# Finding Integers Satisfying The Equations

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.

• For $p=2,\ q=3$ you need to find solutions (if they exist) to $\frac{m(m+1)(2m+1)}{6}=\frac{n^2(n+1)^2}{4}$ – Keith Backman Nov 11 '18 at 16:58
• @KeithBackman They do not exist, except for $n=m=1$. – Dietrich Burde Nov 11 '18 at 17:14

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.
• The question asks for the impossible. I don't think this is known in general (as usual for such problems). But, as you wrote in your question, "considering the smaller values of $p$ and $q$ like $2,3$" is already interesting and can be solved. – Dietrich Burde Nov 12 '18 at 15:41