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.

  • 3
    $\begingroup$ For any $f(x)\in \mathbb Z[x]$ of degree at least 2, by Hilbert's irreducibility theorem (or technically the strengthening of it) you can find an integer $s$ so that $f(x)+s+i$ is irreducible for all $1\le i\le t$. Then you want to show that there are infinitely many primes $p$ for which all $f_i$ are rootless mod p (or even more: remain irreducible mod p) which should follow from Chebotarev's theorem. $\endgroup$ – Gjergji Zaimi Apr 30 '18 at 4:47
  • 1
    $\begingroup$ In the second part of the argument, is it generally true that if $f_1,\ldots, f_t$ are irreducible then there should exist infinitely many primes $p$ where they are all rootless modulo $p$? The reason I am skeptical about this is the example $f_1(x)=x^2-2,f_2(x)=x^2-3$, and $f_3(x)=x^2-6$. They are each irreducible, but at least one of them is reducible mod any prime. $\endgroup$ – Marco Apr 30 '18 at 11:04
  • $\begingroup$ You're absolutely right, I don't know if this argument is fixable. $\endgroup$ – Gjergji Zaimi Apr 30 '18 at 17:27

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.

  • $\begingroup$ Thank you. This sounds like a correct proof. Unfortunately I am not smart or knowledgable enough to judge it. I will award the 50 points when it gets upvoted more. I really think this proof needs to appear in print, especially because I need it :D $\endgroup$ – Marco May 11 '18 at 17:28

(@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.

  • 1
    $\begingroup$ You have posted a new answer. This is not a crime, but you should click the edit button to modify an existing answer instead. You can also delete your answer. If you do not see these buttons nor the flag button to flag your answer and ask for help, you can email moderators@mathoverflow.net. However, try the buttons first. Gerhard "You Have Control Over Answers" Paseman, 2018.05.09. $\endgroup$ – Gerhard Paseman May 9 '18 at 15:17
  • $\begingroup$ Thanks. I'm newish to MOF so I'm still learning the rope. Case in point: If I edited/deleted my incorrect initial answer, wouldn't it make other people's comments (e.g. yours) unintelligible? $\endgroup$ – W Sao May 9 '18 at 19:38
  • $\begingroup$ Depends. If the comment is attached to the answer, the comment goes away with the answer. While I am willing to help, I recommend reading more of the online help and asking these questions on meta.mathoverflow, where the answers can help others. Gerhard "Searching Meta Is Rather Smart" Paseman, 2018.05.09. $\endgroup$ – Gerhard Paseman May 9 '18 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.