Let $p\_1, \ldots, p\_n$ be a finite sequence of nonconstant polynomials with integer coefficients. Does there exist a finite sequence of integers $x\_1, \ldots, x\_n$ such that the integers $p\_1(x\_1), \ldots, p\_n(x\_n)$ have a nontrivial common divisor? 



Noncoprime polynomial values
Here's an argument that might be a bit too high-brow (it uses Chebotarev's density theorem).


**Reduction 1.** Your question is equivalent to: for $f\_1, \ldots, f\_n \in \mathbb Z[x]$, does there exist a prime $p$ such that all $f\_i$ have a root in $\mathbb F\_p$. Indeed, taking lifts $x\_i \in \mathbb Z$ of roots $\bar x\_i \in \mathbb F\_p$ of $\bar f\_i \in \mathbb F\_p[x]$ gives
$$f\_i(x\_i) \equiv 0 \pmod p$$
for all $i$, so the $f\_i(x\_i)$ have $p$ as common divisor.


**Reduction 2.** We may assume that all $f\_i$ are irreducible. Indeed, a root (mod $p$) of a factor of $f\_i$ gives a root of $f\_i$ itself.


*Proof of the assertion.* Let $K\_i = \mathbb Q[x]/(f\_i)$ be the field extension defined by adjoining a root of $f\_i$. Let $K$ be the Galois closure of the compositum of the $K\_i$ (inside $\bar {\mathbb Q}$). By the Chebotarev density theorem, there exists a prime $p$ such that $(p)$ is completely split in $K$. Thus, all the $f\_i$ have a root modulo $p$. $\square$


**Remark.** This actually proves more: there are infinitely many primes $p$ with this property (and their density is $1/[K:\mathbb Q]$). Moreover, there are many lifts $x\_i$ of a root of $\bar{f}\_i$, so we actually get many examples of such $x\_i$. One can ask how many there are (I will not attempt this right now).


