Existence of a Linear Recurrence Modulo $p$ with Period $p^2 - 1$

Question

Let $p$ be a prime number. Prove there is an integer $c$ and an integer sequence $0 \leq a_1, a_2, a_3, \dots < p$ with period $p^2 - 1$ satisfying the recurrence $$ a_{n+2} \equiv a_{n+1} - c \cdot a_n \pmod{p}. $$

Attempt

We can take $a_n=0$ for all $n$, but I assume you mean minimal period. For $p=2$ we can take $c=1$ hence assume $p$ is odd. The characteristic equation is $x^2-x+c=0$ with roots $a,b={{1\pm\sqrt{1-4c}}\over 2}$. The solution is then of the form $a_n=ka^n+lb^n$. Then we want $a,b$ to have multiplicative order $p^2-1$. Hence we want to find an $a$ that generates $F_{p^2}^*$ and such that $(2a-1)^2\in F_p$. I don't know yet how to solve that.

PS. If $1-4c$ is a square mod $p$, then the sequence has period $p-1$ (or a divisor).

Answer 1

$\def\ord{\mathrm{ord}}\def\FF{\mathbb{F}}\def\CC{\mathbb{C}}$I don't think you'll find an easy proof of this. As already explained in the original question, let $\alpha$ and $\beta$ be the roots of $x^2-x+c=0$ in $\mathbb{F}_{p^2}$, the goal is then to make $\alpha$ (and equivalently, $\beta$) have order $p^2-1$ in $\mathbb{F}^{\times}_{p^2}$; in other words, we want $\alpha$ to generate the cyclic group $\mathbb{F}^{\times}_{p^2}$. Given any $\alpha$ in $\mathbb{F}_{p^2}$, $\alpha$ obeys the quadratic polynomial $x^2 - \text{Tr}(\alpha) x + \text{Norm}(\alpha)=0$. So our goal is to find $\alpha \in \mathbb{F}_{p^2}$ which generates $\mathbb{F}^{\times}_{p^2}$ and has trace $1$.

The main result of "Primitive Roots in the Quadratic Extension of a Finite Field", Cohen, Journal of the London Math Society (1983) is that, for any prime $p$ and any nonzero $t \in \mathbb{F}_p$, there is a generator of $\mathbb{F}^{\times}_{p^2}$ with trace $t$. As I understand the history, this result was already proved for $p$ sufficiently large in "On Primitive Roots in Finite Fields", Davenport, The Quarterly Journal of Mathematics (1937) and Cohen's accomplishment was to prove the result for all $p$. A modern exposition can be found in Chapter 14 of Topics in Galois Fields, Hackenberger and Jungnickel, Springer Verlag (2020).

I skimmed the sources that I cited, and they all involve complicated estimates of character sums. I don't think you'll find a simple proof.

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I thought a bit about the character sum proofs, and I have an exposition which I like, so I'll put it here.

Let $p$ be an odd prime and let $q = p^2$. Let $E$ be the set of elements in $\FF_q$ with trace $1$; this is an affine line in the plane $\FF_p^2$, so $|E| = p$.

Lemma 1: Let $a \in \FF_q$. The number of solutions to $ax=y$ with $x$, $y \in E$ is $$\begin{cases} p & a=1 \\ 0 & a \in \FF_p,\ a \neq 1 \\ 1 & a \not\in \FF_p . \end{cases}$$

Proof sketch: If $a=0$, then $y$ must be $0$, but $0 \not \in E$, so the count is $0$. Assume $a \neq 0$ from now on. We want $x$ to lie in $E \cap a^{-1} E$; these are two affine lines in $\FF_p^2$. In the first case, we have two copies of the same line; in the second case, we have parallel lines, in the third case, we have non-parallel lines. $\square$

Lemma 2: Let $\chi : \FF_q^{\ast} \to \CC^{\ast}$ be a character. If $\chi=1$, then $\sum_{x \in E} \chi(x)=p$. Otherwise, $$ \left| \sum_{x \in E} \chi(x) \right| = \begin{cases} 1 & \chi|_{\FF_p} = 1,\ \chi \neq 1 \\ \sqrt{p} & \chi|_{\FF_p} \neq 1 . \end{cases}.$$

