2. Overview of the method

For concreteness we consider the sequence $a_n$ defined by

\begin{align*}a_n = 5^n + (3 + \sqrt{2})^n + (3 - \sqrt{2})^n, n = 1, 2, \ldots\end{align*}

The characteristic polynomial $(x-5)(x - (3 + \sqrt{2}))(x - (3 - \sqrt{2}))$ is reducible and thus not of the form of Theorem 1·1, but we only use this example to demonstrate the idea. (The proof of Theorem 1·1 can be adapted to this sequence, though.)

In the case when 2 is a quadratic residue modulo p the period of the sequence $(a_n)$ modulo p divides $p-1$ . We are unable to say anything nontrivial about whether such primes are prime divisors of $(a_n)$ or not.

The case when 2 is a quadratic nonresidue modulo p, however, turns out to be accessible. Write $n = (p+1)k + r, k, r \in \mathbb{Z}$ . We may view the numbers $3 \pm \sqrt{2}$ as elements of $\mathbb{F}_{p^2}$ , and by norms we have $(3 \pm \sqrt{2})^{p+1} = 7$ in this finite field. Hence

(2·1) \begin{equation}a_{(p+1)k + r} \equiv 5^{2k+r} + 7^k\left((3 + \sqrt{2})^r + (3 - \sqrt{2})^r\right) \pmod{p}.\end{equation}

The equation $a_{(p+1)k + r} \equiv 0 \pmod{p}$ may thus be written, for $p > 7$ , as

(2·2) \begin{equation}\left(\frac{5^2}{7}\right)^k \equiv -\left(\frac{3 + \sqrt{2}}{5}\right)^r - \left(\frac{3 - \sqrt{2}}{5}\right)^r \pmod{p}.\end{equation}

Artin’s primitive root conjecture states that a given rational number a is a primitive root modulo p for infinitely many primes p as long as a is not $-1$ or a square. Under GRH one can prove this in a quantative form: the set of such primes has a positive density (as long as its infinite) [ Reference Hooley6 ]. This density is often quite large. For example, for $a = 2$ the density is roughly 37 percent.

It turns out that the order of a rational number a modulo primes is almost always almost maximal assuming $a \not\in \{-1, 0, 1\}$ (under GRH). More precisely, the density of primes p with $\mathrm{ord}_p(a) \ge (p-1)/C$ goes to 1 as $C \to \infty$ . (See [ Reference Wagstaff24 , section 5] or Lemma 3·2 below.)

Note also that if $\mathrm{ord}_p(a) = (p-1)/h$ , then the function sending integers x to $a^x \pmod{p}$ attains all non-zero hth powers modulo p as its values.

In this light, to prove that (2·2) is solvable for almost any prime p it suffices to show that the equation

(2·3) \begin{equation}x^h \equiv -\left(\frac{3 + \sqrt{2}}{5}\right)^r - \left(\frac{3 - \sqrt{2}}{5}\right)^r \pmod{p}\end{equation}

has a solution (x, r) with $x \neq 0$ for almost any prime p.

Note the right-hand side of (2·3) satisfies a linear recurrence $(b_r)$ . We aim to prove that the sequence $b_1, b_2, \ldots$ almost always attains a hth power as its value modulo p.

By using results from Galois theory and the Chebotarev density theorem, it is not very hard to show that this is true, for example, if there is an infinite subsequence of $(b_r)$ whose elements are distinct primes.

Of course, we cannot guarantee that a linear recurrence has infinitely many prime values. However, there are only a very few cases where such an idea does not work. To name one, if $b_r$ is always three times a square, then if 3 is a quadratic nonresidue modulo p (which happens for a positive density of primes), the sequence $b_r$ may avoid all squares modulo p.

In general, the only obstructions arise when the values of the linear recurrence are almost perfect powers. By applying Zannier’s result on Pisot’s dth root conjecture [ Reference Zannier25 ] we reduce our problem to determining whether or not a linearly recursive sequence arising in the proof is the power of another recurrence. From here on only elementary observations are needed.

As we already mentioned, the sequence $a_n$ considered here does not satisfy the conditions of Theorem 1·1, and the general case is more complicated. There are two notable differences.

