The second answer to this question says the following:

"I believe that sieve methods (in particular methods similar to Chen's result on "almost" Goldbach and twin primes) should give infinitely primes $p$ such that $\frac{p-1}{2}$ is a product of two primes".

Is this claim indeed true and why?

Notice that by Dirichlet's theorem we can obtain the following claim: Given two arbitrary primes $Q$ and $R$, there are infinitely many primes $p$ of the form $1+2nQR$, then $\frac{p-1}{2}=nQR$, so the given product of the two primes, $QR$, divides $\frac{p-1}{2}$.

Thank you very much!

  • $\begingroup$ What is the question? $\endgroup$ – abiessu Jun 26 '18 at 23:12
  • $\begingroup$ The question is: Is the quoted claim true? If it is true, then how to prove it? (at least a short sketch of proof will be ok). $\endgroup$ – user237522 Jun 26 '18 at 23:14
  • $\begingroup$ I think you’ve started with one too many primes... you might consider a prime $P$ with primes in the arithmetic progression $1+2nP$... $\endgroup$ – abiessu Jun 26 '18 at 23:15
  • $\begingroup$ Can we find such $n \in \mathbb{N}$ which is prime? $\endgroup$ – user237522 Jun 26 '18 at 23:16
  • 1
    $\begingroup$ It would already be sufficient that infinite many primes $q>3$ exist such that $6q+1$ is prime. This follows from the generalized Bunyakovsky conjecture. Since the statement is much weaker there might be a proof of it. $\endgroup$ – Peter Jun 28 '18 at 11:29

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