Are there infinitely many $n$ such that $n!−1$ and $n!+1$ are prime numbers?
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9$\begingroup$ If true (which I doubt), this would considerably strengthen the Twin Pirme Conjecture, which is still open. $\endgroup$– Alessandro Della CorteCommented Aug 16, 2021 at 12:31
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$\begingroup$ math.stackexchange.com/questions/1794286/… $\endgroup$– Loreno HeerCommented Aug 16, 2021 at 12:36
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10$\begingroup$ On another note, we do not even know whether $n!+1$ OR $n!-1$ is prime infinitely often! $\endgroup$– Christian BernertCommented Aug 16, 2021 at 12:36
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2$\begingroup$ Given that the underlying mathematical problem seems to be a difficult one, why does this question collect so many negative votes? $\endgroup$– Alex M.Commented Aug 16, 2021 at 14:01
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6$\begingroup$ @AlexM. Although I answered the question, I think I can explain. (I have not voted on it either way). The question is one which if one thinks about for a small amount of time should be likely to be very difficult. And the OP did not indicate any attempt at thinking about the question. If for example, the OP had mentioned that this is a sparse family of twin primes, or if the OP had indicated they had done some small numerical checking it would likely have been substantially better received. $\endgroup$– JoshuaZCommented Aug 16, 2021 at 14:14
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1 Answer
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Almost certainly not. But note that if it were true, proving it would be extremely tough. We can't even prove now that there are infinitely many primes of the form $n!+1$ or that there are infinitely many primes of the form $n!-1$, and we can't prove that there are infinitely many twin primes, that is prime pairs which are two away from each other.
However, while we suspect that there are infinitely many in each of those three categories, we should strongly doubt that there are infinitely many $n$ where both $n!-1$ and $n!+1$ are prime.
Heuristically, the Prime Number Theorem tells that the "chance" a number x is prime should be roughly $\frac{1}{\log x}$. By Stirling's formula, $\frac{1}{\log n!}$ is very close to $\frac{1}{n \log n}$. So the expected number of such prime pairs should be $$\sum_{n=1}^{\infty} \frac{1}{\log (n!+1)} \frac{1}{\log (n!-1)} \approx \sum \frac{1}{(n \log n)^2},$$ which is a convergent series, so we should expect only finitely many such. I'd in fact go out on a limb and strongly suspect that $n=3$ is the largest $n$ where both $n!+1$ and $n!-1$ are prime. Using the relevant OEIS lists here and here one can see at a glance that there are no examples after $n=3$ below $n=150209$.
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$\begingroup$ It seems not very rare that $n!^2-1$ is a product of three primes. $\endgroup$ Commented Aug 16, 2021 at 13:45
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$\begingroup$ @მამუკაჯიბლაძე Do you know such $n$ which is greater than 38? $\endgroup$– Stefan Kohl ♦Commented Aug 16, 2021 at 15:51
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$\begingroup$ @StefanKohl You got me :) $\endgroup$ Commented Aug 16, 2021 at 16:11
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3$\begingroup$ @მამუკაჯიბლაძე I would suspect that for any fixed k, the set of $n$ where $n!^2 -1$ has $k$ prime factors is finite. A similar heuristic to the above will agree. $\endgroup$– JoshuaZCommented Aug 16, 2021 at 16:21