Understand the problem

For a polynomial $f(x)$ with integer coefficients and degree no less than $1$, prove that there are infinitely many primes $p$ which satisfies the following. There exists an integer $n$ such that $f(n) \not= 0$ and $|f(n)|$ is a multiple of $p$.
Source of the problem
Korea Junior MO Problem 7
Topic
Number Theory
Difficulty Level
8/10
Suggested Book
Elementary Number Theory by David Burton

Start with hints

Do you really need a hint? Try it first!

Well, remember the proof that the set of prime numbers is infinite? We started with the assumption that let there be a finite number of prime numbers and then reached a contradiction that there needs to be another extra prime number given that set. Hence, the set of prime numbers is infinite. This problem is also famously known as Schur’s Theorem. Observe that the problem can be restated as every nonconstant polynomial p(x) with integer coefficients if S is the set of all nonzero values,
then the set of primes that divide some member of S is infinite. Let us start by assuming that the set is indeed finite. Let $A$ this set of primes $p$ such that $\exists n$ such than $f(n)\ne 0$ and $p|f(n)$. Let |A| be finite.
 
If $f(0)=0$ the result is immediate since $p|f(p^n)$ $\forall p$ (just choose $n$ such that $f(p^n)\ne 0$ and so any prime $p\in A$. Now let’s take the case when f(0) is non-zero. Let’s take \( f(x) = a_n.x^n + … a_1.x + f(0)\).  Now, \( f(c.f(0)) = a_n.{c.f(0)}^n + … a_1.f(0) + f(0) = f(0).( a_n.c.{cf(0)}^{n-1} + … + a_2.c^2.f(0) + a_1.c + 1 )\). Can you give some appropiate  c to show that another prime must exist?    
Take c = product of all the primes in A.  Prove that it implies some other prime must exist which is not in A.
Now, \( f(c.f(0)) = a_n.{c.f(0)}^n + … a_1.f(0) + f(0) = f(0).( a_n.c.{cf(0)}^{n-1} + … + a_2.c^2.f(0) + a_1.c + 1 )\). Observe that if we take c as mentioned then, i.e. c = product of all the primes in A. Then all f(c.f(0)) must be coprime to all the primes in A. Therefore, it must have a prime factor other than those in A. Hence, a contradiction in the finiteness in A. QED.  

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