An explicit formula for the discrete logarithm

I guess, but i am not sure, many people don't know that there is an explicit formula for the discrete logarithm in finite fields $\mathbb{F}^*_p$. I remember that i was quite surprised when i learned the formula many years ago.

Theorem [Explicit DL Formula] Let $p$ be a prime number $>2$ and $g$ be a primitive root in $\mathbb{F}^*_p$. Then the discrete logarithm of the element $a \in \mathbb{F}^*_p$ is equal to \begin{equation} \mathsf{dlog}_{p,g}(a) = \sum^{p-2}_{k=1} \frac{a^k}{1-g^k}\pmod{p}\end{equation}

The proof is actually not that complicated and can be found in Niederreiter [1]. I will give here a new proof for you:



Proof: To prove this, we rewrite the sum in the following way:
\begin{align*}
-\mathsf{dlog}_{p,g}(a) & = \sum^{p-2}_{k=1} \frac{a^k}{g^k-1}\pmod{p} \\
& \equiv \sum^{p-2}_{k=1}a^k(g^k-1)^{p-2} \pmod{p}
\end{align*}
Next, we apply the Binomial Theorem:
\begin{align*}
-\mathsf{dlog}_{p,g}(a) \equiv \sum^{p-2}_{k=1}a^k\sum^{p-2}_{i=0}\binom{p-2}{i}g^{ki}(-1)^{p-2-i}\pmod{p}
\end{align*}

As already shown in the post Agoh-Guiga Conjecture, it is $$\binom{p-2}{i} \equiv (-1)^i(i+1)\pmod{p}$$ hence

\begin{align*}-\mathsf{dlog}_{p,g}(a)& \equiv \sum^{p-2}_{k=1}a^k\sum^{p-2}_{i=0}\binom{p-2}{i}g^{ki}(-1)^{p-2-i}\pmod{p}\\
& \equiv \sum^{p-2}_{k=1}a^k\sum^{p-2}_{i=0}(-1)^i(i+1)g^{ki}(-1)^{p-2-i}\pmod{p}\\
& \equiv -\sum^{p-2}_{k=1}a^k\sum^{p-2}_{i=0}(i+1)g^{ki}\pmod{p}\\
\end{align*}
Removing the minus-sign and switching the sums
\begin{align*}\mathsf{dlog}_{p,g}(a) & \equiv \sum^{p-2}_{k=1}a^k\sum^{p-2}_{i=0}(i+1)g^{ki}\pmod{p} \\
& \equiv \sum^{p-2}_{i=0}(i+1)\sum^{p-2}_{k=1}(ag^{i})^k\pmod{p} \\
\end{align*}
It is easy to see that (Note, we do not apply the explicit power sum formula)
\begin{equation}
\sum^{p-2}_{k=1}(ag^i)^k \equiv
\begin{cases}
-2 & ag^i \equiv 1\pmod{p}\\
-1 & ag^i \not\equiv 1\pmod{p}\\
\end{cases}

\end{equation}
Note that $ag^i\equiv 1\pmod{p}$ is equivalent to $i = p-1-\mathsf{dlog}_{p,g}(a)$. And this $i$ value is counted twice, since it yields a $-2$ instead of a $-1$. So (using the indicator function $\mathsf{1}_S$ that is $1$ if $S$ is true and $0$ otherwise)
\begin{align*}\mathsf{dlog}_{p,g}(a) & \equiv \sum^{p-2}_{i=0}(i+1)\left(-1 - \mathsf{1}_{i = p-1-\mathsf{dlog}_{p,g}(a)}\right) \\
& \equiv -\sum^{p-2}_{i=0}(i+1) - \sum^{p-2}_{i=0}(i+1)\mathsf{1}_{i = p-1-\mathsf{dlog}_{p,g}(a)}\\ 
& \equiv -\sum^{p-2}_{i=0}(i+1) - (p-1-\mathsf{dlog}_{p,g}(a)+1)\\
& \equiv -\frac{(p-2)(p-1)}{2}-(p-1) + \mathsf{dlog}_{p,g}(a)\\
& \equiv \mathsf{dlog}_{p,g}(a)\\  \end{align*} Q.e.d.

Note that the requirement that $g$ is a primitive root is necessary therewith the denominator never gets zero for any $k$. Somewhere in my old notes, i found a generalisation of the above theorem to non primitive roots, that is:

Theorem [Explicit DL Formula] Let $p$ be a prime number $>2$ and $g$ be an element in $\mathbb{F}^*_p$ with $\mathsf{ord}_p(g) = w$. Then the discrete logarithm of the element $a \in \mathbb{G}_g$ is equal to \begin{equation} \mathsf{dlog}_p(g) =  1+ \frac{w-1}{2} + \sum^{w-1}_{k=1} \frac{a^k}{1-g^k}\pmod{p}\end{equation}

Proof: Analogous.

For $w = p-1$ this turns into the original formula. I find this formula very interesting.

Could this formula perhaps be used, to show some non-trivial relationship between discrete logarithms in different fields $\mathbb{F}^*_p$ and $\mathbb{F}^*_q$ $$\sum^{p-2}_{k=1} \frac{a^k}{1-g^k} \leftrightarrow \sum^{q-2}_{k=1} \frac{a^k}{1-g^k}$$ for $g$ being primitive in both fields?

[1] Harald Niederreiter, A short proof for explicit formulas for discrete logarithms in finite fields, Applicable Algebra in Engineering, Communication and Computing 1990, Volume 1, Issue 1, pp 55-57