# Twisted rational zeros of
linear recurrence sequences

Yuri Bilu
IMB, Université de Bordeaux & CNRS
E-mail<EMAIL_ADDRESS>Florian Luca
School of Maths, Wits, South Africa
CCM, UNAM, Morelia, Mexico
E-mail<EMAIL_ADDRESS>Joris Nieuwveld and Joël Ouaknine
Max Planck Institute for Software Systems, Germany
E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>James Worrell
Department of Computer Science, Oxford University, UK
Email<EMAIL_ADDRESS>
(version of )

###### Abstract

We introduce the notion of a twisted rational zero of a non-degenerate linear
recurrence sequence (LRS). We show that any non-degenerate LRS has only
finitely many such twisted rational zeros. In the particular case of the
Tribonacci sequence, we show that $1/3$ and $-5/3$ are the only twisted
rational zeros which are not integral zeros.

###### Contents

  1. 1 Introduction
    1. 1.1 Twisted zeros and $p$-adic orders
    2. 1.2 Finiteness
  2. 2 Preliminaries
    1. 2.1 Fields
    2. 2.2 Linear recurrence sequences
  3. 3 Twisted rational zeros and the $p$-adic order
    1. 3.1 $p$-adic analytic functions
    2. 3.2 $p$-adic analytic interpolation of a linear recurrence sequence
    3. 3.3 Proof of Theorems 1.1, 1.4 and 1.5
  4. 4 Finiteness of twisted rational zeros
    1. 4.1 Powers in fields
    2. 4.2 Equations in roots of unity
    3. 4.3 A Kummer property
    4. 4.4 Proof of Theorem 1.9
    5. 4.5 An explicit result for ${\mathbb{Q}}$-valued linear recurrence sequences
  5. 5 Twisted rational zeros of the Tribonacci sequence
    1. 5.1 The roots
    2. 5.2 The denominator
    3. 5.3 Proof of Theorem 1.10
  6. 6 On Question 1.7
    1. 6.1 A density result
    2. 6.2 Proof of Theorem 6.1
    3. 6.3 Concluding remarks

##  1 Introduction

Let ${\mathbb{K}}$ be a field of characteristic $0$. We fix an algebraic
closure $\overline{{\mathbb{K}}}$. By a ${\mathbb{K}}$-valued linear
recurrence sequence (LRS) of order $m$ we mean a map
${U:{\mathbb{Z}}\to{\mathbb{K}}}$ such that for every ${n\in{\mathbb{Z}}}$ we
have

$U(n+m)=a_{m-1}U(n+m-1)+\cdots+a_{0}U(n),$ (1.1)

where ${a_{0},\ldots,a_{m-1}\in{\mathbb{K}}}$, with ${a_{0}\neq 0}$. Sometimes
instead of “${\mathbb{K}}$-valued LRS” we will say “LRS over ${\mathbb{K}}$”.

###  1.1 Twisted zeros and $p$-adic orders

Our initial motivation was the work of Marques and Lengyel [17], who computed
the $2$-adic order of the $n^{\text{th}}$ Tribonacci number $T(n)$. The
Tribonacci numbers is the ${\mathbb{Q}}$-valued LRS of order $3$, defined by

$T(0)=0,\quad T(1)=T(2)=1,\quad T(n+3)=T(n+2)+T(n+1)+T(n).$

Marques and Lengyel proved that

$\nu_{2}(T(n))=\begin{cases}0,&\text{if $n\equiv 1,2\pmod{4}$};\\\ 1,&\text{if
$n\equiv 3,11\pmod{16}$};\\\ 2,&\text{if $n\equiv 4,8\pmod{16}$};\\\
3,&\text{if $n\equiv 7\pmod{16}$};\\\ \nu_{2}(n)-1,&\text{if $n\equiv
0\pmod{16}$};\\\ \nu_{2}(n+4)-1,&\text{if $n\equiv 12\pmod{16}$};\\\
\nu_{2}(n+17)+1,&\text{if $n\equiv 15\pmod{32}$};\\\ \nu_{2}(n+1)+1,&\text{if
$n\equiv 31\pmod{32}$}.\end{cases}$ (1.2)

They conjectured that similar formulas must hold for other primes, not just
for ${p=2}$. This conjecture was refuted in [5], where it is shown that having
formulas like (1.2) is exceptional rather than typical.

As one can see in (1.2), on certain residue classes one has formulas like
${\nu_{2}(T(n))=\nu_{2}(n-a)+{\mathrm{const}}}$, where $a$ is one of the
numbers ${0,-1,-4,-17}$. This not surprising, because these numbers are
exactly the zeros of $T$: for ${n\in{\mathbb{Z}}}$ we have

$T(n)=0\quad\text{if and only if}\quad n\in\\{0,-1,-4,-17\\}.$

(For a proof see, for instance, [18], Example 2 on page 360; in that example
$u_{n}$ corresponds to our $T(-n)$.)

Let $U$ be a ${\mathbb{Q}}$-valued LRS; in particular, the coefficients
${a_{0},\ldots,a_{m-1}}$ of the recurrence relation (1.1) belong to
${\mathbb{Q}}$. We call $p$ a regular prime for $U$ if it does not divide the
denominators of the rational numbers ${a_{0},\ldots,a_{m-1}}$, and the
numerator of $a_{0}$. In other words, ${a_{0},\ldots,a_{m-1}}$ are $p$-adic
integers and $a_{0}$ is a $p$-adic unit.

The following theorem can be easily proved, using $p$-adic analysis, see
Section 3.

######  Theorem 1.1.

Let $a$ be a zero of a ${\mathbb{Q}}$-valued LRS $U$, and $p$ a regular prime
for $U$. Then there exist a positive integer $Q$, a positive integer $\kappa$
and an integer $\tau$ such that

$\nu_{p}(U(n))=\kappa\nu_{p}(n-a)+\tau\qquad\text{when}\quad n\equiv
a\pmod{Q}$ (1.3)

Of course the converse also holds: if for some $p$ there exist $Q,\kappa,\tau$
as above such that (1.3) holds, then $a$ is a zero of $U$.

Now let us ask the following slightly more general question: what would happen
if we take $n$ in (1.3) not from the residue class of $a$ modulo $Q$, but from
a different residue class?

######  Question 1.2.

Let $p$ be a prime number. Assume that there exist

$Q\in{\mathbb{Z}}_{>0},\qquad
a^{\prime}\in\\{0,1,\ldots,Q-1\\},\qquad\kappa\in{\mathbb{Z}}_{>0},\qquad\tau\in{\mathbb{Z}}$

such that

$\nu_{p}(U(n))=\kappa\nu_{p}(n-a)+\tau\qquad\text{when}\quad n\equiv
a^{\prime}\pmod{Q}.$

Assume further that $\nu_{p}(n-a)$ is not bounded on the residue class of
$a^{\prime}$ modulo $Q$. Does it imply that ${a^{\prime}\equiv a\pmod{Q}}$ and
${U(a)=0}$?

As the result of Marques and Lengyel implies, it is indeed the case when
${U=T}$ and ${p=2}$. But is it true in general?

The answer is “no”, as the following example shows.

######  Example 1.3.

Consider ${U(n):=2^{n}+1}$ and let $p$ be a prime number satisfying
${p\equiv\pm 3\pmod{8}}$. Then ${2^{(p-1)/2}+1\equiv 0\pmod{p}}$. Define
${\tau:=\nu_{p}(2^{(p-1)/2}+1)}$. Then

$\nu_{p}(U(n))=\nu_{p}(n)+\tau\quad\text{when}\quad
n\equiv\frac{p-1}{2}\pmod{p-1}.$

However, ${U(0)\neq 0}$, and in fact $U(n)$ does not vanish at all.

The explanation is that $0$ is, in fact, a kind of “hidden” zero of the LRS
${2^{n}+1}$. To give an exact definition, recall that a ${\mathbb{K}}$-valued
LRS $U$ satisfying (1.1) admits the Binet expansion

$U(n)=f_{1}(n)\lambda_{1}^{n}+\cdots+f_{s}(n)\lambda_{s}^{n}\qquad(n\in{\mathbb{Z}}),$
(1.4)

where ${\lambda_{1},\ldots,\lambda_{s}}$ are the distinct roots of the
characteristic polynomial

$X^{m}-a_{m-1}X^{m-1}-\cdots-a_{0},$

and ${f_{1},\ldots,f_{s}}$ are polynomials with coefficients in the field
${\mathbb{K}}(\lambda_{1},\ldots,\lambda_{s})$. (Note that the roots
$\lambda_{i}$ are non-zero, because ${a_{0}\neq 0}$.) We call
${a\in{\mathbb{Z}}}$ a twisted zero of the ${\mathbb{K}}$-valued LRS $U$ if
there exist roots of unity
${\xi_{1},\ldots,\xi_{s}\in\overline{{\mathbb{K}}}}$ such that

$\xi_{1}f_{1}(a)\lambda_{1}^{a}+\cdots+\xi_{s}f_{s}(a)\lambda_{s}^{a}=0.$

For example, $0$ is a twisted zero of the LRS with general term
${2^{n}+1^{n}}$, because

$1\cdot 2^{0}+(-1)\cdot 1^{0}=0.$

In Section 3 we will prove the following theorem, which gives a partial
positive answer to Question 1.2.

######  Theorem 1.4.

Let $U$ be a ${\mathbb{Q}}$-valued LRS, ${a\in{\mathbb{Z}}}$ and $p$ a regular
prime number for $U$. Assume that there exists a sequence of integers
$(n_{k})$ satisfying

$\nu_{p}(U(n_{k}))\to+\infty,\qquad\nu_{p}(n_{k}-a)\to+\infty.$

Then $a$ is a twisted zero of $U$.

There is another phenomenon discovered in [5], again in the context of
Tribonacci numbers. There exist infinitely many prime numbers $p$ such that
${\nu_{p}(T(n))\geq\nu_{p}(n-1/3)}$ for ${n\equiv 1/3\pmod{p-1}}$, and the
same holds true with $1/3$ replaced by $-5/3$; see [5, Theorem 1.5]. The
reason is that $1/3$ and $-5/3$ can be viewed as “rational zeros” of the LRS
$T$, see [5, Section 2].

Let us give the general definition. We call ${a\in{\mathbb{Q}}}$ a rational
zero of the ${\mathbb{K}}$-valued LRS $U$ with Binet expansion (1.4) if, for
some definition of the rational powers
${\lambda_{1}^{a},\ldots,\lambda_{s}^{a}\in\overline{{\mathbb{K}}}}$, we have

$f_{1}(a)\lambda_{1}^{a}+\cdots+f_{s}(a)\lambda_{s}^{a}=0.$

We call $a$ a twisted rational zero (TRZ) of $U$ if, for some definition of
${\lambda_{1}^{a},\ldots,\lambda_{s}^{a}}$ and some roots of unity
${\xi_{1},\ldots,\xi_{s}}$, we have

$\xi_{1}f_{1}(a)\lambda_{1}^{a}+\cdots+\xi_{s}f_{s}(a)\lambda_{s}^{a}=0.$

Theorem 1.4 remains true assuming that ${a\in{\mathbb{Q}}}$.

######  Theorem 1.5.

Let $U$ be a ${\mathbb{Q}}$-valued LRS, ${a\in{\mathbb{Q}}}$ and $p$ a regular
prime number for $U$. Assume that there exists a sequence of integers
$(n_{k})$ satisfying

$\nu_{p}(U(n_{k}))\to+\infty,\qquad\nu_{p}(n_{k}-a)\to+\infty.$ (1.5)

Then $a$ is a TRZ of $U$.

This theorem is proved in Section 3 as well.

One may ask whether the converse is true; that is, if $a$ is a TRZ, then there
exists a sequence of integers $(n_{k})$ satisfying (1.5). Easy examples show
that the answer is “no” in general.

######  Example 1.6.

If ${p\equiv-1\pmod{8}}$ then ${\nu_{p}(2^{n}+1)=0}$ for all $n$, though $0$
is a twisted zero of the LRS ${2^{n}+1^{n}}$.

One may still hope that, when $a$ is a TRZ, this holds for infinitely many
primes.

######  Question 1.7.

Let $a$ be a TRZ of a non-degenerate ${\mathbb{Q}}$-valued LRS $U$. Are there
infinitely many prime numbers $p$ with the following property: there exists a
sequence of integers $(n_{k})$ satisfying (1.5)?

We show that the answer is “yes” for twisted integral zeros of LRS of order
$2$; in fact, we will show that for them an analogue of Theorem 1.1 holds.
However, we do not know the answer for rational zeros. As for LRS of higher
order, the answer is, in general, “no” even for integral twisted zeros. See
Section 6 for the details.

###  1.2 Finiteness

Call a non-zero LRS with roots ${\lambda_{1},\ldots,\lambda_{s}}$ non-
degenerate if $\lambda_{k}/\lambda_{\ell}$ is not a root of unity for
${k\neq\ell}$. The following statement is the classical Skolem-Mahler-Lech
Theorem.

######  Theorem 1.8 (Skolem-Mahler-Lech).

A non-degenerate linear recurrence sequence $U$ over a field of characteristic
$0$ has at most finitely many zeros:

$\\#\\{n\in{\mathbb{Z}}:U(n)=0\\}<\infty.$

In Section 4 we prove that the same holds true for TRZs.

######  Theorem 1.9.

Let $U$ be a non-degenerate linear recurrence sequence with values in a field
of characteristic zero. Then $U$ admits at most finitely many TRZs.

The proof is a variation of the principal argument of Laurent’s article [14].
The main step is bounding the denominators of the TRZs; moreover, the bound is
effective if ${\mathbb{K}}$ is a number field. After the denominators are
bounded, Theorem 1.9 can be reduced to the Skolem-Mahler-Lech theorem using
the existing results about equations in roots of unity [10, 11, 16].

The Skolem-Mahler-Lech Theorem is, in general, non-effective, and so is our
Theorem 1.9: while we bound effectively the denominators of the TRZs, we
cannot do the same for their numerators. However, the Skolem-Mahler-Lech
Theorem can be made effective in many special cases, and so can be Theorem
1.9. To illustrate this, we prove (see Section 5) the following.

######  Theorem 1.10.

The only TRZs of the Tribonacci sequence $T$ are

$0,-1,-4,-17,\frac{1}{3},-\frac{5}{3}.$

##  2 Preliminaries

In this section we collect some basic facts and conventions that will be used
throughout the article, usually without special reference.

###  2.1 Fields

The letter $p$ denotes a prime number, and blackboard boldface letters
${\mathbb{K}},{\mathbb{L}},{\mathbb{M}}$ etc. denote (unless indicated
otherwise) fields of characteristic $0$. In particular, they can be number
fields or local fields (finite extensions of ${\mathbb{Q}}_{p}$). If
${\mathbb{K}}$ is a number field and $\mathfrak{p}$ is a prime of
${\mathbb{K}}$ then ${\mathbb{K}}_{\mathfrak{p}}$ denotes the
$\mathfrak{p}$-adic completion.

