Are There Infinitly Many $n$ Which $a\times n!+1$ Be Composite? $ax+1$ is a linear polynomial with integral coefficients.
Are there infinitly many $n$ which $a\times n!+1$ be composite?
As I know this problem is true for polynomials with degree greater that 1, and now my question is about linear polynomials.
 A: Ilya has already answered the question, but maybe it's useful to point out that more general results appear in a paper of Schinzel: On composite integers of the form $c(ak+b)!\pm 1$, Norsk Matematisk Tidskrift 10 (1962), 8--10. This paper was recently reprinted in volume II of Schinzel's selected works (European Math. Society). The method is similar to Ilya's. Here's a link to the Zentralblatt review: https://zbmath.org/?q=an:03175346
A: I do not have a complete answer to this question. However, let me at least prove that there are infinitely many $a \in \mathbb{Z}$ such that $an!+1$ is composite for infinitely many $n$.
In fact, by straightforward calculations based on Wilson's theorem, it is easy to show that for all $k \in \mathbb{N}$ and for all prime numbers $p$ we have $$(2k)![(p-(2k+1)]!+1 \equiv 0 \quad (\textrm{mod }p).$$
Hence for $k=0, \, 1, \, 2,  \ldots$ and for all prime numbers $p$ we have  $$(p-1)!+1 \equiv 0 \quad (\textrm{mod }p), $$ $$2(p-3)!+1 \equiv 0 \quad (\textrm{mod }p),$$ $$24(p-5)!+1 \equiv 0 \quad (\textrm{mod }p),$$ $$\cdots$$ and so on.
See F. G. Elston, A Generalization of Wilson's Theorem,  Mathematics Magazine 30, No. 3 (1957), pp. 159-162. 
A: $\let\eps\varepsilon$Assume that for all $n>N$ the number $an!+1$ is prime.
To start, take any odd $k>2a$ and set $S=k!+a$. Then the number $S/a-1=k!/a$ is divisible by all primes not exceeding $k$. Thus $S/a$ is divisible by some prime $p>k$.
Now we have 
$$
  a(p-k-1)!+1=\frac{a(p-1)!}{(p-1)\dots(p-k)}+1
  \equiv \frac{-a}{-k!}+1=\frac{a+k!}{k!}\equiv 0\pmod p.
$$
If we choose $k$ so that $p-k-1>N$ and $a(p-k-1)!+1>p$, we get a contradiction. The first goal is easily achievable since there are infinitely many $k$ such that all the numbers $k+1.k+2.\dots,k+N,k+N+1$ are composite; moreover, there is an infinite arithmetical progression of such $k$.
Assume that for every such $k$ we have $a(p-1-k)!+1=p$. We have $p=k+o(k)$ (otherwise $p-k\geq \eps k$ for infinitely many $k$, so $p-k\geq \frac{\eps}{1+\eps}p$, and $a(p-1-k)!+1$ is much greater than $p$), So there are two distinct $k$ providing the same value of $p-k-1$. This is absurd, since then we have $p_1=a(p_1-k_1-1)!+1=a(p_2-k_2-1)!+1=p_2$ and hence $k_1=p_1-1-(p_1-k_1-1)=p_2-1-(p_2-k_2-1)=k_2$.
