Theorem Let $P(X) \in \mathbb Z[X]$ be an irreducible polynomial of degree $n > 1$. There exist infinitely many primes $q$ such that $P(X)$ has no roots mod $q$.
I had worked on this a bit but my ideas did not apply. I'm pretty stuck!
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1First line of body says no roots mod $q$, but title and rest of body seem to discuss having roots modulo a prime. So which is it? Please edit your question for consistency. – Gerry Myerson Jun 14 '20 at 10:25
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@GerryMyerson My apologies, I have corrected it. Appreciate you spotting that! – Jun 14 '20 at 11:31
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2Not sure I follow the argument fully, but maybe asnwer by mercio in Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$? can help? It mentions that "Then Cebotarev's theorem says that there are infinitely many primes $p$ for which the polynomial doesn't have a linear factor over $\Bbb F_p$.", see also discussion in comments there. – Sil Jun 14 '20 at 12:33