$ax+1$ 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.



Are There Infinitly Many $n$ Which $a\times n!+1$ Be Composite?
$\let\eps\varepsilon$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 $k\_1=p\_1-1-(p\_1-k\_1-1)=p\_2-1-(p\_2-k\_2-1)=k\_2$.


