Consecutive polynomial non-residues modulo a prime 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. 
 A: 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. 
A: (@Paseman is right; my $\delta/\epsilon$ are off.   I can't fit the correction within the comment section so I'll post a new answer.  Moderator:  Feel free to edit as needed (or tell me how I should edit this).
Lemma. Let $f∈Z[x]$ be a polynomial of degree $n\ge 2$, and let t be a fixed, positive integer. Then for all sufficiently large primes p, there are t consecutive integers not in the range of f (mod p).
Proof. For X>0, define
$V(X):= \{0<k<X:  k=f(m) \text{for some integer m} \}.$
Since $n\ge 2$, there exists a constant C(f)>0 depending only on f such that
$\#V(X)<C(f) \sqrt{X} \text{ for all $X>0$}. $      (*)
Now, let $p>\max(4^6C(f)^6, 8t^3)$ be a prime. Set $q=⌈\frac{1}{2} p^{1/3}⌉$, and consider the pairwise disjoint, consecutive intervals of size q: $[1,q],[q+1,2q],...,[q^2(q−1)+1,q^3]$. The total lenght of these intervals is $q^3 < p$ and there are $q^2$ of them. So if every interval contains at least one integer value of f, then
$\#V(p) > \#V(q^3)\ge q^2 \ge \frac{1}{4} p^{2/3} = \frac{1}{4}p^{1/6}  \cdot \sqrt{p} \ge C(f) \sqrt{p},$
contradicting (*). Thus at least one interval contains no integer value of f. But the number of integers in each interval is $q\ge \frac{1}{2} p^{1/3} > t$, so we are done.
