# What is the best upper bound for “how often” a number n is a primitive root modulo a prime p?

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.

• It's probably $\frac{\varphi(p - 1)}{p - 1}$. This is just the number of nonzero residues $\bmod p$ which are primitive roots. – Qiaochu Yuan Apr 27 at 2:04
• 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 $p$? – Alexander Gruber Apr 27 at 2:13
• See the edited question. – Improve Apr 27 at 2:47
• 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. – Furlo Roth Apr 27 at 10:24
• 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. – Furlo Roth Apr 27 at 10:27