For every positive integer $m$ we fix a primitive root of unity of order $m$
and denote it $\zeta_{m}$. We denote by $\mu_{m}$ the group of roots of unity
of order $m$. Given a field ${\mathbb{K}}$, we denote by $\mu_{\mathbb{K}}$
the group of roots of unity belonging to ${\mathbb{K}}$.

The following lemma, which is Theorem 9.1 in [13, Chapter VI], will be used in
the article on several occasions.

######  Lemma 2.1.

Let ${\mathbb{K}}$ be a field of characteristic $0$ and
${\alpha\in{\mathbb{K}}^{\times}}$. Let $m$ be a positive integer. Assume that

$\displaystyle\text{for all ${p\mid m}$ we have
${\alpha\notin{\mathbb{K}}^{p}}$},$ (2.1) $\displaystyle\text{when ${4\mid m}$
we have ${\alpha\notin-4{\mathbb{K}}^{4}}$}.$ (2.2)

Then the polynomial ${X^{m}-\alpha}$ is irreducible in ${\mathbb{K}}[X]$.

######  Remark 2.2.

If ${\sqrt{-1}\in{\mathbb{K}}}$ then assumption (2.2) can be omitted, because
in this case ${-4\in{\mathbb{K}}^{4}}$ and (2.2) follows from (2.1).

###  2.2 Linear recurrence sequences

Let $U$ be an LRS with values in a field ${\mathbb{K}}$. We call $m$ the
minimal order of $U$ if $U$ admits a linear recurrence relation of order $m$,
but not of order strictly smaller than $m$. By convention, the minimal order
of the identically zero LRS is set to be $0$.

Let $m$ be the minimal order of a (non-zero) LRS $U$ with values in
${\mathbb{K}}$. Then the coefficients ${a_{0},\ldots,a_{m-1}}$ of the
recurrence relation

$U(n+m)=a_{m-1}U(n+m-1)+\cdots+a_{0}U(n)$

are well-defined and belong to the field ${\mathbb{K}}$. Fix an algebraic
closure $\overline{{\mathbb{K}}}$, and let
${\lambda_{1},\ldots,\lambda_{s}\in\overline{{\mathbb{K}}}}$ be the distinct
roots of the characteristic polynomial

$X^{m}-a_{m-1}X^{m-1}-\cdots-a_{0}.$ (2.3)

Then $U$ admits the Binet expansion

$U(n)=f_{1}(n)\lambda_{1}^{n}+\cdots+f_{s}(n)\lambda_{s}^{n},$

where ${f_{1},\ldots,f_{s}}$ are polynomials with coefficients in the field
${\mathbb{K}}(\lambda_{1},\ldots,\lambda_{s})$, such that the order of
$\lambda_{i}$ as a root of the characteristic polynomial (2.3) is equal to
${\deg f_{i}+1}$. In particular, the polynomials $f_{i}$ are all non-zero, and

$\sum_{i=1}^{s}(\deg f_{i}+1)=m.$

Unless the contrary is stated explicitly, in this article, when referring to
an LRS of order $m$, we will assume that $m$ is the minimal order of this LRS.

It is important to note the following: if $U$ is non-degenerate then it does
not vanish identically on any residue class; that is, for any positive integer
$N$ and any ${\ell\in\\{0,\ldots,N-1\\}}$, the function ${n\mapsto
U(\ell+Nn)}$ is not identically zero. Indeed, assuming non-degeneracy of $U$,
the numbers $\lambda_{1}^{N},\ldots,\lambda_{s}^{N}$ are all distinct. Hence

$U(\ell+Nn)=\sum_{i=1}^{s}h_{i}(n)\theta_{i}^{n},\quad\text{where}\quad
h_{i}(T):=\lambda_{i}^{\ell}f_{i}(\ell+NT),\quad\theta_{i}:=\lambda_{i}^{N}.$

This implies that $U(\ell+Nn)$ is an LRS of the same minimal order as $U$; in
particular, it is not identically zero.

##  3 Twisted rational zeros and the $p$-adic order

In this section we prove Theorems 1.1, 1.4 and 1.5 from the Introduction. The
proofs rely on Skolem’s $p$-adic interpolation of LRS, briefly recalled in
Subsections 3.1 and 3.2.

###  3.1 $p$-adic analytic functions

In this subsection we recall some very basic facts about $p$-adic analytic
functions. Most of them are quite standard. All missing proofs, unless
indicated otherwise, can be found in any standard text like [12] or [19].

Let $p$ be a prime number. We fix an algebraic closure
$\overline{{\mathbb{Q}}_{p}}$, and extend the standard $p$-adic absolute value
${|\cdot|_{p}}$ to $\overline{{\mathbb{Q}}_{p}}$, so that ${|p|_{p}=p^{-1}}$.
We will also use the additive valuation $\nu_{p}$ defined by
${\nu_{p}(z)=-\log|z|_{p}/\log p}$ for
${z\in\overline{{\mathbb{Q}}_{p}}^{\times}}$, with the convention
${\nu_{p}(0)=+\infty}$. All algebraic extensions of ${\mathbb{Q}}_{p}$
occurring below will be viewed as subfields of this fixed
$\overline{{\mathbb{Q}}_{p}}$.

Let ${\mathbb{K}}$ be a finite extension of ${\mathbb{Q}}_{p}$. For
${a\in{\mathbb{K}}}$ and ${r>0}$ we denote $\mathcal{D}(a,r)$ and
$\overline{\mathcal{D}}(a,r)$ (or $\mathcal{D}_{\mathbb{K}}(a,r)$,
$\overline{\mathcal{D}_{\mathbb{K}}}(a,r)$, if we want to indicate that the
disk is in the field ${\mathbb{K}}$) the open and the closed disks with center
$a$ and radius $r$:

$\mathcal{D}(a,r)=\\{z\in{\mathbb{K}}:|z-a|_{p}<r\\},\qquad\overline{\mathcal{D}}(a,r)=\\{z\in{\mathbb{K}}:|z-a|_{p}\leq
r\\}.$

It might be worth noting that every open disk in ${\mathbb{K}}$ is also
closed, and any closed disk is open. That is, for every ${r>0}$ there exist
${r^{\prime},r^{\prime\prime}>0}$ such that

$\mathcal{D}(a,r)=\overline{\mathcal{D}}(a,r^{\prime}),\qquad\overline{\mathcal{D}}(a,r)=\mathcal{D}(a,r^{\prime\prime}).$

Another useful observation is that every point of a disk serves as its center:
if ${b\in\mathcal{D}(a,r)}$ then ${\mathcal{D}(a,r)=\mathcal{D}(b,r)}$, and
similarly for the closed disks.

We denote by $\mathcal{O}_{\mathbb{K}}$, or simply by $\mathcal{O}$ if this
does not lead to a confusion, the ring of integers of ${\mathbb{K}}$:

$\mathcal{O}=\\{z\in{\mathbb{K}}:|z|_{p}\leq
1\\}=\overline{\mathcal{D}}(0,1).$

Let $D$ be a disk in ${\mathbb{K}}$ (open or closed), and ${\mathbb{L}}$ a
finite extension of ${\mathbb{K}}$. We call ${g:D\to{\mathbb{L}}}$ an analytic
function if for some ${a\in D}$ we have

$g(z)=\sum_{n=0}^{\infty}\alpha_{n}(z-a)^{n}\qquad(z\in D),$ (3.1)

where ${\alpha_{0},\alpha_{1},\alpha_{2},\ldots\in{\mathbb{L}}}$. In
particular, the infinite sum on the right converges for all ${z\in D}$.

Here are some simple properties of analytic functions, to be used below
without special reference.

  1. 1.

The coefficients ${\alpha_{0},\alpha_{1},\alpha_{2},\ldots}$ are well-defined
as soon as $g$ and $a$ are given. In particular, if the coefficients are not
all $0$, then $g$ is a non-zero function.

  2. 2.

The analytic function $g$ admits a power series expansion around any other
${b\in D}$. Specifically, for any ${b\in\mathcal{O}}$ we have

$g(z)=\sum_{k=0}^{\infty}\beta_{k}(z-b)^{k},$ (3.2)

where

$\beta_{k}=\frac{g^{(k)}(b)}{k!}=\sum_{n=k}^{\infty}\binom{n}{k}\alpha_{n}(b-a)^{n-k}.$

  3. 3.

An analytic function on $D$ is bounded. Indeed, set

$r:=\max\\{|z-w|_{p}:z,w\in D\\}.$

Then ${D=\overline{\mathcal{D}}(a,r)}$, and convergence in (3.1) is equivalent
to ${|\alpha_{n}|_{p}r^{n}\to 0}$. In particular, the sequence
$|\alpha_{n}|_{p}r^{n}$ is bounded. Hence $\alpha_{n}(z-a)^{n}$ is bounded
uniformly in ${z\in D}$. It follows that $g$ is bounded.

  4. 4.

A non-zero analytic function on a disk $D$ may have at most finitely many
zeros in $D$; this is because $D$ is compact and the zeros are isolated. A
quantitative version is given by the classical Theorem of Strassmann; see, for
instance, [8, Theorem 4.1].

######  Theorem 3.1.

Let ${\mathbb{K}}$ be a finite extension of ${\mathbb{Q}}_{p}$ of ramification
index $e$, and let ${g:{\mathbb{Z}}_{p}\to{\mathbb{K}}}$ be an analytic
function, not identically $0$. Denote by $\mathcal{A}$ the (finite) set of
zeros of $g$. Then there exists a positive integer $k$ such that for every
${i\in\\{0,1,\ldots,p^{k}-1\\}}$ we have one of the following two options.

  1. (C)

There exists ${\tau_{i}\in{\mathbb{Z}}}$ such that for
${z\in{\mathbb{Z}}_{p}}$ satisfying ${z\equiv i\pmod{p^{k}}}$ we have
${\nu_{p}(g(z))=e^{-1}\tau_{i}}$; in other words, $\nu_{p}(g(z))$ is constant
on the residue class ${z\equiv i\pmod{p^{k}}}$.

  2. (L)

There exist

$a_{i}\in\mathcal{A},\qquad\tau_{i}\in{\mathbb{Z}},\qquad\kappa_{i}\in{\mathbb{Z}}_{>0}$

such that ${a_{i}\equiv i\pmod{p^{k}}}$, and for ${z\in{\mathbb{Z}}_{p}}$
satisfying ${z\equiv i\pmod{p^{k}}}$ we have

$\nu_{p}(g(z))=\kappa_{i}\nu_{p}(z-a_{i})+e^{-1}\tau_{i}.$

###### Proof.

This is Theorem 3.2 from [5]. ∎

We denote

$\rho:=p^{-1/(p-1)}.$ (3.3)

Let us recall the definition and the basic properties of the $p$-adic
exponential and logarithmic function.

  1. 1.

For ${z\in\mathcal{D}(0,\rho)}$ we define

$\exp(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}.$

For ${z,w\in\mathcal{D}(0,\rho)}$ we have

$|\exp(z)-1|_{p}=|z|_{p},\quad\exp(z+w)=\exp(z)\exp(w),\quad\exp^{\prime}(z)=\exp(z).$

  2. 2.

For ${z\in\mathcal{D}(1,1)}$ we define

$\log(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(z-1)^{n}}{n}.$

For ${z,w\in\mathcal{D}(1,1)}$ we have

$\log(zw)=\log(z)+\log(w),\quad\log^{\prime}(z)=\frac{1}{z}.$

  3. 3.

For ${z\in\mathcal{D}(1,\rho)}$ we have

$|\log(z)|_{p}=|z-1|_{p},\quad\exp(\log(z))=z.$

  4. 4.

For ${z\in\mathcal{D}(0,\rho)}$ we have ${\log(\exp(z))=z}$.

###  3.2 $p$-adic analytic interpolation of a linear recurrence sequence

The contents of this subsection is very classical and goes back to Skolem.
Still, we prefer to include some proofs for the reader’s convenience.

Let $U$ be a non-zero LRS of (minimal) order $m$ with values in a number field
${\mathbb{K}}$. We write its recurrence relation as

$U(n+m)=a_{m-1}U(n+m-1)+\cdots+a_{0}U(n),$

where ${a_{0},\ldots,a_{m-1}\in{\mathbb{K}}}$.

We call a prime $\mathfrak{p}$ of ${\mathbb{K}}$ regular for $U$ if
${a_{1},\ldots,a_{m}}$ are $\mathfrak{p}$-adic integers and $a_{0}$ is a
$\mathfrak{p}$-adic unit.

Let $\mathfrak{p}$ be a regular prime for $U$, and
${\mathbb{K}}_{\mathfrak{p}}$ the $\mathfrak{p}$-adic completion of
${\mathbb{K}}$. It is a finite extension of ${\mathbb{Q}}_{p}$, where $p$ is
the rational prime number below $\mathfrak{p}$.

######  Proposition 3.2 (Skolem).

There exists a positive integer $N$ and analytic functions

$g_{0},\ldots,g_{N-1}:{\mathbb{Z}}_{p}\to\mathcal{O}_{{\mathbb{K}}_{\mathfrak{p}}}$

such that

$u(\ell+Nn)=g_{\ell}(n)\qquad(\ell\in\\{0,\ldots,N-1\\},\quad
n\in{\mathbb{Z}}).$ (3.4)

Moreover, if the LRS $U$ is non-degenerate, then none of the functions
$g_{\ell}$ vanishes identically.

###### Proof.

Denote by ${\mathbb{L}}$ the splitting field over
${\mathbb{K}}_{\mathfrak{p}}$ of the characteristic polynomial
${X^{m}-a_{m-1}X^{m-1}-\cdots-a_{0}}$. Then $U$ admits the Binet expansion

$U(n)=f_{1}(n)\lambda_{1}^{n}+\cdots+f_{s}(n)\lambda_{s}^{n},$

where ${\lambda_{1},\ldots,\lambda_{s}}$ are the distinct roots of the
characteristic polynomial, and ${f_{1},\ldots,f_{s}}$ are non-zero polynomials
with coefficients in ${\mathbb{L}}$. Since $p$ is a regular prime,
${\lambda_{1},\ldots,\lambda_{s}\in\mathcal{O}_{\mathbb{L}}^{\times}}$.

Let $\rho$ be defined as in (3.3). Since ${\rho<1}$, the disk
$\mathcal{D}_{\mathbb{L}}(1,\rho)$ is a multiplicative group. It is a finite
index subgroup of $\mathcal{O}_{\mathbb{L}}^{\times}$, because
$\mathcal{O}_{\mathbb{L}}^{\times}$ is compact and
$\mathcal{D}_{\mathbb{L}}(1,\rho)$ is open. Hence there exists a positive
integer $N$ such that ${x^{N}\in\mathcal{D}_{\mathbb{L}}(1,\rho)}$ for every
${x\in\mathcal{O}_{\mathbb{L}}^{\times}}$. Note that we have

$x^{Nn}=\exp(n\log(x^{N}))\qquad(x\in\mathcal{O}_{\mathbb{L}}^{\times},\quad
n\in{\mathbb{Z}}).$ (3.5)

Now define, for ${\ell=0,1,\ldots,N-1}$ and ${z\in{\mathbb{Z}}_{p}}$,

$g_{\ell}(z):=\sum_{i=1}^{s}\lambda_{i}^{\ell}f_{i}(\ell+Nz)\exp(z\log(\lambda_{i}^{N})).$

Note that, a priori, ${g_{\ell}(z)\in{\mathbb{L}}}$, but we will see that
${g_{\ell}(z)\in{\mathbb{K}}_{\mathfrak{p}}}$ in a while.

