3
$\begingroup$

Let $n$ be an integer and $n=p_1^{a_1}\dots p_s^{a_s}$ be its factorization into primes. Denote by $\Omega(n)$ the sum of $a_i$. Does there exist a constant $k$ such that there are infinitely many primes $p$ such that $\Omega(p^2-1)\leqslant k$?

$\endgroup$
3
  • 1
    $\begingroup$ Any sieve method will show this for some value of $k$ -- e.g. $k=20$ would probably be pretty easy to show (and probably something like $k=10$ would be known with some effort). $\endgroup$
    – Lucia
    May 25, 2016 at 15:27
  • 2
    $\begingroup$ As $p^2-1$ is going to be divisible by 24 for large $p$, one should reduce $k$ by $4$ and work with $(p+1)(p-1)/24$. In view of arxiv.org/abs/1205.5021 it might be possible to get $k$ down to 6+4=10 with current methods, though in that paper one cannot specify one of the three linear forms $L_1(n), L_2(n), L_3(n)$ to be prime, so probably one has to make do with something slightly worse than 10. In any event Maynard's result certainly shows that $\Omega(n(n^2-1)) \leq 11$ infinitely often, with $n$ coprime to any fixed finite number of primes. $\endgroup$
    – Terry Tao
    May 25, 2016 at 16:28
  • $\begingroup$ Actually the discussion on page 2 of the Maynard paper in my previous comment suggests that one can take k=11 for the original problem. $\endgroup$
    – Terry Tao
    May 25, 2016 at 16:35

1 Answer 1

3
$\begingroup$

This will follow with very small $k$ from Schinzel's hypothesis H or as @Gerry Myerson points out Dickson's conjecture.

Let $p$ be of the form $12n+1$. There are no congruence obstructions $p$ and $p_1=(p-1)/12=n$ and $p_2=(p+1)/2=6n+1$ to be simultaneously prime. By the above conjecture, they are prime infinitely often answering with very small $k$ since $p^2-1=12\cdot2\cdot p_1 p_2$

$\endgroup$
1
  • $\begingroup$ Full strength of Schinzel not needed here – Dickson's conjecture will do. $\endgroup$ May 25, 2016 at 13:32

Your Answer

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

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