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Let $\chi$ be the quadratic character on a finite field $F$ of odd characteristic. I'm looking for a reference for a result of this form:

If $\lvert F\rvert > N$ (for some explicit small $N$) then there exists $x \in F$ such that $\chi(x-1) = \chi(x) = \chi(x+1)$.

I'm also interested in other patterns of length 3, such as $\chi(x-1) = \chi(x) = - \chi(x+1)$.

I had a look in the Handbook of Finite Fields but was surprised to not find it there — the handbook didn't even seem to have results on patterns of length 2, although I did find counts for them elsewhere. I could use those to prove the result I want, but I don't see the point in reinventing the wheel when it must be out there.

I found some early papers on runs of residues in prime-order fields, but I need something for all fields.

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  • $\begingroup$ You'll want $F$ of odd characteristic for this, I guess. It seems for Hasse-Weil to work it suffices to have $(\sqrt{|F|}-1)^2>8$ i.e. $|F|>11$ for patterns of length $3$. I checked $\mathbb F_{11}$ has all four patterns manually but it's easy to see $\mathbb F_9$ doesn't have solutions to $\chi(x-1) = \chi(x)= \chi(x+1)$. So you'll want to take $N = 9$. I don't know a reference for this. $\endgroup$
    – Will Sawin
    Commented Oct 20, 2021 at 1:29
  • $\begingroup$ Yes, odd characteristic intended (now included in question). $\endgroup$
    – Ian Wanless
    Commented Oct 20, 2021 at 1:36
  • $\begingroup$ If you fix a pattern for three consecutive character values (all $1$ or all $-1$), then in kconrad.math.uconn.edu/blurbs/ugradnumthy/… see Theorem 2.1 for $F$ of prime order and the last line of Section 2 for general $F$ of odd order. I didn't work out there how large $q$ could be for the count associated to a 3-term consecutive character pattern to be $0$ (using $r = 3$), but if a 3-term pattern is $0$ then $f(q) := q - 16\sqrt{q} - 12 < 0$, so $q \leq 277$ (the first prime power after $277$ is $281$, and $f(281) > 0$). It's a crude bound, but explicit. $\endgroup$
    – KConrad
    Commented Oct 20, 2021 at 2:37
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    $\begingroup$ An earlier MO question on this kind of theme: mathoverflow.net/questions/161271/… $\endgroup$
    – KConrad
    Commented Oct 20, 2021 at 3:22

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