To perform the “norm-trick” and to arrive to an equation of the form (2·2) we need to control the norms of the roots of the characteristic polynomial in finite fields. To do so, in the situation of Theorem 1·1 we consider those primes p for which P factorises modulo p as the product of two irreducibles of degree 1, respectively $d-1$ .

From here we are able to reduce to an equation similar to (2·3), though this time the right hand side is not necessarily a linear recurrence of integers but of algebraic numbers. By taking norms we reduce to the integer case, the same idea can be implemented and we are able to show that the reduction of at least one term of the sequence of algebraic numbers to $\mathbb{F}_p$ is almost always an hth power.

In Section 3 we state the GRH-conditonal result mentioned earlier. In Section 4 we reduce the problem to a polynomial equation of type (2·3) in a similar manner as above. We note that the equation is solvable for a set of primes of density 1 if certain field extensions are linearly disjoint. We present the tool to handle such questions in Section 5. In order to apply it we have to check that no subsequence $(x_{An+B})$ of a linear recurrence $(x_n)$ appearing in our proof consists only of perfect powers, which we do in Section 6. We wrap up the proof in Section 7.

3. Orders of reductions of algebraic numbers

The following lemma is used when transforming our problem into a polynomial equation. This lemma is the only part of the proof that relies on GRH.

Note that we do not say anything about which root f(p) of P modulo p we choose if there are several of them. The result is true no matter how the choices are made.

This result is equivalent to the following algebraic number theoretic formulation.

Proof. (Cf. [ Reference Wagstaff24 , section 5].) Note that almost all prime ideals of $O_K$ belong to T. For $k \in \mathbb{Z}_+$ let $T'_{\!\!k} = T_k \setminus T_{k-1}$ . The results of Lenstra [ Reference Lenstra9 ] imply that $T'_{\!\!k}$ has a density for all k. This density is given by

\begin{align*}d(T'_{\!\!k}) = \sum_{t = 1}^{\infty} \frac{\mu(t)}{[K(\zeta_{kt}, \alpha^{1/kt}) \;:\; K]},\end{align*}

where the sum is absolutely convergent. Let $f(n) = 1/[K(\zeta_n, \alpha^{1/n}) \;:\; K]$ . Now rearranging gives

\begin{align*}\sum_{k = 1}^{\infty} \sum_{t = 1}^{\infty} \mu(t)f(kt) = \sum_{K = 1}^{\infty} f(K) \sum_{d \mid K} \mu(d) = f(1) = 1,\end{align*}

from which the result follows.

4. Reduction to a polynomial equation

We consider the setup of Theorem 1·1. The proof also works in the situation of Theorem 1·2.

Let P and $(a_n)$ be as in Theorem 1·1. Assume $a_n \neq 0$ for all n, as otherwise we are done. Let S denote the set of primes p such that P factorises as the product of polynomials of degree 1 and $d-1$ modulo p. By the assumption and the Chebotarev density theorem, the relative density of S is $1/(d-1)$ . We prove that the density of primes of S which are prime divisors of $(a_n)$ is one (relative to S).

We write

\begin{align*}a_n = \gamma_1\alpha_1^n + \cdots + \gamma_d\alpha_d^n,\end{align*}

where $\alpha_1, \ldots , \alpha_n$ are the roots of P and $\gamma_1, \ldots , \gamma_d$ are constants. For further purposes we define additionally

\begin{align*}b_n = b_{h,\, n} \;:\!=\; \alpha_1^{hn}\!\left(-\frac{a_n}{\gamma_1\alpha_1^n} + 1\right)\end{align*}

and

\begin{align*}c_n = c_{h,\, n} \;:\!=\; N_{F/\mathbb{Q}}(b_n) = (\!-\!1)^dN^{hn}\prod_{i = 1}^d \sum_{j \neq i} \frac{\gamma_j\alpha_j^n}{\gamma_i\alpha_i^n},\end{align*}

where F is the splitting field of P, $N \;:\!=\; N_{F/\mathbb{Q}}(\alpha_1)$ and $h \in \mathbb{Z}_+$ is a parameter fixed later. For an algebraic number field K we let $O_K$ denote its ring of integers.

Note that $\gamma_i \in \mathbb{Q}(\alpha_i) \setminus \{0\}$ (see e.g. [ Reference Roskam19 , section 2]).

