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% are $\pm 1$. Hence, if $|D_K|>3$ we may take $\mu=\pm 1$ and get immediately
% the following corollary.

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\theoremstyle{remark}
\newtheorem*{acknowledgement}{Acknowledgement}


\newcommand{\exceptt}{(-1\pm 3\sqrt{-3})/2}
\newcommand{\toappearin}{To appear in }

\title[A Family of Relative Cubic Thue Equations]{A Family of Relative Cubic Thue Equations over\\Imaginary Quadratic
  Number Fields}
\author{Clemens Heuberger}
\thanks{The first and the third author were supported by grant S8307-MAT of the
  Austrian Science Fund.}
\address[C.\ Heuberger and R.~F.\ Tichy]{Institut f\"ur Mathematik\\
Technische Universit\"at Graz\\
Steyrergasse 30\\
A-8010 Graz\\
Austria}
\email{clemens.heuberger@tugraz.at, tichy@tugraz.at}

\author{Attila Peth\H o}
\address[A. Peth\H o]{Mathematical Institute\\
University of Debrecen\\
P.O. Box 12\\
H-4010 Debrecen\\
Hungary}
\email{pethoe@math.klte.hu}
\thanks{The second author was partially supported by Hungarian National
  Foundation for Scientific Research Grants No 29330 and 38225.}

\author{Robert F.\ Tichy}
% \address{Institut f\"ur Mathematik\\
% Technische Universit\"at Graz\\
% Steyrergasse 30\\
% A-8010 Graz\\
% Austria}
\dedicatory{Dedicated to Professor H.~G.\ Zimmer on his
  65\textsuperscript{th} birthday}




\begin{document}
\begin{abstract}
  We consider the family of relative Thue equations 
  \begin{equation*}
    x^3-(t-1)x^2y-(t+2)xy^2-y^3=\mu,  
  \end{equation*}
  where the parameter $t$, the root of unity $\mu$ and the solutions $x$ and
  $y$ are integers in the same imaginary quadratic number field.

  We prove that there are only trivial solutions (with $\betrag x$, $\betrag
  y\le 1$), if $\betrag{t}$ is large enough or if the discriminant of the quadratic
  number field is large enough or if $\myRe t=-1/2$ (there are a few more
  solutions in this case which are explicitly listed). In the case $\myRe t=-1/2$, 
  an algebraic method is used, in the general case, Baker's method yields the result.
\end{abstract}
\maketitle


\section{Introduction}
Let $F\in\Z[x,y]$ be an irreducible form of degree at least $3$ and $m$ be a
nonzero integer. Then the Diophantine equation
$$F(x,y)=m$$
is called a \emph{Thue} equation in honor of A.~Thue~\cite{Thue:1909} who
proved that it has only finitely many solutions $x$, $y\in \Z$. Upper bounds for the solutions
have been given using A.~Baker's \cite{Baker:1968} theory on linear forms in logarithms of
algebraic numbers, cf.~Bugeaud and Gy\H ory~\cite{Bugeaud-Gyoery:1996}. 
Algorithms to solve single Thue equations have
been developed, we refer to Bilu and Hanrot~\cite{Bilu-Hanrot:1996}.

In the last decade, several families of parametrized Thue equations have been
investigated; a survey containing further references is given in
\cite{Heuberger:2000:nt98}. Furthermore, we refer to the more recent results
\cite{Wakabayashi:ta:cubic-thue-negative-discriminant},
\cite{Thomas:1993:solut-complex-thue}, 
\cite{Togbe:2000}, \cite{Dujella-Jadrijevic:2002:thue},
\cite{Jadrijevic:ta:two-parametric-thue}, \cite{Dujella-Jadrijevic:thue-inequality}, \cite{Gaal-Lettl:2000:param-famil}, and
\cite{Heuberger:2001:tho-eff}.

Relative Thue equations, i.~e.\ Thue equations with coefficients in an algebraic
number field, where solutions come from the ring of integers of the same number
field, have also been successfully solved, we refer to Ga\'al and
Pohst~\cite{Gaal-Pohst:2002:thue} for further references.

