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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!

user237522
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  • What is the question? – abiessu Jun 26 '18 at 23:12
  • 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). – user237522 Jun 26 '18 at 23:14
  • I think you’ve started with one too many primes... you might consider a prime $P$ with primes in the arithmetic progression $1+2nP$... – abiessu Jun 26 '18 at 23:15
  • Can we find such $n \in \mathbb{N}$ which is prime? – user237522 Jun 26 '18 at 23:16
  • Lemma 1 of this paper by Heath Brow gives the answer to your question: https://academic.oup.com/qjmath/article-abstract/37/1/27/1515517?redirectedFrom=PDF – Sungjin Kim Jun 27 '18 at 00:14
  • Heath Brown - a typo.. – Sungjin Kim Jun 27 '18 at 00:20
  • Please, could you quote Lemma 1? I have access to the first page only (you can write your comment as an answer, if you like). – user237522 Jun 27 '18 at 00:20
  • Let $k=1,2$ or $3$, and put $K=2^k$. Let $u$ and $v$ be coprime integers such that $K|u-1$, $16|v$ and $((u-1)/K, v)=1$. Then there exists $\alpha\in(1/4,1/2]$, possibly depending on $k$, $u$ and $v$ such that $$# {p\leq x: p\equiv u (\mathrm{mod} \ v), (p-1)/K = P_2(\alpha)} \gg x (\log x)^{-2},$$ where the implied constant may depend on $k$, $u$, $v$ and $\alpha$. – Sungjin Kim Jun 27 '18 at 00:22
  • Here, $n=P_2(\alpha)$ means that $n$ is either prime or is the product of at most $r$ primes $p_i$ each of which satisfies $p_i\geq n^{\alpha}$. – Sungjin Kim Jun 27 '18 at 00:23
  • Actually, it does not answer your question fully. In fact, your question with product of at most 2 primes. – Sungjin Kim Jun 27 '18 at 00:25
  • Please, could you write your last three comments as an answer? – user237522 Jun 27 '18 at 00:27
  • Thanks for the offer, but I'd let them stay as comments until someone comes up with full solution to your question. – Sungjin Kim Jun 27 '18 at 00:30
  • Did you mean that $r=k$ (a typo)? Have you taken $k=1$? – user237522 Jun 27 '18 at 00:41
  • If you claim that Lemma 1 proves that there are infinitely many primes $p$ such that $\frac{p-1}{2}$ is prime, then this is great. – user237522 Jun 27 '18 at 00:46
  • After receiving an answer to https://math.stackexchange.com/questions/2833165/infinitely-many-primes-p-such-that-fracp-12-is-prime?noredirect=1&lq=1, I guess that you did not claim that Lemma 1 proves that there exist infinitely many primes $p$ such that $\frac{p-1}{2}$ is prime. I still do not understand what is $r$ in your comments ($r \leq 2$?). – user237522 Jun 27 '18 at 02:12
  • Yes, in the comment about $P_2(\alpha)$, the number $r$ should be $2$. – Sungjin Kim Jun 27 '18 at 02:37
  • Thanks. In light of the answer to https://math.stackexchange.com/questions/2833165/infinitely-many-primes-p-such-that-fracp-12-is-prime, your answer is probably the best answer I can get now (unless someone has solved https://en.wikipedia.org/wiki/Sophie_Germain_prime). – user237522 Jun 27 '18 at 14:26
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    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. – Peter Jun 28 '18 at 11:29
  • Thank you very much! Sounds interesting. – user237522 Jun 28 '18 at 17:55

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