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Fix $\epsilon>0$. As $q$ becomes large, is it true that the number of primes less than $q^{1+\epsilon}$ congruent to $1$ modulo $q$ will tend to infinity?

A conjecture of Montgomery says that the number of primes congruent to $a$ mod $q$ should tend to $\pi(x)/\varphi(q)$ when $q<x^{1-\epsilon}$, which gives an extremely strong version of what I want. I don't need the count of primes to be close to $\pi(x)/\varphi(q)$ though, just infinite. Moreover, I know that often these sorts of computations are easier for the residue class $1$, so perhaps that will help here.

I've done some searching online, but I can't find anything. If anybody knows of such a result, that would be fantastic!

Eric Wofsey
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Milo Moses
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  • What would be predicted by the random model for the primes? – reuns Jan 05 '23 at 22:18
  • @reuns I'm not sure exactly, but certainly the infinitude I desire should be implied; Hence Montgomery's conjecture – Milo Moses Jan 05 '23 at 22:21
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    Linnik's theorem is closely related - Linnik has proven that the least prime in any arithmetic progression modulo $q$ is $O(q^d)$ for a universal constant $d$. It is known that $d=5$ works, and GRH implies $d=2$ works (up to some log factors). I don't know if restricting to just the arithmetic progression $1+nq$ improves these results, but I doubt you can get $d=1+\varepsilon$ unconditionally. – Wojowu Jan 05 '23 at 23:29
  • @Wojowu Thank you for your answer! I find it highly unlikely that the arithmetic progression $1+nq$ would lead to $d=1+\epsilon$ unconditionally as well, so this seems like the end of the road for me. – Milo Moses Jan 06 '23 at 00:06
  • I agree with @Wojowu's comment, and furthermore I can confirm that I don't know any reason the problem is any easier for any prescribed residue class, including 1 (mod $q$). – Greg Martin Jan 24 '23 at 02:30

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