From (3.5) we deduce that (3.4) holds. In particular,
${g_{\ell}(z)\in{\mathbb{K}}}$ when ${z\in{\mathbb{Z}}}$. Since ${\mathbb{Z}}$
is dense in ${\mathbb{Z}}_{p}$, we have
${g_{\ell}(z)\in{\mathbb{K}}_{\mathfrak{p}}}$ for all
${z\in{\mathbb{Z}}_{p}}$.

When $U$ is non-degenerate, the function $g_{\ell}$ does not vanish
identically, because the function ${n\mapsto U(\ell+Nn)}$ does not, see
Subsection 2.2. ∎

As a by-product, we prove the Theorem of Skolem-Mahler-Lech (see Theorem 1.8
above) for $\overline{{\mathbb{Q}}}$-valued LRS.

######  Corollary 3.3 (Skolem-Mahler-Lech).

Let $U$ be a non-degenerate $\overline{{\mathbb{Q}}}$-valued LRS. Then the
equation ${U(n)=0}$ has at most finitely many solutions in
${n\in{\mathbb{Z}}}$.

###### Proof.

Pick some prime $\mathfrak{p}$ regular for $U$. Then each of the analytic
functions $g_{\ell}$ has at most finitely many zeros on ${\mathbb{Z}}_{p}$,
hence on ${\mathbb{Z}}$. ∎

Actually, the Skolem-Mahler-Lech Theorem, as stated in Theorem 1.8, applies to
LRS over an arbitrary field of characteristic $0$. To extend it to this
generality, one more ingredient is needed, the Lech-Cassels Specialization
Theorem, see [7].

###  3.3 Proof of Theorems 1.1, 1.4 and 1.5

Let $U$ be an LRS taking values in a number field ${\mathbb{K}}$, and let
$\mathfrak{p}$ be a prime of ${\mathbb{K}}$ regular for $U$, see Subsection
3.2. We denote by $p$ the rational prime below $\mathfrak{p}$. In this section
we prove the following two theorems, which generalize Theorems 1.1, 1.4 and
1.5 from the Introduction.

######  Theorem 3.4.

Let $a$ be a zero of $U$. Then there exist a positive integer $Q$, a positive
integer $\kappa$ and an integer $\tau$ such that
${\nu_{\mathfrak{p}}(U(n))=\kappa\nu_{p}(n-a)+\tau}$ for ${n\equiv
a\pmod{Q}}$.

######  Theorem 3.5.

Let ${a\in{\mathbb{Q}}}$ be such that there exists a sequence of integers
$(n_{k})$ satisfying

$\nu_{\mathfrak{p}}(U(n_{k}))\to+\infty,\qquad\nu_{p}(n_{k}-a)\to+\infty.$

Then $a$ is a TRZ of $U$.

Theorem 3.4 is more general than Theorem 1.1, while Theorem 3.5 is more
general than Theorem 1.5 (and, a fortiori, than Theorem 1.4).

###### Proof of Theorem 3.4.

Let the integer $N$ and the functions $g_{0},\ldots,g_{N-1}$ be as in
Proposition 3.2. Let ${\ell\in\\{0,\ldots,N-1\\}}$ be such that
${a\equiv\ell\pmod{N}}$. Write ${a=\ell+bN}$ with ${b\in{\mathbb{Z}}}$. In the
sequel, we denote ${g:=g_{\ell}}$. We have ${g(b)=0}$.

Theorem 3.1 implies that there exist positive integers $k$ and
$\tau^{\prime}$, and an integer $\kappa^{\prime}$ such that for ${z\equiv
b\pmod{p^{k}}}$ we have

$\nu_{p}(g(z))=\kappa^{\prime}\nu_{p}(z-b)+e^{-1}\tau^{\prime}.$ (3.6)

Now set

$Q:=Np^{k},\qquad\kappa:=e\kappa^{\prime},\qquad\tau:=\tau^{\prime}-\kappa\nu_{p}(N).$

Let ${n\equiv a\pmod{Q}}$. Then ${n\equiv\ell\pmod{N}}$, and for
${m:=(n-\ell)/N}$ we have

$m\equiv b\pmod{p^{k}},\qquad\nu_{p}(m-b)=\nu_{p}(n-a)-\nu_{p}(N),\qquad
g(m)=U(n).$

Applying (3.6) with ${z=m}$, we obtain

$\nu_{\mathfrak{p}}(U(n))=e\nu_{p}(g(m))=e\kappa^{\prime}\nu_{p}(m-b)+\tau^{\prime}=\kappa\nu_{p}(n-a)+\tau.$

The theorem is proved. ∎

###### Proof of Theorem 3.5.

Once again, let $N$ and $g_{0},\ldots,g_{N-1}$ be as in Proposition 3.2. Let
${\ell\in\\{0,\ldots,N-1\\}}$ be such that ${n_{k}\equiv\ell\pmod{N}}$ holds
for infinitely many $k$. By taking a subsequence, we may assume that this
holds for all $k$. We denote ${g:=g_{\ell}}$.

Set ${b:=(a-\ell)/N}$ and ${m_{k}:=(n_{k}-\ell)/N}$. Then, ${m_{k}\to b}$ and
${g(m_{k})\to 0}$ in the $p$-adic topology. Hence ${g(b)=0}$.

(Note that, unlike in the proof of Theorem 3.4, we do not, in general, have
${g(b)=U(\ell+Nb)=U(a)}$; this would only be true if ${b\in{\mathbb{Z}}}$. But
this is not true in general: $b$ is merely a rational number, not necessarily
an integer.)

Recall that

$g(z)=g_{\ell}(z)=\sum_{i=1}^{s}\lambda_{i}^{\ell}h_{i}(z),\quad\text{where}\quad
h_{i}(z):=f_{i}(\ell+Nz)\exp(z\log(\lambda_{i}^{N})).$

Let $A$ be the denominator of the rational number $a$. Then

$(h_{i}(b))^{AN}=\bigl{(}\lambda_{i}^{\ell}f_{i}(a)\bigr{)}^{AN}\exp(ANb\log(\lambda_{i}^{N})).$

Since ${ANb\in{\mathbb{Z}}}$, we have
${\exp(ANb\log(\lambda_{i}^{N}))=\lambda_{i}^{AN^{2}b}}$. Hence

$(h_{i}(b))^{AN}=\lambda_{i}^{AN\ell+AN^{2}b}f_{i}(a)^{AN}=\bigl{(}\lambda_{i}^{a}f_{i}(a)\bigr{)}^{AN},$

where we pick some definition for the rational power $\lambda_{i}^{a}$. Thus,

$h_{i}(b)=\xi_{i}\lambda_{i}^{a}f(a),$

where $\xi_{i}$ is a root of unity.

We have proved that

$0=g(b)=\sum_{i=1}^{s}\xi_{i}\lambda_{i}^{a}f(a),$

which exactly means that $a$ is a TRZ of $U$. The theorem is proved. ∎

##  4 Finiteness of twisted rational zeros

In this section we prove Theorem 1.9. Throughout this section, unless the
contrary is stated explicitly, ${\mathbb{K}}$ is a field of characteristic
$0$. We fix an algebraic closure $\overline{{\mathbb{K}}}$.

For a positive integer $n$ we fix ${\zeta_{n}\in\overline{{\mathbb{K}}}}$, a
primitive $n^{\text{th}}$ root of unity. Recall that we denote by $\mu_{n}$
the group of $n^{\text{th}}$ roots of unity, and by $\mu_{\mathbb{K}}$ the
group of roots of unity in ${\mathbb{K}}$. We denote by ${\mathbb{K}}^{n}$ the
set of $n^{\text{th}}$ powers in ${\mathbb{K}}$:

${\mathbb{K}}^{n}:=\\{\alpha^{n}:\alpha\in{\mathbb{K}}\\}.$

We denote by ${\mathbb{K}}_{\mathrm{ab}}$ the maximal abelian subfield of
${\mathbb{K}}$; that is, the maximal subfield of ${\mathbb{K}}$ which is an
abelian extension of ${\mathbb{Q}}$.

###  4.1 Powers in fields

The following result is due to Chevalley [9] and Bass [1]. The proofs can be
also found in [20] and [2].

######  Theorem 4.1.

[Chevalley, Bass] Let ${\mathbb{K}}$ be a finitely generated field of
characteristic $0$ (in particular, ${\mathbb{K}}_{\mathrm{ab}}$ is a finite
extension of ${\mathbb{Q}}$). Then there exists a positive integer $\Lambda$,
depending only on the degree ${d:=[{\mathbb{K}}_{\mathrm{ab}}:{\mathbb{Q}}]}$,
such that for every positive integer $n$ the following holds: if
${\alpha\in{\mathbb{K}}}$ is a $\Lambda n^{\text{th}}$ power in
${\mathbb{K}}(\zeta_{\Lambda n})$, then $\alpha$ is an $n^{\text{th}}$ power
in ${\mathbb{K}}$. In symbols:

${\mathbb{K}}(\zeta_{\Lambda n})^{\Lambda
n}\cap{\mathbb{K}}\subset{\mathbb{K}}^{n}\qquad(n=1,2,3,\ldots).$ (4.1)

The smallest positive integer $\Lambda$ satisfying (4.1) will be called the
Chevalley-Bass number of the field ${\mathbb{K}}$; see [2, Section 6].

It might not be easy to determine the Chevalley-Bass number of a given field
${\mathbb{K}}$, but it is easy to estimate it in terms of
${d:=[{\mathbb{K}}_{\mathrm{ab}}:{\mathbb{Q}}]}$. For instance, it is shown in
[2, Section 6.1] that, when ${d\geq 3}$, the Chevalley-Bass number $\Lambda$
satisfies ${\Lambda\leq\exp(d^{2/\log\log d})}$.

It will be convenient to introduce the following notion. For
${\alpha\in{\mathbb{K}}^{\times}}$ we define the Kummer exponent of $\alpha$
in ${\mathbb{K}}$ as the biggest positive integer $n$ such that
${\alpha\xi\in{\mathbb{K}}^{n}}$ for some root of unity
${\xi\in{\mathbb{K}}}$. In symbols:

$\varrho_{\mathbb{K}}(\alpha):=\max\\{n:\alpha\in\mu_{\mathbb{K}}{\mathbb{K}}^{n}\\}$

(recall that $\mu_{\mathbb{K}}$ denotes the group of roots of unity in
${\mathbb{K}}$). Clearly, ${\varrho_{\mathbb{K}}(\alpha)=\infty}$ if $\alpha$
is a root of unity, and, when ${\mathbb{K}}$ is a finitely generated field,
$\varrho_{\mathbb{K}}(\alpha)$ is finite if $\alpha$ is not a root of unity.

######  Proposition 4.2.

Let ${\alpha\in{\mathbb{K}}}$ be such that $\varrho_{\mathbb{K}}(\alpha)$ is
finite, and $n$ a positive integer.

  1. 1.

We have ${\alpha\in\mu_{\mathbb{K}}{\mathbb{K}}^{n}}$ if and only if
${n\mid\varrho_{\mathbb{K}}(\alpha)}$.

  2. 2.

Let ${\alpha^{1/n}\in\overline{{\mathbb{K}}}}$ be some determination of the
$n^{\text{th}}$ root, and ${\xi\in\overline{{\mathbb{K}}}}$ a root of unity.
Then the degree ${[{\mathbb{K}}(\alpha^{1/n}\xi):{\mathbb{K}}]}$ is a multiple
of ${n/\gcd(\varrho_{\mathbb{K}}(\alpha),n)}$.

###### Proof.

Item 1 follows immediately from the definition. To prove item 2, define

${\mathbb{L}}:={\mathbb{K}}(\alpha^{1/n}\xi),\qquad
m:=[{\mathbb{L}}:{\mathbb{K}}],\qquad\rho:=\varrho_{\mathbb{K}}(\alpha),\qquad
d:=\gcd(m,n).$

All conjugates of $\alpha^{1/n}$ over ${\mathbb{K}}$ are equal to
$\alpha^{1/n}$ times a root of unity. Hence
${\beta:=\mathcal{N}_{{\mathbb{L}}/{\mathbb{K}}}(\alpha^{1/n}\xi)}$ is
$\alpha^{m/n}$ times a root of unity. Let ${r,s\in{\mathbb{Z}}}$ be such that
${mr+ns=d}$. Then ${\gamma:=\alpha^{s}\beta^{r}}$ is $\alpha^{d/n}$ times a
root of unity. Since $\gamma^{n/d}$ is $\alpha$ times a root of unity, we have
${n/d\mid\rho}$. Hence ${n/d\mid\gcd(\rho,n)}$. It follows that
${n/\gcd(\rho,n)}$ divides $d$. Hence it divides $m$. ∎

###  4.2 Equations in roots of unity

The following result is due to Dvornicich and Zannier [11, 24], who improved
on the previous work of Mann [16] and of Conway and Jones [10]. In this
subsection ${\mathbb{K}}$ is a finitely generated field of characteristic $0$.

######  Theorem 4.3.

[Dvornicich, Zannier] Let ${\alpha_{1},\ldots,\alpha_{s}}$ be non-zero
elements of ${\mathbb{K}}$, and
${\xi_{1},\ldots,\xi_{s}\in\overline{{\mathbb{K}}}}$ roots of unity. Assume
that

$\alpha_{1}\xi_{1}+\cdots+\alpha_{s}\xi_{s}=1,$

and no proper sub-sum of the sum on the left vanishes: ${\sum_{i\in
I}\alpha_{i}\xi_{i}\neq 0}$ when ${\varnothing\subsetneq
I\subsetneq\\{1,\ldots,s\\}}$. Then the order of the multiplicative group
generated by ${\xi_{1},\ldots,\xi_{s}}$ is effectively bounded in terms of
${d=[{\mathbb{K}}_{\mathrm{ab}}:{\mathbb{Q}}]}$ and $s$.

In fact, Dvornicich and Zannier prove that, denoting by $r$ the order of the
group generated by ${\xi_{1},\ldots,\xi_{s}}$, we have the following
properties:

  * •

if ${p^{a+1}\mid r}$ for some positive integer $a$ then ${p^{a}\mid 2d}$;

  * •

${\displaystyle
s+1\geq\dim_{{\mathbb{K}}}({\mathbb{K}}+{\mathbb{K}}\xi_{1}+\cdots+{\mathbb{K}}\xi_{s})\geq
1+\sum_{p\|r}\left(\frac{p-1}{\gcd(d,p-1)}-1\right)}$.

