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Artin's primitive root conjecture asserts that if an integer $a \ne -1$ is not a perfect square then $a$ is a primitive root mod $p$ for infinitely many primes $p$. This conjecture is still open.

An easier question is the following: given $a$ as above and a non-zero integer $b$, does $b$ belong to the multiplicative group $a$ generates mod $p$ for infinitely many $p$? According to paragraph 9.4.2 of this survey on Artin's conjecture, this was proved by Pólya.

I'm interested in a strengthening of this result:

Question 1. Given $a$ and $b$ as above, does $b$ belong to the $p$-adic closure of the subgroup of $\mathbf{Q}_p^{\times}$ generated by $a$ for infinitely many $p$?

In fact, I'd like an even stronger statement:

Question 2. Given $a$ as above and a non-zero element $b$ in a number field $K$, are there infinitely many prime ideals $\mathfrak{p}$ of $K$ such that $b$ belongs to the closure of the subgroup of $K_{\mathfrak{p}}^{\times}$ generated by $a$?

I'd appreciate any thoughts or pointers to the literature!

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  • $\begingroup$ I'd look up Polya's proof and see if I could tweak it to give a positive answer to Q1. In any case: $b \pmod{p} \in \left\langle a \pmod{p} \right\rangle$ does not imply $b \in \overline{\left\langle a \right\rangle}$, but maybe the implication does hold when $a^{p-1} \not\equiv 1 \pmod{p^2}$ (which means that some power $a^k$ will generate $1+p\mathbb{Z}_p \subset \mathbb{Z}_p^\ast$ topologically), for $p>2$ at least. It is worth checking if there is any room in Polya's argument to ensure this last condition holds for infinitely many $p$. $\endgroup$
    – R.P.
    Commented Mar 25, 2022 at 7:58

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