What is the root of $\sum _{i=1}^{m-1} i^k=m^k$ given integer k?

Question

Let $m'$ be the root of $\sum _{i=1}^{m-1} i^k=m^k$ given integer $k$ (solving for $m$). What is $m'$?

Answer 1

Comment; May be this idea can be useful:

Suppose $(i, k)=1$ then due to FLT:

$i^k\equiv i\mod k$ and we may write:

$$1^k+2^k+3^k+4^k+ . . . (m-1)^k\equiv(1+2+3+4+ . . .m-1) \mod k$$

Or:

$$\begin{equation} \sum^{m-1}_{i=1}i^k\equiv\frac{m(m-1)}2 \mod k =k.t+\frac{m(m-1)}2=m^k\end{equation}$$

Note that in the case i=1 , i and k are not coprime but with some manipulation we can get similar result.

$$m(2m^{k-1}-m+1)=2kt$$

Solution of this equation can give m and k. For example to ease the solution we may let m=k for a particular case, then:

$$2m^{m-1}-m+1=2t$$

m must be odd, let $m=2s+1$ we have:

$$2(2s+1)^{2s}-2s=2t$$

This Diophantine equation may have integer solutions , hence we may find m and k.