However, to our knowledge, no infinite family of relative Thue equations has
been solved up to now. The aim of this paper is to solve such a family.
As the ground field, we take an imaginary quadratic number field. 
We consider the form $x^3-(t-1)x^2y-(t+2)xy^2-y^3$. If $t\in\Z$, it
generates Shanks'~\cite{Shanks:1974} simplest cubic number fields. The
corresponding family of absolute Thue
equations was the first family of positive discriminant which has been solved,
cf.~Thomas~\cite{Thomas:1990} and Mignotte~\cite{Mignotte:1993}. We consider
the same form with an imaginary quadratic integral parameter $t$.
This setting has been chosen in order to avoid the unit rank to increase.

Our main result is the following:

\newcommand{\tmin}{3\,023\,195\,238}
\begin{theorem}\label{th:result}
Let $t$ be an integer in an imaginary quadratic number field, 
$\Z_{\Q(t)}$ be the ring of integers of $\Q(t)$ and $\mu$ be a root of unity in $\Z_{\Q(t)}$.

If $\betrag{t}\ge \tmin$, then the only solutions
$x$, $y$ in $\Z_{\Q(t)}$ to
the family of relative Thue equations
\begin{equation}\label{eq:thue}
x^3-(t-1)x^2y-(t+2)xy^2-y^3=\mu  
\end{equation}
are listed in Table~\ref{tab:small-solutions-all-t}.
\end{theorem}

\begin{table}[htbp]
  \begin{center}
    \begin{equation*}
      \begin{array}{rrr}
        x & y & \mu\\\hline
        0 & 1 & -1\\
        -1 & 0 & -1\\
        1 & -1 & -1\\
        0 & -1 & 1\\
        -1 & 1 & 1\\
        1 & 0 & 1\\
        0 & -\complexI & -\complexI\\
        -\complexI & \complexI & -\complexI\\
%      
      \end{array}
\qquad
      \begin{array}{rrr}
        x & y & \mu\\\hline
%
        \complexI & 0 & -\complexI\\
        0 & \complexI & \complexI\\
        -\complexI & 0 & \complexI\\
        \complexI & -\complexI & \complexI\\
        0 & -{{\omega}_3} & -1\\
        0 & -1+{{\omega}_3} & -1\\
        -{{\omega}_3} & {{\omega}_3} & -1\\
        1-{{\omega}_3} & 0 & -1\\
%      
      \end{array}
\qquad
      \begin{array}{rrr}
        x & y & \mu\\\hline
%
        -1+{{\omega}_3} & 1-{{\omega}_3} & -1\\
        {{\omega}_3} & 0 & -1\\
        0 & 1-{{\omega}_3} & 1\\
        0 & {{\omega}_3} & 1\\
        -{{\omega}_3} & 0 & 1\\
        1-{{\omega}_3} & -1+{{\omega}_3} & 1\\
        -1+{{\omega}_3} & 0 & 1\\
        {{\omega}_3} & -{{\omega}_3} & 1
      \end{array} 
    \end{equation*}
    \caption{Solutions (if contained in $\Q(t)$) to \eqref{eq:thue} for all $t$, where
        $\omega_3=(1+\sqrt{-3})/2$.}
    \label{tab:small-solutions-all-t}
  \end{center}
\end{table}

For $t\in\Z^{\ge 0}$, Theorem~\ref{th:result} has been proved by
Thomas~\cite{Thomas:1990}. Writing $F_t(X,Y)=X^3-(t-1)X^2Y-(t+2)XY^2-Y^3$ and
noting that $F_{-t-1}(Y,X)=-F_t(X,Y)$, his result is easily generalized to
negative rational integers $t$. We note that Thomas~\cite{Thomas:1990} and
Mignotte~\cite{Mignotte:1993} solved the rational integer case completely;
they found some extra solutions for $\betrag{t}\le 4$. Therefore, we will
assume $t\notin \Z$ in the remainder of this paper.

