4

As the title says, I want to know if for every odd prime $p$, there is another odd prime $q$ such that $p$ is a primitive root modulo $q^m$ for all $m\ge1$. For small $p$ such as $p=3,5,7$, I could check explicitly that $q=5,7,11$, respectively, do the job by showing that $p$ is a primitive root modulo $q$ and $p^{q-1} \not\equiv 1 \mod q^2$. But I am not sure how to show that such a $q$ exists for all $p$. I would appreciate any ideas on how to proceed.

A related question has been asked before: Is every non-square integer a primitive root modulo some odd prime?, but the answers there don't address my question.

Also related is Artin's conjecture on primitive roots, but it asks for infinitely many prime $q$ such that $p$ is a primitive root modulo $q$ rather than a single odd prime $q$ such that $p$ is a primitive root modulo all powers of $q$.

Pranay
  • 199
  • 1
    I think it should follow by the fact that for each $a$ in the range $1 \le a <q$ we have by Dirichlet's theorem that there are infinitely many primes of the form $p=a+bq$. – Merosity Sep 25 '22 at 16:54
  • 3
    @Merosity I do not understand how Dirichlet's theorem implies the statement I want to prove. Can you please explain? – Pranay Sep 25 '22 at 18:55
  • Am I misremembering/misthinking, or isn't it true that if $p$ is a root mod $q$, then $p$ is a root mod $q^k$ for all $k$ already? So really you're looking for "Is every $p$ a root mod some prime $q$?" – Eric Snyder Sep 26 '22 at 04:02
  • 1
    @EricSnyder That's not true. The correct statement is: if $p$ is a primitive root mod $q$ and $p^{q-1} \not\equiv 1 \mod q^2$, then $p$ is a primitive root mod $q^k$ for all $k$. – Pranay Sep 26 '22 at 04:22
  • 1
    Yeah, it's a bit weird, apparently the implication is from "$p$ is a PR mod $q^2$" to "$p$ is a PR mod $q^k$, including $k=1$." Though there is also "$p$ is a PR mod $q$" implies "$p$ or $p+q$ is a PR mod $q^2$." But so far as I know Artin's conjecture hasn't been proven for even $n=1$, i.e., there may be (very large) primes that are not PR of another prime. – Eric Snyder Sep 26 '22 at 06:10
  • 2
    No idea whether this variant of Artin's conjecture is solved or not but I guess that the existence of a SINGLE prime has been established upto a very large limit. Does someone have a refernece how large ? – Peter Sep 26 '22 at 15:29
  • 2
    Maybe , it is worth mentioning that Artin's conjecture is false for at most two primes (see the wikipedia article). And if there are infinite many primes doing the job , it would be more than a mircale if none satisfies the additional condition (generalized Wieferich primes are very rare) , although we probably cannot completely rule it out. – Peter Sep 26 '22 at 15:42
  • 1
    @Peter I imagine the limit is pretty high, certainly in the $10^9$ range, though I have no good source. – Eric Snyder Sep 26 '22 at 21:18
  • 1
    For $p\le 7.8\cdot 10^9$ , there is a prime as desired. – Peter Sep 27 '22 at 10:16
  • @Peter Thanks! That’s great! Do you mean there is a $q$ such that $p$ is a PR mod all powers of $q$, or just $q$? Could you please give me a reference for that upper limit on $p$? – Pranay Sep 27 '22 at 11:33
  • 1
    I checked this with PARI/GP and also considered the condition $p^{q-1}\ne 1\mod q^2$ making $p$ a primitive root mod every $q^k$. I have not saved the routine, but I can reprogram it , if desired , and maybe even post it as an answer. – Peter Sep 27 '22 at 13:27
  • 1
    @Peter Thanks, your comment is good enough for me. – Pranay Sep 27 '22 at 15:58
  • A related interesting question has unfortunately already been asked here eight years ago , so I will formulate it here : CONJECTURE : For every odd prime $p$ there is a prime $q$ smaller than $p$ that is a primitive root mod $p$. Without this restriction Dirichtlet's theorem establishes the existence. Is this conjecture open ? – Peter Oct 02 '22 at 08:17

0 Answers0