# Infinitely many primes in particular progressions

I'm faced with the following problem on primes. Does someone have any clue? Is it (a reformulation of) an open problem?

Let $$d$$ be a positive integer, $$d\geq 2$$. By Dirichlet's theorem, there is an infinite set $$\mathcal{P}$$ of primes congruent to $$1$$ modulo $$d$$.

Consider the set $$N$$ of integers $$n=1+kd$$ where all prime divisors of $$k$$ belong to $$\mathcal{P}$$.

Could we expect that $$N$$ contains infinitely many primes? (we need at least $$d$$ to be even).

• If $d=2$, the primes in $N$ are those that are $3$ mod $4$. If $d=4$, the primes in $N$ are a subset of primes of the form $x^2+y^2+1$. – zabroshenie goroda Feb 8 '19 at 13:57
• From the work of Landau and Ramanujan, and surveyed by P. Moree, the set $N$ has the property $|\{n<x: n\in N\}| \asymp x/(\log x)^{1-1/\varphi(d)}$. – zabroshenie goroda Feb 8 '19 at 14:16

## 2 Answers

Yes, we should expect it. For any even $$d \ge 2$$, Dickson's conjecture implies that there are infinitely many primes $$p \equiv 1 \bmod d$$ such that $$1 + p d$$ is prime.

Of course, expecting and proving are very different matters.

Chen's theorem says that there are infinitely many numbers $$k$$ such that $$k-2$$ is prime and $$k$$ is either prime or the product of two primes ("$$k$$ is a $$P_2$$ number").

This theorem can be modified relatively easily to prove that for any fixed $$d$$, there are infinitely many numbers $$k$$ such that $$kd+1$$ is primes and $$k$$ is a $$P_2$$ number.

I suspect that the proof of Chen's theorem could be modified, without too much trouble, to prove that there are infinitely numbers $$k\equiv1\pmod d$$ such that $$k-2$$ is prime and $$k$$ is either prime or the product of two primes that are both $$\equiv1\pmod d$$.

Combining these two modifications together would yield an unconditional proof that your $$N$$ contains infinitely many primes.

• Can that really be done? I thought the sieve in Chen's theorem produced numbers with $p+2$ not having any prime factors below $p^{1/3}$. How could one also control the larger primes and put them in a progression? – Lucia Feb 9 '19 at 8:16
• Hmm, you could be right. – Greg Martin Feb 9 '19 at 23:19