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How could I find all the pairs $(n, k)$ for this equation. The most obvious pair solution that I can see is $(1, 1)$.
Using summation identity, I have:

$\frac{n(n+1)}{2} = \frac{k(k + 1)(2k + 1)}{6}$

Then I thought of using cubic formula for $k$-equation, but it involved many variables. Any idea?

Thanks,
Chan

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    @RossMillikan It sounds like you may be thinking of Darmon and Granville's work on the [Fermat-Catalan conjecture](https://en.wikipedia.org/wiki/Fermat–Catalan_conjecture).2017-02-01

2 Answers 2

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There are only two variables involved. If you want to search, you can write it as a quadratic in $n$, just try values of $k$, solve for $n$, and see if it comes out integral. I find k=5, n=10, k=6, n=13 and k=85, n=645 as solutions as well with no more under k=200. Then OEIS has no more and asserts the series is finite. There are references for this claim in A053611

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    Interesting, I never thought of Excel spreadsheet. Thanks for this great idea ;)2011-02-19
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Fix the variable $k$. Let k' = \dfrac{k(k+1)(2k+1)}{6}. Then you get the quadratic equation n^2+n-2k' = 0 with the solutions n_{1/2} = -\dfrac{1}{2} \pm \sqrt{(\dfrac{1}{2})^2+2k'}. Now you can generate your solution pairs.