# On families of cubic split Thue equations parametrised by linear recurrence
sequences

Tobias Hilgart  Department of Mathematics
University of Salzburg
A-5020 Salzburg, Hellbrunnerstraße 34
Austria<EMAIL_ADDRESS>
###### Abstract.

Let
$\left(A_{n}\right)_{n\in\mathbb{N}},\left(B_{n}\right)_{n\in\mathbb{N}}\in\mathbb{Z}^{\mathbb{N}}$
be two linear-recurrent sequences that meet a dominant root condition and a
few more technical requirements. We show that the split family of Thue
equations

$\left|X\left(X-A_{n}Y\right)\left(X-B_{n}Y\right)-Y^{3}\right|=1$

has but the trivial solutions
$\pm\left\\{\left(0,1\right),\left(1,0\right),\left(A_{n},1\right),\left(B_{n},1\right)\right\\}$,
if the parameter $n$ is larger than some effectively computable constant.

###### keywords:

parametrised Thue equations, linear recurrence sequences

###### 1991 Mathematics Subject Classification:

11D25

††issue: 3††issue: 4††issue: 2

##  1\. Introduction

Thue equations are Diophantine equations $F(X,Y)=m$, where
$F\in\mathbb{Z}[X,Y]$ is an irreducible form of degree at least $3$ and $m\neq
0$ a fixed integer. A. Thue showed that such equations only have finitely many
solutions [Thu1909]. But the methods he used were ineffective and not suitable
to derive bounds on the number of solutions for concrete equations. It was
Baker [Bak1968] that could give such effective bounds using his theory on
linear forms in logarithms. These bounds have been used and refined many times
since.

E. Thomas studied so-called split Thue equations,

$X\prod_{i=2}^{N}\left(X-p_{i}(a)Y\right)-Y^{N}=\pm 1,$

that are parametrised by monic polynomials $p_{i}\in\mathbb{Z}[a]$. He
conjectured that, given some technical conditions for the $p_{i}$, the
equation would only have trivial solutions immediately apparent by its form,
provided the parameter $a\geq a_{0}$ is sufficiently large. The bound $a_{0}$
would also be effectively computable. The conjecture has since been proven,
first by Thomas [Tho1990] in the cubic and Heuberger [Heu2001] in the general
case.

The methods involved rely heavily on Baker’s theory of linear forms in
logarithms but not so much on the parameter functions being polynomials.
Hence, the authors in [Hil2022] and [Vuk2021] made attempts using other
parameter functions, namely linear-recurrent sequences. This paper, mimicking
Thomas’ work on polynomial-parametrised split Thue equations, strives to proof
the cubic case for when the parameter functions are linear-recurrent sequences
in as much of a general setting as possible.

Let
$\left(A_{n}\right)_{n_{\in}\mathbb{N}},\left(B_{n}\right)_{n\in\mathbb{N}}$
be two integer sequences that satisfy some linear recurrences with some
starting conditions. We denote the characteristic polynomials of the
recurrence sequences as

$\left(X-\alpha\right)^{d_{A}}\prod_{i=1}^{m_{A}}\left(X-\alpha_{i}\right)^{d_{A,i}},\text{
and
}\left(X-\beta\right)^{d_{B}}\prod_{i=1}^{m_{B}}\left(X-\beta_{i}\right)^{d_{B,i}}$

respectively to give the algebraic multiplicity of each characteristic root.
We will consider the sequences almost exclusively via their explicit formulae

$A_{n}=c_{A}(n)\alpha^{n}+c_{A,1}(n)\alpha_{1}^{n}+\cdots+c_{A,m_{A}}(n)\alpha_{m_{A}}^{n}\\\
$ (1)
$B_{n}=c_{B}(n)\beta^{n}+c_{B,1}(n)\beta_{1}^{n}+\cdots+c_{B,m_{B}}(n)\alpha_{m_{B}}^{n}$
(2)

that always exist for linear recurrence sequences.

Both are to admit to a dominant root condition, i.e.
$\left|\alpha\right|>\max_{i}\left|\alpha_{i}\right|$ and
$\left|\beta\right|>\max_{i}\left|\beta_{i}\right|$. We assume without loss of
generality that $\left|\alpha\right|\leq\left|\beta\right|$. All
characteristic roots must be algebraic over $\mathbb{Q}$.

While we cannot expect the remaining terms, say $c_{A}$, to be constant, they
are polynomials in the splitting field of the respective characteristic
polynomial. In addition, the degree is bounded by the related multiplicity of
the root, e.g. $\deg\left(c_{A}\right)\leq d_{A}-1$. We also define

$\displaystyle d_{1}$
$\displaystyle:=\min\left\\{\deg\left(c_{A}\right),\deg\left(c_{A,i}\right),\deg\left(c_{B}\right),\deg\left(c_{B,i}\right)\right\\},$
$\displaystyle d_{2}$
$\displaystyle:=\max\left\\{\deg\left(c_{A}\right),\deg\left(c_{A,i}\right),\deg\left(c_{B}\right),\deg\left(c_{B,i}\right)\right\\}.$
(3)

In this setting, we prove the following result:

###### Theorem 1

The split family of Thue equations

$\left|X(X-A_{n}Y)(X-B_{n}Y)-Y^{3}\right|=1$ (4)

has only the trivial solutions
$\left\\{\pm\left(0,1\right),\pm\left(A_{n},1\right),\pm\left(B_{n},1\right),\left(\pm
1,0\right)\right\\}$ for sufficiently large $n$, if $A_{n}$ and $B_{n}$ meet a
dominant root condition, where at least one dominant root has absolute value
larger than $1$, and either

  * •

$1\leq A_{n}\leq B_{n}-2$ or

  * •

$-1\geq A_{n}\geq B_{n}+3$.

In case the dominant roots have equal absolute value, we also need that for
sufficiently large $n$, both $\left|c_{B}(n)\right|\neq\left|c_{A}(n)\right|$
and one of the following conditions holds:

  * •

$\left|c_{B}(n)\right|>1$ and
$\left|c_{B}(n)-c_{A}(n)\right|>\frac{1}{\left|c_{B}(n)\right|}$

  * •

$0<\left|c_{B}(n)\right|<1$ and $0<\left|c_{B}(n)-c_{A}(n)\right|<1$

  * •

$\left|c_{B}(n)\right|=1$ and
$\left|c_{B}(n)-c_{A}(n)\right|\not\in\left\\{0,1\right\\}$

##  2\. Preliminaries

Even if the terms $c_{B}(n),c_{A}(n)$ are polynomials in $n$ in general, we
will forego to note the dependency every time for the sake of readability and
simply write $c_{B}\;\left(=c_{B}(n)\right)$ and
$c_{A}\;\left(=c_{A}(n)\right)$.

The conditions $1\leq A_{n}\leq B_{n}-2$ or $-1\geq A_{n}\geq B_{n}+3$
translate to $1\leq\left|\alpha\right|,\left|\beta\right|$, with
$\left|\alpha\right|\leq\left|\beta\right|$ we further get that
$\left|\beta\right|>1$. We need at least one of the dominant roots to be
strictly larger than $1$ so that the corresponding sequence has exponential
growth. This is not really a restriction, since
$\left|\alpha\right|=\left|\beta\right|=1$ would mean that both sequences have
at most polynomial growth and that we are essentially in the case which Thomas
had already studied and solved.

Also note that in the case of $\left|\alpha\right|=\left|\beta\right|$, since
$\left|c_{A}\right|\neq\left|c_{B}\right|$ at least for sufficiently large
$n$, we can conclude that $\left|c_{A}\right|\leq\left|c_{B}\right|$, as
neither of the inequalities $1\leq A_{n}\leq B_{n}-2$ or $-1\geq A_{n}\geq
B_{n}+3$ could hold otherwise.

