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?

  • $\begingroup$ 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. $\endgroup$
    – joriki
    Oct 20, 2011 at 11:30
  • $\begingroup$ $n=m=1$ is also a solution $\endgroup$ Oct 20, 2011 at 11:39
  • $\begingroup$ @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. $\endgroup$ Oct 20, 2011 at 11:41
  • 3
    $\begingroup$ @MartinSleziak: I did say non-trivial, though one person's trivial may be another person's important. $\endgroup$ Oct 20, 2011 at 11:43
  • $\begingroup$ @Peter I can send you the papers by email. Have a look at my profile to find my webpage and there also my address. $\endgroup$ Oct 20, 2011 at 19:47

1 Answer 1


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:

Online resources:

  • $\begingroup$ Link to W. S. Anglin's solution. I think you should add it to your answer. $\endgroup$
    – user236182
    Jan 1, 2016 at 12:55
  • $\begingroup$ @user236182 I have removed the paper from that website. (I should not have put it there in the first place.) $\endgroup$ Jan 8, 2016 at 13:17

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