The condition $\betrag{t}\ge \tmin$ is certainly fulfilled if the discriminant
of the base field is sufficiently large (in absolute value). 
It is well known that if $ D_K<-4$ then the only roots of unity in $K$
are $\pm 1$. Hence, if $|D_K|>4$ we may take $\mu=\pm 1$ and get immediately
the following corollary.

\begin{corollary}Let $k$ be an imaginary quadratic number field with
        discriminant $\ge 3.66\cdot 10^{19}$ and $t\in k\setminus \Z$ be
        an algebraic integer. Then the only solutions $x$, $y\in\Z_k$ to
        \begin{equation*}
          x^3-(t-1)x^2y-(t+2)xy^2-y^3=\pm 1
        \end{equation*}
        are $\pm \{(0,1), (-1,0), (1,-1)\}$.
\end{corollary}

The proof of Theorem~\ref{th:result} is based on Baker's theory on linear forms
in logarithms of algebraic numbers. This yields the rather large constant
$3.03\cdot 10^9$. Since the parameter $t$ is complex, a complete enumeration of
the remaining cases seems to be hopeless. However, if
$\myRe t=-1/2$, we can use an algebraic argument and solve the
equation completely for these values of the parameter.


\begin{theorem}\label{th:result:re-t--1-2}
  Let $t\neq\exceptt$ be an integer in an imaginary quadratic number field with
  $\myRe t=-1/2$ and $\mu$ be a root of unity in $\Q(t)$. 
  Then the only solutions $x$, $y\in\Z_{\Q(t)}$ to \eqref{eq:thue}
  are listed in Table~\ref{tab:small-solutions-all-t} or in Table~\ref{tab:small-solutions-re-t--1-2}.
\end{theorem}


\begin{table}[htbp]
  \begin{center}
    \begin{equation*}
      \begin{array}{r|>{\pm\{}l<{\}}}
        t&\multicolumn{1}{l}{(x,y)\in}\\\hline
                                %
        \vbox to 12pt{}
        -\omega_3& 
        (3 - 6\omega_3,  -1 + 3\omega_3), 
         (-2 + 3\omega_3,  3 - 6\omega_3), 
         (-1 + 3\omega_3,  -2 + 3\omega_3)\\
                                %
        -1 + \omega_3& 
        (-3 + 6\omega_3,  2 - 3\omega_3), 
        ( 1 - 3\omega_3,  -3 + 6\omega_3), 
        ( 2 - 3\omega_3,  1 - 3\omega_3)\\
                                %
        -\omega_7&
        (-1 + 2\omega_7,  -\omega_7), 
        (-\omega_7,  1 - \omega_7), 
        ( 1 - \omega_7,  -1 + 2\omega_7)\\
                                %
        -1 + \omega_7&
        (1 - 2\omega_7,  -1 + \omega_7), 
        (-1 + \omega_7,  \omega_7), 
        (\omega_7,  1 - 2\omega_7)\\
                                %
        -\omega_{19}&
        (-1 + 2\omega_{19},  -1 - \omega_{19}), 
        (-1 - \omega_{19},  2 - \omega_{19}), 
        (2 - \omega_{19},  -1 + 2\omega_{19})\\
                                %
        -1 + \omega_{19}&
        (1 - 2\omega_{19},  -2 + \omega_{19}), 
        ( -2 + \omega_{19},  1 + \omega_{19}), 
        ( 1 + \omega_{19},  1 - 2\omega_{19})
      \end{array}
    \end{equation*}
    \caption{All solutions for $\myRe t=-1/2$ to \eqref{eq:thue}, where $\omega_D=(1+\sqrt{-D})/2$.}
    \label{tab:small-solutions-re-t--1-2}
  \end{center}
\end{table}

The case $t=\exceptt$ has to be excluded because the form $F_t$ is the cube of a
linear polynomial in that case.

The remainder of this paper is organized as follows: In
Section~\ref{sec:elementary-properties}, we collect elementary properties and
asymptotic expressions for the algebraic numbers generating the splitting field
of $F_t$. Small solutions are dealt with in
Section~\ref{sec:small-solutions}. The Galois structure of the family will be
determined in Section~\ref{sec:galois}. It will turn out that the case
$\myRe t=-1/2$ has a special Galois group. The unit structure of the relevant
order will be discussed in Section~\ref{sec:unit-group}. Approximation
properties of solutions are the topic of
Section~\ref{sec:Approximation-Properties}. ``Stable growth'' will be proved
in Section~\ref{sec:stable-growth}, which will exclude all medium sized
solutions. The large solutions are excluded in
Section~\ref{sec:lower-bound-lambda} using a corollary of
Mignotte~\cite{Mignotte:1998:lauren-mignot-nester} of the lower bound
for linear forms in two logarithms by Laurent, Mignotte, and
Nesterenko~\cite{Laurent-Mignotte-Nesterenko:1995}.
Finally, we will prove Theorem~\ref{th:result:re-t--1-2} in 
Section~\ref{sec:proof-re-t--1-2}.

