Given a polynomial $f(x)\in \mathbb{Z}[x]$ of degree at least 2 and a positive integer $t$, does there always exist infinitely many primes $p$ such that the range of $f(x)$ modulo $p$ does not contain the $t$ consecutive numbers $s+1,\ldots, s+t$ for some integer $s$ (where $s$ can depend on $p$)? 


This is known to be true if $f(x)=x^n$, $n\geq 2$. I am wondering if anything is known beyond this example. 



Consecutive polynomial non-residues modulo a prime
Let $d = \deg f$. Consider the covering of $\mathbb A^1\_\mathbb Q$, with coordinate $s$, defined by $f(x\_1) =s+1 ,\dots, f(x\_t) =s +t$. The monodromy representation of this covering gives a map from the etale fundamental group of $\mathbb A^1$ minus the finite set of critical values to $\prod\_{i=1}^t S\_d$. 


The image of this representation has a normal subgroup which is the image of the geometric fundamental group, and a quotient corresponding to some Galois extension of $\mathbb Q$. 


Then any sufficiently large prime $p$ which splits in this quotient has a sequence of $t$ consecutive non-residues.


First, because $p$ splits, the arithmetic monodromy group of the covering over $\mathbb F\_p$ is contained in the geometric monodromy group in characteristic zero. Then because $p$ is sufficiently large, the geometric monodromy group in characteristic $p$ matches the one in characteristic zero, so these are equal.


Then we may apply the function field Chebotarev theorem, which says that as long as $p$ is sufficiently large (with regards to the degree and genus of this covering, say) then for each conjugacy class $\sigma$ in the image of $\pi\_1$, we can find a prime where $\operatorname{Frob}\_q$ acts on the fiber by $\sigma$.


We take $\sigma$ to be a generator of the local monodromy at $\infty$, which for all $i$ acts on the roots of $f(x\_i)= s+i$ by a $d$-cycle, i.e. it is a tuple of $t$ $d$-cycles in $\prod\_{i=1}^t S\_d$. In particular, because $d>1$, it has no fixed points. Hence the Frobenius element has no fixed points when acting on the roots of $f(x\_i)=s+i$ for any $i$ from $1$ to $t$, and thus none of the roots lie in $\mathbb F\_p$, as desired. 


