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

Question

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?

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
@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
@Peter I can send you the papers by email. Have a look at my profile to find my webpage and there also my address. — Martin Sleziak, Oct 20 '11 at 19:47
That may or may not be possible,However the sum of two squares can be a perfect square if all the three roots are Pythagorean triplets — , Mar 05 '16 at 04:03
Sorry if this is a stupid question, but how do we find the solution (24, 70) in the first place? Is it obvious? — Noppawee Apichonpongpan, Oct 13 '16 at 05:51
@NoppaweeApichonpongpan, No I don't think it is obvious. When I wrote this question I think I had just read somewhere that $(24,70)$ is the only solution. It isn't too difficult to try finding a solution by pencil and paper or calculator, and it's easy with a computer search. — Peter Phipps, Oct 13 '16 at 15:38
Now there is also a MathOverflow question about this problem: Squares in the set ${\sum_{j=1}^m j^2: m\in\mathbb{N}}$. — Martin Sleziak, Apr 13 '19 at 23:32

Martin Sleziak · Accepted Answer · 2019-04-13T23:41:42.867

10

This problem is sometimes called cannonball problem or square pyramid problem.

Maybe you may have a look at MathWorld or Wikipedia and some references therein.

Elementary solution is given in W. S. Anglin: The Square Pyramid Puzzle, The American Mathematical Monthly, Vol. 97, No. 2 (Feb., 1990), pp. 120-124.

Further useful references:

Henri Cohen: Number Theory: Tools and diophantine equations, Section 6.8.2
S. C. Althoen and C. B. Lacampagne: Tetrahedral Numbers as Sums of Square Numbers, Mathematics Magazine, Vol. 64, No. 2 (Apr., 1991), pp. 104-108

Online resources:

edited Apr 13 '19 at 23:41

answered Oct 20 '11 at 11:32

Martin Sleziak

53,687

Link to W. S. Anglin's solution. I think you should add it to your answer. – user236182 Jan 01 '16 at 12:55
@user236182 I have removed the paper from that website. (I should not have put it there in the first place.) – Martin Sleziak Jan 08 '16 at 13:17

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

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