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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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    Do you want all solutions or just one?2011-02-19
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    @Moron: I found references (that I didn't check) in OEIS that claim there are only four [or five if you count (0,0)]2011-02-19
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    @Ross: Yes I noticed. Apparently this is called Thomas' problem and the supposed proof: http://linkinghub.elsevier.com/retrieve/pii/0022314X72900364. I got this by following your OEIS link.2011-02-19
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    I vaguely remember there is a theorem that says since the densities are 1/n^2 and 1/n^3 and 1/2+1/3<1, you should expect only finitely many solutions. Maybe somebody will be prompted to cite it.2011-02-19
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    @Moron: unfortunately I don't have free access2011-02-19
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    @Ross: I don't either :-( Just thought it might be useful. And if someone has access, they can perhaps post an answer with the idea.2011-02-19
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    @Ross: Isn't this just Mordell's Theorem that there are a finite number of integral points on any (nontrivial) elliptic curve? Unfortunately a bit of searching for effective bounds turned up nothing that would be specifically useful here, but that was the first thing that jumped into my mind at least...2011-02-19
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    @Steven: maybe so. I was thinking of a result that was cast in number theoretic terms, though I can see how the elliptic curves could fit into this. But I know nothing about elliptic curves.2011-02-19
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    @Steven: I believe the result in the paper is stronger. It tells you exactly what the solutions are.2011-02-19
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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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    How could you find those solutions? Did you use a program to generates k, n or did you solve mathematically?2011-02-19
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    @Chan: I just made an Excel spreadsheet that did what I said. Fill series will make the numbers from 1-200, then copy down is a powerful tool. It works well when you want up to 2000 or so, then becomes cumbersome, but I find debugging much easier than with a Python program.2011-02-19
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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.