For each (unramified) prime $p \in S$ there exists a homomorphism $\varphi \;:\; O_F \to \mathbb{F}_{p^{d-1}}$ mapping one of the roots $\alpha_i$ , say $\alpha_1$ , to an element of $\mathbb{F}_p$ , and the other roots to elements of $\mathbb{F}_{p^{d-1}}$ whose degrees over $\mathbb{F}_p$ are $d-1$ . Let $K = \mathbb{Q}(\alpha_1)$ .

Note that for $n = k(p^{d-1} - 1)/(p-1) + r$ one has

\begin{align*}\varphi(\alpha_1)^n = \varphi(\alpha_1)^{kd + r},\end{align*}

\begin{align*}\varphi(\alpha_i)^n = N_{\mathbb{F}_{p^{d-1}}/\mathbb{F}_p}(\varphi(\alpha_i))^k\varphi(\alpha_i)^r, \quad 2 \le i \le d\end{align*}

and

\begin{align*}N_{\mathbb{F}_{p^{d-1}}/\mathbb{F}_p}(\alpha_i) = \varphi(\alpha_2) \cdots \varphi(\alpha_d) = \frac{N}{\varphi(\alpha_1)}, 2 \le i \le d.\end{align*}

(Clearly $\varphi(\alpha_1) \neq 0$ for all but finitely many p.) Hence

(4·1) \begin{equation}\varphi(a_n) = \varphi(\gamma_1)\varphi(\alpha_1)^{kd + r} + \left(\frac{N}{\varphi(\alpha_1)}\right)^k\left(\varphi(\gamma_2)\varphi(\alpha_2)^r + \cdots + \varphi(\gamma_d)\varphi(\alpha_d)^r\right)\end{equation}

so $\varphi(a_n) = 0$ if and only if

\begin{align*}\varphi(\alpha_1^{d+1}/N)^k = -\frac{\varphi(\gamma_2)\varphi(\alpha_2)^r + \cdots + \varphi(\gamma_d)\varphi(\alpha_d)^r}{\varphi(\gamma_1)\varphi(\alpha_1)^r}.\end{align*}

We then note that $\alpha_1^{d+1}/N$ is not a root of unity, for otherwise the conjugates $\alpha_i^{d+1}/N$ are roots of unity as well. Hence the product

\begin{align*}\prod_{1 \le i \le d} \alpha_i^{d+1}/N = N\end{align*}

is a root of unity. Hence $N = \pm 1$ , and thus the numbers $\alpha_i$ are roots of unity. This contradicts the non-degeneracy of P.

We may hence apply Lemma 3·1 to $\alpha_1^{d+1}/N$ . It suffices to show that for any $h \in \mathbb{Z}_+$ the equation

\begin{align*}x^h = b_r\end{align*}

is solvable in $\mathbb{F}_p \setminus \{0\}$ for almost all primes $p \equiv 1 \pmod{h}$ . This may be reformulated as follows: almost all prime ideals of $O_{K(\zeta_h)}$ split in at least one of the fields $K_r \;:\!=\; K(\zeta_h, b_r^{1/h}),$ r = 1, 2, …

5. A linear disjointness result

We use the following basic result from Galois theory [ Reference Garrett4 ].

Lemma 5·1. Let h be a positive integer, let K be a number field containing a hth root of unity, and let $a_1, a_2, \ldots , a_k$ be a sequence of integers. Assume that for any integers $0 \le e_i < h$ , not all zero, one has $a_1^{e_1/h} \cdots a_k^{e_k/h} \not\in K.$ Then

\begin{align*}\left[K\left(a_1^{1/h}, \ldots , a_k^{1/h}\right) \;:\; K\right] = h^k.\end{align*}

Proof. Note that the condition implies that $(x_n)$ has infinitely many prime divisors – otherwise choose $A = 1, B = 0$ , $D = 2$ and K to be $\mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots , \sqrt{p_n})$ , where $p_1, \ldots , p_n$ are the prime divisors of $(x_n)$ . Note also that for all except finitely many primes p, say for all p not belonging to T, the sequence $(x_n)$ is periodic modulo $p^k$ for all $k \in \mathbb{Z}_+$ .