Proof: In the trivial case, we just have a sum of $p$ ones, so the sum is $p$. We now turn to the other cases. Writing $\overline{z}$ for the complex conjugate of $z$, we have: $$\left| \sum_{x \in E} \chi(x) \right|^2 = \left( \sum_{x \in E} \chi(x) \right) \left( \overline{\sum_{y \in E} \chi(y)} \right) = \sum_{x,y \in E} \chi(x) \overline{\chi}(y).$$ Since $\chi$ is a character of a finite group, we have $$\chi(x) \overline{\chi}(y) = \chi(x) \chi(y^{-1}) = \chi(x y^{-1}).$$ Using Lemma 1, we see that $$\left| \sum_{x \in E} \chi(x) \right|^2 = \sum_{a \in \FF_q} \chi(a) \begin{cases} p & a=1 \\ 0 & a \in \FF_p,\ a \neq 1 \\ 1 & a \not\in \FF_p \end{cases}$$ $$=\sum_{a \in \FF_q} \chi(a) - \sum_{b \in \FF_p} \chi(b) + p \chi(1).$$ This is $(p^2-1)-(p-1)+p=p^2$ if $\chi=1$, it is $0-(p-1)+p=1$ if $\chi|_{\FF_p}=1$ but $\chi \neq 1$, and it is $0-0+p=p$ if $\chi|_{\FF_p} \neq 1$. $\square$

We now use a key identity which is in all of the sources I cite, and where I can't improve on their exposition. See, for example, Proposition 10.2.5 in Topics in Galois Fields:

Vinogradov's identity: Given a character $\chi : \FF_q^{\ast} \to \CC^{\ast}$, define $\ord(\chi)$ to be the minimal positive integer $d$ such that $\chi^d=1$. So $d|q-1$. For $x \in \FF_q$, we have $$\sum_{\chi} \frac{\mu(\ord(\chi))}{\phi(\ord(\chi))} \chi(x) = \begin{cases} \frac{q-1}{\phi(q-1)} & x \ \text{primitive} \\ 0 & \text{otherwise} \end{cases}.$$

So we want to show that $$\sum_{x \in E} \sum_{\chi} \frac{\mu(\ord(\chi))}{\phi(\ord(\chi))} \chi(x) \qquad (\ast)$$ is positive.

Interchange summation so the sum on $\chi$ is on the outside. The trivial character produces $\tfrac{\mu(1)}{\phi(1)} \sum_{x \in E} 1 = p$; so we need to show that all the other terms produce less than $p$.

We have $|\mu(d)|=1$ if $d$ is square free, and $|\mu(d)|=0$ otherwise. We have $\chi|_{\FF_p}=1$ iff $\ord(\chi)|p-1$. Finally, for each $d|q-1$, the number of characters of order $d$ is $\phi(d)$.

So we obtain the lower bound $$(\ast) \geq p - \sum_{d|p-1,\ d\neq 1,\ d \ \text{square free}} \frac{\phi(d)}{\phi(d)} - \sqrt{p} \sum_{d|q-1,\ d\not| p-1,\ d \ \text{square free}} \frac{\phi(d)}{\phi(d)}.$$ The $\phi(d)/\phi(d)$ terms just cancel, so we are just getting $1$ for each relevant $d$.

The number of square free divisors of $N$ is $2^{\omega(N)}$ where $\omega$ is the prime $\omega$ function. So our bound is $$p - (2^{\omega(p-1)}-1) - \sqrt{p} (2^{\omega(q-1)} - 2^{\omega(p-1)}) \qquad (\ast \ast)$$ Unfortunately $(\ast \ast)$ is negative for many $p$'s. The first $p$ for which it is positive is $2797$ and, out of the first $2000$ primes, this bound is only positive $109$ times.

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Asymptotic analysis suggests we will eventually win, but only for $p$ very large. Asymptotically, we might as well work with the simpler $$p - \sqrt{p} 2^{\omega(p^2-1)}. $$ So we want $$p > \sqrt{p} 2^{\omega(p^2-1)}$$ or, in other words, $$\frac{1}{2} \log p \geq \omega(p^2-1) \log 2.$$

We have the asymptotic bound $$\omega(N) \leq (1+o(1)) \frac{\log N}{\log\log N}$$ and concrete bound $$\omega(N) \leq 1.39 \frac{\log N}{\log\log N}.$$ So $$\omega(p^2-1) \leq (1+o(1)) \frac{\log (p^2-1)}{\log \log (p^2-1)} \leq (1+o(1)) \frac{2 \log p}{\log \log p}$$ and we want $$\frac{1}{2} \log p \geq (1+o(1)) \frac{2 \log p}{\log \log p} \log 2$$ or $$\log \log p \geq 2 \log 2 (1+ o(1)) \approx 1.39+o(1).$$ If we use the concrete bound instead of the asymptotic bound, we need $$\log \log p \geq (2 \log 2)(1.39) \geq 1.92.$$ I don't know why the constant in the concrete bound and $2 \log 2$ are both $\approx 1.39$, they are not the same number!

We now run into a problem -- $\log \log p$ grows very slowly. To get $\log \log p > 1.39$, we need $p > 10^7$ and, if we want a hard bound with no error term, then we want $\log \log p > 1.92$ which forces $p > 1.6 \times 10^{20}$.

So it isn't too hard to understand why this works for $p$ large enough, but it is also clear why some serious work was needed to push through to all $p$.