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Assume that $Q(x)$ is a polynomial with integer coefficients. Is there a prime number p such that the equation $Q(x)=0$ has all its root in the finite field $\mathbb{Z}/p\mathbb{Z}$?

I asked this question in RG, too.

https://www.researchgate.net/post/Can_a_given_polynomial_with_integer_coefficients_be_splitted_in_some_Z_p

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Yes, that's true by Chebotarev's Density Theorem. A more elementary proof works as follows: Let $\beta$ be a primitive element of the splitting field of $Q(x)$, and $F(x)$ be the minimal polynomial of $\beta$. Then every root of $Q(x)$ is a polynomial in $\beta$ with rational coefficients. Thus it suffices to show that there are infinitely many primes $p$ modulo which $F(x)$ factors into linear factors. As $\mathbb Q(\beta)/\mathbb Q$ is Galois, the Galois group of $F(x)$ acts regularly on the roots of $F(x)$. By Dedekind now, whenever $F(X)$ has root modulo $p$, $F(X)$ splits into linear factors modulo $p$, and we are done. (For all of this, one has to exclude finitely many primes dividing the denominators of certain rational coefficients, or modulo which $F(x)$ has multiple roots.)

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    $\begingroup$ There is an article by Lenstra and Stevenhagen in which they also discuss the easier Frobenius density theorem and give examples of how frequently a fixed polynomial factoring partitions certain ways over different primes. $\endgroup$ – Will Jagy Feb 22 '16 at 19:20

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