We will inductively choose the indices $n_i$ and the primes $p_i, q_i$ , additionally requiring that $p_i, q_i \not\in T$ . Assume we have already choosen some $n_1, \ldots , n_k, p_1, \ldots , p_k, q_1, \ldots , q_k$ satisfying the conditions. We now pick $n_{k+1}, p_{k+1}, q_{k+1}$ . Let $S = \{p_1, q_1, \ldots , p_k, q_k\} \cup T$ .

Pick some prime divisor $p_{k+1} \not\in S$ , let $v_{p_{k+1}}(x_{n_0}) = t > 0$ for some $n_0$ such that $x_{n_0} \neq 0$ . By periodicity modulo $p_{k+1}^{t+1}$ , there exists an arithmetic progression $An + B, n = 1, 2, \ldots$ such that $v_{p_{k+1}}(x_{An+B}) = t$ for all n

If there exist some prime $q_{k+1} \not\in S \cup \{p_{k+1}\}$ and $n \in \mathbb{Z}_+$ such that $\gcd(v_{q_{k+1}}(x_{An+B}), t) = 1$ , then we are done. Assume this is not the case.

If for any prime $q_{k+1} \not\in S$ and any n we had $\gcd(v_{q_{k+1}}(x_{An+B}), t) = t$ , then we could write

\begin{align*}|x_{An+B}| = f(n)^t\prod_{s \in S} s^{f_s(n)},\end{align*}

for some functions $f, f_s \;:\; \mathbb{Z}_+ \to \mathbb{Z}_{\ge 0}$ . Then $x_{An+B}$ would always be a perfect tth power in a number field containing the tth roots of all primes of S and a 2tth root of unity, contrary to the assumption.

Hence there exist a prime $q_{k+1} \not\in S$ and $n \in \mathbb{Z}_+$ such that $t' \;:\!=\; \gcd(v_{q_{k+1}}(x_{An+B}), t) < t$ . Repeat the above argument with $q_{k+1}$ in place of $p_{k+1}$ and t ^′ in place of t. The value of t decreases. It must happen that for some value of $p_{k+1}$ and t we find a prime $q_{k+1}$ with $\gcd(v_{q_{k+1}}(x_{An+B}), t) = 1$ .

Lemma 5·3. Let $h \in \mathbb{Z}_+$ and let F be a number field containing a hth root of unity. Let $(x_n)$ be a linearly recursive sequence of integers satisfying the assumption of Lemma 5·2. Then there exists a subsequence $x_{n_1}, x_{n_2}, \ldots$ of $(x_n)$ such that the extensions

\begin{align*}F\left(x_{n_1}^{1/h}\right)/F, F\left(x_{n_2}^{1/h}\right)/F, \ldots\end{align*}

are linearly disjoint and of degree h.

Proof. Note first that there exists a constant c (depending on F) such that if x is an integer with $x^{1/h} \in F$ , then all prime divisors of x which are greater than c have multiplicity divisible by h. (Indeed: If p is a prime with for which h does not divide $v_p(x)$ , then p is ramified in the $\mathbb{Q}(x^{1/h})$ . If $\mathbb{Q}(x^{1/h}) \subset F$ , then p is ramified in F, and only finitely many primes ramify in F.)

Let $(x_{n_i})$ be a subsequence constructed in Lemma 5·2. We may assume that the corresponding primes $p_i, q_i$ are larger than c. We prove that this subsequence works by showing that

\begin{align*}\left[F\left(x_{n_1}^{1/h}, x_{n_2}^{1/h}, \ldots , x_{n_k}^{1/h}\right) \;:\; F\right] = h^k\end{align*}

for all k. Apply Lemma 5·1. Assume that

(5·1) \begin{equation}x_{n_1}^{e_1/h} \cdots x_{n_k}^{e_k/h} \in F, 0 \le e_i < h.\end{equation}

By the choice of c, (5·1) implies that the prime divisors of $x_{n_1}^{e_1} \cdots x_{n_k}^{e_k}$ which are larger than c have multiplicity divisible by h. By the choice of the primes $p_i, q_i$ this implies $h \mid e_i$ for all i.

6. Linear recurrences and perfect powers

In this section we show that Lemma 5·3 may be applied to the sequence $(c_n)$ . Assume not, so $c_{An+B}$ is always a Dth power of an element in a fixed number field.

