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