Is there a small finite (perhaps of cardinality two or three) collection of cubic polynomials $p_1, \dotsc, p_k \in \mathbb{Z}[x]$ such that for every prime $p$ at least one of these is irreducible?
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2$\begingroup$ Just a remark that the negative answer to this this also follows from the primitive element theorem, which immediately implies that there exist polynomials $q_1,\ldots,q_k \in \mathbb{Z}[x]$ such that $p_1 \circ q_1, \cdots, p_k \circ q_k$ have a common non-constant factor $H \in \mathbb{Z}[x]$, and Euclid's argument for the polynomial $H$. $\endgroup$– Vesselin DimitrovAug 31, 2015 at 12:52
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No. It follows from Chebotarev's density theorem that for any polynomials $p_1,\dots,p_k\in\mathbb{Z}[x]$ there are infinitely many primes that split all these polynomials. Simply, apply this theorem for the number field $K$ generated by the roots of $p_1,\dots, p_k$, and note that if an unramified prime completely splits in $K$ then each $p_i$ splits into linear factors modulo that prime.