By a result of Zannier [ Reference Zannier25 ], the only case when a linear recurrence is always a Dth power is when it is the Dth power of a linear recurrence. We may hence write

\begin{align*}c_{An+B} = d_n^D,\end{align*}

where d is a linearly recursive sequence (whose elements are not necessarily integers). Write d as an exponential polynomial

\begin{align*}d_n = \sum_{m = 0}^{t} \left(n^t \sum_{k = 1}^{u_m} e_{m, k}\sigma_{m, k}^n\right),\end{align*}

where the coefficients $e_{m, k}$ are non-zero, $\sigma_{m, 1}, \sigma_{m, 2}, \ldots , \sigma_{m, u_m}$ are non-zero and pairwise distinct, and $i_m > 0$ for all $m \le t$ .

We first show that $t = 0$ , i.e. that the characteristic polynomial of $(d_n)$ has no repeated roots. Assume not. Now one sees that $d_n^D$ may be written as

\begin{align*}n^{Dt} \!\left(\sum_{k = 1}^{u_t} e_{t, k}\sigma_{t, k}^n\right)^D + f_n,\end{align*}

where $f_n$ is an exponential polynomial whose polynomial terms have degree less than Dt. Since the representation of a linear recurrence as an exponential polynomial is unique, one sees that $c_{An+B} = d_n^D$ cannot hold for all $n \in \mathbb{Z}$ .

We may thus write

\begin{align*}d_n = e_1\sigma_1^n + \cdots + e_u\sigma_u^n.\end{align*}

Every $\sigma_i$ can be written in the form $\alpha_1^{x_{i, 1}}\alpha_2^{x_{i, 2}} \cdots \alpha_d^{x_{i, d}}$ , where the $x_{i, j}$ are rational numbers [ Reference Ritt18 ]. Let M be a positive integer such that $x_{i, j}M$ is an integer for all i, j. Write the equation $c_{AMn+B} = d_{Mn}^D$ as

\begin{align*}(\!-\!1)^dN^{h(AMn + B)}\prod_{i = 1}^d \sum_{j \neq i} \frac{\gamma_j\alpha_j^{B}\alpha_j^{AMn}}{\gamma_i\alpha_i^B\alpha_i^{AMn}} = \left(\sum_{i = 1}^u e_i \prod_{j = 1}^d \alpha^{x_{i, j}Mn}\right)^D\end{align*}

Note then that if $\prod_{i = 1}^d \alpha_i^{f_i} \in \mathbb{Q}$ for some integers $f_i$ , then $f_i = f_j$ for all i, j. Indeed: since the Galois group of P is $S_d$ , we have $\prod_{i = 1}^d \alpha_i^{f_i'} = \prod_{i = 1}^d \alpha_i^{f_i}$ for any permutation $f_i'$ of $f_i$ . Hence $\alpha_i^{f_i}\alpha_j^{f_j} = \alpha_i^{f_j}\alpha_j^{f_i}$ , so $(\alpha_i/\alpha_j)^{f_i - f_j} = 1$ , which by non-degeneracy of P implies $f_i = f_j$ .

Hence, if $N \neq \pm 1$ , one has $\prod_{i = 1}^d \alpha_i^{f_i} = 1$ only if $f_i = 0$ for all i. By basic results on linear recurrences, this implies that for any $Q \in \mathbb{C}[X_1^{\pm 1}, \ldots , X_d^{\pm 1}]$ in d variables we have

\begin{align*}Q(\alpha_1^{n}, \alpha_2^{n}, \ldots , \alpha_d^{n}) = 0\end{align*}

for all integers n if and only if Q is identically zero. Hence

\begin{align*}(\!-\!1)^dN^{hB}(X_1 \cdots X_d)^{hAM}\prod_{i = 1}^d \sum_{j \neq i} \frac{\gamma_j\alpha_j^BX_j^{AM}}{\gamma_i\alpha_i^BX_i^{AM}} = \left(\sum_{i = 1}^u e_i\prod_{j = 1}^d X_j^{x_{i, j}M}\right)^D\end{align*}

identically as elements of $\mathbb{C}[X_1^{\pm 1}, \ldots , X_d^{\pm 1}]$ .

