Okay, so I have these equations:
$$ x = 1^p + 2^p + 3^p +... m^p $$ $$ x = 1^q + 2^q + 3^q +... n^q $$
How many possible values of positive integers $(m, n, p, q) $ are there, if any, such that the equations hold with $p,q,m, n> 1$ where $p \ne q$?
I tried approaching the problem by first considering the smaller values of $p$ and $q$ like $2,3$. and found no solution over the positive integers. I have no idea how to proceed with it. Any help will be appreciated.
The question is not known in general, I think, but is known for special cases like $p=2$, $q=3$, see the comments:
By the Cannonball problem, $$ 1^2+2^2+3^2+\cdots +m^2=\frac{m(m+1)(2m+1)}{6}=N^2 $$ is only solvable for $m=1$ and $m=24$ with $N=1$, and $N=70$. In order to equal a sum of cubes, $$ 1^3+2^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2=70^2, $$ we obtain $70=n(n+1)/2$, a contradiction. The trivial solution $m=n=1$ is excluded.