Clearly, using these properties, it is easy to bound $r$ explicitly in terms
of $d$ and $s$.

###  4.3 A Kummer property

As before, ${\mathbb{K}}$ is a finitely generated field of characteristic $0$.
Recall that we denote by ${\mathbb{K}}_{\mathrm{ab}}$ the maximal abelian
subfield of ${\mathbb{K}}$. Let $\Gamma$ be the division group of the
multiplicative group ${\mathbb{K}}^{\times}$:

$\Gamma:=\\{a\in\overline{{\mathbb{K}}}^{\times}:\text{$a^{n}\in{\mathbb{K}}^{\times}$
for some positive integer~{}$n$}\\}.$

The following key proposition is, essentially, due to Laurent [14].

######  Proposition 4.4.

Let ${\alpha_{1},\ldots,\alpha_{s}\in\Gamma}$ be such that
${\alpha_{1}+\cdots+\alpha_{s}=1}$, and no proper sub-sum of the sum
${\alpha_{1}+\cdots+\alpha_{s}}$ vanishes. Let $\Lambda$ be the Chevalley-Bass
number of ${\mathbb{K}}$ (see Section 4.1). Then there exist roots of unity
${\xi_{1},\ldots,\xi_{s}\in\overline{{\mathbb{K}}}}$ such that

$\alpha_{i}^{\Lambda}\xi_{i}\in{\mathbb{K}}\qquad(i=1,\ldots,m),$ (4.2)

###### Proof.

We follow Laurent [14, Section 2.2], but we replace the cohomological argument
by a reference to Theorem 4.1.

Let $n$ be a positive integer such that
${\alpha_{1}^{n},\ldots,\alpha_{s}^{n}\in{\mathbb{K}}}$. Denote by $G$ the
Galois group
${\operatorname{Gal}(\overline{{\mathbb{K}}}/{\mathbb{K}}(\zeta_{n}))}$. For
${i\in\\{1,\ldots,s\\}}$ let ${\chi_{i}:G\to{\mathbb{K}}(\zeta_{n})}$ be the
character of $G$ defined by ${\sigma\mapsto\sigma(\alpha_{i})/\alpha_{i}}$.
Since ${\sigma(\alpha_{1})+\cdots+\sigma(\alpha_{s})=1}$ for every ${\sigma\in
G}$, we have ${\alpha_{1}\chi_{1}+\cdots+\alpha_{s}\chi_{s}=1}$.

We claim that the characters ${\chi_{1},\ldots,\chi_{s}}$ are all trivial:

$\chi_{1}=\cdots=\chi_{s}=1.$ (4.3)

Indeed, defining ${\alpha_{0}:=-1}$ and ${\chi_{0}:=1}$, we have
${\alpha_{0}\chi_{0}+\cdots+\alpha_{s}\chi_{s}=0}$. After renumbering the
characters ${\chi_{1},\ldots,\chi_{s}}$, we may assume that for some ${r\geq
0}$ the characters ${\chi_{0},\ldots,\chi_{r}}$ are distinct, and each of the
remaining ${\chi_{r+1},\ldots,\chi_{s}}$ is equal to one of
${\chi_{0},\ldots,\chi_{r}}$. For ${k=0,\ldots,r}$ define
${I_{k}:=\\{i:\chi_{i}=\chi_{k}\\}}$. Then

$\sum_{k=0}^{r}\chi_{k}\sum_{i\in I_{k}}\alpha_{i}=0.$

Artin’s Theorem on Characters (see, for instance, Theorem 4.1 in [13, Chapter
VI]) implies that ${\sum_{i\in I_{k}}\alpha_{i}=0}$ for every $k$. Since no
proper sub-sum of ${\alpha_{1}+\cdots+\alpha_{s}}$ vanishes, this is possible
only if ${r=0}$, which proves (4.3).

It follows from (4.3) that
${\alpha_{1},\ldots,\alpha_{s}\in{\mathbb{K}}(\zeta_{n})}$. Hence each
$\alpha_{i}^{\Lambda n}$ is an $\Lambda n^{\text{th}}$ power in
${\mathbb{K}}(\zeta_{n})$. Theorem 4.1 implies that $\alpha_{i}^{\Lambda n}$
is an $n^{\text{th}}$ power in ${\mathbb{K}}$. It follows that
${\alpha_{i}^{\Lambda}\xi_{i}\in{\mathbb{K}}}$ for some root of unity
$\xi_{i}$, as wanted. ∎

###  4.4 Proof of Theorem 1.9

Let $U$ be a non-degenerate LRS with values in a field of characteristic $0$
and with Binet expansion

$U(n)=f_{1}(n)\lambda_{1}^{n}+\cdots+f_{s}(n)\lambda_{s}^{n}.$ (4.4)

Let ${\mathbb{K}}$ be a finitely generated field, containing
${\lambda_{1},\ldots,\lambda_{s}}$ and the coefficient of the polynomials
$f_{1},\ldots,f_{s}$. Recall that we denote by ${\mathbb{K}}_{\mathrm{ab}}$
the maximal abelian subfield of ${\mathbb{K}}$. Since ${\mathbb{K}}$ is
finitely generated, ${\mathbb{K}}_{\mathrm{ab}}$ is a finite extension of
${\mathbb{Q}}$, and we denote by $d$ its degree over ${\mathbb{Q}}$.

Recall that we denote by $\varrho_{\mathbb{K}}(\alpha)$ the Kummer exponent of
${\alpha\in{\mathbb{K}}}$, see Subsection 4.1. Since the $U$ is non-degenerate
and the field ${\mathbb{K}}$ is finitely generated, we have
${\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{j})<\infty}$ when ${i\neq j}$. We
set

$\rho:=\gcd\\{\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{j}):1\leq i<j\leq
s\\}.$ (4.5)

Let $a$ be a TRZ of $U$. Recall that this means the following: there exist
roots of unity ${\xi_{1},\ldots,\xi_{s}\in\overline{{\mathbb{K}}}}$ such that
for some determinations of
${\lambda_{1}^{a},\ldots,\lambda_{s}^{a}\in\overline{{\mathbb{K}}}}$ we have

$\xi_{1}f_{1}(a)\lambda_{1}^{a}+\cdots+\xi_{s}f_{s}(a)\lambda_{s}^{a}=0.$
(4.6)

We call the TRZ $a$ primitive if no proper sub-sum of the sum in (4.6)
vanishes; that is, if ${\varnothing\subsetneq I\subsetneq\\{1,\ldots,s\\}}$
then ${\sum_{i\in I}\xi_{i}f_{i}(a)\lambda_{i}^{a}\neq 0}$.

######  Proposition 4.5.

Let ${\mathbb{K}}$ be as above (that is, a finitely generated field,
containing ${\lambda_{1},\ldots,\lambda_{s}}$ and the coefficient of the
polynomials $f_{1},\ldots,f_{s}$), and $\Lambda$ the Chevalley-Bass number of
${\mathbb{K}}$. Let $a$ be a primitive TRZ of $U$, and
${\xi_{1},\ldots,\xi_{s}}$ roots of unity satisfying (4.6). Assume that
${s\geq 2}$. Then the denominator of the rational number $a$ divides
$\Lambda\rho$, where $\rho$ is defined in (4.5); in particular, the
denominator is bounded effectively in terms of
${d:=[{\mathbb{K}}_{\mathrm{ab}}:{\mathbb{Q}}]}$ and $\rho$. Moreover, the
orders of the roots of unity ${\xi_{i}/\xi_{j}}$ are effectively bounded in
terms of $d$, $\rho$ and $s$.

###### Proof.

Since ${s\geq 2}$ and no proper sub-sum of the sum in (4.6) vanishes, we have
${f_{i}(a)\neq 0}$ for ${i=1,\ldots,s}$. There will be no loss of generality
to assume that ${\xi_{s}=1}$; so, instead of (4.6) we have

$\xi_{1}f_{1}(a)\lambda_{1}^{a}+\cdots+\xi_{s-1}f_{s-1}(a)\lambda_{s-1}^{a}+f_{s}(a)\lambda_{s}^{a}=0.$
(4.7)

Applying Proposition 4.4 to the relation

$\sum_{i=1}^{s-1}-\xi_{i}\frac{f_{i}(a)}{f_{s}(a)}\left(\frac{\lambda_{i}}{\lambda_{s}}\right)^{a}=1,$
(4.8)

we obtain the following: there exist roots of unity
${\eta_{1},\ldots,\eta_{s-1}}$ such that

$\left({\lambda_{i}}/{\lambda_{s}}\right)^{\Lambda
a}\eta_{i}\in{\mathbb{K}}\qquad(i=1,\ldots,s-1).$

Write ${a=k/\ell}$, where $k$ and $\ell$ are co-prime integers. We want to
show that ${\ell\mid\Lambda\rho}$. We have
${\varrho_{\mathbb{K}}\bigl{(}(\lambda_{i}/\lambda_{s})^{\Lambda
k}\bigr{)}=\Lambda k\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{s})}$.
Proposition 4.2 implies that the quotient ${\ell/\gcd\bigl{(}\ell,\Lambda
k\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{s})\bigr{)}}$ divides the degree
${[{\mathbb{K}}((\lambda_{i}/\lambda_{s})^{\Lambda
a}\eta_{i}):{\mathbb{K}}]}$. But this degree is $1$, which implies that
${\ell\mid\Lambda k\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{s})}$. Since
$\ell$ and $k$ are coprime, this implies that
${\ell\mid\Lambda\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{s})}$ for
${i=1,\ldots,s-1}$.

We have clearly
${\rho=\gcd\\{\varrho_{\mathbb{K}}(\lambda_{i}/\lambda_{s}):i=1,\ldots,s-1\\}}$.
It follows that ${\ell\mid\Lambda\rho}$, which proves the first statement of
the proposition.

Now let us bound the orders of the roots of unity $\xi_{i}$. Let
${\mathbb{L}}$ be the field, generated over ${\mathbb{K}}$ by
${(\lambda_{1}/\lambda_{s})^{a},\ldots,(\lambda_{s-1}/\lambda_{s})^{a}}$.
Since the denominator of $a$ divides $\Lambda\rho$, we have

$[{\mathbb{L}}_{\mathrm{ab}}:{\mathbb{K}}_{\mathrm{ab}}]\leq[{\mathbb{L}}:{\mathbb{K}}]\leq(\Lambda\rho)^{s-1}.$

Hence ${[{\mathbb{L}}_{\mathrm{ab}}:{\mathbb{Q}}]\leq d(\Lambda\rho)^{s-1}}$;
in particular, ${[{\mathbb{L}}_{\mathrm{ab}}:{\mathbb{Q}}]}$ is effectively
bounded in terms of $d$, $s$ and $\rho$.

Applying Theorem 4.3 to relation (4.8), we bound the orders of the roots of
unity $\xi_{i}$ in terms of ${[{\mathbb{L}}_{\mathrm{ab}}:{\mathbb{Q}}]}$ and
$s$. Hence it is bounded in terms of $d$, $s$ and $\rho$, as wanted. The
proposition is proved. ∎

Combining this proposition with the Skolem-Mahler-Lech Theorem, we obtain the
following consequence.

######  Corollary 4.6.

Let $U$ be a non-degenerate LRS over a field of characteristic $0$. Then $U$
admits at most finitely many primitive TRZs. More precisely, if (4.4) is the
Binet expansion of $U$, then there exist at most finitely many $s$-tuples
$(a,\xi_{1},\ldots,\xi_{s-1})$ such that $a$ is a rational number,
${\xi_{1},\ldots,\xi_{s-1}}$ are roots of unity, and (4.7) holds.

###### Proof.

If ${s=1}$ then ${f_{1}(a)=0}$, which is possible only for finitely many $a$.

From now on we assume that ${s\geq 2}$. Proposition 4.5 implies that
${a=n/\Lambda\rho}$, where ${n\in{\mathbb{Z}}}$, and that there are at most
finitely many choices for $(\xi_{1},\ldots,\xi_{s-1})$ in (4.7). Pick some
determinations ${\theta_{i}:=\lambda_{i}^{1/\Lambda\rho}}$, so that
${\lambda_{i}^{a}=\theta_{i}^{n}\eta_{i}}$, where $\eta_{i}$ are
$\Lambda\rho^{\text{th}}$ roots of unity. Then

$\xi_{1}\eta_{1}g_{1}(n)\theta_{1}^{n}+\cdots+\xi_{s-1}\eta_{s-1}g_{s-1}(n)\theta_{s-1}^{n}+\eta_{s}g_{s}(n)\theta_{s}^{n}=0,$
(4.9)

where ${g_{i}(t):=f_{i}(t/\Lambda\rho)}$.

The left-hand side of (4.9) is a non-degenerate LRS, and the Skolem-Mahler-
Lech Theorem implies that there can be at most finitely many $n$ for every
fixed choice of the roots of unity $\xi_{i}$ and $\eta_{i}$. Since there are
at most finitely many choices for $(\xi_{1},\ldots,\xi_{s-1})$ and for
$(\eta_{1},\ldots,\eta_{s})$, the result follows. ∎

Now we are ready to complete the proof of Theorem 1.9. Let $a$ be a TRZ of
$U$, so that (4.6) holds for some choice of roots of unity $\xi_{i}$. Let $I$
be a minimal non-empty subset of ${\\{1,\ldots,s\\}}$ such that ${\sum_{i\in
I}\xi_{i}f_{i}(a)\lambda_{i}^{a}=0}$. Then $a$ is a primitive TRZ of the LRS
$U_{I}$, defined by

$U_{I}(n):=\sum_{i\in I}f_{i}(n)\lambda_{i}^{n}.$ (4.10)

Corollary 4.6 tells us that $U_{I}$ may have at most finitely many primitive
TRZs. Since there are finitely many possible $I$, Theorem 1.9 is proved.

###  4.5 An explicit result for ${\mathbb{Q}}$-valued linear recurrence
sequences

Let $U$ be an LRS over a field of characteristic $0$ with Binet expansion
(4.4), and let $a$ be a rational number. If $a$ is a common root of the
polynomials ${f_{1},\ldots,f_{s}}$, then it is, clearly, a rational zero of
$U$. Such rational zeros will be called trivial; there are only finitely many
of them, and in many interesting cases (for instance, if at least one of
${f_{1},\ldots,f_{s}}$ is constant) there are none.

Arguing as at the end of Subsection 4.4, we obtain the following: the
denominator of a non-trivial TRZ $a$ of $U$ is bounded in terms of $d$, $\rho$
and $s$. Indeed, if ${f_{i}(a)\neq 0}$ for some $i$, then there exists a set
${I\subset\\{1,\ldots,s\\}}$ having at least $2$ elements such that the LRS
$U_{I}$, defined in (4.10), has $a$ as a primitive TRZ. Now Proposition 4.5
implies that the denominator of $a$ is bounded in terms of $d$, $\rho$ and
$s$.

Note that ${s\leq m}$, where $m$ denoted the order of the LRS $U$. Hence the
denominator of $a$ is bounded in terms of $d$, $\rho$ and $m$.