\begin{acknowledgement}
  We are grateful to Peter Kirschenhofer for raising this interesting problem,
  which he posed during an oral discussion.
\end{acknowledgement}




\section{Elementary Properties}\label{sec:elementary-properties}
Let $t\in\C$ and
\begin{align}
  F_t(X,Y)&:=X^3-(t-1)X^2Y-(t+2)XY^2-Y^3,\\
  f_t(X)&:=F_t(X,1).
\end{align}



It is straightforward to verify the identities
\begin{equation}\label{eq:elem:polynomial-identities}
  F_t\bigl(Y,-(X+Y)\bigr)=F_t\bigl(-(X+Y),X\bigr)=F_t(X,Y) \qquad\text{and}\qquad  f_t\left(-1-\frac{1}{X}\right)=\frac{f_t(X)}{X^3}.
\end{equation}

Let $\alpha\conj 1=\alpha=\alpha(t)$ be a root of $f_t(X)=0$. Then
\eqref{eq:elem:polynomial-identities} implies that
\begin{equation}
  \label{eq:elem:galois-action}
  \alpha\conj2:=-1-\frac{1}{\alpha} \qquad\text{and}\qquad
  \alpha\conj3:=-1-\frac{1}{\alpha\conj 2}=-\frac1{\alpha+1}
\end{equation}
are roots of $f_t$. 
The roots $\alpha$, $\alpha\conj 2$, and $\alpha\conj 3$ are pairwise distinct
for $t\neq\exceptt$. Indeed, assuming $\alpha=\alpha\conj 2$ implies
$\alpha^2+\alpha+1=f_t(\alpha)=0$, which is possible only if $t=\exceptt$.
If $t=\exceptt$, we have $f_t(X)=(X-(t-1)/2)^3$.


Assume that $\alpha\in\R$. Then we get $0=f_t(\alpha)-\cc{f_t}(\alpha)=
-(t-\cc t)(\alpha^2+\alpha)$, which implies $t\in\R$.

We will use the following variant of the usual $O$-notation:
For two functions $g(t)$ and $h(\betrag t)$ and a positive number
$t_0$ we will write
$g(t)=L_{t_0}\bigl(h(\betrag t)\bigr)$ if $\betrag{g(t)}\le h(\betrag{t})$ for
all $t$ with absolute value at least $t_0$. We will use this notation in the
middle of an expression in the same way as it is usually done with the $O$-notation.

\begin{lemma}\label{le:elem:root-estimate-alpha-1}
  Let $t\in\C$. Then there is a root $\alpha$ of $f_t$ such that
  \begin{equation}\label{eq:le:elem:root-estimate-alpha-1}
    \alpha=t+\frac{2}{t}-\frac{1}{t^2}-\frac{3}{t^3}+L_{6}\left(\frac{5}{{\betrag
    t}^{7/2}}\right).
  \end{equation}
\end{lemma}

\begin{proof}
  Let $h(z)=f(z+t+2t^{-1}-t^{-2}-3t^{-3})$ and $h_1(z):=h(z)-h(0)$. 
  For $\betrag{t}\ge 6$ and $\betrag{z}=5\betrag{t}^{-7/2}$, we obtain
  \begin{equation*}
    \betrag{h(z)-h_1(z)}=\betrag{h(0)}<\frac{7.38}{\betrag{t}^2}
    <\frac{5}{\betrag{t}^{3/2}}-\frac{10.26}{\betrag{t}^{5/2}}
    <\betrag{h_1(z)}.
  \end{equation*}
  By Rouch\'e's theorem, $h$ and $h_1$ have the same number of zeros
  in the disc $\{z: \betrag{z}<5\betrag{t}^{-7/2}\}$. Since $h_1(0)=0$, this
  implies the theorem.
\end{proof}

Similarly, we will use
\begin{equation}
  \label{eq:elem:root-estimate-alpha-1-order-8}
  \alpha= 
  t +\frac{2}{ t} -\frac{1}{ t^ {2}} -\frac{3}{ t^ {3}} +\frac{5}{ t^ {4}} +\frac{7}{ t^ {5}} -\frac{26}{ t^ {6}} -\frac{10}{ t^ {7}}+L_{20}\left(\frac{33}{ \betrag{t}^ {15/2}}\right).
\end{equation}