We first consider the case where a solution $(x,y)$ has small
$\left|y\right|$.

###### Lemma 1

If a solution $(x,y)$ to the Thue-Equation (4), for sufficiently large $n$,
has $y\in\left\\{-1,0,1\right\\}$, then it is a trivial solution.

###### Proof.

For $y=0$, it follows immediately that $x^{3}=\pm 1$, thus the only solutions
with $y=0$ are trivial. If instead $y=\pm 1$, then either $X\left(X\pm
A_{n}\right)\left(X\pm B_{n}\right)=0$ or $X\left(X\pm A_{n}\right)\left(X\pm
B_{n}\right)=\pm 2$.

The first case immediately gives $x\in\left\\{0,\mp A_{n},\mp B_{n}\right\\}$
and thus that $(x,y)$ is a trivial solution. The second case is solvable for
$x$ only if $\left|B_{n}-A_{n}\right|\leq 2$, which can hold at most for small
$n$: Since either $1\leq A_{n}\leq B_{n}-2$ or $-1\geq A_{n}\geq B_{n}+3$, we
also have $\left|B_{n}-A_{n}\right|\geq 2$ and thus equality, which is
impossible for $\left|\alpha\right|<\left|\beta\right|$ and precluded by
$\left|c_{B}\right|>\left|c_{A}\right|$ for
$\left|\alpha\right|=\left|\beta\right|$. ∎

###### Lemma 2

Assume $x\in\mathbb{C}$ such that $\left|x-1\right|\leq 1/2$, then
$\left|\log\left|x\right|\right|\leq 2\left|x-1\right|$.

###### Proof.

Use the Taylor series expansion of $\log x$ at $1$. ∎

######  Theorem 2 (Thomas, [Tho1979])

Let $F_{\lambda}$ be the field generated by $\lambda$, where either

$\lambda\left(\lambda-r\right)\left(\lambda-s\right)+1=0,\;\;\;1\leq r\leq
s-3$

or

$\lambda\left(\lambda-r\right)\left(\lambda-s\right)-1=0,\;\;\;1\leq r\leq
s-2.$

Then $\left\\{\lambda,\lambda-r\right\\}$ is a fundamental system of units for
the order generated by $\left\\{1,\lambda,\lambda^{2}\right\\}$.

######  Theorem 3 (Bugeaud, Győry, [BuGy1996])

Let $B\geq\max\left\\{\left|m\right|,e\right\\}$, $\alpha$ be a root of
$F(X,1)$, $K:=\mathbb{Q}(\alpha)$, $R:=R_{K}$ the regulator of $K$ and $r$ the
unit rank of $K$. Let $H\geq 3$ be an upper bound for the absolute values of
the coefficients of $F$ and $N$ its degree.

Then all solutions $(x,y)\in\mathbb{Z}^{2}$ of the Thue equation $F(X,Y)=m$
satisfy

$\max\left\\{\log\left|x\right|,\log\left|y\right|\right\\}\leq C(r,N)\cdot
R\cdot\max\left\\{\log R,1\right\\}\cdot\left(R+\log(HB)\right),$

where $C(r,N)=3^{r+27}(r+1)^{7r+19}N^{2N+6r+14}$.

For any algebraic number $\gamma$ of degree $d$ with conjugates
$\gamma^{(1)},\dots,\gamma^{(d)}$, we define the absolute logarithmic height
$h$ of $\gamma$ as

$h\left(\gamma\right)=\frac{1}{d}\left(\log\left|a\right|+\sum_{i=1}^{d}\log\max\left\\{\left|\gamma^{(i)}\right|,1\right\\}\right),$

where $a$ is the leading coefficient of the minimal polynomial of $\gamma$
over $\mathbb{Z}$.

######  Theorem 4 (Baker, Wüstholz, [BaWh1993])

Let $\gamma_{1},\dots,\gamma_{t}$ be algebraic numbers, not $0$ or $1$, in
$K=\mathbb{Q}(\gamma_{1},\dots,\gamma_{t})$ of degree $D$, let
$b_{1},\dots,b_{t}\in\mathbb{Z}$, and let

$\Lambda=b_{1}\log\gamma_{1}+\cdots+b_{t}\log\gamma_{t}$

be non-zero. Then

$\log\left|\Lambda\right|\geq-18(t+1)!t^{t+1}(32D)^{t+2}\log(2tD)h_{1}\cdots
h_{t}\log B,$

where

$B\geq\max\left\\{\left|b_{1}\right|,\dots,\left|b_{t}\right|\right\\},$

and

$h_{i}\geq\max\left\\{h(\gamma_{i}),\left|\log\gamma_{i}\right|D^{-1},0.16D^{-1}\right\\}\text{
for }1\leq i\leq t.$

##  3\. Approximating the roots

We start by giving good approximations for the roots of

$f_{n}(X)=X(X-A_{n})(X-B_{n})-1=\left(x-\lambda_{1}\right)\left(x-\lambda_{2}\right)\left(x-\lambda_{3}\right).$
(5)

The polynomial is irreducible over $\mathbb{Q}$ as per the rational root
theorem, at least for sufficiently large $n$: The only candidates for a
rational root are $1$ and $-1$. When we plug them in and look at either
$f_{n}(1)=0$ or $f_{n}(-1)=0$, we get that $A_{n}\cdot B_{n}=A_{n}+B_{n}$,
which can hold at most for small $n$.

To describe the quality of the approximation, we use the $L$-notation, where
for two complex functions $f,g$, we write $f=L(g)$ if
$\left|f(x)\right|\leq\left|g(x)\right|$ for all $x\in\mathbb{C}$. This way,
we can consider constants that we would lose in the more classical
$O$-notation.

###### Lemma 3

We have, for sufficiently large $n$:

$\displaystyle\lambda_{1}$ $\displaystyle=B_{n}+L\left(B_{n}^{-1}\right)$
$\displaystyle\lambda_{2}$
$\displaystyle=A_{n}+L\left(A_{n}^{-1}\right)=A_{n}+\frac{1}{A_{n}\left(A_{n}-B_{n}\right)}+L\left(\frac{1}{A_{n}^{2}\left(A_{n}-B_{n}\right)^{2}}\right)$
$\displaystyle\lambda_{3}$
$\displaystyle=\frac{1}{A_{n}B_{n}}+L\left(\frac{1}{A_{n}^{2}B_{n}^{2}}\right)$

###### Proof.

We start with the approximation for $\lambda_{1}$. To that end, we first note
that $f_{n}(B_{n})=-1<0$. We then take a look at

$f_{n}(B_{n}\pm B_{n}^{-1})=\left(B_{n}\pm
B_{n}^{-1}\right)\left(B_{n}-A_{n}\pm B_{n}^{-1}\right)\left(\pm
B_{n}^{-1}\right)-1,$

where we plug in $B_{n}+B_{n}^{-1}$ in case $1\leq A_{n}\leq B_{n}-2$ and
$B_{n}-B_{n}^{-1}$ in the other.

Now, since either $B_{n}-A_{n}\geq 2$ or $B_{n}-A_{n}\leq-3$, we have either

$f_{n}(B_{n}+B_{n}^{-1})\geq
1+B_{n}^{-1}\left(1+2B_{n}^{-1}+B_{n}^{-2}\right)>0$

or

$f_{n}(B_{n}-B_{n}^{-1})\geq
2-B_{n}^{-1}\left(1+2B_{n}^{-1}-B_{n}^{-2}\right)>0$

for sufficiently large $n$. By the intermediate value theorem, a root
$\lambda_{1}$ must lie between $B_{n}-B_{n}^{-1}$ and $B_{n}+B_{n}^{-1}$, thus
$\lambda_{1}=B_{n}+L\left(B_{n}^{-1}\right)$.

