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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.

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

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  • $\begingroup$ The question does not ask to solve only for 2,3, but it asks if the solution exists for any positive integers m,n. And whether it is possible to find it or not. $\endgroup$ – Manish Kundu Nov 12 '18 at 15:15
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Nov 12 '18 at 15:41

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