Let $R = \Bbb{Z}/5\Bbb{Z}$ be the ring of integers modulo $5$. Then $4^x = 1, 4, 1, 4, 1, 4, \dots$ as $x$ ranges over $\Bbb{Z}$. Thus $4^x = 3$ and $4^x = 2$ have no solution $x \in \Bbb{Z}$.
Now, $R^{\times}$ the group in which the repeated multiplication resides (exponentiation) is a $\Bbb{Z}$-module. I'm wondering if there's a way to extend the ring of exponents $\Bbb{Z}$ formally to $\Bbb{Z}[x]$ where $x$ is the formal symbol devised so that $4^x = 3 \pmod 5$. Similarly to the way in which $R(i) = \Bbb{C}$ and $i$ solves $x^2 + 1 = 0$. Here, we're wanting to solve $4^x - 3 = 0$ i.e. exponentiation instead of constant powering.
Is there a theory already developed for this case?
Is it even possible? I.e. can we have a ring $S \geqslant \Bbb{Z}$ such that $4^x = 3$ has a solution in $S$? And correspondingly an $S$-module that extends the $\Bbb{Z}$-module $R^{\times}$?
I want to somehow apply this to special cases of the discrete logarithm problem. Since given a prime $p$, some solutions $a^x = b \pmod p$ do not exist, I would like to force them to exist, again similarly to the imaginary unit $i \in \Bbb{C}$.
Two related previous posts:
The group $R^{\times}$ is isomorphic to the additive group $\mathbb{Z}/4\mathbb{Z}$ via the map $f$ where $f(1)=0$, $f(2)=1$, $f(3)=3$, and $f(4)=2$.
Now, working with the additive group that we are more confortable with, the problem translates to asking whether there is an endomorphism $\varphi$ of $\mathbb{Z}/4\mathbb{Z}$ for which $\varphi(2)=3$. The answer is that there is no such endomorphism, for $\varphi(1)+\varphi(1)=\varphi(2)=3$, but the equation $a+a=3$ has no solution in $\mathbb{Z}/4\mathbb{Z}$.
Back to the multiplicative group, the element $4$ is a square, but $3$ is not a square, so there is no endomorphism of the multiplicative group that sends $4$ to $3$. Hence, it is impossible to extend the ring of exponents to the polynomials over the integers in such a way that $4^x=3$.