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

pairsof 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.

anyproof, right? $\endgroup$ – Pete L. Clark Jun 30 '10 at 22:191more comment