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 $\Omega(p^2-1)\leqslant k$?
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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). $\endgroup$– LuciaCommented May 25, 2016 at 15:27
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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 TaoCommented May 25, 2016 at 16:28
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$\begingroup$ 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 TaoCommented May 25, 2016 at 16:35
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1 Answer
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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 $p^2-1=12\cdot2\cdot p_1 p_2$
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$\begingroup$ Full strength of Schinzel not needed here – Dickson's conjecture will do. $\endgroup$ Commented May 25, 2016 at 13:32