TITLE: Composite pairs of the form n!-1 and n!+1
QUESTION [25 upvotes]: It's well known that the numbers of the form $n!\pm1$ are not always prime. Indeed, Wilson's Theorem guarantees that $(p-2)!-1$ and $(p-1)!+1$ are composite for every prime number $p > 5$. 

Is there a proof, preferably an elementary proof, that there are infinitely many composite pairs of the form $n!\pm1$?

The motivation for this question comes from my answer to this recent question. There, I show that every nonstandard model of Peano Arithmetic has a $\mathbb{Z}$-chain consisting entirely of composite numbers. The example I gave is that of a $\mathbb{Z}$-chain contained in the infinite interval $[N!+2,N!+N]$, where $N$ is any nonstandard natural number. I wonder if I could have picked some $\mathbb{Z}$-chain centered at $N!$ instead. A positive answer to the above question would mean that this is indeed possible. Note that it is important in this context that the proof is elementary, but I will also accept beautiful analytic arguments.
Andrey Rekalo pointed out that $(N!)^3 \pm 1$ are both composite. This means that, if $N$ is a nonstandard integer, then the $\mathbb{Z}$-chain centered at $(N!)^3$ has only composite numbers all but two have standard factors. I don't know if it's possible to find a $\mathbb{Z}$-chain all of whose elements have a standard factor.

REPLY [3 votes]: Explicit constructions of infinitely many examples seem to be difficult. Looking at a table of factorizations of $N! \pm 1$ I noticed the following pattern (and now I see that this is essentially what Dror suggested in his comment):
Assume that $q \equiv 3 \bmod 4$ and $p = \frac{q+3}2$ are prime numbers. Then for $n = p-2$, we have $p \mid n!-1$ and $q \mid n!+1$ if $h(-q) \equiv 1 \bmod 4$, where $h(m)$ denotes the class number of ${\mathbb Q}(\sqrt{m})$. Probabilistically, the class number of $h(-p)$ should
be $\equiv 1 \bmod 4$ in half the cases.