In particular, the left hand side is a perfect Dth power in $\mathbb{C}[X_1^{\pm 1}, \ldots , X_d^{\pm 1}]$ . Perform a suitable transformation of the form $X_i \to c_iX_i$ , clear out constants and simplify. One obtains that

(6·1) \begin{equation}(X_1 \cdots X_d)^{(h-1)A'}\prod_{i = 1}^d \left(X_1^{A'} + \cdots + X_d^{A'} - X_i^{A'}\right)\end{equation}

is a Dth power of a polynomial, where $A' \;:\!=\; AM$ . This is clearly impossible if D does not divide $(h-1)A'$ . Otherwise we may drop the term $(X_1 \cdots X_d)^{(h-1)A'}$ . One sees that the polynomials

\begin{align*}X_1^{A'} + \cdots + X_d^{A'} - X_i^{A'}\end{align*}

are pairwise coprime, and hence each of them must be a Dth power. This is not the case, as can be seen, for example, by considering the partial derivative of $X_2^{A'} + \cdots + X_d^{A'}$ with respect to $X_2$ at $(X_2, X_3, \ldots , X_d) = (x_0, 1, \ldots , 1)$ , where $x_0^{A'} + (d-2) = 0$ .

The case $N = \pm 1$ is handled similarly: For any polynomial $Q \in \mathbb{C}[X_1^{\pm 1}, \ldots , X_{d-1}^{\pm 1}]$ we have

\begin{align*}Q(\alpha_1^n, \alpha_2^n, \ldots , \alpha_{d-1}^n) = 0\end{align*}

for all integers n only if Q is zero. Proceeding as before, we have that

\begin{align*}\prod_{i = 1}^d (X_1^{A'} + \cdots + X_d^{A'} - X_i^{A'})\end{align*}

is a Dth power of a polynomial in the variables $X_1, \ldots , X_{d-1}$ (with possibly negative exponents in the monomials), where $X_d$ is shorthand for $N/X_1 \cdots X_{d-1}$ . Note that the term

\begin{align*}X_1^{A'} + \cdots + X_{d-1}^{A'}\end{align*}

is coprime with all the other terms of the product, and, as before, this is not a Dth power of a polynomial.

7. Concluding the proof

We aim to prove that almost all primes of $K(\zeta_h)$ split in at least one of the fields $K(\zeta_h, b_n^{1/h})$ . By the Chebotarev density theorem it suffices to construct a subsequence $b_{n_1}, b_{n_2}, \ldots$ of $(b_n)$ such that the fields

\begin{align*}K\!\left(\zeta_h, b_{n_1}^{1/h}\right)/K(\zeta_h), K\!\left(\zeta_h, b_{n_2}^{1/h}\right)/K(\zeta_h), \ldots\end{align*}

are linearly disjoint

In Section 6 we checked that we may apply Lemma 5·3 to the norm sequence $(c_n)$ . Let $c_{n_1}, c_{n_2}, \ldots$ denote a subsequence given by the lemma with the base field $F(\zeta_h)$ , so

\begin{align*}F\!\left(\zeta_h, c_{n_i}^{1/h}\right)/F(\zeta_h), i = 1, 2, \ldots\end{align*}

are of degree h and linearly disjoint. We claim that this implies that

\begin{align*}F\!\left(\zeta_h, b_{n_i}^{1/h}\right)/F(\zeta_h), i = 1, 2, \ldots\end{align*}

are of degree h and linearly disjoint, too.

By Lemma 5·1 it suffices to show that

\begin{align*}b_{n_1}^{e_1/h} \cdots b_{n_k}^{e_k/h} \in F(\zeta_h), 0 \le e_i < h\end{align*}

implies $e_i = 0$ for all i. But if $b_{n_1}^{e_1} \cdots b_{n_k}^{e_k}$ is an hth power in $F(\zeta_h)$ , the norm $c_{n_1}^e \cdots c_{n_k}^{e_k}$ is a hth power in $F(\zeta_h)$ , too. By the choice of $c_{n_i}$ this happens only if $e_i = 0$ for all i.

Finally, note that linear disjointness over $F(\zeta_h)$ implies linear disjointness over $K(\zeta_h)$ , so $K(\zeta_h, b_{n_1}^{1/h})/K(\zeta_h), K(\zeta_h, b_{n_2}^{1/h})/K(\zeta_h), \ldots$ are linearly disjoint, as desired.