We prove the first approximation for $\lambda_{2}$ analogously by showing that
either $f_{n}(A_{n}-A_{n}^{-1})>0$ o r $f_{n}(A_{n}+A_{n}^{-1})>0$, depending
on the sign of $A_{n}$.

The second approximation for $\lambda_{2}$ follows analogously to the
approximation for $\lambda_{3}$: We first write

$f_{n}\left(\frac{1}{A_{n}B_{n}}+\kappa\right)=\left(\frac{1}{A_{n}B_{n}}+\kappa\right)\left(\frac{1}{A_{n}B_{n}}-A_{n}+\kappa\right)\left(\frac{1}{A_{n}B_{n}}-B_{n}+\kappa\right)-1$

and then expand the product and group by powers of $\kappa$. This then becomes

$\displaystyle f_{n}\left(\frac{1}{A_{n}B_{n}}+\kappa\right)=$
$\displaystyle\kappa\left(A_{n}B_{n}-\frac{2}{A_{n}}-\frac{2}{B_{n}}+\frac{3}{A_{n}^{2}B_{n}^{2}}\right)$
$\displaystyle+$
$\displaystyle\kappa^{2}\left(-\left(A_{n}+B_{n}\right)+\frac{3}{A_{n}B_{n}}\right)$
$\displaystyle+$ $\displaystyle\kappa^{3}$ $\displaystyle-$
$\displaystyle\left(\frac{1}{A_{n}^{2}B_{n}}+\frac{1}{A_{n}B_{n}^{2}}\right)+\frac{1}{A_{n}^{3}B_{n}^{3}}.$

From this expansion, we see that the highest-order term is $\kappa
A_{n}B_{n}$, as long as $\kappa$ is of order strictly greater than
$\frac{1}{A_{n}^{2}B_{n}^{2}}\max\left\\{\frac{1}{A_{n}},\frac{1}{B_{n}}\right\\}$.
Thus, choosing $\kappa=\pm\frac{1}{A_{n}^{2}B_{n}^{2}}$ gives alternating
signs for $f_{n}\left(\frac{1}{A_{n}B_{n}}+\kappa\right)$ and thus the claimed
form for $\lambda_{3}$ by the intermediate value theorem. ∎

Their logarithms play as important a role as the roots themselves. We thus
explicitly state their approximations too.

###### Lemma 4

We have, up to an exponentially decreasing term
$L\left(C\,n^{d_{2}}\varepsilon^{n}\right)$,

$\displaystyle\log\left|\lambda_{1}\right|$
$\displaystyle=n\log\left|\beta\right|+\log\left|c_{B}\right|$
$\displaystyle\log\left|\lambda_{1}-A_{n}\right|$
$\displaystyle=\begin{cases}n\log\left|\beta\right|+\log\left|c_{B}\right|\;\;&\text{
if }\left|\alpha\right|<\left|\beta\right|\\\
n\log\left|\beta\right|+\log\left|c_{B}-c_{A}\right|\;\;&\text{ if
}\left|\alpha\right|=\left|\beta\right|\end{cases}$
$\displaystyle\log\left|\lambda_{2}\right|$
$\displaystyle=n\log\left|\alpha\right|+\log\left|c_{A}\right|$
$\displaystyle\log\left|\lambda_{2}-A_{n}\right|$
$\displaystyle=\begin{cases}-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}\right|\;\;&\text{
if }\left|\alpha\right|<\left|\beta\right|\\\
-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}-c_{A}\right|\;\;&\text{
if }\left|\alpha\right|=\left|\beta\right|\end{cases}$
$\displaystyle\log\left|\lambda_{3}\right|$
$\displaystyle=-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}\right|$
$\displaystyle\log\left|\lambda_{3}-A_{n}\right|$
$\displaystyle=n\log\left|\alpha\right|+\log\left|c_{A}\right|$

###### Proof.

The claims follow immediately from the approximations for the roots by Lemma
3. We demonstrate for $\lambda_{1}$:

$\log\left|\lambda_{1}\right|=\log\left|B_{n}+L\left(B_{n}^{-1}\right)\right|=\log\left|B_{n}\right|+\log\left|1+L\left(B_{n}^{-2}\right)\right|.$

For sufficiently large $n$, we have $\left|B_{n}\right|^{-2}\leq\frac{1}{2}$
and thus, by Lemma 2,

$\log\left|\lambda_{1}\right|=\log\left|B_{n}\right|+L\left(2\left|B_{n}^{-2}\right|\right).$

We expand the term $\log\left|B_{n}\right|$ next:

$\displaystyle\log\left|B_{n}\right|$
$\displaystyle=\log\left|c_{B}\beta^{n}+c_{B,1}\beta_{1}^{n}+\cdots\right|$
$\displaystyle=n\log\left|\beta\right|+\log\left|c_{B}\right|+\log\left|1+\sum_{i}\frac{c_{B,i}}{c_{B}}\left(\frac{\beta_{i}}{\beta}\right)^{n}\right|.$

The $c_{B},c_{B,i}$ are polynomials in $n$ of degree at most $d_{2}$, as
defined in Equation (1). The first terms in the sum are thus of order at most
$n^{d_{2}}$. All the while, the second terms decrease exponentially because of
$\left|\beta\right|>\max_{i}\left|\beta_{i}\right|$. Thus, the sum will have
an absolute value of at most $\frac{1}{2}$ for sufficiently large $n$, and we
can use Lemma 2 again.

We thus have proved the form of $\log\left|\lambda_{1}\right|$, up to an error
of order

$L\left(2\sum_{i}\left|\frac{c_{B,i}}{c_{B}}\right|\left|\frac{\beta_{i}}{\beta}\right|^{n}\right)+L\left(2\left|B_{n}^{-2}\right|\right).$

The first $L$-term is of strictly higher order. We can thus ignore the second
$L$-term if we take the first one $3$ times instead. We then bound the
rational functions $\left|\frac{c_{B,i}}{c_{B}}\right|\leq c_{1}n^{d_{2}}$,
where $c_{1}$ depends only on the coefficients of the $c_{B},c_{B,i}$. We also
have
$\left|\frac{\beta_{i}}{\beta}\right|^{n}\leq\max_{i}\left|\frac{\beta_{i}}{\beta}\right|^{n}$
trivially and thus arrive at the $L$-term

$L\left(n^{d_{2}}\max_{i}\left|\frac{\beta_{i}}{\beta}\right|^{n}\cdot
3c_{1}\sum_{i}1\right)=L\left(3c_{1}m_{B}\cdot
n^{d_{2}}\max_{i}\left|\frac{\beta_{i}}{\beta}\right|^{n}\right),$

which decreases exponentially.

We analogously get

$\displaystyle\log\left|\lambda_{2}\right|$
$\displaystyle=\log\left|A_{n}\right|+L\left(2\left|A_{n}\right|^{-2}\right)$
$\displaystyle=n\log\left|\alpha\right|+\log\left|c_{A}\right|+L\left(3c_{2}m_{A}\cdot
n^{d_{2}}\max_{i}\left|\frac{\alpha_{i}}{\alpha}\right|^{n}\right),$

where we bound $\left|\frac{c_{A,i}}{c_{A}}\right|\leq c_{2}n^{d_{2}}$.

