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Let $n$ be a non-square positive number. The Artin Conjecture states that there are infinitely primes $p$ for which $n$ is a primitive root.

Question: Given a number $n$, what is the best upper bound on the proportion of primes for which $n$ is a primitive root.

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  • $\begingroup$ It's probably $\frac{\varphi(p - 1)}{p - 1}$. This is just the number of nonzero residues $\bmod p$ which are primitive roots. $\endgroup$ – Qiaochu Yuan Apr 27 at 2:04
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    $\begingroup$ Do you mean given an $n$, what proportion of primes $p$ will $n$ be a primitive root mod $p$? Or do you mean given a $p$, how many numbers will be primitive roots mod $p$? $\endgroup$ – Alexander Gruber Apr 27 at 2:13
  • $\begingroup$ See the edited question. $\endgroup$ – Improve Apr 27 at 2:47
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    $\begingroup$ The Artin Conjecture says that $n$ will be a primitive root (assuming it it neither a square nor $-1$) for a positive proportion of primes. The proportionality constant is a rational multiple of $\prod_p \left(1 - \frac{1}{p(p-1)}\right)$ which depends on the factorization of $n$ (for example, when $n = 2$, the constant is $1$). Asymptotically, it is easy to show that the upper bound holds unconditionally, so one upper bound is $C_n + o(1)$ where $C_n$ is what is predicted by the conjecture. $\endgroup$ – Furlo Roth Apr 27 at 10:24
  • $\begingroup$ To see this, ask for the proportion of primes so that $n$ is not an $\ell$th power mod $p$ for $\ell < X$, and apply Chebotarev to get an upper bound $C_n(X) + o(1)$ where $C_n(X)$ involves a product over all primes $p < X$. Then let $X \rightarrow \infty$. I guess using explicit forms of Chebotarev on could make the $o(1)$ term more explicit but it might be a bit painful. $\endgroup$ – Furlo Roth Apr 27 at 10:27

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