In the most interesting special case when $U$ is a ${\mathbb{Q}}$-valued LRS
of order $m$, we can take ${\mathbb{K}}$ in Proposition 4.5 as the splitting
field of the characteristic polynomial of $U$. With this choice of
${\mathbb{K}}$, the degree $d$ is bounded in terms of $m$. Hence the
denominator of $a$ is bounded in terms of $m$ and $\rho$. We are going to make
it totally explicit.

######  Proposition 4.7.

Let $U$ be a ${\mathbb{Q}}$-valued LRS of order $m$ with Binet expansion
(4.4), ${\mathbb{K}}$ the splitting field of the characteristic polynomial of
$U$, and $\rho$ as in (4.5). Let $a$ be a non-trivial TRZ of $U$. Then the
denominator of $a$ does not exceed ${\rho\exp\exp(m/\log m)}$.

###### Proof.

As we have just seen, if $a$ is a non-trivial TRZ of $U$, then there exists
${I\subset\\{1,\ldots,s\\}}$ with ${\\#I\geq 2}$ such that $a$ is a primitive
TRZ of $U_{I}$. Proposition 4.5 implies that the denominator of $a$ is bounded
by $\rho\Lambda$. And we have ${\Lambda\leq\exp\exp(m/\log m)}$, see [2,
Proposition 6.5]. This completes the proof. ∎

##  5 Twisted rational zeros of the Tribonacci sequence

In this section we prove Theorem 1.10. As in the proof of Theorem 1.9, the
principal part is bounding the denominators of the TRZ, see Proposition 5.2.
Instead of adapting the general argument of Section 4, using the Theorems of
Chevalley-Bass and of Dvornicich-Zannier, we use an elementary ad hoc
argument.

In this section we denote by ${\overline{{\mathbb{Q}}}\subset{\mathbb{C}}}$
the field of all complex algebraic numbers, and by ${x\mapsto\overline{x}}$
the complex conjugation.

###  5.1 The roots

Let ${\lambda_{1},\lambda_{2},\lambda_{3}\in\overline{{\mathbb{Q}}}}$ be the
complex roots of the characteristic polynomial

${P(X):=X^{3}-X^{2}-X-1}.$

Then

$T(n)=\alpha_{1}\lambda_{1}^{n}+\alpha_{2}\lambda_{2}^{n}+\alpha_{3}\lambda_{3}^{n},\qquad\alpha_{i}:=P^{\prime}(\lambda_{i})^{-1}\lambda_{i}.$

One of the roots ${\lambda_{1},\lambda_{2},\lambda_{3}}$ is real and the other
two are complex conjugate. We will assume that ${\lambda_{1}\in{\mathbb{R}}}$
and ${\lambda_{3}=\overline{\lambda_{2}}}$.

We denote ${{\mathbb{K}}_{i}:={\mathbb{Q}}(\lambda_{i})}$ and
${{\mathbb{L}}:={\mathbb{Q}}(\lambda_{1},\lambda_{2})}$, the splitting field
of $P$. We have ${[{\mathbb{K}}_{i}:{\mathbb{Q}}]=3}$ and
${[{\mathbb{L}}:{\mathbb{Q}}]=6}$. The maximal abelian subfield
${\mathbb{L}}_{\mathrm{ab}}$ is ${\mathbb{Q}}(\sqrt{-11})$, because the
discriminant of $P$ is $-44$. The $6$ numbers

$\lambda_{i}/\lambda_{j}\qquad(1\leq i\neq j\leq 3)$ (5.1)

form a full Galois orbit over ${\mathbb{Q}}$; in particular, for ${i\neq j}$
we have ${{\mathbb{L}}={\mathbb{Q}}(\lambda_{i}/\lambda_{j})}$.

In the following proposition we collect some less obvious properties of the
roots $\lambda_{i}$, to be used later.

######  Proposition 5.1.

  1. 1.

For ${i\neq j}$, the quotient
${P^{\prime}(\lambda_{i})/P^{\prime}(\lambda_{j})}$ is not a Dirichlet unit.

  2. 2.

Let $\lambda$ be one of ${\lambda_{1},\lambda_{2},\lambda_{3}}$ and
${{\mathbb{K}}:={\mathbb{Q}}(\lambda)}$. Then $\lambda$ is not an
$m^{\text{th}}$ power in ${\mathbb{K}}$ for any integer ${m>1}$, and neither
is ${-\lambda}$. In terms of the Kummer exponent, defined in Subsection 4.1,
this can be stated as ${\varrho_{{\mathbb{K}}}(\lambda)=1}$.

  3. 3.

The quotients $\lambda_{i}/\lambda_{j}$ are cubes in ${\mathbb{L}}$. More
precisely, for ${1\leq i,j\leq 3}$ there exists a unique
${\theta_{ij}\in{\mathbb{L}}}$ such that
${\lambda_{i}/\lambda_{j}=\theta_{ij}^{3}}$.

###### Proof.

If, say, ${P^{\prime}(\lambda_{1})/P^{\prime}(\lambda_{2})}$ is a unit, then
so is every ${P^{\prime}(\lambda_{i})/P^{\prime}(\lambda_{j})}$. It follows
that the $3$ algebraic integers $P^{\prime}(\lambda_{i})$ generate the same
principal ideal in $\mathcal{O}_{\mathbb{L}}$. This ideal must divide the sum
$\sum_{i=1}^{3}P^{\prime}(\lambda_{i})=4$, which implies that the product
$\prod_{i=1}^{3}P^{\prime}(\lambda_{i})$ must be a power of $2$. But

$\prod_{i=1}^{3}P^{\prime}(\lambda_{i})=44,$ (5.2)

a contradiction. This proves item 1.

In the proof of item 2 we will use the notion of the absolute logarithmic
height $\operatorname{h}(\cdot)$ of an algebraic number. The definition can be
found in many sources, say, in [6, Section 1.5.7]. We will need only the
following properties: if $\gamma$ is an algebraic integer of degree $d$ with
conjugates ${\gamma_{1},\ldots,\gamma_{d}\in{\mathbb{C}}}$, and $m$ is a
positive integer, then
$d\operatorname{h}(\gamma)=\sum_{i=1}^{d}\max\\{\log|\gamma_{i}|,0\\}$, and
${\operatorname{h}(\gamma^{m})=m\operatorname{h}(\gamma)}$. In particular,

$3\operatorname{h}(\lambda)=\log\lambda_{1}<0.61,$

because ${\lambda_{1}>1}$ and ${|\lambda_{2}|=|\lambda_{3}|<1}$.

Now assume that ${\lambda=\gamma^{m}}$ for some ${\gamma\in{\mathbb{K}}}$ and
${m>1}$. Then ${\operatorname{h}(\lambda)=m\operatorname{h}(\gamma)}$. On the
other hand, the famous result of Smyth [21] implies that

$3\operatorname{h}(\gamma)\geq\log\vartheta>0.28.$

where $\vartheta$ is the real root of the polynomial ${X^{3}-X-1}$. Hence

$m\leq\frac{\log\lambda}{\log\vartheta}<2.2.$

It follows that ${m=2}$. Hence $\gamma$ is a root of the polynomial
${P(X^{2})}$. However, this polynomial is irreducible over ${\mathbb{Q}}$,
which means that $\gamma$ is of degree $6$, a contradiction. In a similar
fashion one shows that $-\lambda$ is not a proper power in ${\mathbb{K}}$.
This proves item 2.

In item 3 uniqueness is clear, because ${\zeta_{3}\notin{\mathbb{L}}}$, so we
only have to prove existence. Using PARI [22] (or another similar tool), we
calculate the $X$-resultant of the polynomials $P(X)$ and $P(XY)$. It is a
polynomial in $Y$ of degree $9$, whose roots are exactly the $9$ quotients
$\lambda_{i}/\lambda_{j}$. It has a root $1$ of multiplicity $3$, which
corresponds to the $3$ quotients $\lambda_{i}/\lambda_{i}$, and it factors as
${(Y-1)^{3}R(Y)}$, where

$R(Y):=Y^{6}+4Y^{5}+11Y^{4}+12Y^{3}+11Y^{2}+4Y+1$

is the irreducible polynomial whose roots are the quotients (5.1).

Polynomial $R(Y^{3})$ is reducible over ${\mathbb{Q}}$: it has an irreducible
factor of degree $6$

$Q(Y):=Y^{6}+Y^{5}+2Y^{4}+3Y^{3}+2Y^{2}+Y+1$

and another irreducible factor of degree $12$. Let $\theta$ be a root of $Q$.
Then the field ${\mathbb{Q}}(\theta)$ is of degree $6$, and $\theta^{3}$ is
one of the quotients (5.1); in particular, ${\mathbb{Q}}(\theta)$ contains
${\mathbb{L}}$. By equality of degrees we obtain
${{\mathbb{Q}}(\theta)={\mathbb{L}}}$, which completes the proof of item 3. ∎

###  5.2 The denominator

In this subsection we prove that the denominator of a TRZ divides $3$.

######  Proposition 5.2.

Let $a$ be a TRZ of the Tribonacci LRS. Then ${3a\in{\mathbb{Z}}}$.

We will use the following very simple lemma.

######  Lemma 5.3.

Let ${\gamma\in{\mathbb{C}}}$ be a complex number such that
${\gamma/\overline{\gamma}}$ is not a root of unity, and
${\delta\in{\mathbb{R}}}$ a real number. Then the equation
${\gamma\eta+\overline{\gamma}\eta^{\prime}=\delta}$ may have at most one
solution in roots of unity $(\eta,\eta^{\prime})$. This solution satisfies
${\eta\in{\mathbb{Q}}(\gamma,\overline{\gamma},\delta)}$ and
${\eta^{\prime}=\overline{\eta}}$.

###### Proof.

If ${\eta,\eta^{\prime}}$ are roots of unity such that
${\gamma\eta+\overline{\gamma}\eta^{\prime}\in{\mathbb{R}}}$, then the complex
numbers $\overline{\gamma}\eta^{\prime}$ and
$\overline{\gamma}\overline{\eta}$ have the same imaginary part:
${\overline{\gamma}\eta^{\prime}-\gamma\overline{\eta^{\prime}}=\overline{\gamma}\overline{\eta}-\gamma\eta}$.
If ${\eta\neq\overline{\xi}}$ then

$\frac{\gamma}{\overline{\gamma}}=\frac{\eta^{\prime}-\overline{\eta}}{\overline{\eta^{\prime}}-\eta}=-\frac{\eta^{\prime}}{\eta},$

contradicting the hypothesis that ${\gamma/\overline{\gamma}}$ is not a root
of unity. Hence ${\eta^{\prime}=\overline{\eta}}$.

Thus, $\eta$ is a root of the polynomial

$F(X):=X^{2}-(\delta/\gamma)X+\overline{\gamma}/\gamma\in{\mathbb{Q}}(\gamma,\overline{\gamma},\delta)[X].$

Since the free term $\overline{\gamma}/\gamma$ is not a root of unity, the
other root of $F$ cannot be a root of unity. This proves that there may exist
only one possible $\eta$ for given $\gamma$ and $\delta$.

Furthermore, if $F$ is irreducible over
${\mathbb{Q}}(\gamma,\overline{\gamma},\delta)$ then its other root is a root
of unity as well, which is impossible, as we just saw. Hence $F$ is reducible,
which implies that ${\eta\in{\mathbb{Q}}(\gamma,\overline{\gamma},\delta)}$.
The lemma is proved. ∎

###### Proof of Proposition 5.2.

Let $m$ be the denominator of $a$; that is, ${a=n/m}$, where
${m,n\in{\mathbb{Z}}}$ are coprime and ${m>0}$. Let $\lambda_{1}^{1/m}$ be the
positive real $m^{\text{th}}$ root, $\lambda_{2}^{1/m}$ some complex
$m^{\text{th}}$ root, and we define $\lambda_{3}^{1/m}$ as the complex
conjugate of $\lambda_{2}^{1/m}$. With this choice of $m^{\text{th}}$ roots we
have

$\lambda_{1}^{1/m}\lambda_{2}^{1/m}\lambda_{3}^{1/m}=1.$ (5.3)

Once the $m^{\text{th}}$ roots are defined, the rational powers
$\lambda_{i}^{a}$ are well-defined as $(\lambda_{i}^{1/m})^{n}$. Note that
${\lambda_{3}^{a}=\overline{\lambda_{2}^{a}}}$.

Since $a$ is a TRZ, we have
${\alpha_{1}\lambda_{1}^{a}\eta_{1}+\alpha_{2}\lambda_{2}^{a}\eta_{2}+\alpha_{2}\lambda_{3}^{a}\eta_{3}=0}$
for some roots of unity ${\eta_{1},\eta_{2},\eta_{3}}$. We may clearly assume
that ${\eta_{1}=1}$, in which case Lemma 5.3 implies that
${\eta_{3}=\overline{\eta_{2}}}$. In the sequel we write $\eta_{2}$ as $\eta$
and $\eta_{3}$ as $\overline{\eta}$; that is, we have

$\alpha_{1}\lambda_{1}^{a}+\alpha_{2}\lambda_{2}^{a}\eta+\alpha_{3}\lambda_{3}^{a}\overline{\eta}=0.$
(5.4)

Recall that we denote ${{\mathbb{K}}_{1}={\mathbb{Q}}(\lambda_{1})}$. Fix an
element $\sigma$ in the absolute Galois group
${G_{1}:=\operatorname{Gal}(\overline{{\mathbb{Q}}}/{\mathbb{K}}_{1})}$. Then
${(\lambda_{1}^{1/m})^{\sigma}=\lambda_{1}^{1/m}\xi_{1}}$ for some
${\xi_{1}\in\mu_{m}}$. The group
${H:=\operatorname{Gal}(\overline{{\mathbb{Q}}}/{\mathbb{L}})}$ is an index
$2$ subgroup of $G_{1}$. If ${\sigma\in H}$ then
${\lambda_{2}^{\sigma}=\lambda_{2}}$ and ${\lambda_{3}^{\sigma}=\lambda_{3}}$,
in which case there exist ${\xi_{2},\xi_{3}\in\mu_{m}}$ such that

$(\lambda_{2}^{1/m})^{\sigma}=\lambda_{2}^{1/m}\xi_{2},\qquad(\lambda_{3}^{1/m})^{\sigma}=\lambda_{3}^{1/m}\xi_{3}.$
(5.5)

If ${\sigma\notin H}$ then ${\lambda_{2}^{\sigma}=\lambda_{3}}$ and
${\lambda_{3}^{\sigma}=\lambda_{2}}$; in this case there exist
${\xi_{2},\xi_{3}\in\mu_{m}}$ such that

$(\lambda_{2}^{1/m})^{\sigma}=\lambda_{3}^{1/m}\xi_{3},\qquad(\lambda_{3}^{1/m})^{\sigma}=\lambda_{2}^{1/m}\xi_{2}.$
(5.6)

Let us apply $\sigma$ to equalities (5.3) and (5.4). We obtain

$\displaystyle
1=1^{\sigma}=(\lambda_{1}^{1/m}\lambda_{2}^{1/m}\lambda_{3}^{1/m})^{\sigma}=\lambda_{1}^{1/m}\lambda_{2}^{1/m}\lambda_{3}^{1/m}\xi_{1}\xi_{2}\xi_{3}=\xi_{1}\xi_{2}\xi_{3},$
(5.7)
$\displaystyle\alpha_{1}^{\sigma}\lambda_{1}^{a}\xi_{1}^{n}+\alpha_{2}\lambda_{2}^{a}\xi_{2}^{n}\eta^{\prime}+\alpha_{3}\lambda_{3}^{a}\xi_{3}^{n}\eta^{\prime\prime}=0,\qquad(\eta^{\prime},\eta^{\prime\prime})=\begin{cases}(\eta^{\sigma},\overline{\eta}^{\sigma}),&\sigma\in
H,\\\ (\overline{\eta}^{\sigma},\eta^{\sigma}),&\sigma\notin H.\end{cases}$
(5.8)

Note that ${\alpha_{1}^{\sigma}=\alpha_{1}}$ because
${\alpha_{1}\in{\mathbb{K}}_{1}}$, and that
${\eta^{\prime}\eta^{\prime\prime}=(\eta\overline{\eta})^{\sigma}=1}$.

