A summation of powers defined by an equation over finite fields Let $p$ be an odd prime and let $k\mid q$ for some positive integer numbers $k$ and $q$. Suppose that $r \in \mathbb{F}_{p^q}$ has multiplicative order $p^k-1$. 
For each $1\leq u \leq p^k-3$, the subfield $\mathbb{F}_p(r) \cong \mathbb{F}_{p^k}$ of $\mathbb{F}_{p^q}$ contains $r^u + r^{u-1} + \cdots + r+1 \not=0$. Hence there exists a positive integer number $m_u$ such that the following equation holds. 
$$
r^{m_u}+r^u+r^{u-1}+\cdots+r+1=0
$$

Is there a systematic method to get the value of $\sum_{u=1}^{p^k-3}m_u$?

 A: The bad news is that the sum depends on the choice of $r$. We can do the following, reducing the problem to that of finding the discrete logarithm of $r-1$ (base $r$, of course).
Consider the sum
$$
1+r+r^2+\cdots+r^u=\frac{r^{u+1}-1}{r-1}.
$$
The exponent $u+1$ in the numerator takes all the values in the range $2\le u+1\le p^k-2$. Consequently $r^{u+1}$ ranges over the $p^k$ elements of the field $K=\Bbb{F}_{p^k}$ with the three exceptions $r^{u+1}\notin\{0,1,r\}$.
This means that
$$
r^{m(u)}=-\frac{r^{u+1}-1}{r-1}
$$
ranges over the set $K\setminus\{0,-1,1/(r-1)\}.$
Therefore the exponent $m(u)$ otherwise takes all the possible values of the logarithm, $\{0,1,2,\ldots,p^k-2\}$, but omits
$$\log_r(-1)=(p^k-1)/2\quad\text{and}\quad\log_r(1/(r-1))=p^k-1-\log_r(r-1).$$
Given all this, your sum equals
$$
\begin{aligned}
\sum_{u=1}^{p^k-3}m(u)&=\sum_{\ell=0}^{p^k-2}\ell-\frac{p^k-1}2-(p^k-1-\log_r(r-1))\\
&=\frac12(p^k-1)(p^k-2)-\frac{3(p^k-1)}2+\log_r(r-1)\\
&=\frac12(p^k-1)(p^k-5)+\log_r(r-1).
\end{aligned}
$$
But this is all we can say. We don't have a formula for the discrete logarithm $\log_r(r-1)$. The discrete logarithm problem is difficult in general, and in most cases the answer depends on the choice of $r$.
