$$ When $n=24$ the left hand side is $24\times 25 \times 49$ and there are two things that make it work as a solution: $$7^2+1=2\times5^2$$ $$7^2-1=12\times2^2.$$ We can write these algebraically as $$x^2=2y^2-1$$ $$x^2=12z^2+1$$ and solve them simultaneously (with $x,y,z\in \mathbb N$).

These are instances of Pell's equation and each individually has an infinite number of solutions.
How do we show there is only one (non-trivial) value of $x$ that is common to the solutions of both equations?

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You mean $y=5$? If your edit answers your question, it would be preferable to write it as an answer and accept it, so that the question doesn't remain unanswered. However, it's not clear to me how $z=2$ and $y=5$ follows directly. – joriki Oct 20 '11 at 11:30
$n=m=1$ is also a solution – Martin Sleziak Oct 20 '11 at 11:39
@joriki: Thanks for pointing that out. After I'd posted the equestion I had a momentary panic and added my Edit but hadn't thought it through. I have removed the edit. – Peter Phipps Oct 20 '11 at 11:41
@MartinSleziak: I did say non-trivial, though one person's trivial may be another person's important. – Peter Phipps Oct 20 '11 at 11:43
Unfortunately, I don't have access to JSTOR so the info below is of little use. If someone could write a few words I'll be grateful. – Peter Phipps Oct 20 '11 at 19:11