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I encountered this fact yesterday: $1$ and $4900$ are the only squares as the sum of $1+4+9+\ldots +n^2$. I was trying to solve this problem using my knowledge of elementary number theory. I reduce it to the point:

Show that $(a,b,c)=(2,5,7)$ is the only positive integer solution to $$ \left\lbrace \begin{array}{} 6\times a^2+1=b^2 \\ 12\times a^2+1=c^2 \end{array} \right. $$ (then let $n=6\times a^2$, you get $4900= 24\times25\times49/6$)

I recognize these as Pell's equations, but I don't know how to proceed.

Gary
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Chao H
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1 Answers1

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You are exploring what has been called the Cannonball problem. Édouard Lucas formulated the cannonball problem as a Diophantine equation $$\sum_{n = 1}^N n^2 = M^2$$ or $$\frac{1}{6} N(N + 1)(2N + 1) = \frac{2N^3 + 3N^2 + N}{6} = M^2.$$

Lucas conjectured that the only solutions are $N=1, M=1$, and $N=24, M=70$, using either $1$ or $4900$ cannon balls (as it seems you have observed). It was not until 1918 that G. N. Watson proved it, using a method that is far beyond on the scope of elementary number theory. However it was proved in $1990$ in just over $4$ pages, if you are interested in seeing a simpler proof.

Clyde Kertzer
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  • The last link is broken, I think you may be logged in to an university account when copying the link. – Andrew Feb 08 '23 at 03:21
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    Yes, the only place to access the paper is on JSTOR, so you need a login. – Clyde Kertzer Feb 08 '23 at 03:27
  • It’s not that it’s on JSTOR that’s the issue. Click on the link while in incognito mode. – Andrew Feb 08 '23 at 04:25
  • If you don't have access to JSTOR, I don't know any other way to access the paper. Let me know if you find another link that doesn't require login. – Clyde Kertzer Feb 08 '23 at 04:32
  • The issue that the paper is found on JSTOR is NOT the issue. https://www-jstor-org.colorado.idm.oclc.org/stable/2323911?origin=crossref You linked to your universities access page. Try accessing the link on incognito, off your university wifi. It will just bring you to a ucolorado login page. I guess changing the link like this actually works: https://www.jstor.org/stable/2323911?origin=crossref – Andrew Feb 08 '23 at 04:57
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    Ah, gotchya. Good catch, I just changed the link. – Clyde Kertzer Feb 08 '23 at 05:06
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    This is why, instead of "a paper", the link text should be a proper citation with title, author's name, year, and journal name. That way, people can find the paper even if something is wrong with the link. – MJD Feb 08 '23 at 05:36