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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).

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    $\begingroup$ 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 $x^2+y^2+1$. $\endgroup$
    – zabroshenie goroda
    Commented Feb 8, 2019 at 13:57
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    $\begingroup$ From the work of Landau and Ramanujan, and surveyed by P. Moree, the set $N$ has the property $|\{n<x: n\in N\}| \asymp x/(\log x)^{1-1/\varphi(d)}$. $\endgroup$
    – zabroshenie goroda
    Commented Feb 8, 2019 at 14:16

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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.

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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.

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    $\begingroup$ 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? $\endgroup$
    – Lucia
    Commented Feb 9, 2019 at 8:16
  • $\begingroup$ Hmm, you could be right. $\endgroup$
    – Greg Martin
    Commented Feb 9, 2019 at 23:19

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