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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.

Since this appears to be harder than at first sight, let me ask a slightly different question which has a better chance of having an answer using known methods.

Is it true that for every positive integer $B$ there is a positive integer $N$ such

Andrey Rekalo pointed out that $N$ is divisible by all primes up to $B$, and $N \pm 1$ are both composite?

If $n \geq B$ and $n! (N!)^3 \pm 1$ are both composite. This means that, then if $N = n!$ N$ is as required. Again, elementary arguments are preferreda nonstandard integer, then the $\mathbb{Z}$-chain centered at $(N!)^3$ has only composite numbers all but any proof will be accepted.

Redactedtwo have standard factors. Andrey Rekalo pointed out that this modified question has the easy answer I don't know if it's possible to find a $N = (B!)^3$, for example\mathbb{Z}$-chain all of whose elements have a standard factor.
show/hide this revision's text 4 redaction

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.

Since this appears to be harder than at first sight, let me ask a slightly different question which has a better chance of having an answer using known methods.

Is it true that for every positive integer $B$ there is a positive integer $N$ such that $N$ is divisible by all primes up to $B$, and $N \pm 1$ are both composite?

If $n \geq B$ and $n! \pm 1$ are both composite, then $N = n!$ is as required. Again, elementary arguments are preferred, but any proof will be accepted.

Redacted. Andrey Rekalo pointed out that this modified question has the easy answer $N = (B!)^3$, for example.
show/hide this revision's text 3 addendum

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.

Since this appears to be harder than at first sight, let me ask a slightly different question which has a better chance of having an answer using known methods.

Is it true that for every positive integer $B$ there is a positive integer $N$ such that $N$ is divisible by all primes up to $B$, and $N \pm 1$ are both composite?

If $n \geq B$ and $n! \pm 1$ are both composite, then $N = n!$ is as required. Again, elementary arguments are preferred, but any proof will be accepted.
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