When is $\mathbb{F}_{p^k}^{\times} \cong \mathbb{Z}_m^{\times}$ for $k>1$?

Question

Let $\mathbb{F}_{p^k}^{\times}$ denote the group of units of a finite field, and let $\mathbb{Z}_m^{\times}$ denote the group of units modulo some integer $m$. I'm curious about when $\mathbb{F}_{p^k}^{\times} \cong \mathbb{Z}_m^{\times}$. Two relevant and well known results are:

1. The group of units of a finite field is cyclic.
2. $\mathbb{Z}_m^{\times}$ is cyclic iff $m=2, 4, q^k$ or $2q^k$ where $q$ is an odd prime and $k \geq 1$.

For $k=1$, we get a solution whenever $|\mathbb{Z}_m^{\times}| = \varphi(m) = p-1$ and $\mathbb{Z}_m^{\times}$ is cyclic (by the first result and since cyclic groups of the same order are isomorphic). Then by the second result, $m=p$ and $m=2p$ are the only such values where $p$ is a prime.

For $k>1$, by the first and second result we need to find $\mathbb{F}_{p^k}^{\times} \cong \mathbb{Z}_{q^s}^{\times} \cong \mathbb{Z}_{2q^s}^{\times}$ for an odd prime $q$. And since cyclic groups of the same order are isomorphic, it suffices to have $|\mathbb{F}_{p^k}^{\times}| =|\mathbb{Z}_{q^s}^{\times}|$. Thus, we get the diophantine equation $p^k - 1 = q^s - q^{s-1}$ which has solutions for odd primes $p,q$ iff $\mathbb{F}_{p^k}^{\times} \cong \mathbb{Z}_{q^s}^{\times} \cong \mathbb{Z}_{2q^s}^{\times}$.

One might suspect that this would not have solutions, but there is this one:

$7^3 - 1 = 19^2 - 19 \implies \mathbb{F}_{7^3}^{\times} \cong \mathbb{Z}_{19^2}^{\times} \cong \mathbb{Z}_{2 \cdot19^2}^{\times}$

I've checked for $p, q, k, s \leq 300$ and this appears to be the only such solution. I have tried to prove this and have a few very partial results. For instance, any solution must satisfy the following conditions:

• $q > p \iff (p-1) \lvert (q-1)$

This follows fairly easily after reducing modulo $p-1$.

• $q^{s-1} < p^k < q^s$

Since $p^k - 1 = q^s - q^{s-1} < q^s$ and we cannot have $p^k = q^s$ for distinct primes $p, q$ so, $p^k < q^s$. For the other inequality, $p^k > p^k - 1 = q^s - q^{s-1} = q^{s-1}(q-1) > q^{s-1}$.

Any suggestions how one can show that $p^k - 1 = q^s - q^{s-1}$ has no solutions with $k > 1$ beyond $(p, q, k, s) = (7, 19, 3, 2)$?

EDIT (22/10/23): This "proof" is currently wrong, but I will keep it here for future reference (see comments) I believe we can show $s=2$ for any solution. Let $s > 1$. Since $(p-1) \lvert (q-1)$, we of course get that $t := \frac{q-1}{p-1}$ is an integer. Writing the equation as,

$$p^k - 1 = q^{s-1}(q - 1)$$

factoring,

$$(p-1)(1 + p + \cdots + p^{k-1}) = q^{s-1}(q-1)$$

$$(1 + p + \cdots + p^{k-1}) = q^{s-1} \frac{q-1}{p-1}$$

$$(1 + p + \cdots + p^{k-1}) = q^{s-1} t = q^{s-2} \cdot qt$$

We want to show that $GCD(1 + p + \cdots + p^{k-1}, q^{s-2}) = 1$. Otherwise, we would get that $1 + p + \cdots + p^{k-1} = q^r$ for $r \leq s-2 \implies p^{k-1} < q^r \leq q^{s-2}$.

We also have, $p^k > p^k - 1 = q^s - q^{s-1} > q^{s-1}$ since $q > 2$.

Putting these together, we get that $p^{k-1} < q^{s-2} < q^{s-1} < p^k$ which is a contradiction since $q > p$.

Thus, $GCD(1 + p + \cdots + p^{k-1}, q^{s-2}) = 1$ and since $1 + p + \cdots + p^{k-1} = q^{s-2} \cdot qt$ we have,

$$q^{s-2} = 1 \implies s = 2$$.

Answer 1

This is an attempt at a partial answer. Assume the abc conjecture holds.

Rewrite the diophantine equation as $(q^s - q^{s-1}) + 1 = p^k$. We will now bound from below the quality

$$\frac{\log(p^k)}{\log(\textrm{rad}((q^s - q^{s-1})p^k))} = \frac{k\log(p)}{\log(\textrm{rad}(q(q - 1)p))} \geq \frac{k\log(p)}{\log(q(q - 1)p)}.$$

If we assume $k$ to be at least 4 (so 5 minimum, as it has to be odd), and $q < p$, then it is clear that $\log(q(q - 1)p) < 3\log(p)$ and $k\log(p) > 4\log(p)$, so the fraction is bounded from below by $\frac{4}{3}$. This means that, unless you are about to disprove the abc conjecture, there can only be finitely many solutions with $k$ above 3 and $q < p$. If $k = 3$, you can only really make the same argument if $q^{1 + \epsilon} < p$ always holds for some $\epsilon > 0$ independent of $p$ and $q$. In your example, $q > p$, so I shall try to prove a similar result for that.

We will again use abc conjecture, with (to adopt your notation in the edit) $a = 1 + p + p^2 + \ldots + p^{\frac{k-1}{2}}$, $b = p^{\frac{k+1}{2}} + \ldots + p^{k-1}$ and $c = q^{s-1}t$. We shall bound the quality from below:

$$\frac{(s-1)\log(q)+\log(t)}{\log(\textrm{rad}(abqt))}$$

Now, $\textrm{rad}(abqt) = \textrm{rad}(paqt)$, because $b = p^{\frac{k+1}{2}}a$. Furthermore, $at \leq (p^{\frac{k+1}{2}} - p^{\frac{k-1}{2}})t = p^{\frac{k-1}{2}}(q - 1)$. This means

$$\frac{(s-1)\log(q)+\log(t)}{\log(\textrm{rad}(abqt))} \geq \frac{(s-1)\log(q)}{\log(p^{\frac{k+1}{2}}q(q-1))}.$$

As $p^k < q^s$, $p^{\frac{k+1}{2}} < q^{\frac{s+1}{2}}$, we can finally write

$$\frac{(s-1)\log(q)}{\log(p^{\frac{k+1}{2}}q(q-1))} \geq \frac{(s-1)\log(q)}{(2 + \frac{s+1}{2})\log(q)} = \frac{2s - 2}{s + 5}.$$

Thus, if $s > 7$, there should only be finitely many solutions.