If we look at $\lambda_{1}-A_{n}$, we get

$\log\left|\lambda_{1}-A_{n}\right|=\log\left|B_{n}-A_{n}\right|+L\left(2\left|B_{n}\right|^{-2}\right)$

and must now differentiate between the cases
$\left|\alpha\right|<\left|\beta\right|$ and
$\left|\alpha\right|=\left|\beta\right|$. In the first case, we factorise
$\left|c_{B}\right|\left|\beta\right|^{n}$ and get the subsequent $L$-term

$3c_{3}\left(m_{B}+m_{A}+1\right)n^{d_{2}}\max_{i}\left\\{\left|\frac{\beta_{i}}{\beta}\right|,\left|\frac{\alpha_{i}}{\beta}\right|,\left|\frac{\alpha}{\beta}\right|\right\\}^{n},$

where we bound
$\left|\max_{i}\left\\{\frac{c_{A}}{c_{B}},\frac{c_{A,i}}{c_{B}},\frac{c_{B_{i}}}{c_{B}}\right\\}\right|\leq
c_{3}n^{d_{2}}$. With $\left|\alpha\right|<\left|\beta\right|$, the term still
decreases exponentially. In the second case, we substitute
$\left|\alpha\right|$ for $\left|\beta\right|$ in $A_{n}$ and factorise
$\left|c_{B}-c_{A}\right|\left|\beta\right|^{n}$ instead. This then leads to
the $L$-term

$3c_{4}\left(m_{B}+m_{A}\right)n^{d_{2}}\max_{i}\left\\{\left|\frac{\beta_{i}}{\beta}\right|,\left|\frac{\alpha_{i}}{\beta}\right|\right\\}^{n},$

with
$\left|\max_{i}\left\\{\frac{c_{A}}{c_{B}-c_{A}},\frac{c_{A,i}}{c_{B}-c_{A}},\frac{c_{B_{i}}}{c_{B}-c_{A}}\right\\}\right|\leq
c_{4}n^{d_{2}}$.

The remaining logarithms follow analogously, as we already know the $L$-terms
that result from
$\log\left|B_{n}\right|,\log\left|A_{n}\right|,\log\left|B_{n}-A_{n}\right|$.
For example,

$\displaystyle\log\left|\lambda_{2}-A_{n}\right|$
$\displaystyle=\log\left|A_{n}^{-1}\left(A_{n}-B_{n}\right)^{-1}+L\left(A_{n}^{-2}\left(A_{n}-B_{n}\right)^{-2}\right)\right|$
$\displaystyle=-\log\left|A_{n}\right|-\log\left|B_{n}-A_{n}\right|+L\left(\left|A_{n}\right|^{-3}\left|B_{n}-A_{n}\right|^{-3}\right),$

and we can simply sum the $L$-terms that result from $\log\left|A_{n}\right|$
and $\log\left|B_{n}-A_{n}\right|$ and increase the constant by $1$ to account
for the extra $L$-term.

We can use the same $L$-term for all approximations by putting

$\displaystyle C$
$\displaystyle=5\max\left\\{c_{1},c_{2},c_{3},c_{4}\right\\}\left(m_{B}+m_{A}+1\right),$
$\displaystyle\varepsilon$
$\displaystyle=\max_{i}\left\\{\left|\frac{\beta_{i}}{\beta}\right|,\left|\frac{\alpha_{i}}{\beta}\right|,\left|\frac{\alpha_{i}}{\alpha}\right|,\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\left|\frac{\alpha}{\beta}\right|\right\\},$

where
$\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)=0$
if $\left|\alpha\right|=\left|\beta\right|$ and $1$ otherwise. This way, we
can use the same exponentially decreasing $L$-term
$L\left(C\,n^{d_{2}}\varepsilon^{n}\right)$ across all approximations. ∎

###### Lemma 5

There are effectively computable constants $c_{5},c_{6}>0$ such that the
difference between two distinct roots $\lambda_{i},\lambda_{i^{\prime}}$ is
bounded by

$\displaystyle
c_{5}\left|\beta\right|^{n}\leq\left|\lambda_{1}-\lambda_{2}\right|\leq
c_{6}\,n^{d_{2}}\left|\beta\right|^{n}$ $\displaystyle
c_{5}\,n^{d_{1}}\left|\beta\right|^{n}\leq\left|\lambda_{1}-\lambda_{3}\right|\leq
c_{6}\,n^{d_{2}}\left|\beta\right|^{n}$ $\displaystyle
c_{5}\,n^{d_{1}}\left|\alpha\right|^{n}\leq\left|\lambda_{2}-\lambda_{3}\right|\leq
c_{6}\,n^{d_{2}}\left|\alpha\right|^{n}$

###### Proof.

By Lemma 3, the difference between two distinct roots is approximately either

$\displaystyle\left|\lambda_{1}-\lambda_{2}\right|$
$\displaystyle\approx\left|B_{n}-A_{n}\right|$
$\displaystyle\approx\left|c_{B}-\chi_{\left\\{\beta\right\\}}(\alpha)c_{A}\right|\left|\beta\right|^{n}$
$\displaystyle\left|\lambda_{1}-\lambda_{3}\right|$
$\displaystyle\approx\left|B_{n}-\frac{1}{A_{n}B_{n}}\right|$
$\displaystyle\approx\left|c_{B}\right|\left|\beta\right|^{n}$
$\displaystyle\left|\lambda_{2}-\lambda_{3}\right|$
$\displaystyle\approx\left|A_{n}-\frac{1}{A_{n}B_{n}}\right|$
$\displaystyle\approx\left|c_{A}\right|\left|\alpha\right|^{n},$

in the sense that we can bound it from below and above by the respective
right-hand term, using effectively computable constants. Doing the same again,
i.e. $n^{d_{1}}\ll\left|c_{A}\right|,\left|c_{B}\right|\ll n^{d_{2}}$ and
$1\ll\left|c_{B}-c_{A}\right|\ll n^{d_{2}}$, we can bound the difference
between two distinct roots as in the statement. ∎

##  4\. Regulator and upper bound for $\log\left|y\right|$

As per Theorem 2, the pairs $\lambda_{i},\lambda_{i}-A_{n}$ are fundamental
units of the order $\mathbb{Z}[\lambda_{i}]$. As they are generators, we can
express the regulator $R$ up to sign as the determinant

$\pm
R=\det\begin{pmatrix}\log\left|\lambda_{i}\right|&\log\left|\lambda_{i}-A_{n}\right|\\\
\log\left|\lambda_{i^{\prime}}\right|&\log\left|\lambda_{i^{\prime}}-A_{n}\right|\end{pmatrix},$

independent of the choice for $i,i^{\prime}$.

###### Lemma 6

For the regulator of the order $\mathbb{Z}[\lambda_{i}]$, we have
$R=\Theta(n^{2})$.

###### Proof.

We calculate the regulator using $(i,i^{\prime})=(1,2)$, thus

$\pm
R=\log\left|\lambda_{1}\right|\log\left|\lambda_{2}-A_{n}\right|-\log\left|\lambda_{2}\right|\log\left|\lambda_{1}-A_{n}\right|.$

With the approximations from Lemma 4, where we omit the $L$-terms for
readability, this yields

$\displaystyle\pm R$
$\displaystyle\approx\left(n\log\left|\beta\right|+\log\left|c_{B}\right|\right)\left(-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}\right|\right)$
$\displaystyle\;-\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)\left(n\log\left|\beta\right|+\log\left|c_{B}\right|\right)$
$\displaystyle=-n^{2}\,\log\left|\beta\right|\left(2\log\left|\alpha\right|+\log\left|\beta\right|\right)+O\left(n\log
n\right)$

if $\left|\alpha\right|<\left|\beta\right|$. If instead
$\left|\alpha\right|=\left|\beta\right|$, we have

$\displaystyle\pm$ $\displaystyle
R\approx\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)\left(-n\left(\log\alpha+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}-c_{A}\right|\right)$
$\displaystyle\;-\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)\left(n\log\left|\beta\right|+\log\left|c_{B}-c_{A}\right|\right)$
$\displaystyle=-n^{2}\,\log\left|\beta\right|\left(2\log\left|\alpha\right|+\log\left|\beta\right|\right)+O\left(n\log
n\right).$

In both cases, the regulator is quadratic in $n$. ∎

If we use the regulator $R$ being quadratic, which is then a quadratic upper
bound to the regulator of the number field, we immediately get from Theorem 3:

###### Lemma 7

We have

$\log\left|y\right|\ll n^{4}\log n.$

###### Proof.

Let $K=\mathbb{Q}(\lambda_{1})$. By Lemma 6, we have $R_{K}\ll n^{2}$. The
unit rank is $r=2$. The highest coefficient of the Thue equation is given by
$A_{n}B_{n}$, which is at most of order $c_{A}\,c_{B}\,\beta^{2n}$. We plug
everything into the upper bound from Theorem 3, and as the highest-order term
is of the form $n^{4}\log n$, the claim follows. ∎

##  5\. Constructing the linear form in logarithms

We now set out to construct a linear form in logarithms, on which we want to
use Baker’s and Wüstholz’s lower bound to then deduce a lower bound for
$\log\left|y\right|$. To that end, we start with the expressions
$x-\lambda_{i}y$ and fix the index $j$, sometimes called the type of the
solution $(x,y)$, such that

$\left|x-\lambda_{j}y\right|\leq\left|x-\lambda_{i}y\right|\;\;\;\forall
i\in\left\\{1,2,3\right\\},$

and denote the other two indices by $k$ and $l$, i.e.
$\left\\{j,k,l\right\\}=\left\\{1,2,3\right\\}$. First, we ascertain that the
$x-\lambda_{i}y$ give units in $\mathbb{Z}[\alpha_{i}]$.

###### Lemma 8

For each $i\in\left\\{1,2,3\right\\}$, we have
$x-\lambda_{i}y\in\left(\mathbb{Z}[\lambda_{i}]\right)^{\times}$.

###### Proof.

We take the Norm $N=N_{\mathbb{Q}(\lambda_{1})/\mathbb{Q}}$, which is the
product of the conjugates, i.e.

$N(x-\lambda_{i}y)=\left(x-\lambda_{j}y\right)\left(x-\lambda_{k}y\right)\left(x-\lambda_{l}y\right).$

Now the right hand side is $y^{3}f_{n}\left(\frac{x}{y}\right)$ and thus
exactly the left hand side of our Thue equation at $(x,y)$. Thus,
$\left|N(x-\lambda_{i}y)\right|=1$, from which the claim follows. ∎

As the $x-\lambda_{i}y$ are units in $\mathbb{Z}[\lambda_{i}]$, we can write
them in terms of the fundamental units $\lambda_{i},\lambda_{i}-A_{n}$. Let
thus

$x-\lambda_{i}y=\pm\lambda_{i}^{b_{1}}\left(\lambda_{i}-A_{n}\right)^{b_{2}}$
(6)

for each $i\in\left\\{1,2,3\right\\}$. We then take a look at Siegel’s
identity,

$\left(x-\lambda_{j}\,y\right)\left(\lambda_{k}-\lambda_{l}\right)+\left(x-\lambda_{l}\,y\right)\left(\lambda_{j}-\lambda_{k}\right)+\left(x-\lambda_{k}\,y\right)\left(\lambda_{l}-\lambda_{j}\right)=0,$

which we rewrite into

$\frac{x-\lambda_{l}y}{x-\lambda_{k}y}\cdot\frac{\lambda_{j}-\lambda_{k}}{\lambda_{j}-\lambda_{l}}=1+\frac{x-\lambda_{j}y}{x-\lambda_{k}y}\cdot\frac{\lambda_{l}-\lambda_{k}}{\lambda_{j}-\lambda_{l}}=:1+\gamma.$

We take the absolute value and use our decomposition into fundamental units on
the first term of the left hand side, thus defining the linear form in
logarithms

$\Lambda:=b_{1}\log\left|\frac{\lambda_{l}}{\lambda_{k}}\right|+b_{2}\log\left|\frac{\lambda_{l}-A_{n}}{\lambda_{k}-A_{n}}\right|+\log\left|\frac{\lambda_{j}-\lambda_{k}}{\lambda_{j}-\lambda_{l}}\right|=\log\left|1+\gamma\right|,$

which is very small, as $\gamma$ is very small. In fact:

###### Lemma 9

We have

$\left|\gamma\right|\leq
2c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}$

and, if
$2c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}\leq\frac{1}{2}$,

$\left|\Lambda\right|\leq
4c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}.$

###### Proof.

As

$\gamma=\frac{x-\lambda_{j}y}{x-\lambda_{k}y}\cdot\frac{\lambda_{l}-\lambda_{k}}{\lambda_{j}-\lambda_{l}},$

we have to bound all four terms. For $i\neq j$, we have

$\left|y\right|\left|\lambda_{i}-\lambda_{j}\right|\leq\left|x-\lambda_{i}y\right|+\left|x-\lambda_{j}y\right|\leq
2\left|x-\lambda_{i}y\right|.$

In combination with $\left|y\right|\geq 2$, we thus have

$\left|x-\lambda_{k}y\right|\geq\left|\lambda_{k}-\lambda_{j}\right|$

and, as the product
$\left(x-\lambda_{j}y\right)\left(x-\lambda_{k}y\right)\left(x-\lambda_{l}y\right)=\pm
1$,

$\left|x-\lambda_{j}y\right|=\left|x-\lambda_{k}y\right|^{-1}\left|x-\lambda_{l}y\right|^{-1}\leq\left|\lambda_{k}-\lambda_{j}\right|^{-1}\left|\lambda_{l}-\lambda_{j}\right|^{-1}.$

We put everything together and get

$\displaystyle\left|\frac{x-\lambda_{j}y}{x-\lambda_{k}y}\right|\cdot\left|\frac{\lambda_{l}-\lambda_{k}}{\lambda_{j}-\lambda_{l}}\right|\leq\frac{\left|\lambda_{k}-\lambda_{j}\right|^{-1}\left|\lambda_{l}-\lambda_{j}\right|^{-1}}{\left|\lambda_{k}-\lambda_{j}\right|}\cdot\frac{\left|\lambda_{l}-\lambda_{j}\right|+\left|\lambda_{k}-\lambda_{j}\right|}{\left|\lambda_{l}-\lambda_{j}\right|},$

where we also used the triangle inequality on the numerator of the second
fraction. The right hand side is now equal to

$\left|\lambda_{k}-\lambda_{j}\right|^{-2}\left|\lambda_{l}-\lambda_{j}\right|^{-1}+\left|\lambda_{k}-\lambda_{j}\right|^{-1}\left|\lambda_{l}-\lambda_{j}\right|^{-2}\leq
2c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}$

as per Lemma 5, and the bound decreases exponentially even if
$\left|\alpha\right|=1$. ∎

We’d like to use the lower bound from Theorem 4 directly on $\Lambda$.
However, as the arguments of some of the logarithms are exponential in $n$,
their logarithmic heights are linear in $n$. This will give
$\log\left|\Lambda\right|\gg-n^{3}\log\left(\frac{1}{n}\log\left|y\right|\right)$,
from which we can only deduce

$\log\left|y\right|\gg n$ (7)

instead of the desired exponential lower bound.

We thus aim to shift the exponential dependency on $n$ from the logarithms
into the coefficients, where an additional factor $n$ will not ruin the
argument.

###  5.1. Transforming the linear form

Let $j=1$ and choose $(k,l)=(3,2)$. We use the approximations from Lemma 4 to
write:

$\displaystyle\log\left|\frac{\lambda_{l}}{\lambda_{k}}\right|$
$\displaystyle+L\left(2C\,n^{d_{2}}\varepsilon^{n}\right)=\log\left|\lambda_{2}\right|-\log\left|\lambda_{3}\right|+L\left(2C\,n^{d_{2}}\varepsilon^{n}\right)$
$\displaystyle=n\left(2\log\left|\alpha\right|+\log\left|\beta\right|\right)+2\log\left|c_{A}\right|+\log\left|c_{B}\right|$
$\displaystyle\log\left|\frac{\lambda_{l}-A_{n}}{\lambda_{k}-A_{n}}\right|$
$\displaystyle+L\left(2C\,n^{d_{2}}\varepsilon^{n}\right)=\log\left|\lambda_{2}-A_{n}\right|-\log\left|\lambda_{3}-A_{n}\right|+L\left(2C\,n^{d_{2}}\varepsilon^{n}\right)$
$\displaystyle=\begin{cases}-n\,\left(2\log\left|\alpha\right|+\log\left|\beta\right|\right)-2\log\left|c_{A}\right|-\log\left|c_{B}\right|\;\;&\text{
if }\left|\alpha\right|<\left|\beta\right|\\\
-n\,\left(2\log\left|\alpha\right|+\log\left|\beta\right|\right)-2\log\left|c_{A}\right|-\log\left|c_{B}-c_{A}\right|\;\;&\text{
if }\left|\alpha\right|=\left|\beta\right|\end{cases}$
$\displaystyle\log\left|\frac{\lambda_{k}-\lambda_{j}}{\lambda_{l}-\lambda_{j}}\right|$
$\displaystyle+L\left(4C\,n^{d_{2}}\varepsilon^{n}\right)=\log\left|\lambda_{3}-\lambda_{1}\right|-\log\left|\lambda_{2}-\lambda_{1}\right|+L\left(4C\,n^{d_{2}}\varepsilon^{n}\right)$
$\displaystyle=\begin{cases}0\;\;&\text{ if
}\left|\alpha\right|<\left|\beta\right|\\\
\log\left|c_{B}\right|-\log\left|c_{B}-c_{A}\right|\;\;&\text{ if
}\left|\alpha\right|=\left|\beta\right|\end{cases}$

We shift the $L$-terms into the upper bound, and get a linear form in at most
$4$ logarithms. The coefficients are gathered in the table below, where
$\chi_{M}$ denotes the characteristic function on the set $M$.

| $j=1$
---|---
$\log\left|\alpha\right|$ | $2n\left(b_{1}-b_{2}\right)$
$\log\left|\beta\right|$ | $n\left(b_{1}-b_{2}\right)$
$\log\left|c_{A}\right|$ | $2\left(b_{1}-b_{2}\right)$
$\log\left|c_{B}\right|$ | $b_{1}-\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)b_{2}+\chi_{\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)$
$\log\left|c_{B}-c_{A}\right|$ | $\chi_{\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)\left(b_{2}+1\right)$

We do the same for the cases $(j,k,l)=(2,3,1)$ and $(j,k,l)=(3,2,1)$ and get:

| $j=2$ | $j=3$
---|---|---
$\log\left|\alpha\right|$ | $n\left(b_{1}-\left(b_{2}-1\right)\right)$ | $n\left(b_{2}-\left(b_{1}-1\right)\right)$
$\log\left|\beta\right|$ | $n\left(2b_{1}+b_{2}-1\right)$ | $n\left(2b_{2}+b_{1}-1\right)$
$\log\left|c_{A}\right|$ | $b_{1}-\left(b_{2}-1\right)$ | $b_{2}+-\left(b_{1}-1\right)$
$\log\left|c_{B}\right|$ | $2b_{1}+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)\left(b_{2}-1\right)$ | $\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)2b_{2}-1$
$\log\left|c_{B}-c_{A}\right|$ | $\chi_{\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)\left(b_{2}-1\right)$ | $\chi_{\left\\{\left|\beta\right|\right\\}}\left(\left|\alpha\right|\right)2b_{2}$

We call the linear form in the logarithms and coefficients as per the tables
above $\xi_{j}$. As we shifted the $L$-terms into the upper bound, we have to
worsen it accordingly, we thus have

$\left|\xi_{j}\right|\leq
4c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}+6C\,n^{d_{2}}\varepsilon^{n}\ll
n^{d_{2}}\varepsilon^{n}.$ (8)

###  5.2. A closer look at the powers

To apply Baker’s and Wüstholz’s lower bound on the linear form $\xi$, we have
to argue $\xi\neq 0$ first. For that, we have to either restrict the
logarithms, such as demanding linear independence or gain further information
on the coefficients, and thus on $b_{1},b_{2}$.

To that end, we return to the equations

$\left|x-\lambda_{i}y\right|=\left|\lambda_{i}^{b_{1}}\left(\lambda_{i}-A_{n}\right)^{b_{2}}\right|\;\;\forall
i\in\left\\{1,2,3\right\\}.$

If we take the logarithm, the equations for $i=k,l$ become

$\begin{pmatrix}\log\left|x-\lambda_{k}y\right|\\\
\log\left|x-\lambda_{l}y\right|\end{pmatrix}=\begin{pmatrix}\log\left|\lambda_{k}\right|&\log\left|\lambda_{k}-A_{n}\right|\\\
\log\left|\lambda_{l}\right|&\log\left|\lambda_{l}-A_{n}\right|\end{pmatrix}\begin{pmatrix}b_{1}\\\
b_{2}\end{pmatrix}.$ (9)

The $2\times 2$ matrix on the right hand side is invertible, as its
determinant is equal to the regulator $R$, which is non-zero as per Lemma 6.
We multiply the equation with the inverse matrix

$\frac{1}{R}\begin{pmatrix}\log\left|\lambda_{l}-A_{n}\right|&-\log\left|\lambda_{k}-A_{n}\right|\\\
-\log\left|\lambda_{l}\right|&\log\left|\lambda_{k}\right|\end{pmatrix}$

to get an explicit formula for $b_{1},b_{2}$. We further break down the
formula by writing

$\displaystyle\log\left|x-\lambda_{i}y\right|$
$\displaystyle=\log\left|x-\lambda_{j}y-\left(\lambda_{i}-\lambda_{j}\right)y\right|$
$\displaystyle=\log\left|y\right|+\log\left|\lambda_{i}-\lambda_{j}\right|+\log\left(1-\frac{x-\lambda_{j}y}{\left(\lambda_{i}-\lambda_{j}\right)y}\right)$

for $i=k,l$. The last term is small, as seen in the proof of Lemma 9. By
writing

$\frac{\left|x-\lambda_{j}\right|}{\left|\lambda_{i}-\lambda_{j}\right|\left|y\right|}\leq\frac{\left|\lambda_{k}-\lambda_{j}\right|^{-1}\left|\lambda_{l}-\lambda_{j}\right|^{-1}}{2\left|\lambda_{i}-\lambda_{j}\right|}\leq\frac{1}{2}c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}$

as per Lemma 5, we thus have

$\log\left|x-\lambda_{i}y\right|=\log\left|y\right|+\log\left|\lambda_{i}-\lambda_{j}\right|+L\left(\frac{1}{2}c_{5}^{3}\,n^{-d_{1}}\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}\right).$

We use this expression to rewrite Equation (9), multiplied with the inverse
matrix, into

$\begin{pmatrix}b_{1}\\\
b_{2}\end{pmatrix}=\frac{1}{R}\begin{pmatrix}\log\left|\lambda_{l}-A_{n}\right|&-\log\left|\lambda_{k}-A_{n}\right|\\\
-\log\left|\lambda_{l}\right|&\log\left|\lambda_{k}\right|\end{pmatrix}\begin{pmatrix}\log\left|y\right|+\log\left|\lambda_{k}-\lambda_{j}\right|\\\
\log\left|y\right|+\log\left|\lambda_{l}-\lambda_{j}\right|\end{pmatrix},$
(10)

up to an error term
$L\left(C^{\prime}(n)\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}\right)$
with

$C^{\prime}(n):=\frac{1}{R}\log\left|\frac{\lambda_{k}\left(\lambda_{l}-A_{n}\right)}{\lambda_{l}\left(\lambda_{k}-A_{n}\right)}\right|c_{5}^{3}n^{-d_{1}},$

and we mainly care that the $L$-term decreases exponentially and that
$C^{\prime}(n)\ll 1$.

In the case that $j=1$ and $\left|\alpha\right|<\left|\beta\right|$ only, we
can actually forego to fiddle around with the linear form and directly derive
an exponential lower bound for $\log\left|y\right|$. To that end, we use a
different set of fundamental units, $\lambda_{i}$ and $\lambda_{i}-B_{n}$.
They still fall under the results of Theorem 2, and Equation (10) holds for
the new set of powers $u_{1},u_{2}$ if we substitute $A_{n}$ for $B_{n}$ in
the matrix.

###### Lemma 10

If $j=1$ and $\left|\alpha\right|<\left|\beta\right|$, then
$\log\log\left|y\right|\gg n$.

###### Proof.

We take a closer look at $\pm R\,u_{1}$, which is

$\left(\log\left|\lambda_{2}-B_{n}\right|-\log\left|\lambda_{3}-B_{n}\right|\right)\log\left|y\right|+t(n)+L\left(C^{\prime}(n)\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}\right),$

where we only care that $\left|t(n)\right|$ at most decreases exponentially
per Lemma 4, so $\left|t(n)\right|\ll c_{7}^{-n}$.

We have
$c_{8}^{-n}\ll\left|\log\left|\lambda_{2}-B_{n}\right|-\log\left|\lambda_{3}-B_{n}\right|\right|\ll
c_{9}^{-n}$ with the same argument. By Equation (7) we also have
$\log\left|y\right|\gg n$, while $C^{\prime}(n)\ll 1$. Thus,
$\left|Ru_{1}\right|\neq 0$, hence $\left|Ru_{1}\right|\geq R$ for
sufficiently large $n$.

This, however, means that

$\log\left|y\right|\geq\frac{R-t(n)-C^{\prime}(n)\left|\alpha\right|^{-2n}\left|\beta\right|^{-n}}{\left|\log\left|\lambda_{2}-B_{n}\right|-\log\left|\lambda_{3}-B_{n}\right|\right|}\gg
c_{10}^{n},$

and if $c_{7},c_{8},c_{9}>1$, so is $c_{10}$. ∎

###### Lemma 11

For $j\in\left\\{2,3\right\\}$ or
$j=1,\,\left|\alpha\right|=\left|\beta\right|$, we have $\xi_{j}\neq 0$.

###### Proof.

Assume $j=1$ and take $(k,l)=(3,2)$. By Equation (10) and Lemma 4, the powers
$b_{1},b_{2}$ are then, up to exponentially decreasing errors,

$\displaystyle R\,b_{1}=$
$\displaystyle\big{(}-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|$
$\displaystyle-\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\big{)}\left(\log\left|y\right|+n\log\left|\beta\right|+\log\left|c_{B}\right|\right)$
$\displaystyle-$
$\displaystyle\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)$
$\displaystyle\left(\log\left|y\right|+n\log\left|\beta\right|+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|-\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\right)$

and

$\displaystyle R\,b_{2}=-$
$\displaystyle\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)\left(\log\left|y\right|+n\log\left|\beta\right|+\log\left|c_{B}\right|\right)$
$\displaystyle+$
$\displaystyle\left(-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}\right|\right)$
$\displaystyle\left(\log\left|y\right|+\log\left|\beta\right|+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|-\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\right).$

If we multiply the terms out and look at the highest order terms – those of
$n^{2}$ and $\log\left|y\right|\cdot n$, then we see that they have negative
sign and are the same for $b_{1}$ and $b_{2}$. The other terms all get
dominated by the factor $\frac{1}{R}$, which decreases like $\frac{1}{n^{2}}$,
as per Lemma 6. Thus, $b_{1}=b_{2}$ for sufficiently large $n$. Note that this
would mean that all coefficients in $\xi_{1}$ are $0$ if
$\left|\alpha\right|<\left|\beta\right|$.

If $\left|\alpha\right|=\left|\beta\right|$ however, $b_{1}=b_{2}$ immediately
implies that $\xi_{1}\neq 0$, except for when $b_{1}=b_{2}=-1$, as per our
conditions on $\left|c_{B}\right|,\left|c_{B}-c_{A}\right|$ in Theorem 1.

Going back to the unit decompositions in Equation (6), we can rewrite them
into

$\left(x-\lambda_{i}y\right)\lambda_{i}^{-b_{1}}\left(\lambda_{i}-A_{n}\right)^{-b_{2}}-1=0$

for $i\in\left\\{1,2,3\right\\}$. With $b_{1}=b_{2}=-1$, this is a polynomial
equation and the $\lambda_{i}$ thus roots of the polynomial

$\left(-yX+x\right)X\left(X-A_{n}\right)-1,$

which must then be divisible by the minimal polynomial
$X\left(X-A_{n}\right)\left(X-B_{n}\right)-1$. The constant terms are equal,
so the leading coefficients must be, too. This gives $y=-1$ and thus the
contradiction to $(x,y)$ being a non-trivial solution. So $j$ cannot be $1$ if
$\left|\alpha\right|=\left|\beta\right|$.

Assume, then, $j=2$ and take $(k,l)=(3,1)$. Equation (10) and Lemma 4 give

$\displaystyle R\,b_{1}=$
$\displaystyle\left(n\log\left|\beta\right|+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|+\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\right)$
$\displaystyle\left(\log\left|y\right|+n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)$
$\displaystyle-$
$\displaystyle\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)$
$\displaystyle\left(\log\left|y\right|+n\log\left|\beta\right|+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|+\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\right)$

and

$\displaystyle R\,b_{2}=-$
$\displaystyle\left(n\log\left|\beta\right|+\log\left|c_{B}\right|\right)\left(\log\left|y\right|+n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)$
$\displaystyle+$
$\displaystyle\left(-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\log\left|c_{B}\right|\right)$
$\displaystyle\left(\log\left|y\right|+n\log\left|\beta\right|+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|+\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\right)$

up to exponentially decreasing errors. Further assume that
$\left|\alpha\right|\neq\left|\beta\right|$ and we look at the logarithms
$\log\left|\alpha\right|,\log\left|\beta\right|$ in $\xi_{2}$. The term in
question is

$n\left(b_{1}-\left(b_{2}-1\right)\right)\log\left|\alpha\right|+n\left(2b_{1}+b_{2}-1\right)\log\left|\beta\right|$

and dominates the other terms. Thus, it suffices to show that its highest
order terms do not cancel each other out, which is easily verified with the
formula derived for $b_{1}$ and $b_{2}$.

If $\left|\alpha\right|=\left|\beta\right|$ instead, the same term simplifies
to $3nb_{1}\log\left|\alpha\right|$ and the formula for $b_{1}$ itself to
$R\,b_{1}=\log\left|y\right|\left(\log\left|c_{B}-c_{A}\right|-\log\left|c_{A}\right|\right)$.
We can no longer claim that $3nb_{1}$ dominates the other coefficients of
$\xi_{2}$ and have to take a closer look.

To that end, we first write $R\,b_{2}$ explicitly in terms of $n^{2},n$ and
$\log\left|y\right|$. We have

$\displaystyle R\,b_{2}$
$\displaystyle=n^{2}\left(-3\left(\log\left|\alpha\right|\right)^{2}\right)+n\cdot
2\log\left|\alpha\right|\left(-\log\left|c_{A}\right|-\log\left|c_{B}\right|-\log\left|c_{B}-c_{A}\right|\right)$
$\displaystyle+\log\left|y\right|\left(-3n\log\left|\alpha\right|-2\log\left|c_{B}\right|-\log\left|c_{A}\right|\right).$

We plug everything into the linear form $\xi_{2}$,

$3nb_{1}\log\left|\alpha\right|+\left(b_{1}-b_{2}+1\right)\log\left|c_{A}\right|+2b_{1}\log\left|c_{B}\right|+\left(b_{2}-1\right)\log\left|c_{B}-c_{A}\right|,$

and group the terms with respect to their order in $n$, differentiating two
cases. Either the term with $\frac{1}{R}\log\left|y\right|\,n$, the whole term
is

$\frac{1}{R}\log\left|y\right|\,n\cdot
3\log\left|\alpha\right|\left(\log\left|c_{A}\right|-\log\left|c_{B}-c_{A}\right|\right),$

is at most constant, equivalently $\log\left|y\right|=O(n)$. In that case,
since $\log\left|c_{A}\right|-\log\left|c_{B}-c_{A}\right|>0$ as per our
conditions on $A_{n},B_{n}$, the whole term equals at most some positive
constant. While the term $\frac{1}{R}n^{2}\cdot
3\log\left|\alpha\right|^{2}\left(\log\left|c_{A}\right|-\log\left|c_{B}-c_{A}\right|\right)$
equals some positive constant, and thus their sum. All the other terms
decrease with increasing $n$, thus the linear form $\xi_{2}$ does not cancel
for sufficiently large $n$. In the other case, the term with
$\frac{1}{R}\log\left|y\right|\cdot n$ dominates all others – including the
one with $\frac{1}{R}n^{2}$ – and we can conclude $\xi_{2}\neq 0$ for
sufficiently large $n$ just by the non-vanishing of the
$\frac{1}{R}\log\left|y\right|\,n$ term.

Finally, assume $j=3$ and take $(k,l)=(2,1)$. We again have, by Equation (10)
and Lemma 4,

$\displaystyle R\,b_{1}=$
$\displaystyle\left(n\log\left|\beta\right|+\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|+\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\right)$
$\displaystyle\left(\log\left|y\right|+n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)$
$\displaystyle-$
$\displaystyle\big{(}-n\left(\log\left|\alpha\right|+\log\left|\beta\right|\right)-\log\left|c_{A}\right|-\chi_{\mathbb{R}\backslash\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}\right|$
$\displaystyle+\chi_{\left\\{\left|\beta\right|\right\\}}(\left|\alpha\right|)\log\left|c_{B}-c_{A}\right|\big{)}\left(\log\left|y\right|+n\log\left|\beta\right|+\log\left|c_{B}\right|\right)$

and

$\displaystyle R\,b_{2}=-$
$\displaystyle\left(n\log\left|\beta\right|+\log\left|c_{B}\right|\right)\left(\log\left|y\right|+n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)$
$\displaystyle+$
$\displaystyle\left(n\log\left|\alpha\right|+\log\left|c_{A}\right|\right)\left(\log\left|y\right|+n\log\left|\beta\right|+\log\left|c_{B}\right|\right)$

up to exponentially decreasing errors. Again, if
$\left|\alpha\right|\neq\left|\beta\right|$, the coefficients of
$\log\left|\alpha\right|,\log\left|\beta\right|$ dominate the others in
$\xi_{3}$, the term in question being

$n\left(b_{2}-\left(b_{1}-1\right)\right)\log\left|\alpha\right|+n\left(2b_{2}+\left(b_{1}-1\right)\right)\log\left|\beta\right|.$

It follows from the formulas for $b_{1},b_{2}$ that its highest order terms do
not cancel each other out and we can thus conclude $\xi_{3}\neq 0$.

If $\left|\alpha\right|=\left|\beta\right|$, this again fails, as the term
simplifies to $3nb_{2}\log\left|\alpha\right|$ and $b_{2}$ to
$R\,b_{2}=\log\left|y\right|\left(\log\left|c_{A}\right|-\log\left|c_{B}\right|\right)$.
We state $R\,b_{1}$ in terms of $n^{2}$, $n$, and $\log\left|y\right|$ and get

$\displaystyle R\,b_{1}$
$\displaystyle=n^{2}\left(3\left(\log\left|\alpha\right|\right)^{2}\right)+n\cdot
2\log\left|\alpha\right|\left(\log\left|c_{A}\right|+\log\left|c_{B}\right|+\log\left|c_{B}-c_{A}\right|\right)$
$\displaystyle+\log\left|y\right|\left(3n\log\left|\alpha\right|+\log\left|c_{A}\right|+2\log\left|c_{B}-c_{A}\right|\right).$

We then plug everything into the linear form $\xi_{3}$,

$3nb_{2}\log\left|\alpha\right|+\left(b_{2}+b_{1}-1\right)\log\left|c_{A}\right|-\log\left|c_{B}\right|+2b_{2}\log\left|c_{B}-c_{A}\right|,$

and conclude analogously to the case $j=2$ that $\xi_{3}\neq 0$. It should be
noted that $\log\left|c_{A}\right|-\log\left|c_{B}\right|$ is non-zero, but
could be either positive or negative. However, the factor occurs in both the
terms with $\frac{1}{R}n^{2}$ and $\frac{1}{R}\log\left|y\right|\,n$ and is
the only one of questionable sign, thus both terms have the same sign and do
not cancel with the same argument as for the case $j=2$. ∎

With $\xi_{j}\neq 0$, we can apply Theorem 4 and derive a lower bound for
$\log\left|\xi\right|$. The coefficients of $\xi_{j}$ can be bounded by
$n\cdot\max\left\\{b_{1},b_{2}\right\\}$. By Equation (10), in combination
with $R$ being quadratic by Lemma 6 and $\log\left|y\right|$ being at most of
order $n^{4}\log n$ by Lemma 7, we have $B\ll n^{4}\log n$ in Theorem 4. The
terms $c_{A},c_{B}$ are at most polynomial in $n$, thus their logarithmic
heights are of order at most $\log n$, i.e.
$h\left(\left|c_{A}\right|\right),h\left(\left|c_{B}\right|\right),h\left(\left|c_{B}-c_{A}\right|\right)\ll\log
n$, and the logarithmic heights of $\left|\alpha\right|,\left|\beta\right|$
are constants.

Everything put together, Theorem 4 gives

$\log\left|\xi_{j}\right|\gg-\left(\log n\right)^{3}\cdot\log\left(n^{4}\log
n\right).$

If we take the logarithm of the upper bound from Equation (8), we get

$-n\gg-\left(\log n\right)^{3}\cdot\log\left(n^{4}\log n\right),$

which gives us a contradiction for sufficiently large $n$, or in turn an
effective upper bound for $n$. We thus conclude our proof of Theorem 1.

Acknowledgement

The author wants to thank Volker Ziegler for the initial idea as well as many
helpful discussions and feedback.

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