Is either $n! + 1$ or $n! - 1$ not prime for all $n$? I was looking at an article about factorial primes, and I noticed that both $n!+1$ and $n!-1$ were not prime. (As in, there are no numbers $n$ such that both $n!+1$ and $n!-1$ are prime). I think that for any $n$, both $n!+1$ and $n!-1$ cannot be prime. Is this an easy thing to prove? If so, how? Would Wilson's theorem be applicable in some way?
This is just a conjecture that I am asking out of curiosity. I would love some thoughts on how one might approach such a problem as this one.
 A: The smallest prime factors of $n!^2 - 1$ for the first few values of $n$ seem to me to be bizarrely small but I only know how to "explain" some of them. Wilson's theorem gives, for a prime $p$, the following:
$$(p-1)! \equiv -1 \bmod p$$
$$(p-2)! \equiv 1 \bmod p$$
$$\left( \frac{p-1}{2} \right)! \equiv \pm 1 \bmod p, p \equiv 3 \bmod 4$$
(the last one is a nice exercise). More generally we have
$$(p-k)! \equiv (-1)^k (k-1)! \bmod p$$
which will explain one mystery a bit later. Now, applying the first three facts, we have
$$3! \equiv 1 \bmod 5, -1 \bmod 7$$
$$4! \equiv -1 \bmod 5$$
$$5! \equiv 1 \bmod 7, -1 \bmod 11$$
$$6! \equiv -1 \bmod 7$$
The first one I don't know how to explain is
$$7! \equiv -1 \bmod 71$$
but it's striking that $71 \equiv 1 \bmod 7$. Also we have
$$8! \equiv 1 \bmod 23, -1 \bmod 61$$
which I also don't know how to explain, but it's again striking that $23 \equiv -1 \bmod 8$. Then we have easy Wilson cases again,
$$9! \equiv 1 \bmod 11, -1 \bmod 19$$
$$10! \equiv -1 \bmod 11$$
$$11! \equiv 1 \bmod 13, 23$$
$$12! \equiv -1 \bmod 13$$
and then the fairly mysterious
$$13! \equiv -1 \bmod 83$$
(here we have $83 \equiv 5 \bmod 13$ which is a square root of $-1 \bmod 13$, what's up with that) and the somewhat more explainable
$$14! \equiv (23-9)! \equiv -8! \equiv -1 \bmod 23.$$
Next is a round of Wilson's theorem again:
$$15! \equiv 1 \bmod 17$$
$$16! \equiv -1 \bmod 17$$
$$17! \equiv 1 \bmod 19$$
$$18! \equiv -1 \bmod 19$$
and then
$$19! \equiv -1 \bmod 71$$
(recall that we saw above that $7! \equiv -1 \bmod 71$, and I didn't include it above but we also have $9! \equiv -1 \bmod 71$). Up until this point the smallest prime factor was at most $2$ digits which I personally think is wacky, but now I am defeated: for $n = 20$ the smallest prime factor is
$$20! \equiv 1 \bmod 124769$$
so whatever's been powering our luck it's run out. There are some other coincidences I don't know how to explain: for example, $61$ divides not only $8!+1$ but also $16!+1$ and $18!+1$, and $661$ divides not only $8!+1$ but also $17!+1$. Very strange.
A: The OEIS entry on factorial primes currently states that this is an open problem:

Conjecture: 3 is the intersection of A002981 and A002982.

The two referenced sequences consist of the natural numbers $n$ such that $n!+1$ is prime, and the $n$ such that $n!-1$ is prime, respectively. Their intersection is exactly the numbers $n$ you are looking for, where both $n!+1$ and $n!-1$ are prime.
