1
$\begingroup$

I have the following question:

Given an integer $n \ge 1 $ which is not a square, does there exist a prime number $p$ for which $n$ is a primitive root modulo $p?$

It is easy to see that if $n$ is a square, then it is not a primitive root modulo any odd prime. Artin conjectures that for any such $n$ there are infinitely many such primes, but is it known that at least one such prime always exists?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Not an expert at all, but I find it hard to imagine how such an argument would go. Algebraic techniques seem kind of hopeless to me on this type of problem, and analytic techniques often produce infinitely many primes while they're at it. (I guess that leaves combinatorial techniques, aka "being really clever".) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 10, 2021 at 4:13
  • 4
    $\begingroup$ If I’m not mistaken, this is regarded as pretty much as hard as the general infinitely-many-primes version of the conjecture (for example, assume there are finitely many Fermat primes, then if $n$ is a prim. root mod $p$ and $p$ is suff. large, write $p-1 = 2^e \cdot k$ with $k > 1$ odd and then apply the existence-of-one-prime version of the conjecture to $n^k$ to produce a second sufficiently large prime for which $n$ is a primitive root, etc.). That’s the case for most conj.s/thm.s in analytic num. theory —- producing one example is often thought to be as hard as producing infinitely many. $\endgroup$
    – alpoge
    Commented Feb 10, 2021 at 6:20
  • 1
    $\begingroup$ To illustrate the previous comment with another conjecture, see mathoverflow.net/questions/226794/… $\endgroup$
    – KConrad
    Commented Feb 10, 2021 at 13:25

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .