# Solutions to a Pillai–type equation involving Tribonacci numbers and
$S-$units

Herbert Batte1,∗ and Florian Luca2

###### Abstract

Let $\\{T_{n}\\}_{n\geq 0}$ be the sequence of Tribonacci numbers. In this
paper, we study the exponential Diophantine equation $T_{n}-2^{x}3^{y}=c$, for
$n,x,y\in\mathbb{Z}_{\geq 0}$. In particular, we show that there is no integer
$c$ with at least six representations of the form $T_{n}-2^{x}3^{y}$.

Keywords and phrases: Tribonacci numbers; $S-$units; linear forms in
logarithms; $p-$adic numbers; Pillai’s problem.

2020 Mathematics Subject Classification: 11B39, 11D61, 11D45, 11Y50.

∗ Corresponding author

##  1 Introduction

###  1.1 Background

We consider the sequence of Tribonacci numbers $\\{T_{n}\\}_{n\geq 0}$ defined
by $T_{0}=0$, $T_{1}=T_{2}=1$ and the recurrence relation

$T_{n+3}=T_{n+2}+T_{n+1}+T_{n},$

for all $n\geq 0$. The first few terms of this sequence are given by

$0,\;1,\;1,\;2,\;4,\;7,\;13,\;24,\;44,\;81,\;149,\;274,\;504,\ldots.$

The Diophantine equation

$a^{x}-b^{y}=c,$ (1.1)

where fixed integers $a>1$, $b>1$ and $c$ are involved, is known as the Pillai
equation. Originating from Pillai’s work in [24], this equation’s potential to
produce multiple nonnegative integer solutions $(x,y)$ has been a subject of
interest. Pillai’s significant contribution was the demonstration that, for
positive, coprime integers $a$ and $b$, and when $|c|>c_{0}(a,b)$, there is no
more than at most a single solution $(x,y)$.

Further explorations of the Pillai problem have focused on the cases where $a$
is fixed as $2$, $3$, or a particular prime $p$, replacing the sequence of
powers of powers of $b$ with other exponentially growing sequences of positive
integers, such as Pell numbers, Fibonacci numbers, Tribonacci numbers, and
more complex $k$–generalized Fibonacci numbers for an integer parameter $k\geq
2$. These studies, see for example [2], [8], [11], [13], and [14], have
largely confirmed the conclusion of the original Pillai problem: all integers
are uniquely represented in the above way, except for finitely many outliers.

In a number field ${\mathbb{K}}$ with ring of integers
${\mathcal{O}}_{\mathbb{K}}$ and a finite set of prime ideals $S$, an element
$x\in{\mathbb{K}}$ is an $S$–unit if its principal fractional ideal is a
product of primes in $S$. For rational numbers, an $S$–unit is a rational
number whose numerator and denominator are divisible only by primes in $S$.
More research on the variation of (1.1) are found in [3] and [26]. In [26], in
equation (1.1) Fibonacci numbers replaced the powers of $a$, and $S$–units
replaced the powers of $b$, while in the similar study [3], equation (1.1) was
investigated with Lucas numbers and $S$–units. In this note, we revisit (1.1)
instead with Tribonacci numbers while retaining the $S$–units with
$S=\\{2,3\\}$.

Therefore, we investigate the exponential Diophantine equation

$\displaystyle T_{n}-2^{x}3^{y}=c,$ (1.2)

for $n,x,y\in\mathbb{Z}_{\geq 0}$.

We discard the situation $n=1$ since $T_{1}=T_{2}=1$. Thus, we always assume
that $n=0$, $n\geq 2$. The main results of this paper are the following
theorems.

###  1.2 Main Results

######  Theorem 1.1.

The Diophantine equation (1.2) has in the case that $c=0$, exactly five
solutions, namely

$(n,x,y)=(2,0,0),(3,1,0),(4,2,0),(7,3,1),(9,0,4).$

Furthermore, this representation is given by

$\displaystyle 0$
$\displaystyle=T_{2}-2^{0}3^{0}=T_{3}-2^{1}3^{0}=T_{4}-2^{2}3^{0}=T_{7}-2^{3}3^{1}=T_{9}-2^{0}3^{4}.$

######  Theorem 1.2.

Let $c\in\mathbb{N}$ such that the Diophantine equation (1.2) has at least
four solutions $(n,x,y)\in\mathbb{Z}_{\geq 0}^{3}$. Then,

$c=1.$

Furthermore, this representation is given by

$\displaystyle 1$
$\displaystyle=T_{3}-2^{0}3^{0}=T_{4}-2^{0}3^{1}=T_{5}-2^{1}3^{1}=T_{6}-2^{2}3^{1}.$

######  Theorem 1.3.

Let $c\in-\mathbb{N}$ such that the Diophantine equation (1.2) has at least
five solutions $(n,x,y)\in\mathbb{Z}_{\geq 0}^{3}$. Then,

$c\in\\{-8,-2\\}.$

Furthermore, these representations are given by

$\displaystyle-8$
$\displaystyle=T_{0}-2^{3}3^{0}=T_{2}-2^{0}3^{2}=T_{4}-2^{2}3^{1}=T_{7}-2^{5}3^{0}=T_{12}-2^{9}3^{0},$
$\displaystyle-2$
$\displaystyle=T_{0}-2^{1}3^{0}=T_{2}-2^{0}3^{1}=T_{3}-2^{2}3^{0}=T_{4}-2^{1}3^{1}=T_{5}-2^{0}3^{2}.$

##  2 Methods

###  2.1 Preliminaries

The Tribonacci sequence $(T_{n})_{n\geq 0}$ has a characteristic polynomial
given by

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

This polynomial $\Psi(X)$ is irreducible in $\mathbb{Q}[X]$, and has a
positive real zero

$\alpha=\frac{1}{3}\left(1+(19+3\sqrt{33})^{1/3}+(19-3\sqrt{33})^{1/3}\right),$

lying strictly outside the unit circle and two complex conjugate zeros $\beta$
and $\gamma$ lying strictly inside the unit circle. Furthermore,
$|\beta|=|\gamma|=\alpha^{-1/2}$. According to Dresden and Zu [15], we have

$\displaystyle
T_{n}=C_{\alpha}\alpha^{n-1}+C_{\beta}\beta^{n-1}+C_{\gamma}\gamma^{n-1}\quad\text{for
all }n\geq 0,$ (2.1)

where $C_{X}=(X-1)/(4X-6)$. Dresden and Zu [15], also showed that the
contribution of the complex conjugate zeros $\beta$ and $\gamma$ to the
right–hand side of (2.1) is very small. More precisely,

$\displaystyle\left|T_{n}-C_{\alpha}\alpha^{n-1}\right|<\frac{1}{2}\quad\text{for
all }n\geq 0.$ (2.2)

The minimal polynomial $P(X)$ of $C_{\alpha}$ over the integers is given by

$P(X)=44X^{3}-44X^{2}+12X-1,$ (2.3)

and has zeros $C_{\alpha}$, $C_{\beta}$, $C_{\gamma}$ with $|C_{\alpha}|$,
$|C_{\beta}|$, $|C_{\gamma}|<1$. Numerically, these values are approximated as

$\displaystyle 1.83$ $\displaystyle<\alpha<1.84,$ $\displaystyle 0.73<|\beta|$
$\displaystyle=|\gamma|=\alpha^{-1/2}<0.74,$ $\displaystyle 0.61$
$\displaystyle<|C_{\alpha}|<0.62,$ (2.4) $\displaystyle 0.19<|C_{\beta}|$
$\displaystyle=|C_{\gamma}|<0.20.$

Let $\mathbb{K}:=\mathbb{Q}(\alpha,\beta)$ be the splitting field of the
polynomial $\Psi$ over $\mathbb{Q}$. Then, $[\mathbb{K}:\mathbb{Q}]=6$.
Furthermore, $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$. The Galois group of
$\mathbb{K}$ over $\mathbb{Q}$ is given by

$\mathcal{G}:=\text{Gal}(\mathbb{K}/\mathbb{Q})\simeq\\{(1),(\alpha\beta),(\alpha\gamma),(\beta\gamma),(\alpha\beta\gamma),(\alpha\gamma\beta)\\}\simeq
S_{3}.$

Thus, we identify the automorphisms of $\mathcal{G}$ with the permutations of
the zeros of the polynomial $\Psi$. For instance, the permutation
$(\alpha\gamma)$ corresponds to the automorphism $\sigma:\alpha\mapsto\gamma$,
$\gamma\mapsto\alpha$ and $\beta\mapsto\beta$.

We need additional properties.

######  Proposition 2.1.

The following hold:

  * (i)

There exists a unique $z\in{\mathbb{K}}$ such that $z^{3}=\alpha/\beta$.

  * (ii)

If $m\geq 2$, then $\pm z$ is not an $m$th power of some other element in
${\mathbb{K}}$.

###### Proof.

(i) For the existence of $z$, see Proposition 5.1 (i) in [6]. The uniqueness
of $z$ follows from the fact that ${\mathbb{K}}$ does not contain cubic roots
of unity. The minimal polynomial $Q(X)$ of $z$ is

$Q(X)=X^{6}+X^{5}+2X^{4}+3X^{3}+2X^{2}+X+1.$

(ii) Assume that $\pm z=y^{m}$ for some integer $m\geq 2$. We may assume that
$m=p$ is a prime and write $y=(\pm z)^{1/p}\in{\mathbb{K}}$. Note that $y$ is
an algebraic integer as it is a unit. Also,
$|y|=|z|^{1/p}=|\alpha/\beta|^{1/3p}=|\alpha|^{1/2p}$. Note that all
conjugates of $y$ have absolute values in the set
$\\{|\alpha|^{1/2p},1,|\alpha|^{-1/2p}\\}$. Using the main result of Voutier
in [23], we have that

$|\alpha|^{1/2p}=|y|\geq 1+\frac{1}{2d}\left(\dfrac{\log\log d}{\log
d}\right)^{3},$

where $d$ is the degree of ${\mathbb{K}}$. Using $d=6$ and taking logarithms,
we get

$p\leq\frac{\log\alpha}{2\log\left(1+\dfrac{1}{2\cdot 6}\left(\dfrac{\log\log
6}{\log 6}\right)^{3}\right)}<106.$

Now we used SageMath to check that for prime $p\leq 106$, the polynomial
$Q(X^{p})$ is irreducible. This proves (ii). ∎

We recall one additional simple fact from calculus. This is Lemma 1 in [26].

######  Lemma 2.1 (Lemma 1 in [26]).

If $x\in\mathbb{R}$ satisfies $|x|<\frac{1}{2}$, then
$|\log(1+x)|<\frac{3}{2}|x|$.

To end this section, we present an analytic argument which is Lemma 7 in [16].

######  Lemma 2.2 (Lemma 7 in [16]).

If $m\geq 1$, $T>(4m^{2})^{m}$ and $T>\displaystyle\frac{z}{(\log z)^{m}}$,
then

$z<2^{m}T(\log T)^{m}.$

###  2.2 Linear forms in logarithms

We use several times Baker–type lower bounds for nonzero linear forms in four
or more logarithms of algebraic numbers. There are many such bounds mentioned
in the literature like that of Baker and Wüstholz from [1] or Matveev from
[19]. Before we can formulate such inequalities we need the notion of height
of an algebraic number recalled below.

######  Definition 2.1.

Let $\lambda$ be an algebraic number of degree $d$ with minimal primitive
polynomial over the integers

$a_{0}x^{d}+a_{1}x^{d-1}+\cdots+a_{d}=a_{0}\prod_{i=1}^{d}(x-\lambda^{(i)}),$

where the leading coefficient $a_{0}$ is positive. Then, the logarithmic
height of $\lambda$ is given by

$h(\lambda):=\dfrac{1}{d}\Big{(}\log
a_{0}+\sum_{i=1}^{d}\log\max\\{|\lambda^{(i)}|,1\\}\Big{)}.$

In particular, if $\lambda$ is a rational number represented as $\lambda:=p/q$
with coprime integers $p$ and $q\geq 1$, then
$h(\lambda)=\log\max\\{|p|,q\\}$. The following properties of the logarithmic
height function $h(\cdot)$ will be used in the rest of the paper without
further reference:

$\displaystyle h(\lambda_{1}\pm\lambda_{2})$ $\displaystyle\leq
h(\lambda_{1})+h(\lambda_{2})+\log 2;$ $\displaystyle
h(\lambda_{1}\lambda_{2}^{\pm 1})$ $\displaystyle\leq
h(\lambda_{1})+h(\lambda_{2});$ $\displaystyle h(\lambda^{s})$
$\displaystyle=|s|h(\lambda)\quad{\text{\rm valid for}}\quad s\in\mathbb{Z}.$

A linear form in logarithms is an expression

$\Lambda:=b_{1}\log\lambda_{1}+\cdots+b_{t}\log\lambda_{t},$ (2.5)

where for us $\lambda_{1},\ldots,\lambda_{t}$ are positive real algebraic
numbers and $b_{1},\ldots,b_{t}$ are nonzero integers. We assume, $\Lambda\neq
0$. We need lower bounds for $|\Lambda|$. We write
${\mathbb{L}}:={\mathbb{Q}}(\lambda_{1},\ldots,\lambda_{t})$ and $D$ for the
degree of ${\mathbb{L}}$. We start with the general form due to Matveev [19].

######  Theorem 2.1 (Matveev, [19]).

Put $\Gamma:=\lambda_{1}^{b_{1}}\cdots\lambda_{t}^{b_{t}}-1=e^{\Lambda}-1$.
Assume $\Gamma\neq 0$. Then

$\log|\Gamma|>-1.4\cdot 30^{t+3}\cdot t^{4.5}\cdot D^{2}(1+\log D)(1+\log
B)A_{1}\cdots A_{t},$

where $B\geq\max\\{|b_{1}|,\ldots,|b_{t}|\\}$ and
$A_{i}\geq\max\\{Dh(\lambda_{i}),|\log\lambda_{i}|,0.16\\}$ for
$i=1,\ldots,t$.

We also employ a $p$–adic variation of Laurent’s result as established by
Bugeaud and Laurent in [9], Corollary 1. Prior to outlining their result, we
require a few additional notations.

######  Definition 2.2.

Let $p$ be a prime number. The $p$-adic valuation of an integer $x$, denoted
by $\nu_{p}(x)$, is defined by

$\nu_{p}(x):=\begin{cases}\max\\{k\in\mathbb{N}:p^{k}\mid x\\},&\text{if
}x\neq 0;\\\ \infty,&\text{if }x=0.\end{cases}$

Additionally, if $x=a/b$ is a rational number and $a,~{}b$ are integers, then
we put

$\nu_{p}(x):=\nu_{p}(a)-\nu_{p}(b).$

The formula for $\nu_{p}(x)$ when $x$ is rational given by Definition 2.2 does
not depend on the representation of $x$ as a ratio of integers $a/b$. It
follows easily from Definition 2.2 that if $x$ is rational, then

$\nu_{p}(x)=\text{ord}_{p}(x),$

where $\text{ord}_{p}(x)$ is the exponent of $p$ in the factorization of $x$.
For example, $\nu_{2}(9/8)=-3$. Now for the algebraic number $\lambda$ in
Definition 2.1, we define

$\nu_{p}(\lambda):=\frac{\nu_{p}(a_{d}/a_{0})}{d},$

that is, it is the $p$–adic valuation of the rational number $a_{d}/a_{0}$
divided by the degree $d$ of $\lambda$. For example, when $x$ is a rational
number, we write it as $x=a_{d}/a_{0}$ with coprime integers $a_{d}$ and
$a_{0}\geq 1$, then its minimal polynomial is $f(X)=a_{0}X-a_{d}$ has degree
$1$ so

$\nu_{p}(x)=\nu_{p}(a_{d}/a_{0}),$

consistent with Definition 2.2. The $p$–adic valuation gives rise to the
conventional absolute value. When the rational numbers $\mathbb{Q}$ are
completed with the standard absolute value, the outcome is the set of real
numbers, $\mathbb{R}$. Conversely, employing the $p$-adic absolute value for
the completion of $\mathbb{Q}$ yields the $p$-adic numbers, represented as
$\mathbb{Q}_{p}$.

Similarly to the preceding context, let $\lambda_{1}$ and $\lambda_{2}$ be
algebraic numbers over $\mathbb{Q}$, treated as elements of the field
${\mathbb{K}}_{p}:=\mathbb{Q}_{p}(\lambda_{1},\lambda_{2})$, with
$D:=[\mathbb{Q}_{p}(\lambda_{1},\lambda_{2}):\mathbb{Q}_{p}]$. Similar to the
situation in Theorem 2.1 above, we must employ an adjusted height function.
Specifically, we express it as follows:

$h^{\prime}(\lambda_{i})\geq\max\left\\{h(\lambda_{i}),\dfrac{\log
p}{D}\right\\},~{}~{}\text{for }~{}i=1,2.$

######  Theorem 2.2 (Bugeaud and Laurent, [9]).

Let $b_{1}$, $b_{2}$ be positive integers and suppose that $\lambda_{1}$ and
$\lambda_{2}$ are multiplicatively independent algebraic numbers such that
$\nu_{p}(\lambda_{1})=\nu_{p}(\lambda_{2})=0$. Put

$E^{\prime}:=\dfrac{b_{1}}{h^{\prime}(\lambda_{2})}+\dfrac{b_{2}}{h^{\prime}(\lambda_{1})},$

and

$E:=\max\left\\{\log E^{\prime}+\log\log p+0.4,10,10\log p\right\\}.$

Then

$\nu_{p}\left(\lambda_{1}^{b_{1}}\lambda_{2}^{b_{2}}-1\right)\leq\frac{24pg}{(p-1)(\log
p)^{4}}E^{2}D^{4}h^{\prime}(\lambda_{1})h^{\prime}(\lambda_{2}),$

where $g>0$ is the smallest integer such that
$\nu_{p}\left(\lambda_{i}^{g}-1\right)>0$.

Applying Theorem 2.1 and 2.2, we get upper bounds on our variables.

However, such upper bounds are too large, thus there is need to reduce them.
In this paper, we use the following result related with continued fractions
(see Theorem 8.2.4 in [20]).

######  Lemma 2.3 (Legendre).

Let $\mu$ be an irrational number, $[a_{0},a_{1},a_{2},\ldots]$ be the
continued fraction expansion of $\mu$. Let
$p_{i}/q_{i}=[a_{0},a_{1},a_{2},\ldots,a_{i}]$, for all $i\geq 0$, be all the
convergents of the continued fraction of $\mu$, and $M$ be a positive integer.
Let $N$ be a non-negative integer such that $q_{N}>M$. Then putting
$a(M):=\max\\{a_{i}:i=0,1,2,\ldots,N\\}$, the inequality

$\bigg{|}\mu-\frac{r}{s}\bigg{|}>\dfrac{1}{(a(M)+2)s^{2}},$

holds for all pairs $(r,s)$ of positive integers with $0<s<M$.

However, since there are no methods based on continued fractions to find a
lower bound for linear forms in more than two variables with bounded integer
coefficients, we use at some point a method based on the LLL–algorithm. We
next explain this method.

###  2.3 Reduced Bases for Lattices and LLL–reduction methods

Let $k$ be a positive integer. A subset $\mathcal{L}$ of the $k$–dimensional
real vector space $\mathbb{R}^{k}$ is called a lattice if there exists a basis
$\\{b_{1},b_{2},\ldots,b_{k}\\}$ of $\mathbb{R}^{k}$ such that

$\displaystyle\mathcal{L}=\sum_{i=1}^{k}\mathbb{Z}b_{i}=\left\\{\sum_{i=1}^{k}r_{i}b_{i}\mid
r_{i}\in\mathbb{Z}\right\\}.$

In this situation we say that $b_{1},b_{2},\ldots,b_{k}$ form a basis for
$\mathcal{L}$, or that they span $\mathcal{L}$. We call $k$ the rank of
$\mathcal{L}$. The determinant $\text{det}(\mathcal{L})$, of $\mathcal{L}$ is
defined by

$\displaystyle\text{det}(\mathcal{L})=|\det(b_{1},b_{2},\ldots,b_{k})|,$

with the $b_{i}$ being written as column vectors. This is a positive real
number that does not depend on the choice of the basis (see [10] Sect 1.2).

Given linearly independent vectors $b_{1},b_{2},\ldots,b_{k}$ in
$\mathbb{R}^{k}$, we refer back to the Gram–Schmidt orthogonalization
technique. This method allows us to inductively define vectors $b^{*}_{i}$
(with $1\leq i\leq k$) and real coefficients $\mu_{i,j}$ (for $1\leq j\leq
i\leq k$). Specifically,

$\displaystyle b^{*}_{i}$
$\displaystyle=b_{i}-\sum_{j=1}^{i-1}\mu_{i,j}b^{*}_{j},~{}~{}~{}\mu_{i,j}=\dfrac{\langle
b_{i},b^{*}_{j}\rangle}{\langle b^{*}_{j},b^{*}_{j}\rangle},$

where $\langle\cdot,\cdot\rangle$ denotes the ordinary inner product on
$\mathbb{R}^{k}$. Notice that $b^{*}_{i}$ is the orthogonal projection of
$b_{i}$ on the orthogonal complement of the span of $b_{1},\ldots,b_{i-1}$,
and that $\mathbb{R}b_{i}$ is orthogonal to the span of
$b^{*}_{1},\ldots,b^{*}_{i-1}$ for $1\leq i\leq k$. It follows that
$b^{*}_{1},b^{*}_{2},\ldots,b^{*}_{k}$ is an orthogonal basis of
$\mathbb{R}^{k}$.

######  Definition 2.3.

The basis $b_{1},b_{2},\ldots,b_{n}$ for the lattice $\mathcal{L}$ is called
reduced if

$\displaystyle\|\mu_{i,j}\|$
$\displaystyle\leq\frac{1}{2},\quad\text{for}\quad 1\leq j<i\leq
n,~{}~{}\text{and}$ $\displaystyle\|b^{*}_{i}+\mu_{i,i-1}b^{*}_{i-1}\|^{2}$
$\displaystyle\geq\frac{3}{4}\|b^{*}_{i-1}\|^{2},\quad\text{for}\quad 1<i\leq
n,$

where $\|\cdot\|$ denotes the ordinary Euclidean length. The constant
$\frac{3}{4}$ above is arbitrarily chosen, and may be replaced by any fixed
real number $\mathcal{P}$ with $\frac{1}{4}<\mathcal{P}<1$, (see [17] Sect 1).

Let $\mathcal{L}\subseteq\mathbb{R}^{k}$ be a $k-$dimensional lattice with
reduced basis $b_{1},\ldots,b_{k}$ and denote by $B$ the matrix with columns
$b_{1},\ldots,b_{k}$. We define

$l\left(\mathcal{L},v\right)=\left\\{\begin{array}[]{c}\min_{u\in\mathcal{L}}||u-v||\quad;~{}~{}v\not\in\mathcal{L}\\\
\min_{0\neq
u\in\mathcal{L}}||u||\quad;~{}~{}v\in\mathcal{L}\end{array}\right.,$

where $||\cdot||$ denotes the Euclidean norm on $\mathbb{R}^{k}$. It is well
known that, by applying the LLL-algorithm, it is possible to give in
polynomial time a lower bound for $l\left(\mathcal{L},v\right)\geq c_{1}$ (see
[25], Sect. V.4).

######  Lemma 2.4.

Let $v\in\mathbb{R}^{k}$ and $z=B^{-1}v$ with $z=(z_{1},\ldots,z_{k})^{T}$.
Furthermore,

  1. (i)

if $v\not\in\mathcal{L}$, let $i_{0}$ be the largest index such that
$z_{i_{0}}\neq 0$ and put $\sigma:=\\{z_{i_{0}}\\}$, where $\\{\cdot\\}$
denotes the distance to the nearest integer.

  2. (ii)

if $v\in\mathcal{L}$, put $\sigma:=1$.

Finally, we have

$c_{1}:=\max\limits_{1\leq j\leq
k}\left\\{\dfrac{||b_{1}||}{||b_{j}^{*}||}\right\\}\qquad\text{and}\qquad
c_{2}:=c_{1}^{-1}\sigma||b_{1}||.$

In our application, we are given real numbers
$\eta_{0},\eta_{1},\ldots,\eta_{k}$ which are linearly independent over
$\mathbb{Q}$ and two positive constants $c_{3}$ and $c_{4}$ such that

$\displaystyle|\eta_{0}+a_{1}\eta_{1}+\cdots+a_{k}\eta_{k}|\leq
c_{3}\exp(-c_{4}H),$ (2.6)

where the integers $a_{i}$ are bounded as $|a_{i}|\leq A_{i}$ with $A_{i}$
given upper bounds for $1\leq i\leq k$. We write $A_{0}:=\max\limits_{1\leq
i\leq k}\\{A_{i}\\}$.

The basic idea in such a situation, from [12], is to approximate the linear
form (2.6) by an approximation lattice. So, we consider the lattice
$\mathcal{L}$ generated by the columns of the matrix

$\mathcal{A}=\begin{pmatrix}1&0&\ldots&0&0\\\ 0&1&\ldots&0&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&\ldots&1&0\\\ \lfloor
M\eta_{1}\rfloor&\lfloor M\eta_{2}\rfloor&\ldots&\lfloor
M\eta_{k-1}\rfloor&\lfloor M\eta_{k}\rfloor\end{pmatrix},$

where $M$ is a large constant usually of the size of about $A_{0}^{k}$ . Let
us assume that we have an LLL–reduced basis $b_{1},\ldots,b_{k}$ of
$\mathcal{L}$ and that we have a lower bound $l\left(\mathcal{L},v\right)\geq
c_{1}$ with $v:=(0,0,\ldots,-\lfloor M\eta_{0}\rfloor)$. Note that $c_{2}$ can
be computed by using the results of Lemma 2.4. Then, with these notations the
following result is Lemma VI.1 in [25].

######  Lemma 2.5 (Lemma VI.1 in [25]).

Let $S:=\displaystyle\sum_{i=1}^{k-1}A_{i}^{2}$ and
$T:=\dfrac{1+\sum_{i=1}^{k}A_{i}}{2}$. If $c_{2}^{2}\geq T^{2}+S$, then
inequality (2.6) implies that we either have $a_{1}=a_{2}=\cdots=a_{k-1}=0$
and $a_{k}=-\dfrac{\lfloor M\eta_{0}\rfloor}{\lfloor M\eta_{k}\rfloor}$, or

$H\leq\dfrac{1}{c_{4}}\left(\log(Mc_{3})-\log\left(\sqrt{c_{2}^{2}-S}-T\right)\right).$

SageMath 9.5 is used to perform all the computations in this work.

###  2.4 Equations in $\alpha,\beta,\gamma$

In this section, we prove the following results which will be helpful and
recalled later.

######  Lemma 2.6.

The equation

$\displaystyle\dfrac{1-\gamma^{v}}{1-\gamma^{u}}=\dfrac{1-\alpha^{v}}{1-\alpha^{u}}.$
(2.7)

has no integer solutions $u,v$, with $u>v\geq 1$.

###### Proof.

Let $u,v\in\mathbb{Z}$, with $u>v\geq 1$. Then we can write (2.7) as

$\displaystyle\alpha^{u}-\alpha^{v}$
$\displaystyle=\gamma^{u}-\gamma^{v}+\gamma^{v}\alpha^{u}-\gamma^{u}\alpha^{v};$
$\displaystyle\alpha^{u}-\alpha^{v}=\left|\alpha^{u}-\alpha^{v}\right|$
$\displaystyle=\left|\gamma^{u}-\gamma^{v}+\gamma^{v}\alpha^{u}-\gamma^{u}\alpha^{v}\right|$
$\displaystyle\leq|\gamma|^{u}+|\gamma|^{v}+\alpha^{u-0.5v}+\alpha^{v-0.5u},$

where we have used the fact that $|\gamma|=\alpha^{-1/2}$. Assume for a moment
that $v\geq 5$ which implies that $u\geq 6$ since $u>v$. Then the above
inequality implies

$\displaystyle\alpha^{u-2}+\alpha^{u-3}$
$\displaystyle=\alpha^{u}-\alpha^{u-1}\leq\alpha^{u}-\alpha^{v}\leq
1+\alpha^{u-0.5v}+\alpha^{v-0.5u}$ $\displaystyle\leq
1+\alpha^{u-2.5}+\alpha^{u-1-0.5\cdot 6}=1+\alpha^{u-2.5}+\alpha^{u-4}$
$\displaystyle<1+\alpha^{u-2.5}+\alpha^{u-3}.$

Therefore, $\alpha^{u-2}<1+\alpha^{u-2.5}$. From this, we see that
$\alpha^{u-2}\left(1-\alpha^{-0.5}\right)<1$. This implies that

$\displaystyle\alpha^{u-2}$
$\displaystyle<\dfrac{\sqrt{\alpha}}{\sqrt{\alpha}-1}<\dfrac{\sqrt{1.84}}{\sqrt{1.83}-1}<3.85,$

so that $u<4.24$. This contradicts the assumption that $u\geq 6$. Therefore
(2.7) has no solutions with $v\geq 5$. This means that (2.7) may have
solutions only for $1\leq v\leq 4$. We go back and rewrite (2.7) as

$\displaystyle\alpha^{u}-\alpha^{v}$
$\displaystyle=\gamma^{u}-\gamma^{v}+\gamma^{v}\alpha^{u}-\gamma^{u}\alpha^{v};$
$\displaystyle\alpha^{u}$
$\displaystyle=\gamma^{u}-\gamma^{v}+\gamma^{v}\alpha^{u}-\gamma^{u}\alpha^{v}+\alpha^{v}$
$\displaystyle\leq|\gamma|^{u}+|\gamma|^{v}+\alpha^{u-0.5v}+\alpha^{v-0.5u}+\alpha^{v}$
$\displaystyle<2+\alpha^{u-0.5}+\alpha^{3.5}+\alpha^{4},$

so that $\alpha^{u}\left(1-\alpha^{-0.5}\right)<22$, which gives $u\leq 7$.
With help of SageMath, we check for $v\in\\{1,2,3,4\\}$ and
$u\in\\{2,3,4,5,6,7\\}$ that satisfy (2.7) and find no solutions. ∎

######  Lemma 2.7.

Assume that the algebraic numbers

$\displaystyle\lambda_{1}:=\dfrac{\alpha}{\beta}\qquad{\text{\rm
and}}\qquad\lambda_{2}:=\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}\right),$

are multiplicatively dependent for $n>n_{1}>n_{2}$. Then $\lambda_{2}=\pm
z^{t}$ for some integer $t$ with $|t|<80n+244$ and $z$ is defined in
Proposition 2.1.

###### Proof.

Suppose $\lambda_{1}$ and $\lambda_{2}$ are multiplicatively dependent. This
means that there exists integers $A$ and $B$, not both zero, such that
$\lambda_{1}^{A}=\lambda_{2}^{B}$. Since $\lambda_{1}=z^{3}$, we have

$\lambda_{2}^{B}=z^{3A}.$

Let $d:=\gcd(3A,B)$ and write $3A=A_{1}d,~{}B=B_{1}d$. Then
$\lambda_{2}^{B_{1}}=\zeta_{d}z^{A_{1}}$ where $\zeta_{d}$ is a root of unity
of order dividing $d$. Since ${\mathbb{K}}$ does not contain roots of unity
other than $\pm 1$, it follows that $\lambda_{2}^{B_{1}}=\pm z^{A_{1}}$. We
show that $B_{1}=1$. Indeed, since $\gcd(A_{1},B_{1})=1$, we get that
$A_{1}p+B_{1}q=1$ for some integers $p,~{}q$. Thus,

$z=z^{A_{1}p+B_{1}q}=(\pm 1)^{p}(\lambda_{2}^{p}z^{q})^{B_{1}}.$

The above calculation shows that $\pm z$ is a power of exponent $B_{1}$ of
some other number in ${\mathbb{K}}$, so by (ii) of Proposition 2.1, we get
$B_{1}=1$. Thus, $\lambda_{2}=\pm z^{t}$ with $t=A_{1}$, and we need to bound
$t$. Note that $|t|=|\log|\lambda_{2}||/|\log z|$. We estimate

$\displaystyle|\log|\lambda_{2}||$ $\displaystyle=$
$\displaystyle\left|\log\left|\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}\right)\right|\right|$
$\displaystyle\leq$
$\displaystyle|\log|C_{\alpha}||+|\log|C_{\beta}||+\left|\log\left|\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right|\right|$
$\displaystyle+\left|\log\left|\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right|\right|.$

The first two terms above are bounded as $|\log|C_{\alpha}||<0.5$ and
$|\log|C_{\beta}||<1.67$. The next two terms are absolute values of logarithms
of absolute values of differences of complex numbers. These can be very small
or can be very large. In the “large” part it is easy. Since $\alpha>1$ and
$n>n_{1}>n_{2}$, $|\beta|=|\gamma|=\alpha^{-1/2}$, we have by the absolute
value inequality and the fact that $|1-\alpha^{n_{i}-n}|=1-\alpha^{n_{i}-n}<1$
for $i=1,2$, that

$\displaystyle|(1-\alpha^{n_{1}-n})(1-\gamma^{n_{2}-n})-(1-\alpha^{n_{2}-n})(1-\gamma^{n_{1}-n})|<|1-\gamma^{n_{2}-n}|+|1-\gamma^{n_{1}-n}|$
(2.8) $\displaystyle\leq$ $\displaystyle
1+\alpha^{(n-n_{1})/2}+1+\alpha^{(n-n_{2})/2}<2\alpha^{(n-n_{2})/2+2}<\alpha^{n+4},$

where we used $\alpha^{3}>\alpha+1$ and $2<\alpha^{2}$. Furthermore,

$\displaystyle|(1-\beta^{n_{1}-n})(1-\gamma^{n_{2}-n})-(1-\beta^{n_{2}-n})(1-\gamma^{n_{1}-n})|$
(2.9) $\displaystyle<$
$\displaystyle|1-\beta^{n_{1}-n}||1-\gamma^{n_{2}-n}|+|(1-\beta^{n_{2}-n}||1-\gamma^{n_{1}-n}|<2(\alpha^{(n-n_{1})/2}+1)(\alpha^{(n-n_{2})/2}+1)$
$\displaystyle<$ $\displaystyle
2(\alpha^{(n-n_{2})/2+2})^{2}=2\alpha^{n-n_{2}+4}<\alpha^{n+5}.$

The challenge is computing lower bounds. Let

$u:=1-\beta^{n_{1}-n},\quad v:=1-\gamma^{n_{2}-n}.$

Then we need to compute a lower bound for

$|(uv)^{(\sigma)}-(uv)^{(\tau)}|$

over all the distinct conjugates of $uv$. For example, taking
$\sigma={\text{\rm id}}$ and $\tau$ to be the map that swaps $\beta$ with
$\gamma$, we get a lower bound on the absolute value the expression appearing
in the denominator of $\lambda_{2}$ in the statement of Lemma 2.7 (ignoring
$C_{\beta}$), while taking $\sigma$ to be the map that swaps $\alpha$ with
$\beta$ and $\tau$ the three cycle $\beta\mapsto\gamma\mapsto\alpha$ we get a
lower bound on the numerator appearing in the expression of $\lambda_{2}$ in
the statement of Lemma 2.7 (ignoring $C_{\alpha}$). For this, we use a
well–known root separation result of Mahler [18]. In our particular case, it
says that if

$R(X)=\prod_{i=1}^{d}(X-z_{i})\in{\mathbb{Z}}[X]$

is a monic polynomial with integer coefficients and distinct roots then
writing

$M(R):=\prod_{i=1}^{d}\max\\{1,|z_{i}|\\}$

for its Mahler measure, we have

$|z_{i}-z_{j}|>\frac{1}{d^{d/2}M(P)^{d-1}}.$

For us,

$R(X)=\prod_{\sigma\in S_{3}}(X-(uv)^{(\sigma)})$

has degree $d=6$. Writing down the 6 conjugates $(uv)^{(\sigma)}$, we get a
total of $12$ factors namely two occurrences of each of $1-\delta^{n_{s}-n}$
for each $s\in\\{1,2\\}$ and $\delta\in\\{\alpha,\beta,\gamma\\}$. Ignoring
the contributions of $1-\alpha^{n_{s}-n}$ (as they are in $(0,1)$) and using
$|1-\delta^{n_{s}-n}|\leq 1+\alpha^{(n-n_{s})/2}<\alpha^{(n-n_{s}+4)/2}$ as
before twice for each of $s\in\\{1,2\\}$ and $\delta\in\\{\beta,\gamma\\}$, we
get that

$M(P)<(\alpha^{(n-n_{1}+4)/2})^{4}(\alpha^{(n-n_{2}+4)/2})^{4}=\alpha^{4n-2n_{2}-2n_{1}+16}\leq\alpha^{4n+10}.$

Thus,

$|(uv)^{(\sigma)}-(uv)^{(\tau)}|>\frac{1}{6^{3}(\alpha^{4n+10})^{5}},$

showing that if the left–hand side above is $<1$, then

$|\log|((uv)^{(\sigma)}-(uv)^{(\tau)})||<\log(6^{3}\alpha^{20n+50})<(20n+50)\log\alpha+3\log
6.$

Comparing the above with (2.8) and (2.9), we conclude that the last bound
above holds in all cases. Putting everything together, we get

$|\log|\lambda_{2}||<0.5+1.67+2((20n+50)\log\alpha+3\log
6)<(40n+100)\log\alpha+13.$

We thus get

$|t|\leq\frac{|\log|\lambda_{2}||}{\log|z|}\leq\frac{(40n+100)\log\alpha+13}{0.5\log\alpha}<80n+244.$

∎

######  Lemma 2.8.

For any positive integer $t$, we have

  * (i)

$\nu_{2}(z^{t}-1)<1+\nu_{2}(n),\quad\text{and}$

  * (ii)

$\nu_{3}(z^{t}-1)\leq 1+\nu_{3}(n).$

###### Proof.

For a prime ideal $\pi$ of ${\mathcal{O}}_{\mathbb{K}}$ and an element
$x\in{\mathbb{K}}$, we write $\nu_{\pi}(x)$ for the exponent at which $\pi$
appears in the factorization of the principal fractional ideal
$x{\mathcal{O}}_{\mathbb{K}}$. We shall use the following well-known fact. Let
$p\in{\mathbb{Z}}$ be the prime such that $\pi$ sits above $p$ (so, $\pi\mid
p$) and let $e_{\pi}:=\nu_{\pi}(p)$ is the ramification index of $\pi$. If
$\zeta\in{\mathbb{K}}$ is such that

$\nu_{\pi}(\zeta-1)\geq\frac{e_{\pi}}{p-1},\qquad{\text{\rm
then}}\qquad\nu_{\pi}(\zeta^{t}-1)=\nu_{\pi}(\zeta-1)+\nu_{\pi}(n)$ (2.10)

(see Lemma 1 in [22] or Lemma 4.4 in [5]). We start with $p=2$. Then
$2{\mathcal{O}}_{\mathbb{K}}=\pi^{3}$, where $\pi$ is a prime ideal of
${\mathcal{O}}_{\mathbb{K}}$. In fact, $\pi$ is principal and generated by
$1+z+z^{2}$, as it can be seen by rewriting the equation

$Q(z)=0\qquad{\text{\rm as}}\qquad(z^{2}+z+1)^{3}=2z(z+1)^{2}(z^{2}+1),$

and using that $z+1,~{}z^{2}+1$ are units in ${\mathcal{O}}_{\mathbb{K}}$. Now
${\mathcal{O}}_{\mathbb{K}}/\pi$ is a field with $4$ elements. Since
$\nu_{2}(z-1)=\nu_{2}(z^{2}-1)=0$, we get that the order of $z$ modulo $\pi$
is $3$. Thus, $\nu_{\pi}(z^{t}-1)=0$ unless $3\mid t$ and
$\nu_{\pi}(z^{3}-1)=1$. For $t=3k$, we get that

$\nu_{\pi}(z^{3k}-1)=\nu_{\pi}(z^{3}-1)((z^{3})^{k-1}+\cdots+1)=\nu_{\pi}(z^{3}-1)=1,$

provided $k$ is odd. This proves (i) when $n$ is odd. Since
$z^{3}+1=z^{3}-1+2\equiv{z^{3}-1}\pmod{\pi^{3}}$, we get that
$\nu_{\pi}(z^{3}+1)=1$. Thus, $\nu_{\pi}(z^{6}-1)=2$. Now
$z^{6}+1=z^{6}-1+2\equiv z^{6}-1\pmod{\pi^{3}}$, so also
$\nu_{\pi}(z^{6}+1)=2$. Thus,

$\nu_{\pi}(z^{12}-1)=\nu_{\pi}((z^{6}-1)(z^{6}+1))=\nu_{\pi}(z^{3}-1)+\nu_{\pi}(z^{3}+1)=4.$

The above calculations show that $\nu_{\pi}(z^{t}-1)<4$ unless $12\mid t$. In
particular, unless $12\mid t$, we get
$\nu_{2}(z^{t}-1)=\nu_{\pi}(z^{t}-1)/3\leq 1$, again confirming (i) in this
case. However, if $12\mid t$, then setting $k:=t/12$, then since

$\nu_{\pi}(z^{12}-1)=4>\frac{e_{\pi}}{p-1}(=3),$

we get by formula (2.10) that

$\nu_{\pi}(z^{t}-1)=\nu_{\pi}((z^{12})^{k}-1)=\nu_{\pi}(z^{12}-1)+\nu_{\pi}(k)=4+(\nu_{\pi}(n)-6)<\nu_{\pi}(n)=3\nu_{2}(n),$

and since $\nu_{2}(z^{t}-1)=\nu_{\pi}(z^{t}-1)/3$, we get the conclusion (i).
Conclusion (ii) is even easier. Indeed, $\pi=3{\mathcal{O}}_{\mathbb{K}}$ is
prime. Further, the order of $z$ in ${\mathcal{O}}_{\mathbb{K}}/\pi$ is
exactly $13$ and $\nu_{3}(z^{13}-1)=1>1/12$. Thus, formula (2.10) applies and
gives that $\nu_{3}(z^{t}-1)=0$ unless $13\mid t$ and if $13\mid t$, then
$\nu_{3}(z^{t}-1)=1+\nu_{3}(n)$. ∎

###  2.5 Bounds for solutions to $S-$unit equations

It is also a well known fact from [8] that

$\displaystyle\alpha^{n-2}<T_{n}<\alpha^{n-1}\quad\text{holds for all }n\geq
1.$ (2.11)

Now, if $c\geq 0$ in (1.2), then

$\displaystyle 2^{x}3^{y}=T_{n}-c\leq T_{n}<\alpha^{n-1}.$

This implies that $x\log 2+y\log 3<(n-1)\log\alpha<(n-1)\log 1.84$. Hence,

$\displaystyle x,y<n.$ (2.12)

On the other hand, if $c<0$ in (1.2), then

$\displaystyle\alpha^{n-2}<T_{n}=2^{x}3^{y}+c<2^{x}3^{y}.$

This implies that $\alpha^{n-2}<2^{x}3^{y}$, or simply,

$\displaystyle(n-2)\log\alpha<2\max\\{x\log 2,y\log 3\\}.$ (2.13)

The purpose of this subsection is to deduce a result from the following lemma.
The following lemma is Proposition 1 from [26].

######  Lemma 2.9 (Proposition 1, [26]).

Let $\Delta>10^{80}$ be a fixed integer and assume that

$\displaystyle 2^{x}3^{y}-2^{x_{1}}3^{y_{1}}=\Delta.$ (2.14)

Then $2^{x}3^{y}<\Delta(\log\Delta)^{60\log\log\Delta}$.

We now state and prove the following consequence of Lemma 2.9.

######  Lemma 2.10.

Assume that $(n,n_{1},x,x_{1},y,y_{1})$ is a solution to
$T_{n}-2^{x}3^{y}=T_{n_{1}}-2^{x_{1}}3^{y_{1}}$, with $n>310$ and $n>n_{1}$.
Then, for $X=x\log 2+y\log 3$, we have

$0.08\alpha^{n}<\exp(X)<\alpha^{n}\left(n\log\alpha\right)^{60\log(n\log\alpha)},$

and

$n\log\alpha-3<X<n\log\alpha+60\left(\log(n\log\alpha)\right)^{2}.$

###### Proof.

If $n>310$, then

$\Delta=2^{x}3^{y}-2^{x_{1}}3^{y_{1}}=T_{n}-T_{n_{1}}\geq
T_{n-2}>T_{308}>\alpha^{306}>1.83^{306}>10^{80}.$

So, we apply Lemma 2.9, with $\Delta=T_{n}-T_{n_{1}}<T_{n}<\alpha^{n-1}$, by
(2.11). This yields

$\displaystyle\exp(X)$
$\displaystyle<\alpha^{n-1}\left(\log\alpha^{n-1}\right)^{60\log\log\alpha^{n-1}}<\alpha^{n}\left(\log\alpha^{n}\right)^{60\log\log\alpha^{n}}=\alpha^{n}\left(n\log\alpha\right)^{60\log(n\log\alpha)}.$
(2.15)

On the other hand,

$\displaystyle 0.08\alpha^{n}<\alpha^{n-4}<T_{n-2}\leq
T_{n}-T_{n_{1}}=2^{x}3^{y}-2^{x_{1}}3^{y_{1}}<2^{x}3^{y}=\exp(X).$ (2.16)

Combining (2.15) and (2.16), we get

$0.08\alpha^{n}<\exp(X)<\alpha^{n}\left(n\log\alpha\right)^{60\log(n\log\alpha)},$

and taking logarithms both sides gives

$\displaystyle
n\log\alpha-3<X<n\log\alpha+60\left(\log(n\log\alpha)\right)^{2},$

which proves Lemma 2.10. ∎

##  3 Proof of Theorem 1.1

We go back to (1.2) and treat the case $c=0$. This reduces the Diophantine
equation to

$T_{n}=2^{x}3^{y}.$ (3.1)

Now, an investigation on the largest prime factor of $k-$generalized Fibonacci
numbers was made in [7]. Specifically, when $k=3$, it was shown in [7] that
the only solutions to equation (3.1) have $n\leq 9$. A verification by hand in
this range yields the corresponding solutions from Theorem 1.1. ∎

##  4 Proof of Theorem 1.2

###  4.1 An absolute upper bound on $n$

For technical reasons, we assume $n>310$ and we determine an upper bound on
$n$. Let $(n,x,y)$, $(n_{1},x_{1},y_{1})$, $(n_{2},x_{2},y_{2})$ and
$(n_{3},x_{3},y_{3})$ be elements from $\mathbb{Z}_{\geq 0}^{3}$ such that

$\displaystyle
T_{n}-2^{x}3^{y}=T_{n_{1}}-2^{x_{1}}3^{y_{1}}=T_{n_{2}}-2^{x_{2}}3^{y_{2}}=T_{n_{3}}-2^{x_{3}}3^{y_{3}}.$
(4.1)

Without loss of generality, we assume $n>n_{1}>n_{2}>n_{3}$.

######  Lemma 4.1.

Let $c\geq 1$ be such that the Diophantine equation (1.2) has at least two
representations as

$c=T_{n}-2^{x}3^{y}=T_{n_{1}}-2^{x_{1}}3^{y_{1}}.$

Then

$n-n_{1}<9.6\cdot 10^{15}\log n.$

###### Proof.

Since $c\geq 1$, we have $T_{n}>2^{x}3^{y}$ and
$T_{n_{1}}>2^{x_{1}}3^{y_{1}}$. We go back to equation (1.2) and rewrite it as

$\displaystyle T_{n}-2^{x}3^{y}$ $\displaystyle=T_{n_{1}}-2^{x_{1}}3^{y_{1}}$
$\displaystyle
C_{\alpha}\alpha^{n-1}+C_{\beta}\beta^{n-1}+C_{\gamma}\gamma^{n-1}-2^{x}3^{y}$
$\displaystyle=C_{\alpha}\alpha^{n_{1}-1}+C_{\beta}\beta^{n_{1}-1}+C_{\gamma}\gamma^{n_{1}-1}-2^{x_{1}}3^{y_{1}}$
$\displaystyle C_{\alpha}\alpha^{n-1}-2^{x}3^{y}$
$\displaystyle=C_{\alpha}\alpha^{n_{1}-1}+C_{\beta}\beta^{n_{1}-1}+C_{\gamma}\gamma^{n_{1}-1}-C_{\beta}\beta^{n-1}-C_{\gamma}\gamma^{n-1}-2^{x_{1}}3^{y_{1}}$
$\displaystyle<C_{\alpha}\alpha^{n_{1}-1}+2<10C_{\alpha}\alpha^{n_{1}-1},$

for all $n>310$. So, we conclude that

$\displaystyle\left|2^{x}3^{y}C_{\alpha}^{-1}\alpha^{-(n-1)}-1\right|<10\alpha^{-(n-n_{1})}.$
(4.2)

We now apply Theorem 2.1 on the left–hand side of (4.2). Let
$\Gamma_{0}:=2^{x}3^{y}C_{\alpha}^{-1}\alpha^{-(n-1)}-1=e^{\Lambda_{0}}-1$.
Notice that $\Lambda_{0}\neq 0$ otherwise $\Gamma_{0}=0$ and we would have
$C_{\alpha}\alpha^{n-1}=2^{x}3^{y}\in\mathbb{Z}$. Conjugating this relation by
any automorphism that sends $\alpha$ to $\beta$, we get
$C_{\beta}\beta^{n-1}=2^{x}3^{y}$, which is a contradiction because
$|C_{\beta}\beta^{n-1}|<1$ while $2^{x}3^{y}\geq 1$ for all $x,y\geq 0$. We
use the field ${\mathbb{K}}:=\mathbb{Q}(\alpha)$ of degree $D=3$. Here,
$t:=4$,

$\displaystyle\lambda_{1}$
$\displaystyle:=2,~{}~{}\lambda_{2}:=3,~{}~{}\lambda_{3}:=C_{\alpha},~{}~{}\lambda_{4}:=\alpha,$
$\displaystyle b_{1}$
$\displaystyle:=x,~{}~{}b_{2}:=y,~{}~{}b_{3}:=-1,~{}~{}b_{4}:=-(n-1).$

Next, $\max\\{|b_{1}|,|b_{2}|,|b_{3}|,|b_{4}|\\}=\max\\{x,y,1,n-1\\}<n$, so we
can take $B:=n$. The minimal polynomial of $C_{\alpha}$ over the integers is
indicated at (2.3). Since its roots satisfy $|C_{\alpha}|$, $|C_{\beta}|$,
$|C_{\gamma}|<1$, it follows that $h(C_{\alpha})=\frac{1}{3}\log 44$. So, we
can take $A_{1}:=Dh(\lambda_{1})=3\log 2$, $A_{2}:=Dh(\lambda_{2})=3\log 3$,
$A_{3}:=Dh(\lambda_{3})=\log 44$, and
$A_{4}:=Dh(\lambda_{4})=3\cdot\frac{1}{3}\log\alpha=\log\alpha$. Therefore,
Theorem 2.1 gives,

$\displaystyle\log|\Gamma|$ $\displaystyle>-1.4\cdot 30^{7}\cdot 4^{4.5}\cdot
3^{2}(1+\log 3)(1+\log n)(3\log 2)(3\log 3)(\log 44)(\log\alpha)$
$\displaystyle>-5.6\cdot 10^{15}\log n,$ (4.3)

where the last inequality holds for $n>310$. Comparing (4.2) and (4.1), we get

$\displaystyle(n-n_{1})\log\alpha-\log 10<5.6\cdot 10^{15}\log n,$

so that $n-n_{1}<9.6\cdot 10^{15}\log n$. ∎

Next, we state and prove the following result.

######  Lemma 4.2.

Let $c\geq 1$, $X:=x\log 2+y\log 3$ and $X_{1}:=x_{1}\log 2+y_{1}\log 3$. Then

$\displaystyle X-X_{1}<8.5\cdot 10^{30}\left(\log n\right)^{2}.$

###### Proof.

We go back to equation (1.2) and rewrite it as

$\displaystyle C_{\alpha}\alpha^{n-1}-C_{\alpha}\alpha^{n_{1}-1}-2^{x}3^{y}$
$\displaystyle=C_{\beta}\beta^{n_{1}-1}-C_{\beta}\beta^{n-1}+C_{\gamma}\gamma^{n_{1}-1}-C_{\gamma}\gamma^{n-1}-2^{x_{1}}3^{y_{1}};$
$\displaystyle\dfrac{C_{\alpha}\alpha^{n_{1}-1}\left(\alpha^{n-n_{1}}-1\right)}{2^{x}3^{y}}-1$
$\displaystyle=\dfrac{C_{\beta}\left(\beta^{n_{1}-1}-\beta^{n-1}\right)}{2^{x}3^{y}}+\dfrac{C_{\gamma}\left(\gamma^{n_{1}-1}-\gamma^{n-1}\right)}{2^{x}3^{y}}-\dfrac{2^{x_{1}}3^{y_{1}}}{2^{x}3^{y}},$

and taking absolute values, we have

$\displaystyle\left|\dfrac{C_{\alpha}\alpha^{n_{1}-1}\left(\alpha^{n-n_{1}}-1\right)}{2^{x}3^{y}}-1\right|$
$\displaystyle\leq\dfrac{1}{\exp(X)}+\dfrac{1}{\exp(X)}+\dfrac{1}{\exp\left(X-X_{1}\right)}\leq
3\exp\left(-(X-X_{1})\right),$

where we have used the fact that the contribution of the complex conjugate
zeros $\beta$ and $\gamma$ to the right–hand side of the equation above is
very small, see relation (2.1).

Let
$\Gamma_{1}:=C_{\alpha}\alpha^{n_{1}-1}\left(\alpha^{n-n_{1}}-1\right)\cdot
2^{-x}3^{-y}-1$. Then

$\displaystyle|\Gamma_{1}|=\left|\dfrac{C_{\alpha}\alpha^{n_{1}-1}\left(\alpha^{n-n_{1}}-1\right)}{2^{x}3^{y}}-1\right|$
$\displaystyle\leq 3\exp\left(-(X-X_{1})\right).$ (4.4)

Notice that $\Gamma_{1}=e^{\Lambda_{1}}-1\neq 0$, otherwise we would have

$\displaystyle\dfrac{C_{\alpha}\left(\alpha^{n-1}-\alpha^{n_{1}-1}\right)}{2^{x}3^{y}}=1.$

Taking algebraic conjugates, we would get

$\displaystyle
1=\left|\dfrac{C_{\beta}\left(\beta^{n-1}-\beta^{n_{1}-1}\right)}{2^{x}3^{y}}\right|<1,$

a contradiction. Therefore, $\Gamma_{1}\neq 0$ (so, also $\Lambda_{1}\neq 0$).
We again use the field $\mathbb{Q}(\alpha)$ of degree $D=3$. Here, $t:=4$,

$\displaystyle\lambda_{1}$
$\displaystyle:=2,~{}~{}\lambda_{2}:=3,~{}~{}\lambda_{3}:=\alpha,~{}~{}\lambda_{4}:=C_{\alpha}(\alpha^{n-n_{1}}-1),$
$\displaystyle b_{1}$
$\displaystyle:=-x,~{}~{}b_{2}:=-y,~{}~{}b_{3}:=n_{1}-1,~{}~{}b_{4}:=1.$

Next, $\max\\{|b_{1}|,|b_{2}|,|b_{3}|,|b_{4}|\\}=\max\\{x,y,1,n_{1}-1\\}<n$,
so we can take $B:=n$. As before, we can still take $A_{1}:=3\log 2$,
$A_{2}:=3\log 3$, $A_{3}:=\log\alpha$. Moreover, assuming $n-n_{1}\geq 2$, so
$\alpha^{n-n_{1}}-1>1$, we have

$\displaystyle 3h(\lambda_{4})$ $\displaystyle=$ $\displaystyle
3h(C_{\alpha}(\alpha^{n-n_{1}}-1))\leq
3h(C_{\alpha})+3h(\alpha^{n-n_{1}}-1)\leq\log
44+\log(\alpha^{n-n_{1}}-1)+2\log(1.55)$ (4.5) $\displaystyle<$
$\displaystyle(n-n_{1})\log\alpha+\log 44+2\log 1.55<5.7\times 10^{15}\log n.$

where we have used Lemma 4.1. The number $1.55$ above is an upper bound on

$|\beta^{n-n_{1}}-1|\leq 1+|\beta|^{n-n_{1}}|\leq 1+1/\alpha<1.55$

and also on $|\gamma^{n-n_{1}}-1|$. This was if $n-n_{1}\geq 2$, but if
$n-n_{1}=1$, the above bound (4.5) still holds. So, we take $A_{4}:=5.7\cdot
10^{15}\log n$.

Then, by Theorem 2.1,

$\displaystyle\log|\Gamma_{1}|$ $\displaystyle>-1.4\cdot 30^{7}\cdot
4^{4.5}\cdot 3^{2}(1+\log 3)(1+\log n)(3\log 2)(3\log 3)(\log\alpha)(5.7\cdot
10^{15}\log n)$ $\displaystyle>-8.4\cdot 10^{30}\left(\log n\right)^{2}.$
(4.6)

Comparing (4.4) and (4.1), we get

$\displaystyle X-X_{1}<8.5\cdot 10^{30}\left(\log n\right)^{2}.$ (4.7)

This proves Lemma 4.2. ∎

To proceed further, let us write $x_{min}:=\min\\{x,x_{1},x_{2}\\}$,
$y_{min}:=\min\\{y,y_{1},y_{2}\\}$. We state and prove the following result.

######  Lemma 4.3.

Assume that $c\geq 1$. Then either

$x_{min},~{}y_{min}<10^{22}(\log n)^{3},$

or $n<36100$.

###### Proof.

Again, we go back to equation (1.2) and assume it has two solutions $(n,x,y)$,
$(n_{1},x_{1},y_{1})$. We rewrite it as

$\displaystyle
C_{\alpha}\alpha^{n-1}-C_{\alpha}\alpha^{n_{1}-1}+C_{\beta}\beta^{n-1}-C_{\beta}\beta^{n_{1}-1}+C_{\gamma}\gamma^{n-1}-C_{\gamma}\gamma^{n_{1}-1}$
$\displaystyle=2^{x}3^{y}-2^{x_{1}}3^{y_{1}};$ $\displaystyle
C_{\alpha}\alpha^{n-1}\left(1-\alpha^{n_{1}-n}\right)+C_{\beta}\beta^{n-1}\left(1-\beta^{n_{1}-n}\right)+C_{\gamma}\gamma^{n-1}\left(1-\gamma^{n_{1}-n}\right)$
$\displaystyle=2^{x}3^{y}-2^{x_{1}}3^{y_{1}}.$ (4.8)

At this point, if we assume a third solution $(n_{2},x_{2},y_{2})$ to (1.2),
then we can also rewrite a relation analogous to (4.1) as

$\displaystyle
C_{\alpha}\alpha^{n-1}\left(1-\alpha^{n_{2}-n}\right)+C_{\beta}\beta^{n-1}\left(1-\beta^{n_{2}-n}\right)+C_{\gamma}\gamma^{n-1}\left(1-\gamma^{n_{2}-n}\right)$
$\displaystyle=2^{x}3^{y}-2^{x_{2}}3^{y_{2}}.$ (4.9)

We now eliminate $C_{\gamma}\gamma^{n-1}$ from (4.1) and (4.9) to get

$\displaystyle~{}~{}~{}~{}C_{\alpha}\alpha^{n-1}\left[\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right]+$
$\displaystyle~{}~{}~{}~{}C_{\beta}\beta^{n-1}\left[\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right]$
$\displaystyle=\left(2^{x}3^{y}-2^{x_{1}}3^{y_{1}}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(2^{x}3^{y}-2^{x_{2}}3^{y_{2}}\right)\left(1-\gamma^{n_{1}-n}\right).$
(4.10)

We first show that the left–hand side of (4.1) is nonzero. Indeed, if it was
equal to zero, then we would have
$\left(2^{x}3^{y}-2^{x_{1}}3^{y_{1}}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(2^{x}3^{y}-2^{x_{2}}3^{y_{2}}\right)\left(1-\gamma^{n_{1}-n}\right)=0,$
so that

$\displaystyle\dfrac{1-\gamma^{n_{1}-n}}{1-\gamma^{n_{2}-n}}=\dfrac{2^{x}3^{y}-2^{x_{1}}3^{y_{1}}}{2^{x}3^{y}-2^{x_{2}}3^{y_{2}}}.$

Taking algebraic conjugates, we get

$\displaystyle\dfrac{1-\gamma^{n_{1}-n}}{1-\gamma^{n_{2}-n}}=\dfrac{1-\alpha^{n_{1}-n}}{1-\alpha^{n_{2}-n}}.$
(4.11)

Moreover, $n>n_{1}>n_{2}$ implies $n-n_{2}>n-n_{1}$, and Lemma 2.6 tells us
that (4.11) has no integer solutions $n-n_{1}$, $n-n_{2}$, with
$n-n_{2}>n-n_{1}$. Hence, the left–hand side of (4.1) is nonzero. Note also
that the term
$\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)$
appearing on the right–hand side of (4.1) is nonzero. Indeed, if it was zero,
then we would have

$\displaystyle\dfrac{1-\gamma^{n_{1}-n}}{1-\gamma^{n_{2}-n}}=\dfrac{1-\beta^{n_{1}-n}}{1-\beta^{n_{2}-n}},$

a relation which is an algebraic conjugate of (4.11), which cannot hold. Thus,
the above relation is nonzero.

Next, we go back to (4.1) and rewrite it as

$\displaystyle\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}=\dfrac{\left(2^{x}3^{y}-2^{x_{1}}3^{y_{1}}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(2^{x}3^{y}-2^{x_{2}}3^{y_{2}}\right)\left(1-\gamma^{n_{1}-n}\right)}{C_{\beta}\beta^{n-1}\left[\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right]}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}=\dfrac{2^{x_{min}}3^{y_{min}}\cdot
A}{C_{\beta}\beta^{n-1}\left[\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right]},$

where

$A:=\left(2^{x-x_{min}}3^{y-y_{min}}-2^{x_{1}-x_{min}}3^{y_{1}-y_{min}}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(2^{x-x_{min}}3^{y-y_{min}}-2^{x_{2}-x_{min}}3^{y_{2}-y_{min}}\right)\left(1-\gamma^{n_{1}-n}\right).$

Since $\nu_{p}(\beta)=0$, for $p=2,3$ and

$\nu_{p}(C_{\beta})=\begin{cases}-2/3,&\text{if }~{}p=2;\\\ 0,&\text{if
}~{}p=3,\end{cases}$

we have

$\displaystyle\nu_{2}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\quad\quad=x_{min}+\nu_{2}(A)-(-2/3)-\nu_{2}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right).$

Therefore, since $\nu_{2}(A)\geq 0$ because $A$ is an algebraic integer, we
get

$\displaystyle x_{min}$
$\displaystyle\leq\nu_{2}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\quad\quad+\nu_{2}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right).$
(4.12)

In a similar way,

$\displaystyle y_{min}$
$\displaystyle\leq\nu_{3}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\quad\quad+\nu_{3}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right).$
(4.13)

Next, we estimate

$\displaystyle\nu_{p}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right),$
(4.14)

for $p=2,3$. Well, the number shown at (4.14) is an algebraic integer all
whose conjugates are bounded in absolute value by

$2(1+\alpha^{(n-n_{1})/2})(1+\alpha^{(n-n_{2})/2})<\alpha^{(n-n_{1})/2+(n-n_{2})/2+6}\leq\alpha^{n-n_{2}+6}.$

Thus, the last coefficient of the minimal polynomial for the number shown
inside the valuation at (4.14) is at most $\alpha^{6(n-n_{2})+36}$. This shows
that

$\nu_{p}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right)\leq\frac{(6(n-n_{2})+30)\log\alpha}{\log
p}.$

Using Lemma 3.1, we get that the above bounds are at most

$5.7\cdot 10^{16}\log n\quad(p=2)\qquad{\text{\rm and}}\qquad 3.2\cdot
10^{16}\log n\quad(p=3).$

At this point, relations (4.1) and (4.1) become

$\displaystyle x_{min}$
$\displaystyle\leq\nu_{2}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)+5.7\cdot
10^{16}\log n$ (4.15)

and

$\displaystyle y_{min}$
$\displaystyle\leq\nu_{3}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)+3.2\cdot
10^{16}\log n.$ (4.16)

Lastly, we estimate

$\displaystyle\nu_{p}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right).$
(4.17)

Let

$\lambda_{1}:=\dfrac{\alpha}{\beta}\qquad{\text{\rm
and}}\qquad\lambda_{2}:=-\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}\right).$

We distinguish two cases.

Case 1. $\lambda_{1}$ and $\lambda_{2}$ are multiplicatively independent.

We can apply Lemma 2.2 to (4.17) with the field $\mathbb{Q}(\alpha,\beta)$ of
degree $D:=6$. Since $h(\lambda_{1})=3(\log\alpha)/3=\log\alpha$, we can
choose $h(\lambda_{1}):=\log\alpha$ for all cases $p=2,3$ (note that
$\log\alpha>\log p/D$ for $p=2,3$). Further,

$\displaystyle h^{\prime}(\lambda_{2})$ $\displaystyle\leq
2h(C_{\alpha})+4h\left(1-\alpha^{n_{1}-n}\right)+4\log(1-\alpha^{n-n_{2}})+2\log
2$ $\displaystyle<\frac{2\log 44}{3}+2\log
2+\frac{4}{3}\left(\log(1+|\beta|^{n_{1}-n})+\log(1+|\gamma|^{n_{2}-n})\right)$
$\displaystyle<4+\frac{4}{3}\left(\log(\alpha^{(n-n_{1})/2}+1)+\log(\alpha^{(n-n_{2})/2}+1)\right)$
$\displaystyle<4+\frac{4}{3}\left((n-n_{1})/2+2+(n-n_{2})/2+2)\right)\log\alpha$
$\displaystyle<4+\frac{8(n-n_{2}+4)}{3}\log\alpha<1.6\times 10^{16}\log n,$

where we have used Lemma 4.1 and $n>310$. Therefore,

$E^{\prime}=\dfrac{b_{1}}{h^{\prime}(\lambda_{2})}+\dfrac{b_{2}}{h^{\prime}(\lambda_{1})}=\dfrac{n-1}{h^{\prime}(\lambda_{1})}+\dfrac{1}{h^{\prime}(\lambda_{2})}<n.$

We now have

$\displaystyle E$ $\displaystyle=\max\left\\{\log E^{\prime}+\log\log
p+0.4,10,10\log p\right\\}$ $\displaystyle=\max\left\\{\log n+\log\log
p+0.4,10,10\log p\right\\}.$

If $n>36100$, then $E<1+\log n$ in both cases $p=2,3$. Therefore, Lemma 2.2
gives

$\displaystyle\nu_{p}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\leq\dfrac{24pg}{(p-1)(\log
p)^{4}}E^{2}D^{4}h^{\prime}(\lambda_{1})h^{\prime}(\lambda_{2})$
$\displaystyle<\dfrac{24pg}{(p-1)(\log p)^{4}}(\log
n)^{2}\left(1+\frac{1}{\log 36100}\right)^{2}\cdot 6^{4}(\log\alpha)\cdot
1.6\cdot 10^{16}\log n$ $\displaystyle<\dfrac{3.7\cdot 10^{20}pg}{(p-1)(\log
p)^{4}}\left(\log n\right)^{3}.$

Hence, inequalities (4.15) and (4.16) become

$\displaystyle x_{min}$ $\displaystyle<\dfrac{3.7\cdot 10^{20}\cdot 2\cdot
3}{(2-1)(\log 2)^{4}}\left(\log n\right)^{3}+5.7\times 10^{16}\log
n<10^{22}(\log n)^{3}$

and

$\displaystyle y_{min}$ $\displaystyle<\dfrac{3.7\cdot 10^{20}\cdot 3\cdot
13}{(3-1)(\log 3)^{4}}\left(\log n\right)^{3}+3.2\times 10^{16}\log n<5\times
10^{21}(\log n)^{3}.$

In the above, we used $g=3$ when $p=2$ and $g=13$ when $p=3$ as in the proof
of Lemma 2.8. This was in case $\lambda_{1}$ and $\lambda_{2}$ are
multiplicatively independent.

Case 2. $\lambda_{1}$ and $\lambda_{2}$ are multiplicatively dependent.

In this case, $\lambda_{1}=z^{3}$ and $\lambda_{2}=\pm z^{t}$ for some integer
$t$ with $|t|\leq 80n+244$ by Lemma 2.7. Thus,

$\lambda_{1}^{n-1}\lambda_{2}-1=\pm z^{t+3(n-1)}+1\mid z^{2t+6n-2}-1.$

The exponent of $z$ above is in absolute value is at most $2|t|+6n-2\leq
186n+500<n^{2}$. Lemma 2.8 now shows that

$\nu_{p}(\lambda_{1}^{n-1}\lambda_{2}+1)\leq 1+\frac{\log n^{2}}{\log p}<3\log
n$

for $n>36100$ and $p\in\\{2,3\\}$. Comparing this to Case 1, we see that the
bounds from Case 1 still hold. This completes the proof of Lemma 4.3. ∎

Lastly, we consider a fourth solution $(n_{3},x_{3},y_{3})$, with
$n>n_{1}>n_{2}>n_{3}$ and we find an absolute bound for $n$. We prove the
following result.

######  Lemma 4.4.

If $c\geq 1$ and $n>36100$, then

$n<1.2\cdot 10^{37},\qquad x<1.1\cdot 10^{37},\qquad y<6.8\cdot 10^{36}.$

###### Proof.

Lemma 4.3 indicates that within any set of three solutions, the smallest
values of $x$ and $y$ are constrained to be less than $10^{22}(\log n)^{3}$.
Consequently, in a quadrat of solutions, it is possible for at most one
solution to have an $x$–value exceeding $10^{22}(\log n)^{3}$ and similarly,
at most one solution can have a $y$–value surpassing this bound. Therefore, it
follows that at least one solution in each set will have both $x$ and $y$
confined within this limit. This particularly implies that the smallest
solution adheres to these bounds. Hence,

$X_{3}=x_{3}\log 2+y_{3}\log 3<(\log 2+\log 3)\cdot 10^{22}(\log
n)^{3}<1.8\cdot 10^{22}(\log n)^{3}.$

With Lemmas 2.10 and 4.2, we get

$\displaystyle n\log\alpha-3$ $\displaystyle<X$
$\displaystyle=X_{3}+(X_{2}-X_{3})+(X_{1}-X_{2})+(X-X_{1})$
$\displaystyle<1.8\cdot 10^{22}(\log n)^{3}+3\cdot 8.5\cdot 10^{30}\left(\log
n\right)^{2}$ $\displaystyle<2.44\cdot 10^{30}\left(\log n\right)^{3},$

which implies

$\displaystyle\dfrac{n}{(\log n)^{3}}<4.1\cdot 10^{30}.$ (4.18)

We apply Lemma 2.2 to inequality (4.18) above with $z:=n$, $m:=3$,
$T:=4.1\cdot 10^{30}$. Since $T>(4\cdot 3^{2})^{3}=46656$, we get

$n<2^{m}T(\log T)^{m}=2^{3}\cdot 4.1\cdot 10^{30}(\log 4.1\cdot
10^{30})^{3}<1.2\cdot 10^{37}.$

Further, we have by Lemma 2.10 that

$\displaystyle X$
$\displaystyle<n\log\alpha+60\left(\log(n\log\alpha)\right)^{2};$
$\displaystyle x\log 2+y\log 3$ $\displaystyle<1.2\cdot
10^{37}\log\alpha+60\left(\log(1.2\cdot 10^{37}\log\alpha)\right)^{2}<7.4\cdot
10^{36}.$

This gives $x<1.1\cdot 10^{37}$ and $y<6.8\cdot 10^{36}$ and completes the
proof of Lemma 4.4. ∎

###  4.2 Reduction of the upper bound on $n$

Here, we use the LLL–reduction method, the theory of continued fractions and
as well as $p-$adic reduction methods due to [21] to find a rather small bound
for $n$.

To begin, we go back to equation (4.2). Assuming $n-n_{1}\geq 5$, we can write

$\displaystyle|\Lambda_{0}|=\left|\log C_{\alpha}-x\log 2-y\log
3+(n-1)\log\alpha\right|\leq 15\alpha^{-(n-n_{1})},$

where we used Lemma 2.1 with $n-n_{1}\geq 5$ since
$\alpha^{n-n_{1}}\geq\alpha^{5}>2$. So, we consider the approximation lattice

$\mathcal{A}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\ \lfloor M\log 2\rfloor&\lfloor
M\log 3\rfloor&\lfloor M\log\alpha\rfloor\end{pmatrix},$

with $M:=10^{112}$ and choose $v:=\left(0,0,-\lfloor M\log
C_{\alpha}\rfloor\right)$. Now, by Lemma 2.4, we get

$|\Lambda|\geq c_{1}:=10^{-40}\qquad\text{and}\qquad c_{2}:=2.15\cdot
10^{38}.$

Moreover, by Lemma 4.4, we have

$x<A_{1}:=1.1\cdot 10^{37},~{}~{}y<A_{2}:=6.8\cdot
10^{36},~{}~{}n-1<A_{3}:=1.2\cdot 10^{37}.$

So, Lemma 2.5 gives $S=1.7\cdot 10^{74}$ and $T=1.49\cdot 10^{37}$. Since
$c_{2}^{2}\geq T^{2}+S$, then choosing $c_{3}:=15$ and $c_{4}:=\log\alpha$, we
get $n-n_{1}\leq 285$.

Next, we now go back to equation (4.4). Assume that $X-X_{1}\geq 2$. We can
then write

$\displaystyle|\Lambda_{1}|=\left|(n_{1}-1)\log\alpha+\log
C_{\alpha}\left(\alpha^{n-n_{1}}-1\right)-x\log 2-y\log 3\right|$
$\displaystyle<4.5\exp\left(-(X-X_{1})\right),$

where we used Lemma 2.1 together with the fact that
$\exp(X-X_{1})\geq\exp(2)>6$. So, we use the same approximation lattice

$\mathcal{A}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\ \lfloor M\log 2\rfloor&\lfloor
M\log 3\rfloor&\lfloor M\log\alpha\rfloor\end{pmatrix},$

with $M:=10^{113}$ and choose $v:=\left(0,0,-\lfloor M\log
C_{\alpha}\left(\alpha^{n-n_{1}}-1\right)\rfloor\right)$. It turns out that
for all values $1\leq n-n_{1}\leq 285$, the chosen constant $M$ is
sufficiently large, so we can still apply Lemma 2.5. By Lemma 2.4, we maintain
the lower bound $|\Lambda_{1}|\geq c_{1}:=10^{-40}$, $c_{2}:=6.13\cdot
10^{37}$ and by Lemma 4.4, we also have

$x<A_{1}:=1.1\cdot 10^{37},~{}~{}y<A_{2}:=6.8\cdot
10^{36},~{}~{}n-1<A_{3}:=1.2\cdot 10^{37}.$

So, Lemma 2.5 still gives the same values of $S$ and $T$ as before. Since
$c_{2}^{2}\geq T^{2}+S$, we now choose $c_{3}:=4.5$ and $c_{4}:=1$ and we get
$X-X_{1}\leq 174$.

Next, we find reduced bounds on $x_{min}$ and $y_{min}$ using $p-$adic
reduction methods due to [21]. We go back to equation (1.2) and assume it has
two solutions $(n,x,y)$, $(n_{1},x_{1},y_{1})$. We rewrite it as

$c=T_{n}-2^{x}3^{y}=T_{n_{1}}-2^{x_{1}}3^{y_{1}},$

so that we can determine $\nu_{p}\left(T_{n}-T_{n_{1}}\right)$. Notice that
from the above equation,

$\displaystyle\nu_{p}\left(T_{n}-T_{n_{1}}\right)$
$\displaystyle=\nu_{p}\left(2^{x}3^{y}-2^{x_{1}}3^{y_{1}}\right)$
$\displaystyle=\nu_{p}\left(2^{x_{min}}3^{y_{min}}\left(2^{x-x_{min}}3^{y-y_{min}}-2^{x_{1}-x_{min}}3^{y_{1}-y_{min}}\right)\right)$
$\displaystyle=\nu_{p}\left(2^{x_{min}}3^{y_{min}}\right)+\nu_{p}\left(2^{x-x_{min}}3^{y-y_{min}}-2^{x_{1}-x_{min}}3^{y_{1}-y_{min}}\right),$

and thus

$\displaystyle\nu_{p}\left(2^{x_{min}}3^{y_{min}}\right)$
$\displaystyle=\nu_{p}\left(T_{n}-T_{n_{1}}\right)-\nu_{p}\left(2^{x-x_{min}}3^{y-y_{min}}-2^{x_{1}-x_{min}}3^{y_{1}-y_{min}}\right)$
$\displaystyle\leq\nu_{p}\left(T_{n}-T_{n_{1}}\right).$ (4.19)

Now, we determine $\nu_{p}\left(T_{n}-T_{n_{1}}\right)$ for all $n-n_{1}\geq
1$, $n<1.2\cdot 10^{37}$ and $p=2,3$. Here, we show how to do it for $p=2$,
and then we automate the process in SageMath. Since $n<1.2\cdot
10^{37}<2^{124}$, then $n$ has at most 124 binary digits. Let $d:=n-n_{1}\leq
285$, by the results in the reduction above. So, we need an upper bound for

$\nu_{2}\left(T_{n}-T_{n_{1}}\right)=\nu_{2}\left(T_{n_{1}+d}-T_{n_{1}}\right),$

but since $n_{1}<n$, we instead bound

$\nu_{2}\left(T_{n+d}-T_{n}\right),\qquad d\in[1,285],\qquad n<1.2\cdot
10^{37}.$

The Tribonacci sequence is periodic modulo $2^{k}$ with period $2^{k+1}$. In
particular, $T_{n+d}-T_{n}$ is periodic modulo $2^{8}$ with period
$2^{9}<1000$. With a simple program in SageMath, we checked over all
$d\in[1,285]$ for which there is $n\leq 1000$ such that $2^{8}\mid
T_{n+d}-T_{n}$. There are $214$ such $d$’s, namely
$\\{1,3,4,5,7,8,9,11,12,13,15,16,17,19,20,21,23,24,25,27,28,,\ldots,281,283,284,285\\}$.
Some numbers are missing from this list. This means that for the missing
numbers $d$, we have $\nu_{2}(T_{n+d}-T_{n})\leq 7$ always.

Here, we will work out one $d$ only, for explanation. Namely, we take $d=9$.
We calculate $n_{0}(d)\in[1,2^{9}]$ such that for $n=n_{0}(d)$ we have that
$\nu_{2}(L_{n+d}-L_{n})\geq 8$. This is unique in this case and it is
$n_{0}(d)=167$. So, from now on, every $n<1.2\cdot 10^{37}$ such that
$\nu_{2}(T_{n+d}-T_{n})\geq 8$ is of the form $n=167+2^{9}z$, for some
$z\in\mathbb{Z}$. So now, we need to find out $z$ such that
$\nu_{2}(T_{n+9}-T_{n})$ is as large as possible. For this, we go to the Binet
formula and get

$\displaystyle T_{n+9}-T_{n}$
$\displaystyle=\left(C_{\alpha}\alpha^{n+8}+C_{\beta}\beta^{n+8}+C_{\gamma}\gamma^{n+8}\right)-\left(C_{\alpha}\alpha^{n-1}+C_{\beta}\beta^{n-1}+C_{\gamma}\gamma^{n-1}\right)$
$\displaystyle=\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{n-1}+\left(\beta^{9}-1\right)C_{\beta}\beta^{n-1}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{n-1}$
$\displaystyle=\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{167+2^{9}z-1}+\left(\beta^{9}-1\right)C_{\beta}\beta^{167+2^{9}z-1}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{167+2^{9}z-1}$
$\displaystyle=\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{166}\alpha^{2^{9}z}+\left(\beta^{9}-1\right)C_{\beta}\beta^{166}\beta^{2^{9}z}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{166}\gamma^{2^{9}z}$
$\displaystyle=\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{166}\exp_{2}\left(2^{9}z\log_{2}\alpha\right)+\left(\beta^{9}-1\right)C_{\beta}\beta^{166}\exp_{2}\left(2^{9}z\log_{2}\beta\right)$
$\displaystyle~{}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{166}\exp_{2}\left(2^{9}z\log_{2}\gamma\right).$

We can apply a simple cosmetic to the above relation so that we have terms in
$\log\alpha^{4}$, $\log\beta^{4}$ and $\log\gamma^{4}$ inside the respective
exponentials. We write

$\displaystyle T_{n+9}-T_{n}$
$\displaystyle=\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{166}\exp_{2}\left(2^{7}z\log_{2}\alpha^{4}\right)+\left(\beta^{9}-1\right)C_{\beta}\beta^{166}\exp_{2}\left(2^{7}z\log_{2}\beta^{4}\right)$
$\displaystyle~{}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{166}\exp_{2}\left(2^{7}z\log_{2}\gamma^{4}\right).$

Moreover,

$\log_{2}\alpha^{4}=\log_{2}\left(1-\left(1-\alpha^{4}\right)\right)=-\sum_{n\geq
1}\frac{\left(1-\alpha^{4}\right)^{n}}{n},$ (4.20)

and in the right–hand side, $|(1-\alpha^{4})^{n}/n|_{2}\leq 2^{-(n-\log n/\log
2)}<2^{-4n/3+\log n/\log 2},$ where we have used the fact that
$\nu_{2}(\alpha^{4}-1)=\nu_{2}((\alpha-1)(\alpha+1)(\alpha^{2}+1))=4/3$. This
therefore shows that series given in the right–hand side of (4.20) converges.
For the argument in the exponential, we have

$\nu_{2}\left(2^{7}z\log_{2}\alpha^{4}\right)\geq\nu_{2}(2^{8}z)\geq 8,$

so $\left|2^{7}z\log_{2}\alpha^{4}\right|_{2}\leq 2^{-8}<2^{-1}$, therefore
the exponential in this input is convergent $2$–adically. The same arguments
work with $\alpha$ replaced by $\beta$ or $\gamma$. We now stop the argument
of the logarithm at $n=120$, so put

$P:=-\sum_{n=1}^{120}\frac{\left(1-\alpha^{4}\right)^{n}}{n},$ (4.21)

such that

$\log_{2}\alpha^{4}=P-\sum_{n\geq
121}\frac{\left(1-\alpha^{4}\right)^{n}}{n}.$

Since $n-\log_{2}n\geq n-\log n/\log 2$ and the function $n-\log n/\log 2$ is
at least $121$ for all $n\geq 128$, then $n-\nu_{2}(n)\geq 121$ for all $n\geq
121$. Thus, $\log_{2}\alpha^{4}=P+u$, where $\nu_{2}(u)\geq 121$. We therefore
have,

$2^{7}z\log_{2}\alpha^{4}=2^{7}zP+2^{7}zu,$

so that

$\exp_{2}(2^{7}z\log_{2}\alpha^{4})=\exp_{2}(2^{7}zP+2^{7}zu)=\exp_{2}(2^{7}zP)\exp_{2}(2^{7}zu).$

We have

$\exp_{2}(y)=1+y+\frac{y^{2}}{2}+\cdots+\frac{y^{n}}{n!}+\cdots.$

For $\nu_{2}(y)\geq 2$ and $n\geq 2$ we have

$\nu_{2}\left(\frac{y^{n}}{n!}\right)=n\nu_{2}(y)-\nu_{2}(n!)\geq
n\nu_{2}(y)-(n-\sigma_{2}(n))>n(\nu_{2}(y)-1)\geq\nu_{2}(y),$

where the last inequality holds as it is equivalent to $\nu_{2}(y)\geq
n/(n-1)$, which is so since $\nu_{2}(y)\geq 2\geq n/(n-1)$ for all $n\geq 2$.
In the above, $\sigma_{2}(n)$ is the sum of the digits of $n$ in base $2$. It
then follows that $\exp_{2}(y)\equiv 1\pmod{2^{\nu_{2}(y)}},$ provided
$\nu_{2}(y)\geq 2$. Hence, $\exp_{2}(2^{7}zu)\equiv
1\pmod{2^{7+\nu_{2}(u)}}\equiv 1\pmod{2^{128}}.$ Thus

$\exp_{2}(2^{7}z\log_{2}\alpha)\equiv\exp_{2}(2^{7}zP)\pmod{2^{128}}\equiv\sum_{k\geq
0}\frac{(2^{7}zP)^{k}}{k!}\pmod{2^{128}}.$

Indeed,

$\nu_{2}\left(\frac{(2^{7}zP)^{k}}{k!}\right)=k\nu_{2}(2^{7}zP)-\nu_{2}(k!)\geq(7+\nu_{2}(P))k-(k-\sigma_{2}(k))>7k,$

since $\sigma_{2}(k)\geq 1$ and $\nu_{2}(P)\geq 1$, so it follows that the
above numbers are $\geq 7\cdot 19=133>128$ for $k\geq 19$. Thus, we may
truncate the series at $k=18$ and write

$\exp_{2}\left(2^{7}z\log_{2}\alpha^{4}\right)=\sum_{k=0}^{18}\frac{(2^{7}zP)^{k}}{k!}\pmod{2^{128}}.$

The same argument works with $\alpha$ replaced by $\beta$ or $\gamma$, so we
may write

$Q:=-\sum_{n=1}^{120}\frac{\left(1-\beta^{4}\right)^{n}}{n}\qquad\text{and}\qquad
R:=-\sum_{n=1}^{120}\frac{\left(1-\gamma^{4}\right)^{n}}{n},$ (4.22)

so that

$\exp_{2}(2^{7}z\log_{2}\beta^{4})=\sum_{k=0}^{18}\frac{(2^{7}zQ)^{k}}{k!}\pmod{2^{128}}\quad\text{and}\quad\exp_{2}(2^{7}z\log_{2}\gamma^{4})=\sum_{k=0}^{18}\frac{(2^{7}zR)^{k}}{k!}\pmod{2^{128}}.$

Thus,

$\displaystyle
T_{n+9}-T_{n}=\sum_{k=0}^{18}\frac{\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{166}(2^{7}zP)^{k}+\left(\beta^{9}-1\right)C_{\beta}\beta^{166}(2^{7}zQ)^{k}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{166}(2^{7}zR)^{k}}{k!}\pmod{2^{128}}.$

The right–hand side above is a polynomial of degree $18$ in $z$ whose
coefficients are rational numbers which are $2$–adic integers (that is, the
numerators of those rational numbers are always odd). We will show that in our
range the above expression is never $0$ modulo $2^{128}$. This will show that
$\nu_{2}(T_{n+9}-T_{n})<128$ for $n<1.2\cdot 10^{37}.$

We need to find these numbers which is not so easy in SageMath as $P$, $Q$ and
$R$ involve large powers of $\alpha$, $\beta$ and $\gamma$ respectively.
Nevertheless, we can compute $A:=P+Q+R$, $B:=PQ+PR+QR$ and $C:=PQR$. Next, the
coefficients

$\displaystyle
u_{k}:=\left(\alpha^{9}-1\right)C_{\alpha}\alpha^{166}P^{k}+\left(\beta^{9}-1\right)C_{\beta}\beta^{166}Q^{k}+\left(\gamma^{9}-1\right)C_{\gamma}\gamma^{166}R^{k}$
(4.23)

form a linearly recurrence sequence of recurrence

$u_{k+3}=Au_{k+2}-Bu_{k+1}+Cu_{k},\qquad{\text{\rm for}}\qquad k\geq 0,$

with $u_{0}$, $u_{1}$ and $u_{2}$ obtained from (4.23) when $k=0,1,2$
respectively. So, we can compute all the remaining ones iteratively and look
at the polynomial

$f(z):=\sum_{k=0}^{18}(2^{7}z)^{k}\frac{u_{k}}{k!}\pmod{2^{128}}.$

All coefficients $u_{k}/k!$ are 2–adic integers, and we can reduce them modulo
$2^{128}$. As a result, we obtain a polynomial in
$\mathbb{Z}/(2^{128}\mathbb{Z})[z]$. Our goal is to find a $z$ such that this
polynomial $=0\pmod{2^{128}}$. This is achieved incrementally. Specifically,
by reducing $f(z)$ modulo $2^{10},2^{11},2^{12},2^{13},$ and so forth, we
deduce the necessary digit of $z$ in the subsequent power of 2 (either 0 or 1)
to ensure divisibility by higher and higher powers of 2, in accordance with
Hensel’s lemma. This process yields

$z=2^{0}+2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+2^{8}+2^{9}+\cdots,$

and upon extracting digits up to $2^{101}$, we express

$z=2^{0}+2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+2^{8}+2^{9}+\cdots+2^{100}+2^{101}+2^{102}t,$

and reduce $f(z)$ modulo $2^{127}$, obtaining

$2^{123}(8+9t)\pmod{2^{127}}.$

Notice that we should choose $z$ as a multiple of 8, leading to

$n\geq 2^{19}\cdot(\cdots+2^{102}\cdot 8)=2^{124}>1.2\cdot 10^{37}.$

This analysis demonstrates that effectively, $\nu_{2}(T_{n+9}-T_{n})<126$.

A similar analysis was performed for the other values of $d$ as well as for
$p=3$. In the case of $p=3$, the period of $(T_{n})_{n\geq 0}$ modulo
$3^{k+1}$ is $13\cdot 3^{k}$, so we work $3-$adically with
$\log_{3}(\alpha^{13})$, $\log_{3}(\beta^{13})$ and $\log_{3}(\gamma^{13})$.
In all cases, we obtained that $\nu_{p}(T_{n}-T_{n_{1}})<128$. Hence, in all
cases, we conclude from (4.2) that

$x_{min},~{}y_{min}<128.$

Lastly, we find a smaller upper bound for $n$. If we write $c_{X}$ for the
upper bound of $X-X_{1}$, then

$\displaystyle X=X_{2}+(X_{1}-X_{2})+(X-X_{1})$ $\displaystyle<x_{min}\log
2+y_{min}\log 3+2c_{X},$ $\displaystyle x\log 2+y\log 3$
$\displaystyle<128\log 2+128\log 3+2\cdot 174<578.$

Hence, $x<834$ and $y<527$. On the other hand, Lemma 2.10 implies that
$n\log\alpha-3<X<578,$ so that $n<962$.

###  4.3 Conclusion

To finalize with the proof of Theorem 1.2, we note that for $n>310$, the
bounds are $n\leq 962$, $x<834$, and $y<527$. To efficiently handle large
$T_{n}$ values, our SageMath 9.5 code utilized batch processing, iterating
through all $(n,x,y)$ combinations within these ranges. This approach selected
$c\geq 1$ values with at least four representations of the form
$T_{n}-2^{x}3^{y}$, aligning with the solutions in Theorem 1.2. The
computation, performed on an 8GB RAM laptop, was completed in about 3 hours. ∎

##  5 Proof of Theorem 1.3

###  5.1 An absolute upper bound on $n$

We proceed as in Subsection 4.1 and determine an absolute upper bound on $n$.
Without loss of generality, we may assume $n>n_{1}>310$, and continue by
proving a series of results.

######  Lemma 5.1.

Let $c\leq-1$ such that $(n,x,y)$ and $(n_{1},x_{1},y_{1})$ satisfy (1.2) with
$n>n_{1}>310$, then

$X-X_{1}<5.55\cdot 10^{15}\log n.$

###### Proof.

Since $c\leq-1$, then $T_{n}-2^{x}3^{y}\leq-1$ and hence $2^{x}3^{y}-T_{n}>0$.
We go back to (1.2) and rewrite it as

$\displaystyle 0<2^{x}3^{y}-T_{n}$
$\displaystyle=2^{x_{1}}3^{y_{1}}-T_{n_{1}};$ $\displaystyle
2^{x}3^{y}-C_{\alpha}\alpha^{n-1}$
$\displaystyle=2^{x_{1}}3^{y_{1}}-C_{\alpha}\alpha^{n_{1}-1}+C_{\beta}\left(\beta^{n-1}-\beta^{n_{1}-1}\right)+C_{\gamma}\left(\gamma^{n-1}-\gamma^{n_{1}-1}\right)$
$\displaystyle<2^{x_{1}}3^{y_{1}}+C_{\beta}\left(\beta^{n-1}-\beta^{n_{1}-1}\right)+C_{\gamma}\left(\gamma^{n-1}-\gamma^{n_{1}-1}\right);$
$\displaystyle|2^{x}3^{y}-C_{\alpha}\alpha^{n-1}|$
$\displaystyle<2^{x_{1}}3^{y_{1}}+2,$ (5.1)

where the last inequality holds for all $n>n_{1}>310$. So, dividing through
both sides of (5.1) by $2^{x}3^{y}$, we conclude that

$\displaystyle\left|2^{-x}3^{-y}C_{\alpha}\alpha^{n-1}-1\right|\leq
3\exp\\{-(X-X_{1})\\}.$ (5.2)

We now apply Theorem 2.1 on the left–hand side of (5.2). Let
$\Gamma_{2}:=2^{-x}3^{-y}C_{\alpha}\alpha^{n-1}-1=e^{\Lambda_{2}}-1$. Like
before, $\Lambda_{2}\neq 0$ otherwise we would have
$C_{\alpha}\alpha^{n-1}=2^{x}3^{y}\in\mathbb{Z}$. Conjugating this relation by
any automorphism that swaps $\alpha$ and $\beta$, we get
$C_{\beta}\beta^{n-1}=2^{x}3^{y}$, which is a contradiction because
$|C_{\beta}\beta^{n-1}|<1$ while $2^{x}3^{y}\geq 1$ for all $x,y\geq 0$. We
use the field ${\mathbb{K}}:=\mathbb{Q}(\alpha)$ of degree $D=3$. Here,
$t:=4$,

$\displaystyle\lambda_{1}$
$\displaystyle:=2,~{}~{}\lambda_{2}:=3,~{}~{}\lambda_{3}:=C_{\alpha},~{}~{}\lambda_{4}:=\alpha,$
$\displaystyle b_{1}$
$\displaystyle:=-x,~{}~{}b_{2}:=-y,~{}~{}b_{3}:=1,~{}~{}b_{4}:=n-1.$

We have $\max\\{|b_{1}|,|b_{2}|,|b_{3}|,|b_{4}|\\}=\max\\{x,y,1,n-1\\}=n$. So,
we can take $B:=n$. Again, $h(C_{\alpha})=\frac{1}{3}\log 44$. So, we can take
$A_{1}:=3\log 2$, $A_{2}:=3\log 3$, $A_{3}:=\log 44$, and $A_{4}:=\log\alpha$.
Therefore, Theorem 2.1 gives,

$\displaystyle\log|\Gamma_{2}|$ $\displaystyle>-1.4\cdot 30^{7}\cdot
4^{4.5}\cdot 3^{2}(1+\log 3)(1+\log n)(3\log 2)(3\log 3)(\log 44)(\log\alpha)$
$\displaystyle>-5.5\cdot 10^{15}\log n,$ (5.3)

where the last inequality holds for $n>310$. Comparing (5.2) and (5.1), we get

$\displaystyle X-X_{1}<5.55\cdot 10^{15}\log n.$ (5.4)

This proves Lemma 5.1. ∎

Next, we prove the following.

######  Lemma 5.2.

Let $c\leq-1$ such that $(n,x,y)$, $(n_{1},x_{1},y_{1})$ and
$(n_{2},x_{2},y_{2})$ satisfy (1.2) with $n>n_{1}>n_{2}$ and $n_{1}>310$, then

$n-n_{1}<2\cdot 10^{15}(\log n)^{2}.$

###### Proof.

Since we assume a third solution $(n_{2},x_{2},y_{2})$ to (1.2), then Lemma
2.10 also holds for $n_{1}>n_{2}$. By hypothesis, $n_{1}>310$, so

$0.08\alpha^{n_{1}}<\exp(X_{1})=2^{x_{1}}3^{y_{1}}<\alpha^{n_{1}}\left(n_{1}\log\alpha\right)^{60\log(n_{1}\log\alpha)}.$

We can then rewrite (5.1) with the above inequality and get

$\displaystyle\left|2^{x}3^{y}-C_{\alpha}\alpha^{n-1}\right|$
$\displaystyle<2^{x_{1}}3^{y_{1}}+2$
$\displaystyle<\alpha^{n_{1}}\left(n_{1}\log\alpha\right)^{60\log(n_{1}\log\alpha)}+2$
$\displaystyle<1.01\alpha^{n_{1}}\left(n_{1}\log\alpha\right)^{60\log(n_{1}\log\alpha)}.$

Dividing through by $C_{\alpha}\alpha^{n-1}$, we obtain

$\displaystyle\left|2^{x}3^{y}C_{\alpha}^{-1}\alpha^{-(n-1)}-1\right|$
$\displaystyle\leq\dfrac{1.01\alpha}{C_{\alpha}}\alpha^{-(n-n_{1})}\left(n_{1}\log\alpha\right)^{60\log(n_{1}\log\alpha)}$
$\displaystyle<3\alpha^{-(n-n_{1})}\left(n_{1}\log\alpha\right)^{60\log(n_{1}\log\alpha)}.$
(5.5)

Let
$\Gamma_{3}:=2^{x}3^{y}C_{\alpha}^{-1}\alpha^{-(n-1)}-1=e^{\Lambda_{3}}-1$. By
the same arguments and data used in the proof of Lemma 5.1 above, we conclude
by Matveev’s Theorem 2.1 that

$\displaystyle(n-n_{1})\log\alpha-\log
3-60\left[\log(n_{1}\log\alpha)\right]^{2}$ $\displaystyle<5.5\cdot
10^{15}\log n;$ $\displaystyle n-n_{1}$
$\displaystyle<\dfrac{60\left[\log(n_{1}\log\alpha)\right]^{2}+\log 3+5.5\cdot
10^{15}\log n}{\log\alpha}$ $\displaystyle<2\cdot 10^{15}(\log n)^{2},$

where we have used the fact that $n>n_{1}>310$. Hence, Lemma 5.2 is proved. ∎

Next, we retain the notation $x_{min}:=\min\\{x,x_{1},x_{2}\\}$ and
$y_{min}:=\min\\{y,y_{1},y_{2}\\}$ and prove the following result.

######  Lemma 5.3.

Let $c\leq-1$ such that $(n,x,y)$, $(n_{1},x_{1},y_{1})$,
$(n_{2},x_{2},y_{2})$ and $(n_{3},x_{3},y_{3})$ satisfy (1.2) with
$n>n_{1}>n_{2}>n_{3}$ and $n_{1}>310$, then either

$x_{min},~{}y_{\min}<2\cdot 10^{21}\left(\log n\right)^{4},$

or $n<36100$.

###### Proof.

We follow the same arguments as in the proof of Lemma 4.3 but using the bounds
from Lemma 5.2. In particular, we consider again the $p-$adic valuations (4.1)
and (4.1). We have,

$\displaystyle x_{min}$
$\displaystyle\leq\nu_{2}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\quad\quad+\nu_{2}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right),$
(5.6)

and

$\displaystyle y_{min}$
$\displaystyle\leq\nu_{3}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\quad\quad+\nu_{3}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right).$
(5.7)

As before, we first estimate
$\nu_{p}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right),$
for $p=2,3$. Specifically,

$\displaystyle\nu_{p}\left(\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)\right)$
$\displaystyle\leq\frac{(6(n-n_{2})+30)\log\alpha}{\log p}$
$\displaystyle<\frac{\left(6\cdot 2\cdot 10^{15}(\log
n)^{2}+30\right)\log\alpha}{\log p}$ $\displaystyle<\begin{cases}1.1\cdot
10^{16}(\log n)^{2},&\text{if }~{}p=2;\\\ 6.7\cdot 10^{15}(\log
n)^{2},&\text{if }~{}p=3,\end{cases}$

where we have used Lemma 5.2 and $n>310$. At this point, relations (5.1) and
(5.1) become

$\displaystyle x_{min}$
$\displaystyle\leq\nu_{2}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)+1.1\cdot
10^{16}\left(\log n\right)^{2},$ (5.8)

and

$\displaystyle y_{min}$
$\displaystyle\leq\nu_{3}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)+6.7\cdot
10^{15}\left(\log n\right)^{2}.$ (5.9)

Lastly, we estimate

$\displaystyle\nu_{p}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right),$
(5.10)

with

$\lambda_{1}:=\dfrac{\alpha}{\beta}\qquad{\text{\rm
and}}\qquad\lambda_{2}:=-\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}\right),$

as before. We again distinguish two cases.

Case 1. $\lambda_{1}$ and $\lambda_{2}$ are multiplicatively independent.

We apply Lemma 2.2 to (5.10) with the field $\mathbb{Q}(\alpha,\beta)$ of
degree $D:=6$. Since $h(\lambda_{1})=3(\log\alpha)/3=\log\alpha$, we can
choose $h(\lambda_{1}):=\log\alpha$ for all cases $p=2,3$. Like in Subsection
4.1, Case 1 of Lemma 4.3,

$\displaystyle h^{\prime}(\lambda_{2})$
$\displaystyle<4+\frac{8(n-n_{2}+4)}{3}\log\alpha$
$\displaystyle<4+\frac{8\left(2\cdot 10^{15}(\log
n)^{2}+4\right)}{3}\log\alpha$ $\displaystyle<3.3\cdot 10^{15}(\log n)^{2},$

where we have used Lemma 5.2 and $n>310$. Therefore,

$E^{\prime}=\dfrac{b_{1}}{h^{\prime}(\lambda_{2})}+\dfrac{b_{2}}{h^{\prime}(\lambda_{1})}=\dfrac{n-1}{h^{\prime}(\lambda_{1})}+\dfrac{1}{h^{\prime}(\lambda_{2})}<n.$

We have

$\displaystyle E$ $\displaystyle=\max\left\\{\log E^{\prime}+\log\log
p+0.4,10,10\log p\right\\}$ $\displaystyle=\max\left\\{\log n+\log\log
p+0.4,10,10\log p\right\\}.$

If $n>36100$, then $E<1+\log n$ in both cases $p=2,3$. Therefore, Lemma 2.2
gives

$\displaystyle\nu_{p}\left(\dfrac{C_{\alpha}}{C_{\beta}}\left(\dfrac{\alpha}{\beta}\right)^{n-1}\dfrac{\left(1-\alpha^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\alpha^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}{\left(1-\beta^{n_{1}-n}\right)\left(1-\gamma^{n_{2}-n}\right)-\left(1-\beta^{n_{2}-n}\right)\left(1-\gamma^{n_{1}-n}\right)}+1\right)$
$\displaystyle\leq\dfrac{24pg}{(p-1)(\log
p)^{4}}E^{2}D^{4}h^{\prime}(\lambda_{1})h^{\prime}(\lambda_{2})$
$\displaystyle<\dfrac{24pg}{(p-1)(\log p)^{4}}(\log
n)^{2}\left(1+\frac{1}{\log 36100}\right)^{2}\cdot 6^{4}(\log\alpha)\cdot
3.3\cdot 10^{15}(\log n)^{2}$ $\displaystyle<\dfrac{7.6\cdot
10^{19}pg}{(p-1)(\log p)^{4}}\left(\log n\right)^{4}.$

Hence, inequalities (5.8) and (5.9) become

$\displaystyle x_{min}$ $\displaystyle<\dfrac{7.6\cdot 10^{19}\cdot 2\cdot
3}{(2-1)(\log 2)^{4}}\left(\log n\right)^{4}+1.1\cdot 10^{16}\left(\log
n\right)^{2}<2\cdot 10^{21}(\log n)^{4},$

and

$\displaystyle y_{min}$ $\displaystyle<\dfrac{7.6\cdot 10^{19}\cdot 3\cdot
13}{(3-1)(\log 3)^{4}}\left(\log n\right)^{4}+6.7\cdot 10^{15}\left(\log
n\right)^{2}<1.1\times 10^{21}(\log n)^{4}.$

In the above, we used $g=3$ when $p=2$ and $g=13$ when $p=3$.

Case 2. $\lambda_{1}$ and $\lambda_{2}$ are multiplicatively dependent.

Like the explanation done before in Subsection 4.1,

$\nu_{p}(\lambda_{1}^{n-1}\lambda_{2}+1)\leq 1+\frac{\log n^{2}}{\log p}<3\log
n,$

for $n>36100$ and $p\in\\{2,3\\}$. Comparing this to Case 1, we see that the
bounds from Case 1 still hold. This completes the proof of Lemma 5.3. ∎

To conclude this subsection, we consider a fourth solution
$(n_{3},x_{3},y_{3})$, with $n>n_{1}>n_{2}>n_{3}$ and we find an absolute
bound for $n$. We prove the following result.

######  Lemma 5.4.

If $c\leq-1$ and $n>36100$, then

$n<6.1\cdot 10^{29},\qquad x<5.5\cdot 10^{29},\qquad y<3.5\cdot 10^{29}.$

###### Proof.

Lemma 5.3 tells that out of any four solutions, the minimal $x$ and $y$ are
bounded by $2\cdot 10^{21}(\log n)^{4}$. So, out of the five solutions, at
most one of them has $x$ which is not bounded by $2\cdot 10^{21}(\log n)^{4}$
and at most one of them has $y$ which is not bounded by $2\cdot 10^{21}(\log
n)^{4}$. Hence, at least one of the solutions has both $x$ and $y$ bounded by
$2\cdot 10^{21}(\log n)^{4}$, which in particular, shows that,

$X_{4}=x_{4}\log 2+y_{4}\log 3<(\log 2+\log 3)\cdot 2\cdot 10^{21}(\log
n)^{4}<3.6\cdot 10^{21}(\log n)^{4}.$

Now, by Lemmas 2.10 and 5.1, we get

$\displaystyle n\log\alpha-3$
$\displaystyle<X=X_{4}+(X_{3}-X_{4})+(X_{2}-X_{3})+(X_{1}-X_{2})+(X-X_{1})$
$\displaystyle<3.6\cdot 10^{21}(\log n)^{4}+4\cdot 5.55\cdot 10^{15}\log n$
$\displaystyle<3.61\cdot 10^{21}\left(\log n\right)^{4},$

which implies

$\displaystyle\dfrac{n}{(\log n)^{4}}<6\cdot 10^{21}.$ (5.11)

We apply Lemma 2.2 to inequality (5.11) above with $z:=n$, $m:=4$, $T:=6\cdot
10^{21}$. Since $T>(4\cdot 4^{2})^{4}$, we get

$n<2^{m}T(\log T)^{m}=2^{4}\cdot 6\cdot 10^{21}(\log 6\cdot
10^{21})^{4}<6.1\cdot 10^{29}.$

Further, we have by Lemma 2.10 that

$\displaystyle X$
$\displaystyle<n\log\alpha+60\left(\log(n\log\alpha)\right)^{2};$
$\displaystyle x\log 2+y\log 3$ $\displaystyle<6.1\cdot
10^{29}\log\alpha+60\left(\log(6.1\cdot 10^{29}\log\alpha)\right)^{2}<3.8\cdot
10^{29}.$

This gives $x<5.5\cdot 10^{29}$ and $y<3.5\cdot 10^{29}$ and completes the
proof of Lemma 5.4. ∎

###  5.2 Reduction of the upper bound on $n$

Again, we use the LLL–reduction methods, the theory of continued fractions and
as well as $p-$adic reduction methods due to [21] to find a rather small bound
for $n$.

First, we consider (5.2) with the assumption that $X-X_{1}\geq 2$ and take
logarithms, that is

$|\Lambda_{2}|=\left|\log C_{\alpha}+(n-1)\log\alpha-x\log 2-y\log
3\right|<4.5\exp(-(X-X_{1})).$

Like before, we consider the approximation lattice

$\mathcal{A}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\ \lfloor M\log 2\rfloor&\lfloor
M\log 3\rfloor&\lfloor M\log\alpha\rfloor\end{pmatrix},$

with $C:=10^{90}$ and choose $v:=\left(0,0,-\lfloor M\log
C_{\alpha}\rfloor\right)$. Now, by Lemma 2.4, we get
$|\Lambda_{2}|>c_{1}:=9.3\cdot 10^{-32}$ and hence $c_{2}:=5.12\cdot 10^{30}$.
Moreover, by Lemma 5.4, we have

$x<A_{1}=5.5\cdot 10^{29},~{}~{}y<A_{2}=3.5\cdot
10^{29},~{}~{}n-1<n<A_{3}=6.1\cdot 10^{29}.$

So, Lemma 2.5 gives $S=4.25\cdot 10^{59}$ and $T=7.55\cdot 10^{29}$. Since
$c_{2}^{2}\geq T^{2}+S$, then choosing $c_{3}:=4.5$ and $c_{4}:=1$, we get
$X-X_{1}\leq 138$.

Next, we continue with the assumption that $n>n_{1}$, and consider the
inequality

$0<2^{x}3^{y}-2^{x_{1}}3^{y_{1}}=T_{n}-T_{n_{1}}<T_{n}<\alpha^{n-1}.$

Dividing through by $2^{x}3^{y}$ and taking logarithms, we get

$|\tau|:=\left|(x-x_{1})\log 2+(y-y_{1})\log
3\right|<1.5\dfrac{\alpha^{n-1}}{2^{x}3^{y}},$

where we have assumed that $2^{x}3^{y}>\alpha^{n}$ and applied Lemma 2.1.
Next, we divide the above equation by $|y_{1}-y|\log 2$ and get

$\displaystyle\left|\dfrac{\log 3}{\log
2}-\dfrac{x-x_{1}}{y_{1}-y}\right|<\dfrac{1.2\alpha^{n}}{2^{x}3^{y}|y_{1}-y|},$

since $\alpha>1.83$ and $y_{1}$, $y$ are distinct. Note that $y_{1}$ and $y$
are indeed distinct since if they were not, then
$0.5\leq|e^{\tau}-1|=\left|1-2^{x_{1}-x}\right|<\alpha^{n-1}/2^{x}3^{y}<0.5$,
a contradiction. By Lemma 2.3 with $\mu:=\dfrac{\log 3}{\log 2}$ and
$M:=10^{48}$, we have

$\displaystyle\dfrac{1}{(a(M)+2)(y_{1}-y)^{2}}<\left|\dfrac{\log 3}{\log
2}-\dfrac{x-x_{1}}{y_{1}-y}\right|<\dfrac{1.2\alpha^{n}}{2^{x}3^{y}|y_{1}-y|},$

where $a(M)=55$ (in fact, $q_{100}>10^{48}$ and $\max\\{a_{k}:0\leq k\leq
100\\}=55$). Multiplying the above inequality by $|y_{1}-y|\log 2$ gives

$\displaystyle\dfrac{\log 2}{57\cdot 3.5\cdot 10^{29}}<\dfrac{\log
2}{(a(M)+2)|y_{1}-y|}<\left|(x-x_{1})\log 2+(y-y_{1})\log 3\right|,$

so that

$3.4\cdot 10^{-32}<\left|(x-x_{1})\log 2+(y-y_{1})\log
3\right|<1.5\dfrac{\alpha^{n-1}}{2^{x}3^{y}},$

where we have used the upper bound $y<3.5\cdot 10^{29}$. This gives
$2^{x}3^{y}<2.5\cdot 10^{31}\alpha^{n}$, since $\alpha>1.83$. If the
assumption that $2^{x}3^{y}>\alpha^{n}$ is violated, then we would have
$2^{x}3^{y}\leq\alpha^{n}<2.5\cdot 10^{31}\alpha^{n}$, and we are in the same
situation. Now, by assuming a third solution to (1.2), then

$0<2^{x}3^{y}-C_{\alpha}\alpha^{n-1}<2^{x_{1}}3^{y_{1}}+2<2.5\cdot
10^{31}\alpha^{n_{1}}+2<2.51\cdot 10^{31}\alpha^{n_{1}},$

which gives

$\displaystyle\left|2^{x}3^{y}C_{\alpha}^{-1}\alpha^{-(n-1)}-1\right|<7.6\cdot
10^{31}\alpha^{-(n-n_{1})}.$

Assume that $7.6\cdot 10^{31}\alpha^{-(n-n_{1})}<0.5$, which is certainly true
if $n-n_{1}\geq 123$. Taking logarithms gives

$\displaystyle\left|x\log 2+y\log 3-(n-1)\log\alpha-\log C_{\alpha}\right|\leq
1.5\cdot 10^{50}\alpha^{-(n-n_{1})}.$

So, we consider the approximation lattice

$\mathcal{A}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\ \lfloor M\log 2\rfloor&\lfloor
M\log 3\rfloor&\lfloor M\log\alpha\rfloor\end{pmatrix},$

with $M:=10^{90}$ and choose $v:=\left(0,0,-\lfloor M\log
C_{\alpha}\rfloor\right)$. Now, by Lemma 2.4, we maintain
$|\Lambda_{2}|>c_{1}:=9.3\cdot 10^{-32}$ and $c_{2}:=5.12\cdot 10^{30}$.
Still, by Lemma 5.4, we have

$x<A_{1}=5.5\cdot 10^{29},~{}~{}y<A_{2}=3.5\cdot
10^{29},~{}~{}n-1<n<A_{3}=6.1\cdot 10^{29}.$

So, Lemma 2.5 gives $S=4.25\cdot 10^{59}$ and $T=7.55\cdot 10^{29}$, so that
choosing $c_{3}:=1.5\cdot 10^{50}$ and $c_{4}:=\log\alpha$, we get
$n-n_{1}\leq 417$.

To continue, we proceed as in Subsection 4.2 but with different upper bounds
for $n$ and $n-n_{1}$. Notice that in this case, $d:=n-n_{1}\leq 417$ and
$n<6.1\cdot 10^{29}<2^{99}$. We repeat the algorithm described in Subsection
4.2 after relation (4.2) and obtain rather smaller bounds as
$x_{min},y_{min}\leq 152$.

Lastly, we find a smaller upper bound for $n$. If we write $c_{X}$ for the
upper bound of $X-X_{1}$, then

$\displaystyle X=X_{2}+(X_{1}-X_{2})+(X-X_{1})$ $\displaystyle<x_{min}\log
2+y_{min}\log 3+2c_{X};$ $\displaystyle x\log 2+y\log 3$
$\displaystyle<152\log 2+152\log 3+2\cdot 138<549.$

Hence, $x<793$ and $y<500$. On the other hand, Lemma 2.10 implies that
$n\log\alpha-3<X<549,$ so that $n<914$.

###  5.3 Conclusion

Concluding the proof of Theorem 1.3, we note that for $n>310$, the bounds are
$n\leq 913$, $x<793$, and $y<500$. To efficiently handle large $T_{n}$ values,
our SageMath 9.5 code utilized batch processing, iterating through all
$(n,x,y)$ combinations within these ranges. This approach selected $c\leq-1$
values with at least five representations of the form $T_{n}-2^{x}3^{y}$,
aligning with the solutions in Theorem 1.3. The computation, performed on an
8GB RAM laptop, was completed in about 2 hours. ∎

## Acknowledgments

The first author thanks the Eastern Africa Universities Mathematics Programme
(EAUMP) for funding his doctoral studies. His time at Wits University,
Johannesburg, greatly contributed to this paper, and he is thankful for the
university’s hospitality and supportive environment. Both authors appreciate
the CoEMaSS Grant #2024–029–NUM from Wits University, which was crucial for
this research.

## References

  * [1] Baker, A., & Wüstholz, G. Logarithmic forms and group varieties, J. Reine Angew. Math. 442 (1993),19–62.
  * [2] Batte, H., Ddamulira, M., Kasozi, J., & Luca, F. On the multiplicity in Pillai’s problem with Fibonacci numbers and powers of a fixed prime. Glasnik matematic̆ki 57 (2022), 185–201.
  * [3] Batte, H., Ddamulira, M., Kasozi, J., & Luca. On a problem of Pillai involving $S-$units and Lucas numbers, (2024), arXiv:2401.05361.
  * [4] Bilu, Y., Hanrot, G., & Voutier, P. M. Existence of primitive divisors of Lucas and Lehmer numbers. With an appendix by M. Mignotte. J. Reine Angew. Math. 539 (2001), 75–122.
  * [5] YBilu, Y. & Luca, F. Binary polynomial powers sums vanishing at roots of unity. Acta Arith. 198 (2021), 195–217.
  * [6] Bilu, Y., Luca, F., Nieuwveld, J., Ouaknine, J., & Worrell, J. Twisted rational zeros of linear recurrence sequences. (2024), arXiv:2401.06537.
  * [7] Bravo, J. J., & Luca, F. On the largest prime factor of the $k-$Fibonacci numbers. International J. Number Theory 9 (2013), 1351–1366.
  * [8] Bravo, J. J., Luca, F., & Yazán, K. On Pillai’s problem with Tribonacci numbers and Powers of 2. Bull. Korean Math. Soc. 54 (2017),1069–1080.
  * [9] Bugeaud, Y., & Laurent, M. Minoration effective de la distance $p$-adique entre puissances de nombres algb́riques. J. Number Theory 61 (1996), 311–342.
  * [10] Cassels, J. W. S. An introduction to the geometry of numbers. Springer Science & Business Media, 2012.
  * [11] Chim, K. C., Pink, I., & Ziegler, V. On a variant of Pillai’s problem II. J. Number Theory 183 (2018), 269–290.
  * [12] de Weger, B. M. Solving exponential Diophantine equations using lattice basis reduction algorithms. J. Number Theory 26 (1987), 325–367.
  * [13] Ddamulira, M., & Luca, F. On the problem of Pillai with $k$–generalized Fibonacci numbers and powers of 3. Internat. J. Number Theory 16 (2020), 1643–1666.
  * [14] Ddamulira, M., Luca, F., & Rakotomalala, M. On a problem of Pillai with Fibonacci numbers and powers of 2. Proc. Indian Acad. 127 (2017), 411–421.
  * [15] Dresden, G. P., & Du, Z. A Simplified Binet Formula for k-Generalized Fibonacci Numbers. J. Integer Seq. 17 (2015), Article 14.4.7
  * [16] Gúzman, S. S., & Luca, F. Linear combinations of factorials and $S$-units in a binary recurrence sequence. Ann. Math. Québec 38 (2014), 169–188.
  * [17] Lenstra, A. K., Lenstra, H. W., & Lovász, L. Factoring polynomials with rational coefficients. Math. Ann. 261 (1982), 515–534.
  * [18] Mahler, K. An inequality for the discriminant of a polynomial. Michigan Math. J.  11 (1964), 1153–1157.
  * [19] Matveev, E. M. An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers. II. Izvestiya: Mathematics, 64 (2000), 1217–1269.
  * [20] Murty M. R., & Esmonde, J. Problems in Algebraic Number Theory, second edition, Graduate Texts in Mathematics, 190, Springer–Verlag, New York, 2005.
  * [21] Pethö, A., & de Weger, B. M. Products of prime powers in binary recurrence sequences. I. The hyperbolic case, with an application to the generalized Ramanujan-Nagell equation. Math. Comp. 176 (1986), 713–727.
  * [22] Postnikova, L., P. & and Schinzel A., Primitive divisors of the expression $a^{n}-b^{n}$ in algebraic number fields, Math. Sb. (N.S.) 75 (117) (1968), 171–177 (in Russian).
  * [23] Voutier, P. An effective lower bound for the height of algebraic numbers. Acta Arithmetica 74 (1996), 81–95.
  * [24] Pillai, S. S. On $a^{x}-b^{y}=c$. J. Indian Math. Soc.(NS) 2 (1936), 119–122.
  * [25] Smart, N. P. The algorithmic resolution of Diophantine equations: a computational cookbook. Cambridge University Press, 1998.
  * [26] Ziegler, V. (2022). On a variant of Pillai’s problem involving $S-$units and Fibonacci numbers. Bol. Soc. Mat. Mexicana 28 (2022), Paper number 57.

## Addresses

1 School of Mathematics, Wits University, Johannesburg, South Africa and
Department of Mathematics, School of Physical Sciences, College of Natural
Sciences, Makerere University, Kampala, Uganda.

Email<EMAIL_ADDRESS>
2 School of Mathematics, Wits University, Johannesburg, South Africa and
Centro de Ciencias Matemáticas UNAM, Morelia, Mexico.

Email<EMAIL_ADDRESS>