Rewrite (5.8) as

$\alpha_{1}\lambda_{1}^{a}+\alpha_{2}\lambda_{2}^{a}\left(\frac{\xi_{2}}{\xi_{1}}\right)^{n}\eta^{\prime}+\alpha_{3}\lambda_{3}^{a}\left(\frac{\xi_{3}}{\xi_{1}}\right)^{n}\eta^{\prime\prime}=0.$

Comparing this to (5.4), the uniqueness statement in Lemma 5.3 implies that

$\left(\frac{\xi_{2}}{\xi_{1}}\right)^{n}\eta^{\prime}=\eta,\qquad\left(\frac{\xi_{3}}{\xi_{1}}\right)^{n}\eta^{\prime\prime}=\overline{\eta}.$

Multiplying these equalities, we obtain
${(\xi_{2}\xi_{3}/\xi_{1}^{2})^{n}=1}$. Since
${\xi_{1},\xi_{2},\xi_{3}\in\mu_{m}}$ and $m,n$ are coprime, this implies that
${\xi_{2}\xi_{3}/\xi_{1}^{2}=1}$, which, together with (5.7), implies that
${\xi_{1}^{3}=1}$.

We have proved the following: for any ${\sigma\in G_{1}}$ there exist
${\xi_{1}\in\mu_{3}}$ such that
${(\lambda_{1}^{1/m})^{\sigma}=\lambda_{1}^{1/m}\xi_{1}}$. If ${\xi_{1}=1}$
for every ${\sigma\in G_{1}}$ then ${\lambda_{1}^{1/m}\in{\mathbb{K}}_{1}}$.
Now assume that ${\xi_{1}=\zeta_{3}}$ for some $\sigma$. Since
${\mathbb{K}}_{1}$ is a real field, any Galois orbit over ${\mathbb{K}}_{1}$
must be stable under the complex conjugation. Hence the Galois orbit of
$\lambda_{1}^{1/m}$ over ${\mathbb{K}}_{1}$ is
${\lambda_{1}^{1/m},\lambda_{1}^{1/m}\zeta_{3},\lambda_{1}^{1/m}\overline{\zeta_{3}}}$.

Thus, ${[{\mathbb{K}}_{1}(\lambda_{1}^{1/m}):{\mathbb{K}}_{1}]\in\\{1,3\\}}$.
On the other hand, item 2 of Proposition 5.1 implies that $\lambda_{1}$ is not
a $p^{\text{th}}$ power in ${\mathbb{K}}_{1}$ for any $p$, and
${-\lambda_{1}}$ is not a square in ${\mathbb{K}}_{1}$. Hence
${[{\mathbb{K}}_{1}(\lambda_{1}^{1/m}):{\mathbb{K}}_{1}]=m}$ by Lemma 2.1. It
follows that ${m\in\\{1,3\\}}$, as wanted. ∎

We also need to take care of the roots of unity occurring in the definition of
TRZ.

######  Proposition 5.4.

Let $a$ be a TRZ of the Tribonacci LRS. Then, with suitable definitions of the
rational powers $\lambda_{i}^{a}$ we have

$\displaystyle\alpha_{1}\lambda_{1}^{a}+\alpha_{2}\lambda_{2}^{a}+\alpha_{3}\lambda_{3}^{a}$
$\displaystyle=0,$ (5.9)
$\displaystyle\lambda_{1}^{a}\lambda_{2}^{a}\lambda_{3}^{a}$ $\displaystyle=1$
(5.10)

###### Proof.

We have ${a=n/3}$, with ${n\in{\mathbb{Z}}}$. We define $\lambda_{1}^{1/3}$ as
the real cubic root of $\lambda_{1}$, and for ${i=2,3}$ we define
${\lambda_{i}^{1/3}:=\lambda_{1}^{1/3}\theta_{i1}}$, where
${\theta_{ij}\in{\mathbb{L}}}$ are defined in item 3 of Proposition 5.1.

Note that ${\theta_{31}=\overline{\theta_{21}}}$. Indeed,
${\overline{\theta_{21}}^{3}=\overline{\lambda_{2}}/\overline{\lambda_{1}}=\lambda_{3}/\lambda_{1}}$.
Since $\lambda_{3}/\lambda_{1}$ has only one cubic root in ${\mathbb{L}}$, we
must have ${\theta_{31}=\overline{\theta_{21}}}$.

From our definitions it follows that
${\lambda_{1}^{a}\lambda_{2}^{a}\lambda_{3}^{a}}$ is a positive real number.
Since
${(\lambda_{1}^{a}\lambda_{2}^{a}\lambda_{3}^{a})^{3}=(\lambda_{1}\lambda_{2}\lambda_{3})^{n}=1}$,
this proves (5.10), so we are only left with (5.9).

As we have seen in the proof of Proposition 5.2, there exists a root of unity
$\eta$ such that (5.4) holds. This can be rewritten as
${\alpha_{1}+\alpha_{2}\theta_{21}^{n}\eta+\alpha_{3}\theta_{31}^{n}\overline{\eta}=0}$.
Lemma 5.3 implies that
${\eta\in{\mathbb{Q}}(\alpha_{1},\alpha_{2},\alpha_{3},\theta_{21},\theta_{31})={\mathbb{L}}}$.
Since ${\mathbb{L}}$ contains no roots of unity other than ${\pm 1}$, we must
have ${\eta=1}$ or ${\eta=-1}$. In the former case we are done. Now let us
assume that ${\eta=-1}$. In this case

$\alpha_{1}-\alpha_{2}\theta_{21}^{n}-\alpha_{3}\theta_{31}^{n}=0.$ (5.11)

Let ${\sigma\in\operatorname{Gal}({\mathbb{L}}/{\mathbb{Q}})}$ be such that

$\lambda_{1}^{\sigma}=\lambda_{2},\qquad\lambda_{2}^{\sigma}=\lambda_{1},\qquad\lambda_{3}^{\sigma}=\lambda_{3}.$

Then

$\alpha_{1}^{\sigma}=\alpha_{2},\quad\alpha_{2}^{\sigma}=\alpha_{1},\quad\alpha_{3}^{\sigma}=\alpha_{3},\quad\theta_{21}^{\sigma}=\theta_{12}=\theta_{21}^{-1},\quad\theta_{31}^{\sigma}=\theta_{32}=\theta_{31}\theta_{21}^{-1}.$

Applying $\sigma$ to (5.11), we obtain
${\alpha_{2}-\alpha_{1}\theta_{21}^{-n}-\alpha_{3}\theta_{31}^{n}\theta_{21}^{-n}=0}$,
which can be rewritten as
${\alpha_{1}-\alpha_{2}\theta_{21}^{n}+\alpha_{3}\theta_{31}^{n}=0}$.
Comparing this with (5.11), we obtain ${\alpha_{3}\theta_{31}^{n}=0}$, a
contradiction. The proposition is proved. ∎

###  5.3 Proof of Theorem 1.10

Let $W$ be the ${\mathbb{Q}}$-valued LRS with the general term given by

$W(n)=44\alpha_{1}^{3}\lambda_{1}^{n}+44\alpha_{2}^{3}\lambda_{2}^{n}+44\alpha_{3}^{3}\lambda_{3}^{n}-3.$

It is an LRS of order $4$, defined by

$W(0)=W(1)=0,\quad W(2)=2,\quad W(3)=8,\qquad W(n+4)=2W(n+3)-W(n).$

######  Proposition 5.5.

Let ${a\in{\mathbb{Q}}}$ be a TRZ of the Tribonacci LRS. Then $3a$ is a zero
of $W$.

###### Proof.

As we have seen in Proposition 5.2, if $a$ is a TRZ of the Tribonacci LRS,
then ${n:=3a\in{\mathbb{Z}}}$, so we only need to prove that ${W(n)=0}$.

Define the rational powers $\lambda_{i}^{a}$ as in Proposition 5.4, so that
both (5.9) and (5.10) hold. Using (5.2), we also find

$\alpha_{1}\alpha_{2}\alpha_{3}=\frac{\lambda_{1}\lambda_{2}\lambda_{3}}{P^{\prime}(\lambda_{1})P^{\prime}(\lambda_{2})P^{\prime}(\lambda_{3})}=\frac{1}{44}.$
(5.12)

Consider the polynomial

$F(X_{1},X_{2},X_{3})=X_{1}^{3}+X_{2}^{3}+X_{3}^{3}-3X_{1}X_{2}X_{3}\in{\mathbb{Z}}[X_{1},X_{2},X_{3}].$

Then
${44F(\alpha_{1}\lambda_{1}^{a},\alpha_{2}\lambda_{2}^{a},\alpha_{3}\lambda_{3}^{a})=W(n)}$,
because
${\alpha_{1}\lambda_{1}^{a}\alpha_{2}\lambda_{2}^{a}\alpha_{3}\lambda_{3}^{a}=1/44}$,
as follows from (5.10) and (5.12). The polynomial $F$ factors as

$F(X_{1},X_{2},X_{3})=(X_{1}+X_{2}+X_{3})(X_{1}+\zeta_{3}X_{2}+\overline{\zeta_{3}}X_{3})(X_{1}+\overline{\zeta_{3}}X_{2}+\zeta_{3}X_{3}),$

which implies that
${F(\alpha_{1}\lambda_{1}^{a},\alpha_{2}\lambda_{2}^{a},\alpha_{3}\lambda_{3}^{a})=0}$
by (5.9). This completes the proof. ∎

######  Proposition 5.6.

The only zeros of $W$ are ${-51,-12,-5,-3,0,1}$.

###### Proof.

In [3] an algorithm is suggested which, when terminates, produces the full set
of zeros of a given non-degenerate LRS, together with a mathematically
rigorous proof that no other zeros exist. This algorithm is implemented, for
simple111An LRS is called simple if its characteristic polynomial has only
simple roots. non-degenerate ${\mathbb{Q}}$-valued LRS, in the Skolem Tool
[4]. Running the Skolem Tool for the LRS $W$, we obtain the result. ∎

We know (see [5, Section 2]) that $-17,-4,-5/3,-1,0,1/3$ are indeed TRZs of
the Tribonacci LRS. Hence Theorem 1.10 is an immediate consequence of
Propositions 5.5 and 5.6.

##  6 On Question 1.7

In this section we discuss Question 1.7. We will see that the answer is
positive for twisted (integral) zeros of LRS of order $2$, but (in general)
not for higher order LRS. Unless the contrary is stated explicitly, the letter
$p$ in this section denotes a prime number.

For twisted (integral) zeros of LRS of order $2$ we not only answer Question
1.7, but obtain a partial analog of Theorem 3.4.

######  Theorem 6.1.

Let $U$ be a non-degenerate LRS of order $2$ with values in a number field
${\mathbb{K}}$ and ${a\in{\mathbb{Z}}}$ a twisted zero of $U$. Then for
infinitely many primes $\mathfrak{p}$ of ${\mathbb{K}}$ the following holds:
there exist

$Q\in{\mathbb{Z}}_{>0},\qquad
a^{\prime}\in\\{0,1,\ldots,Q-1\\},\qquad\tau\in{\mathbb{Z}}$ (6.1)

such that for every ${n\in{\mathbb{Z}}}$ satisfying ${n\equiv
a^{\prime}\pmod{Q}}$ we have

$\nu_{\mathfrak{p}}(U(n))=\nu_{p}(n-a)+\tau,$

where $p$ is the rational prime below $\mathfrak{p}$. Moreover, ${p\nmid Q}$;
in particular, ${\nu_{p}(n-a)}$ is unbounded on the set of $n$ satisfying
${n\equiv a^{\prime}\pmod{Q}}$.

In fact, we show that this holds true for primes $\mathfrak{p}$ from a set of
positive lower density. Let us recall the definition of density. Denote by
$\pi_{\mathbb{K}}(x)$ the counting function for primes of ${\mathbb{K}}$; that
is, the number of primes $\mathfrak{p}$ of ${\mathbb{K}}$ such that
${\mathcal{N}\mathfrak{p}\leq x}$; here $\mathcal{N}\mathfrak{p}$ denotes the
absolute norm. Let $\mathcal{P}$ be a set of primes of ${\mathbb{K}}$. We
denote its lower density as
$\liminf_{x\to+\infty}{\\#\\{\mathfrak{p}\in\mathcal{P}:\mathcal{N}\mathfrak{p}\leq
x\\}}/{\pi_{\mathbb{K}}(x)}.$

###  6.1 A density result

The proof of Theorem 6.1 relies on a certain Chebotaryov222We prefer the
spelling Chebotaryov, as in [23], because it is more consistent with the
original Russian and Ukrainian pronunciation.-style density result. To state
it, let us introduce some more notation. In this subsection ${\mathbb{K}}$ is
a number field, unless stated otherwise.

Let ${\alpha\in{\mathbb{K}}^{\times}}$ and a prime $\mathfrak{p}$ of
${\mathbb{K}}$ be such that ${\nu_{\mathfrak{p}}(\alpha)=0}$. We denote by
$\operatorname{ord}_{\mathfrak{p}}(\alpha)$ the multiplicative order of
$\alpha$ modulo $\mathfrak{p}$. That is, let
${\mathcal{O}_{\mathfrak{p}}:=\\{x\in{\mathbb{K}}:\nu_{p}(x)\geq 0\\}}$ be the
local ring of $\mathfrak{p}$, and
${\mathcal{O}_{\mathfrak{p}}\to\mathcal{O}_{\mathfrak{p}}/\mathfrak{p}:x\mapsto\overline{x}}$
the reduction map333This is in conflict with the convention of Section 5,
where ${x\mapsto\overline{x}}$ denotes the complex conjugation. However, in
this section we never use complex conjugation, so there is no risk of
confusion. modulo $\mathfrak{p}$. Then
${\overline{\alpha}\in(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}}$,
and $\operatorname{ord}_{\mathfrak{p}}(\alpha)$ is the order of
$\overline{\alpha}$ in the multiplicative group
$(\mathcal{O}\mathfrak{p}/\mathfrak{p})^{\times}$.

