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

Question

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

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

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Added in Edit: Francesco Polizzi presented a simple counterexample to Question 1: $a=4$ and $b\ge 3$. 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 $a,b\in\mathbb N$ and $a\notin\{n^k:n,k\in\mathbb N\setminus\{1\}\}$. Is there a prime number $p\in 1+b\mathbb N$ such that $a+p\mathbb Z$ is a generator of the multiplicative group of the field $\mathbb Z/p\mathbb Z$?

Answer 1

This is a hybrid of Dirichlet's theorem with Artin's conjecture on primitive roots. Artin's primitive root conjecture says that if $a \in \mathbf{Z}$ 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:

• $a \ne -1$;
• $a$ is not an $\ell$-th power for any prime $\ell$ dividing $b$;
• $b$ is not divisible by the integer $\Delta$, where $\Delta$ is defined as follows: write $a = u v^2$, where $u$ is square-free; then $\Delta = u$ if $u = 1 \bmod 4$ and $\Delta = 4u$ otherwise.

(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.)

Answer 2

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 $(\mathbf{Z}_p)^{\times}$ 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 $\frac{p-1}{2}$, in particular it is strictly less than $p-1$.