# Find a pair (n, k) such that $\sum_{i=1}^{n} i = \sum_{i=1}^{k} i^2$?

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

-
Do you want all solutions or just one? – Aryabhata Feb 19 '11 at 1:08
@Moron: I found references (that I didn't check) in OEIS that claim there are only four [or five if you count (0,0)] – Ross Millikan Feb 19 '11 at 1:20
@Ross: Yes I noticed. Apparently this is called Thomas' problem and the supposed proof: linkinghub.elsevier.com/retrieve/pii/0022314X72900364. I got this by following your OEIS link. – Aryabhata Feb 19 '11 at 1:36
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. – Ross Millikan Feb 19 '11 at 1:44
@Moron: unfortunately I don't have free access – Ross Millikan Feb 19 '11 at 1:46

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
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.