A stronger form of the Dirichlet Theorem on prime numbers in arithmetic sequences

Question 1. Let $a,b>1$ be two natural numbers. Is there a prime number $pin 1+bmathbb N$ such that $a+pmathbb Z$ is a generator of the multiplicative group of the field $mathbb Z/pmathbb Z$?

In need this fact for establishing some properties of the Golomb topology on positive integers.

Added in Edit: Francesco Polizzi presented a simple counterexample to Question 1: $a=4$ and $bge 3$. In this case $a$ is a square. So, we can ask a more refined version of Question 1 (which is still sufficient for my purposes).

Question 1'. Let $a,binmathbb N$ and $anotinn^k:n,kinmathbb Nsetminus1$. Is there a prime number $pin 1+bmathbb N$ such that $a+pmathbb Z$ is a generator of the multiplicative group of the field $mathbb Z/pmathbb Z$?

It seems to me that the answer to Question $1'$ is still no in general: for any $a$, if we take $b=4a$, then every prime $p equiv 1pmodb$ is such that $a$ is a square modulo $p$ (a consequence of the law of quadratic reciprocity). As in Francesco's answer, this means that $a+pmathbbZ$ does not generate $mathbbZ/pmathbbZ$.

– so-called friend Don
Sep 8 '18 at 3:57

@so-calledfriendDon is right; I had misquoted Moree's theorem. I have updated my answer with the correct formulation.

– David Loeffler
Sep 8 '18 at 21:32

2 Answers
2

This is a hybrid of Dirichlet's theorem with Artin's conjecture on primitive roots. Artin's primitive root conjecture says that if $a in mathbfZ$ is not a perfect square or $-1$, then there are infinitely many p such that a is a generator mod p (and, further, that the set of such p has some specific non-zero relative density inside the set of primes, depending on a).

The primitive root conjecture is presently open, I'm afraid, so there is no hope of proving this stronger statement. However, if you're content with a conditional proof, then Artin's primitive root conjecture is known to follow from the Generalised Riemann Hypothesis by work of Hooley (whose 90th birthday it is today, incidentally); and Hooley's work has been extended by subsequent authors to consider the case of primes in arithmetic progressions. Here is a paper by Moree which gives an explicit formula (Theorem 4 of the paper) for the density of primes $p$ such that $p = 1 bmod b$ and $a$ is a generator mod $p$, for any non-zero integers $a, b$.

In particular, Moree shows (mid-way down page 14 of the paper) that there are infinitely many primes $p$ with this property if and only if the following are satisfied:

(If the last condition is not satisfied, then $a$ will be a square modulo $p$ for every $p = 1 bmod b$, so it cannot be a generator. The examples by Francesco and Don show that this condition is really needed.)

But I assume from $a$ a bit more: it is not a power of any element (not just that it is not a perfect square). Maybe this will help somehow (using Grunwald-Wang Theorem)? For example $a=8$ is not a square but is a 3-d power.

– Taras Banakh
Sep 7 '18 at 10:59

Great answer—succint, yet says everything that needs to be said!

– Greg Martin
Sep 7 '18 at 18:00

My original answer was a little too succinct, and was actually wrong, as Don's example shows. I have fixed it.

– David Loeffler
Sep 8 '18 at 21:29

I think that, in this generality, the answer is no.

For instance, take $a=4$, $b geq 3$. Then $p geq 7$, so $a=2^2$ is a square mod $p$.

But a generator $x$ of $(mathbfZ_p)^times$ is always a non-square: indeed, if $x=y^2$ then $x^(p-1)/2=y^p-1=1$, so the order of $x$ divides $fracp-12$, in particular it is strictly less than $p-1$.

Thank you for the answer. In fact, for my purposes it suffices to assume that $a$ is not a power of any other number. Then what is the answer in this case?

– Taras Banakh
Sep 7 '18 at 10:13

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