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).

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

2 Answers 2


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.

  • 3
    $\begingroup$ 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
    Feb 9, 2019 at 8:16
  • $\begingroup$ Hmm, you could be right. $\endgroup$ Feb 9, 2019 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.