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I tried solving this so first $1^2+2^2+\cdots+n^2=n(n+1)(2n+1)/6$ so problem is $n(n+1)(2n+1)=6y^2$ where $n$ and $y$ are natural. I tried doing this $n=6k,6k+1....6k+5$, so say for $n=6k$ you get $k(6k+1)(12k+1)=y^2$ so $k(6k+1)$ and $12k+1$ are co-prime which means they are all squares meaning $k=a^2 \quad 6k+1=b^2\quad 12k+1=c^2$ but idk what to do from here?
Whatever I try I can't solve this.

ps. I'm pretty sure $n=24, y=70$ is the only solution besides $1,1$.

Jyrki Lahtonen
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    See art of problem solving. The unique solution is indeed $n=24$ and $y=70$, see also the duplicate (please try to search next time before posting). – Dietrich Burde May 02 '24 at 13:39
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    Or google "cannonball problem". – garondal May 02 '24 at 13:40
  • A better duplicate might be this one, or this one by the way. But there are very many. – Dietrich Burde May 02 '24 at 13:42
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    None of the other questions actually have a solution to the problem, mostly just links to payblocked articles. Which not only violates the "link only answers" standard but does so in a really bad way. The "already answered" question is to a different problem than the one asked here. – DanielV May 02 '24 at 14:35
  • @DietrichBurde you should change the dupe target, as it does not answer the question directly. – D S May 02 '24 at 16:17
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    @DietrichBurde FYI, your "Yes, but I don't know how to change this" implies to me that you're unaware of how to change the list of duplicate questions. If so, then as a "number-theory" gold-tag member, note that in the upper-right hand corner of the blue square, you should see a little pencil type icon, with "Edit" after it. The tool-tip for it says "Edit the list of duplicate questions". Clicking on this replaces the page with a new one where it shows the current list of duplicate questions, along with options to remove from and/or add to the list. – John Omielan May 02 '24 at 16:36
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    @JohnOmielan Thank you very much. In the future I know how to do it. – Dietrich Burde May 02 '24 at 17:05

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