For a positive integer $r$ we denote by $\mathcal{P}_{\mathbb{K}}(\alpha,r)$
the set of ${\mathbb{K}}$-primes $\mathfrak{p}$ such that the multiplicative
order $\operatorname{ord}_{\mathfrak{p}}(\alpha)$ is divisible by $r$. In
symbols:

$\mathcal{P}_{\mathbb{K}}(\alpha,r):=\\{\mathfrak{p}:r\mid\operatorname{ord}_{\mathfrak{p}}(\alpha)\\}.$

######  Proposition 6.2.

Let $r$ be a positive integer with the property

$\text{$\zeta_{p}\in{\mathbb{K}}$ for every ${p\mid r}$}.$ (6.2)

Let ${\alpha\in{\mathbb{K}}^{\times}}$ be not a root of unity. Then the set
$\mathcal{P}_{\mathbb{K}}(\alpha,r)$ is infinite, and, moreover, it is of
positive lower density.

The proof of Proposition 6.2 depends on the following lemma.

######  Lemma 6.3.

Let $p$ be a prime number and ${\mathbb{K}}$ a field of characteristic
distinct from $p$. Define $\ell$ as the biggest integer such that
${\zeta_{p^{\ell}}\in{\mathbb{K}}}$. Assume that ${\ell\geq 1}$. Let
${\alpha\in{\mathbb{K}}}$ be such that
${\alpha^{1/p}\in{\mathbb{K}}(\zeta_{p^{\ell+1}})}$. Then
${\alpha\in\mu_{p^{\ell}}{\mathbb{K}}^{p}}$.

###### Proof.

We may assume that ${\alpha\neq 0}$, since there is nothing to prove
otherwise. We use Kummer’s Theory, as in Theorem 8.1 from [13, Chapter VI].
Let $B$ be the subgroup of the multiplicative group ${\mathbb{K}}^{\times}$
generated by $\alpha$, $\mu_{p^{\ell}}$ and $({\mathbb{K}}^{\times})^{p}$. By
the hypothesis, ${{\mathbb{K}}(B^{1/p})={\mathbb{K}}(\zeta_{p^{\ell+1}})}$.
The above-mentioned theorem implies that

$[B:({\mathbb{K}}^{\times})^{p}]=[{\mathbb{K}}(B^{1/p}):{\mathbb{K}}]=[{\mathbb{K}}(\zeta_{p^{\ell+1}}):{\mathbb{K}}]=p.$

Since
${[\mu_{p^{\ell}}({\mathbb{K}}^{\times})^{p}:({\mathbb{K}}^{\times})^{p}]=p}$,
this proves that ${B=\mu_{p^{\ell}}({\mathbb{K}}^{\times})^{p}}$, which
exactly means that ${\alpha\in\mu_{p^{\ell}}({\mathbb{K}}^{\times})^{p}}$. ∎

###### Proof of Proposition 6.2.

If the statement holds true with ${\mathbb{K}}$ replaced by a bigger field,
then it is true for ${\mathbb{K}}$. Indeed, let ${\mathbb{K}}^{\prime}$ be a
finite extension of ${\mathbb{K}}$. If $\mathfrak{p}$ is a
${\mathbb{K}}$-prime, and $\mathfrak{p}^{\prime}$ is a
${\mathbb{K}}^{\prime}$-prime above $\mathfrak{p}$, then for any
${\alpha\in{\mathbb{K}}^{\times}}$ we have
${\operatorname{ord}_{\mathfrak{p}^{\prime}}(\alpha)=\operatorname{ord}_{\mathfrak{p}}(\alpha)}$;
so, it suffices to to show that the set
$\mathcal{P}_{{\mathbb{K}}^{\prime}}(\alpha,r)$ is of positive lower density.
Thus, we may assume that ${\sqrt{-1}\in{\mathbb{K}}}$.

Let $m$ be the order of the group of roots of unity $\mu_{\mathbb{K}}$.
Condition (6.2) may be re-stated as ${p\mid r\Rightarrow p\mid m}$.

If the conclusion of the proposition holds true with $r$ replaced a multiple
of $r$, then it holds for $r$. Hence we may replace $r$ by ${\prod_{p\mid
r}p^{\max\\{\nu_{p}(r),\nu_{p}(m)+1\\}}}$ and assume in the sequel that

$\nu_{p}(r)>\nu_{p}(m)\qquad(p\mid m);$ (6.3)

in particular, ${p\mid r\Leftrightarrow p\mid m}$.

Next, we may replace $\alpha$ by $\alpha\xi$, where ${\xi\in{\mathbb{K}}}$ is
a root of unity. Indeed, if
${r\mid\operatorname{ord}_{\mathfrak{p}}(\alpha)}$, then
${\nu_{p}\bigl{(}\operatorname{ord}_{\mathfrak{p}}(\xi)\bigr{)}<\nu_{p}\bigl{(}\operatorname{ord}_{\mathfrak{p}}(\alpha)\bigr{)}}$
for every ${p\mid\operatorname{ord}_{\mathfrak{p}}(\xi)}$, by (6.3). Hence
${\operatorname{ord}_{\mathfrak{p}}(\alpha\xi)=\operatorname{ord}_{\mathfrak{p}}(\alpha)}$.

Finally, we may assume that

$\alpha\notin\mu_{\mathbb{K}}{\mathbb{K}}^{p}\qquad(p\mid r).$ (6.4)

Indeed, call $\alpha$ with property (6.4) $r$-reduced. Since $\alpha$ is not a
root of unity, we have ${\alpha\xi=\beta^{N}}$, where $\beta$ is $r$-reduced,
$N$ is composed of primes dividing $r$ and $\xi$ is a root of unity. Now, if
${Nr\mid\operatorname{ord}_{\mathfrak{p}}(\beta)}$ then
${r\mid\operatorname{ord}_{\mathfrak{p}}(\alpha\xi)=\operatorname{ord}_{\mathfrak{p}}(\alpha)}$.
Hence we may assume (6.4), replacing $\alpha$ by $\beta$ and $r$ by $Nr$.

Denote ${{\mathbb{L}}:={\mathbb{K}}(\zeta_{r})}$. Properties (6.2) and (6.3)
imply that

$\operatorname{Gal}({\mathbb{L}}/{\mathbb{K}})\cong\prod_{p\mid
m}{\mathbb{Z}}/p^{\nu_{p}(r/m)}{\mathbb{Z}}.$

In particular, for every ${p\mid r}$, the extension ${\mathbb{L}}$ has exactly
one subfield of degree $p$ over ${\mathbb{K}}$; precisely, it is
${{\mathbb{K}}(\zeta_{p^{\ell_{p}+1}})}$, where ${\ell_{p}:=\nu_{p}(m)}$.

Pick some value of the $r^{\text{th}}$ root $\alpha^{1/r}$. This would define
the roots $\alpha^{1/p}$ for every ${p\mid r}$. We claim that
${\alpha^{1/p}\notin{\mathbb{L}}}$ for any ${p\mid r}$. Indeed, in the
opposite case, ${\mathbb{K}}(\alpha^{1/p})$ would be a subfield of
${\mathbb{L}}$ of degree $p$ over ${\mathbb{K}}$. This would imply
${\alpha^{1/p}\in{\mathbb{K}}(\zeta_{p^{\ell_{p}+1}})}$, which, by Lemma 6.3,
implies that ${\alpha\in\mu_{p^{\ell_{p}}}{\mathbb{K}}^{p}}$, contradicting
(6.4).

Lemma 2.1, together with Remark 2.2, implies now that
${{\mathbb{M}}:={\mathbb{L}}(\alpha^{1/r})}$ is an extension of ${\mathbb{L}}$
of degree $r$. Moreover, since ${\zeta_{r}\in{\mathbb{L}}}$, extension
${\mathbb{M}}/{\mathbb{L}}$ is cyclic, and the map

$\sigma\mapsto\xi_{\sigma}:\frac{(\alpha^{1/r})^{\sigma}}{\alpha^{1/r}}$ (6.5)

defines an isomorphism of the groups
${H:=\operatorname{Gal}({\mathbb{M}}/{\mathbb{L}})}$ and $\mu_{r}$.

Since ${{\mathbb{L}}={\mathbb{K}}(\zeta_{r})}$, extension
${\mathbb{M}}/{\mathbb{K}}$ is Galois, and we denote
${G:=\operatorname{Gal}({\mathbb{M}}/{\mathbb{K}})}$. Then $H$ is the subgroup
of $G$ fixing $\zeta_{r}$.

Let $\mathfrak{p}$ be a ${\mathbb{K}}$-prime not dividing $r$ and satisfying
${\nu_{\mathfrak{p}}(\alpha)=0}$. In particular, $\mathfrak{p}$ does not
ramify in ${\mathbb{M}}$. For an ${\mathbb{M}}$-prime $\mathfrak{P}$ above
$\mathfrak{p}$, we denote by ${\phi_{\mathfrak{P}}\in G}$ the Frobenius of
$\mathfrak{P}$ above ${\mathbb{K}}$. Recall that $\phi_{\mathfrak{P}}$ is the
element of $G$ with the following property: let
${\mathcal{O}_{\mathfrak{P}}:=\\{x\in{\mathbb{M}}:\nu_{\mathfrak{P}}(x)\geq
0\\}}$ be the local ring of $\mathfrak{P}$; then for every
${x\in\mathcal{O}_{\mathfrak{P}}}$ we have
${x^{\mathcal{N}_{\mathfrak{p}}}\equiv
x^{\phi_{\mathfrak{P}}}\pmod{\mathfrak{P}}}$ (as before, $\mathcal{N}(\cdot)$
denotes the absolute norm).

Next, let
${\bigl{(}({\mathbb{M}}/{\mathbb{K}})\big{/}\mathfrak{p}\bigr{)}:=\\{\phi_{\mathfrak{P}}:\mathfrak{P}\mid\mathfrak{p}\\}}$
be the Artin symbol of $\mathfrak{p}$. Note that it is a full conjugacy class
in $G$. Denote by $\Sigma$ the subset of $H$ consisting of the elements of
exact order $r$. In symbols: ${\Sigma:=\\{\sigma\in
H:H=\langle\sigma\rangle\\}}$. Let $\mathfrak{p}$ be such that
${\bigl{(}({\mathbb{M}}/{\mathbb{K}})\big{/}\mathfrak{p}\bigr{)}\subset\Sigma}$.
By the Theorem of Chebotaryov, the set of such $\mathfrak{p}$ is of positive
density444Actually, it is of density $\\#\Sigma/\\#G$, but the exact value of
the density is not relevant to us., so we only have to prove that
${r\mid\operatorname{ord}_{\mathfrak{p}}(\alpha)}$.

We are going to prove that ${r\mid\mathcal{N}\mathfrak{p}-1}$, but
$\overline{\alpha}$ is not a $p^{\text{th}}$ power in
$(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}$ for any ${p\mid r}$.
Since ${\mathcal{N}\mathfrak{p}-1}$ is the order of the cyclic group
$(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}$, the order of
$\overline{\alpha}$ in this group must be divisible by $r$, as wanted.

As before, let $\mathfrak{P}$ be an ${\mathbb{M}}$-prime above $\mathfrak{p}$.
Since ${\phi_{\mathfrak{P}}\in\Sigma\subset H}$, we have
${\phi_{\mathfrak{P}}(\zeta_{r})=\zeta_{r}}$, which means that
${\zeta_{r}^{\mathcal{N}\mathfrak{p}}\equiv\zeta_{r}\pmod{\mathfrak{P}}}$.
Since ${\mathfrak{P}\nmid r}$, this implies that
${\zeta_{r}^{\mathcal{N}\mathfrak{p}}=\zeta_{r}}$, that is,
${r\mid\mathcal{N}\mathfrak{p}-1}$.

We are left with proving that $\overline{\alpha}$ is not a $p^{\text{th}}$
power in $(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}$ for any prime
$p$ dividing $r$. Fix such $p$. Since ${\phi_{\mathfrak{P}}\in\Sigma}$, it is
of exact order $r$ in $H$. Hence ${\xi:=\xi_{\phi_{\mathfrak{P}}}}$ (as
defined in (6.5)) is a primitive $r^{\text{th}}$ root of unity, and
${\eta:=\xi^{r/p}}$ is a primitive $p^{\text{th}}$ root of unity.

Applying (6.5) with ${\sigma=\phi_{\mathfrak{P}}}$, we obtain
${\alpha^{(\mathcal{N}\mathfrak{p}-1)/r}\equiv\xi\pmod{\mathfrak{P}}}$.
Raising this congruence to the power $r/p$, we obtain
${\alpha^{(\mathcal{N}\mathfrak{p}-1)/p}\equiv\eta\pmod{\mathfrak{P}}}$. Since
both sides of this congruence belong to ${\mathbb{K}}$, it is actually a
congruence modulo $\mathfrak{p}$, and it implies the identity
${\overline{\alpha}^{(\mathcal{N}\mathfrak{p}-1)/p}=\overline{\eta}}$ in the
group $(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}$. If
$\overline{\alpha}$ is a $p^{\text{th}}$ power in
$(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}$ then
${\overline{\alpha}^{(\mathcal{N}\mathfrak{p}-1)/p}=1}$, which is impossible
because $\overline{\eta}$ is a primitive $p^{\text{th}}$ root of unity. This
completes the proof of the proposition. ∎

###  6.2 Proof of Theorem 6.1

Besides Proposition 6.2, the proof of Theorem 6.1 relies on the following
lemma. It is very classical and goes back to the work of Lucas [15] or even
earlier. Still, we give a proof for convenience.

######  Lemma 6.4.

Let ${\mathbb{K}}$ be a number field, $\mathfrak{p}$ a prime of ${\mathbb{K}}$
with underlying rational prime $p$ and ${e:=\nu_{\mathfrak{p}}(p)}$ the
ramification index. Let ${\theta\in{\mathbb{K}}}$ be such that
${\nu_{\mathfrak{p}}(\theta-1)>0}$ and ${n\in{\mathbb{Z}}}$. Then we have the
following.

  1. 1.

If ${p\nmid n}$ then
${\nu_{\mathfrak{p}}(\theta^{n}-1)=\nu_{\mathfrak{p}}(\theta-1)}$.

  2. 2.

In general,
${\nu_{\mathfrak{p}}(\theta^{n}-1)\geq\nu_{\mathfrak{p}}(\theta-1)+\nu_{p}(n)\min\\{e,(p-1)\nu_{\mathfrak{p}}(\theta-1)\\}}$.

  3. 3.

If ${\nu_{\mathfrak{p}}(\theta-1)>e/(p-1)}$ then
${\nu_{\mathfrak{p}}(\theta^{n}-1)=\nu_{\mathfrak{p}}(\theta-1)+e\nu_{p}(n)}$.

