Integers mod $p$ "together" algebraically closed Let $f(x)$ be a nonconstant polynomial over $\mathbb Z$.
Must $f$ have a zero in $\mathbb F_p$ for some prime number $p$?
More generally, let $f_1,\dots,f_k$ be such polynomials, must there exist a $p$ such that each $f_i$ has a zero in $\mathbb F_p$?
 A: Let $d$ be the degree of $f$. There are at most $2d$ integers $m$ such that $|f(m)|=1$. Pick any other integer $n$ and then an arbitrary prime factor $p$ of $f(n)$.
A: Easy answer to the first question: for $n \gg 0$, we have $f(n) > 1$ or $f(n) < -1$. Hence $f(n)$ is divisible by some prime number $p$, so $n$ is a root mod $p$.
A: The answer to the first question is positive. We may assume WLOG that $f$ is irreducible.
The Frobenius Density Theorem states that density of primes $p$ such that $f \mod p$ has a given decomposition (i.e. factors into irred. polynomials of prescribed degrees $n_1,\cdots,n_r$) is 
$$\frac{\# \{ \sigma \in \text{Gal}(f) \mid \sigma \text{ has cycle structure }n_1,\cdots, n_r \} }{|\text{Gal}(f)|},$$
where $\text{Gal}(f)$ is the Galois group of $f$. See the discussion in this introductory paper by Sury.
By choosing the decomposition type of linear factors ($n_i = 1$), we see that for infinitely many primes $p$ (their density is $1$ over the size of the Galois group) we have that $f \mod p$ splits into linear factors (and in particular has zeroes). This is because the identity element of the Galois group of $f$ have cycle structure with $n_i=1$.
Remark: If you solely want the existence of infinitely many such primes (and not their density), The Frobenius density theorem may be avoided, see this discussion.
