Special case of [Bunyakovsky conjecture](https://en.wikipedia.org/wiki/Bunyakovsky_conjecture)


Let $f(x)$ be non-constant irreducible polynomial with integer
coefficients, no fixed prime factor and positive
leading coefficient. Let $S$
be a finite set of primes.



> 
> Q1 Is it possible $f(n)$ to be divisible by a member of
>  $S$ for almost all natural $n$?
> 
> 
> Q2 Must $f(n)$ take at least one prime value for
>  natural $n$?
> 
> 
> 



Can a polynomial be almost always divisible by a member of a finite set of primes?
The answer to Q1 is "no". The value of $f(n)$ mod $p$ depends only on the value of $n$, mod $p$.


Let $N$ be the product of the primes in $S$. Assuming "almost all" means "all but finitely many", then that means that for $n$ sufficiently large, the set $n+1,\ldots,n+N$ are all multiples of primes from $S$. But since all this is invariant modulo $N$, that means every number is a multiple of some prime from $S$.


The question is then, can we cover $1,\ldots,N$ by residue classes mod primes from $S$ without using all the residue classes modulo any prime (because if we did we have a fixed prime factor)? And the answer is "no", by an easy application of the Chinese Remainder Theorem: choose a residue class modulo each $p\in S$ which is not used; they give a unique residue mod $N$ which is uncovered.