###### Proof.

Item 1 must be proved only for ${n>0}$ and ${n=-1}$. If ${n>0}$ then

$\theta^{n}-1=(\theta-1)(\theta^{n-1}+\cdots+1).$

Since ${\theta\equiv 1\pmod{\mathfrak{p}}}$ in the local ring
$\mathcal{O}_{\mathfrak{p}}$, we have

$\theta^{n-1}+\cdots+1\equiv n\pmod{\mathfrak{p}}.$

In particular, ${\nu_{\mathfrak{p}}(\theta^{n-1}+\cdots+1)=0}$ if ${p\nmid
n}$. This proves item 1 for ${n>0}$. As for ${n=-1}$, it is obvious that
${\nu_{\mathfrak{p}}(\theta^{-1}-1)=\nu_{\mathfrak{p}}(\theta-1)}$.

Due to item 1, in items 2 and 3 we may assume that ${n=p^{k}}$. Moreover,
using induction in $k$, we may assume that ${n=p}$. Write ${\theta=1+\gamma}$.
Then

$\theta^{p}-1=p\gamma+\sum_{\ell=2}^{p-1}\binom{p}{\ell}\gamma^{\ell}+\gamma^{p}.$

Each of the terms inside the sum has $\mathfrak{p}$-adic valuation strictly
bigger than ${\nu_{\mathfrak{p}}(p\gamma)=\nu_{\mathfrak{p}}(\gamma)+e}$.
Hence

$\nu_{\mathfrak{p}}(\theta^{p}-1)\geq\nu_{\mathfrak{p}}(\gamma)+\min\\{e,(p-1)\nu_{\mathfrak{p}}(\gamma)\\}$
(6.6)

which proves item 2. Finally, when ${\nu_{\mathfrak{p}}(\gamma)>e/(p-1)}$,
inequality (6.6) becomes equality, and the minimum on the right is $e$, which
proves item 3. ∎

Now we are ready to prove Theorem 6.1.

###### Proof of Theorem 6.1.

By shifting, we may assume that ${a=0}$. Thus, we have to prove that, for
infinitely many ${\mathbb{K}}$-primes $\mathfrak{p}$ the following holds:
there exist $Q$, $a^{\prime}$ and $\tau$ as in the statement of the theorem
such that

$\nu_{\mathfrak{p}}(U(n))=\nu_{\mathfrak{p}}(n)+\tau\qquad\text{when}\quad
n\equiv a^{\prime}\pmod{Q}.$ (6.7)

If the statement holds true with ${\mathbb{K}}$ replaced by a bigger number
field, then it holds for ${\mathbb{K}}$. Hence we may assume that the roots of
the characteristic polynomial of $U$ belong to ${\mathbb{K}}$. Multiplying
$U(n)$ by $\beta\theta^{n}$ with suitable
${\beta,\theta\in{\mathbb{K}}^{\times}}$, we may assume that either
${U(n)=n-\beta}$, where ${\beta\in{\mathbb{K}}}$, or
${U(n)=\eta\lambda^{n}-1}$, where ${\eta,\lambda\in K^{\times}}$. If
${U(n)=n-\beta}$ and $0$ is a twisted zero of $U$, then ${\beta=0}$ and
${U(n)=n}$, so there is nothing to prove.

Now let us assume that ${U(n)=\eta\lambda^{n}-1}$. Note that $\lambda$ is not
a root of unity, because $U$ is non-degenerate. Since $0$ is a twisted zero of
$U$, we have ${\eta\lambda^{0}-\xi=0}$ for some root of unity $\xi$. Hence
$\eta$ is a root of unity. Denote by $r$ its order.

Let $\mathcal{P}^{\prime}_{\mathbb{K}}(\lambda,r)$ be the subset of
$\mathcal{P}_{\mathbb{K}}(\lambda,r)$, consisting of
${\mathfrak{p}\in\mathcal{P}_{\mathbb{K}}(\lambda,r)}$, unramified and of
degree $1$ over ${\mathbb{Q}}$, and with underlying prime
${\mathcal{N}\mathfrak{p}>2}$. Proposition 6.2 implies that the set
$\mathcal{P}_{\mathbb{K}}(\lambda,r)$ is of positive lower density. Hence so
is $\mathcal{P}^{\prime}_{\mathbb{K}}(\lambda,r)$: this is because the set of
unramified primes of degree $1$ is of density $1$. We claim that for every
${\mathfrak{p}\in\mathcal{P}^{\prime}_{\mathbb{K}}(\lambda,r)}$, there exist
$Q$, $a^{\prime}$ and $\tau$ as in (6.1) such that (6.7) holds, and ${p\nmid
Q}$.

Thus, fix ${\mathfrak{p}\in\mathcal{P}^{\prime}_{\mathbb{K}}(\lambda,r)}$.
Since $\mathfrak{p}$ is of degree $1$, the cyclic group
${(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}}$ is of order ${p-1}$,
where ${p:=\mathcal{N}\mathfrak{p}}$ is an odd prime number. Since
${r\mid\operatorname{ord}_{\mathfrak{p}}(\lambda)}$, the subgroup
$\langle\overline{\lambda}\rangle$ of
${(\mathcal{O}_{\mathfrak{p}}/\mathfrak{p})^{\times}}$ contains
$\overline{\eta}$. Hence there exists ${s\in\\{1,\ldots,p-1\\}}$ such that
${\eta\lambda^{s}\equiv 1\pmod{\mathfrak{p}}}$. Lemma 6.4 implies that

$\nu_{\mathfrak{p}}\bigl{(}(\eta\lambda^{s})^{m}-1\bigr{)}=\nu_{p}(m)+\nu_{\mathfrak{p}}(\xi\lambda^{s}-1)\qquad(m\in{\mathbb{Z}}).$

Now note that, when ${m\equiv 1\pmod{r}}$, we have
${(\xi\lambda^{s})^{m}-1=U(sm)}$. Hence (6.7) holds with

$Q:=rs,\qquad a^{\prime}:=s,\qquad\tau:=\nu_{\mathfrak{p}}(\xi\lambda^{s}-1).$

Note also that ${p\nmid Q}$ because ${r\mid p-1}$ and ${1\leq s\leq p-1}$. The
theorem is proved. ∎

######  Remark 6.5.

  1. 1.

Our argument can be illustrated with the LRS from Example 1.3. In that case

${\mathbb{K}}={\mathbb{Q}},\qquad\lambda=2,\qquad\eta=-1,\qquad r=2,\qquad
s=\frac{p-1}{2},$

and the set $\mathcal{P}^{\prime}_{\mathbb{Q}}(2,2)$ contains all the primes
satisfying ${p\equiv\pm 3\pmod{8}}$ (and some other primes as well).

  2. 2.

This argument does not extend to rational $a$, because we can no longer do the
shifting and assume that ${a=0}$. We do not know whether Theorem 6.1 can be
extended to twisted rational zeros.

  3. 3.

The non-degeneracy hypothesis cannot be dropped. Indeed, consider

$U(n):=\zeta_{3}^{n}+\overline{\zeta_{3}}^{n}=\begin{cases}2,&\text{if $3\mid
n$},\\\ -1,&\text{if $3\nmid n$}.\end{cases}$

Then $0$ is a twisted zero of $U$, but ${\nu_{p}(U(n))=0}$ for all $n$ and all
${p\neq 2}$.

###  6.3 Concluding remarks

As we already indicated in the introduction, the answer to Question 1.7 is
negative for LRS of order $3$ or higher. Here is an example of order $3$, but
one can easily construct similar examples of any order.

######  Example 6.6.

Let $p$ be a prime number. The LRS ${U(n):=8^{n}+2^{n}+1}$ has a twisted zero
at $0$, because ${1+\zeta_{3}+\overline{\zeta_{3}}=0}$. However, there does
not exist a sequence $(n_{k})$ such that

$\nu_{p}(U(n_{k}))\to+\infty,\qquad\nu_{p}(n_{k})\to+\infty.$

Indeed, let $(n_{k})$ be such a sequence. Let ${\mathbb{K}}$ be the splitting
field of the polynomial ${X^{3}+X+1}$ and $\mathfrak{p}$ a prime of
${\mathbb{K}}$ above $p$. Then, replacing $(n_{k})$ by a subsequence, we have

$\nu_{\mathfrak{p}}(2^{n_{k}}-\alpha)\to+\infty,\qquad\nu_{\mathfrak{p}}(n_{k})\to+\infty$

for some root $\alpha$ of this polynomial. Theorem 3.5 implies that $0$ is a
twisted zero of the LRS ${2^{n}-\alpha}$, a contradiction, because $\alpha$ is
not a root of unity.

To conclude, let us ask one more question. Theorem 1.5 deals with just one
individual prime number. What happens if a sequence $(n_{k})$ satisfying (1.5)
can be found for all but finitely many primes? One may expect that in this
case $a$ is a genuine zero of $U$, not merely a TRZ.

######  Question 6.7.

Let $U$ be an LRS with values in a number field ${\mathbb{K}}$, and
${a\in{\mathbb{Q}}}$. Assume that for every ${\mathbb{K}}$-prime
$\mathfrak{p}$, with finitely many exceptions, there exist a sequence of
integers $(n_{k})$ (depending on $\mathfrak{p}$) such that

$\nu_{\mathfrak{p}}(U(n_{k}))\to+\infty,\qquad\nu_{\mathfrak{p}}(n_{k}-a)\to+\infty.$

Does it imply that either $a$ a trivial TRZ (as defined in Subsection 4.5) or
${a\in{\mathbb{Z}}}$ and ${U(a)=0}$?

Probably, “all but finitely many primes” can be replaced by “primes from a set
of density $1$”. But, as Example 1.3 shows, it is not enough to assume that
this holds for infinitely many primes, or even for primes from a set of
positive lower density.

### Acknowledgments

Yuri Bilu and Florian Luca were supported by the ANR project JINVARIANT. James
Worrell was supported by UKRI Fellowship EP/X033813/1.

While working on this project, Yuri Bilu enjoyed hospitality and support of
MPIM Bonn, where he was visiting in June 2023.

Florian Luca worked on this paper in 2023 during research visits at the MPI-
SWS, Saarbrücken, Germany and the Stellenbosch Institute for Advanced Studies,
Stellenbosch, South Africa. He thanks these institutions for their hospitality
and support.

Joël Ouaknine is also affiliated with Keble College, Oxford as emmy.network
Fellow, and supported by DFG grant 389792660 as part of TRR 248 (see
https://perspicuous-computing.science).

## References

  * [1] H. Bass, _A remark on an arithmetic theorem of Chevalley_ , Proc. Amer. Math. Soc. 16 (1965), 875–878. MR 184925
  * [2] Yuri Bilu, _The Chevalley-Bass Theorem_ , Essays in Analytic Number Theory In Honor of Helmut Maier’s 70th Birthday, to appear, arXiv:2305.05041 (2023).
  * [3] Yuri Bilu, Florian Luca, Joris Nieuwveld, Joël Ouaknine, David Purser, and James Worrell, _Skolem meets Schanuel_ , 47th International Symposium on Mathematical Foundations of Computer Science, LIPIcs. Leibniz Int. Proc. Inform., vol. 241, Schloss Dagstuhl. Leibniz-Zent. Inform., Wadern, 2022, pp. Art. No. 20, 15. MR 4481938
  * [4] Yuri Bilu, Florian Luca, Joris Nieuwveld, Joël Ouaknine, David Purser, and James Worrell, _The Skolem Tool_ , \https://skolem.mpi-sws.org, 2022, [Online; accessed 15-November-2023].
  * [5] Yuri Bilu, Florian Luca, Joris Nieuwveld, Jöel Ouaknine, and James Worrell, _On the p-adic zeros of the Tribonacci sequence_ , Math. Comp. (2023).
  * [6] Enrico Bombieri and Walter Gubler, _Heights in Diophantine geometry_ , New Mathematical Monographs, vol. 4, Cambridge University Press, Cambridge, 2006\. MR 2216774
  * [7] J. W. S. Cassels, _An embedding theorem for fields_ , Bull. Austral. Math. Soc. 14 (1976), 193–198, 479–480. MR 422221
  * [8] , _Local fields_ , London Mathematical Society Student Texts, vol. 3, Cambridge University Press, Cambridge, 1986. MR 861410
  * [9] Claude Chevalley, _Deux théorèmes d’arithmétique_ , J. Math. Soc. Japan 3 (1951), 36–44. MR 44570
  * [10] J. H. Conway and A. J. Jones, _Trigonometric Diophantine equations (On vanishing sums of roots of unity)_ , Acta Arith. 30 (1976), no. 3, 229–240. MR 422149
  * [11] Roberto Dvornicich and Umberto Zannier, _On sums of roots of unity_ , Monatsh. Math. 129 (2000), no. 2, 97–108. MR 1742911
  * [12] Fernando Q. Gouvêa, _$p$ -adic numbers_, Universitext, Springer, Cham, 2020. MR 4175370
  * [13] Serge Lang, _Algebra_ , third ed., Graduate Texts in Mathematics, vol. 211, Springer-Verlag, New York, 2002. MR 1878556
  * [14] Michel Laurent, _Équations diophantiennes exponentielles_ , Invent. Math. 78 (1984), no. 2, 299–327. MR 767195
  * [15] Edouard Lucas, _Theorie des Fonctions Numeriques Simplement Periodiques_ , Amer. J. Math. 1 (1878), no. 4, 289–321. MR 1505176
  * [16] Henry B. Mann, _On linear relations between roots of unity_ , Mathematika 12 (1965), 107–117. MR 191892
  * [17] Diego Marques and Tamás Lengyel, _The $2$-adic order of the Tribonacci numbers and the equation ${T_{n}=m!}$_, J. Integer Seq. 17 (2014), no. 10, Article 14.10.1, 8. MR 3275869
  * [18] M. Mignotte and N. Tzanakis, _Arithmetical study of recurrence sequences_ , Acta Arith. 57 (1991), no. 4, 357–364. MR 1109992
  * [19] W. H. Schikhof, _Ultrametric calculus_ , Cambridge Studies in Advanced Mathematics, vol. 4, Cambridge University Press, Cambridge, 2006, An introduction to $p$-adic analysis, Reprint of the 1984 original [MR0791759]. MR 2444734
  * [20] John H. Smith, _A result of Bass on cyclotomic extension fields_ , Proc. Amer. Math. Soc. 24 (1970), 394–395. MR 257049
  * [21] C. J. Smyth, _On the product of the conjugates outside the unit circle of an algebraic integer_ , Bull. London Math. Soc. 3 (1971), 169–175. MR 289451
  * [22] The PARI Group, Univ. Bordeaux, _PARI/GP version 2.15.4_, 2023, available from http://pari.math.u-bordeaux.fr/.
  * [23] Wikipedia, _Nikolai Chebotaryov_ , https://en.wikipedia.org/wiki/Nikolai_Chebotaryov, 2023, [Online; accessed 15-November-2023].
  * [24] U. Zannier, _Vanishing sums of roots of unity_ , Rend. Sem. Mat. Univ. Politec. Torino 53 (1995), no. 4, 487–495, Number theory, II (Rome, 1995). MR 1452400