# Are there infinitely many primes p such that both p-1 and p+1 have at most 3 prime factors, counted with multiplicity?

This is a subsequence of primes $p$ for which $p^2-1$ has at most 6 prime divisors counted with multiplicity.

This sequence described in the question is the sequence A079153 in OEIS.

I could not find references on the first sequence, but I found a mention of the second sequence in another MO question: Number of prime factors of the order of a finite non-abelian simple group with a reference to a survey article by Solomon from 2001: are there infinitely many primes $p$ such that $|PSL(2,p)|=(p-1)*p*(p+1)/2$ is a product of six prime factors? In the survey article it is said this problem is akin to the twin prime conjecture, but no references are given.

• The statement that $p^2-1$ has at most 6 prime factors (with multiplicity) is weaker than the statement that both $p-1$ and $p+1$ have most 3 prime factors (with multiplicity). For example, if $p-1$ has 2 prime factors and $p+1$ has 4 prime factors (with multiplicity), then the first statement is true, while the second is false. – GH from MO Nov 11 '14 at 17:38
• just to make sure I understand the previous comment with a specific example, say $p=23$ so then $p-1=22=2*11$ and $p+1=24=2*2*2*3$, thus $p^2-1$ has 6 prime divisors (with multiplicity) while $p-1$ has only 2, and $p+1$ has 4. Is that right (and if so, should then the OP question be clarified)? – Mirko Nov 11 '14 at 18:09
• Both of you are of course quite right. I am editing my question. – wishcow Nov 11 '14 at 18:50
• Dickson's conjecture would imply that there are infinitely many members of the subsequence where, e.g., $(p-1)/6$ and $(p+1)/4$ are prime. – Robert Israel Nov 11 '14 at 19:37

## 1 Answer

At the present time, the best (general) result is due to James Maynard, who has proven that any admissible 3-tuple takes at most 7 prime factors infinitely often (see this paper). In particular, (using a certain admissible 3-tuple related to what Robert Israel wrote) we can get that $n(n^2-1)$ has at most 11 prime factors infinitely often.

I don't know if this number can be improved for the specific case you are interested in. However, the general consensus is that Dickson's conjecture is true, so both $p-1$ and $p+1$ should consist of exactly 3 prime factors, for infinitely many $p$. But this conjecture is currently out of reach of any known techniques, and appears to be harder than the twin prime conjecture, since we must control admissible 3-tuples, but the twin prime problem deals with only admissible 2-tuples.

• The result of James Maynard is much stronger than the twin prime conjecture. – Jovan Oct 1 '20 at 11:50
• No, it is not. Why do you say it is? – Wojowu Oct 1 '20 at 11:58
• Because $n \frac {\log (\log n)^4} {\log (n^3)}$ is much less than $\frac n {\log^2 n}$ for $n$ sufficiently large. – Jovan Radenkovic Apr 9 at 13:46