If all primes larger than $n$ divide a polynomial, then is the polynomial reducible?

Question

You are given a non-linear integer polynomial $P$ and an integer $n$ such that any prime above $n$ divides $P(x)$ for some integer $x$. Is it necessary that the polynomial is reducible?

I was trying to solve the problem "show that any non-square integer $a$ is a quadratic non-residue modulo infinitely many primes" and noticed that if we show that the polynomial $x^2-a$ cannot have only finitely many primes not dividing it, we are done. This then generalized to the above statement which I am posing as a question.

It seems intuitively obvious that the above statement is true. Searching for a counterexample has not helped.

Is there an elementary way of tackling this problem?

Thanks in advance.

A similar question has been asked here. The answer by Eric Schneider is a gem.

Isn't this true for $P(x) = x$? Maybe you meant "non-linear polynomial"? — Arthur, Oct 15 '15 at 11:01
@Arthur, yes I was actually looking for non-linear polynomials. Edited,Thanks. — rah4927, Oct 15 '15 at 11:15

score 2 · Answer 1 · answered Oct 15 '15 at 11:25

2

You can use Chebotarev density theorem.If you read carefully,excersise 7.1 is exactly what you want.
An elementary argument could be found I suppose but the answer you seek is in this paper.
Nice question!

answered Oct 15 '15 at 11:25

Konstantinos Gaitanas

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Thanks for the answer. Currently, this is out of my depth, but I am sure countless others will find this useful. – rah4927 Oct 15 '15 at 11:27

If all primes larger than $n$ divide a polynomial, then is the polynomial reducible?

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