Denoting the root calculated in Lemma~\ref{le:elem:root-estimate-alpha-1} by 
$\alpha\conj 1=\alpha$ and using \eqref{eq:elem:galois-action}, 
we get the following estimates for the other roots:

\begin{subequations}
  \label{eq:elem:alpha-2-3-estimate}
  \begin{align}
  \alpha\conj 2&=-1-\frac1{\alpha\conj1}=
-1 -\frac{1}{ t}+L_{6}\left(\frac{102}{5 \betrag{t}^ {5/2}}\right)
,\\
  \alpha\conj 3&=-\frac1{\alpha\conj1+1}=
-\frac{1}{ t} +\frac{1}{ t^ {2}}+L_{6}\left(\frac{183}{10 \betrag{t}^ {5/2}}\right)
.
\end{align}
\end{subequations}

\section{Small Solutions}\label{sec:small-solutions}







\begin{lemma}\label{le:small-solutions}
  Let $\betrag{t}>4$. 
  All solutions $(x,y)\in\Z_{\Q(t)}$  to \eqref{eq:thue} with
  $\betrag{y}<\sqrt 5$, $\betrag{x+y}<\sqrt 5$, or $\betrag{x}<\sqrt 5$ are
  listed in Table~\ref{tab:small-solutions-all-t}.
\end{lemma}
\begin{proof}
  We will first consider solutions with $\betrag{y}<\sqrt{5}$.

  Let some imaginary quadratic root of unity $\mu$ and some imaginary quadratic integer
  $y$ with $\betrag{y}<\sqrt{5}$ be given such that $[\Q(\mu,y):\Q]\le 2$. It is clear
  that there is only a finite number of such pairs $(y,\mu)$. We are
  looking for quadratic integers $x$ and $t$ such that $[\Q(t,x,y,\mu):\Q]\le
  2$ and $F_t(x,y)=\mu$. The latter condition is equivalent to
  \begin{equation}\label{eq:small-solutions-thue-1}
    x(x^2-(t-1)xy-(t+2)y^2)=\mu+y^3.
  \end{equation}

  If $\mu+y^3\neq 0$, we conclude that $N(x)\divides N(\mu+y^3)$, 
  where $N(\,\cdot\,)=N_{\Q(t)/\Q}(\,\cdot\,)$ denotes the norm function.
  This yields
  a finite number of possible values for $x$. We only consider those $x$ such that
  $[\Q(\mu,x,y):\Q]\le 2$ and such that $(\mu+y^3)/x$ is an algebraic integer. 
  
  Then
  \eqref{eq:small-solutions-thue-1} implies
  $t(-xy-y^2)=(\mu+y^3)/x-x^2-xy+2y^2$. If
  $(-xy-y^2)=(\mu+y^3)/x-x^2-xy+2y^2=0$, we found a solution for all values of
  $t$ and list it in Table~\ref{tab:small-solutions-all-t}. Otherwise, if
  $xy+y^2\neq 0$, we calculate an explicit value for $t$.
  We check whether this $t$ is an algebraic integer. It turns out that this process
  only yields values of $t$ with $\betrag{t}\le 4$.
 
  If $\mu+y^3=0$, we may either have $x=0$ and arbitrary $t$ (in the
  appropriate field) and the solution is listed in
  Table~\ref{tab:small-solutions-all-t}, or \eqref{eq:small-solutions-thue-1}
  results in
  \begin{equation*}
    (x+y)(x-yt)=2y^2.
  \end{equation*}
  This implies $N(x+y)\divides N(2y^2)=4$. Therefore, we only have a finite
  number of choices for $x+y$. For each choice, we get a value of $t$. It turns
  out that all those $t$ have absolute value at most $4$.

  Let now $(x,y)$ be a solution of \eqref{eq:thue} with
  $\betrag{x}<\sqrt{5}$. By~\eqref{eq:elem:polynomial-identities},
  $\bigl(-(x+y),x\bigr)$ is a solution contained in
  Table~\ref{tab:small-solutions-all-t}. The case $\betrag{x+y}<\sqrt 5$ is
  dealt with similarly. It can be checked that for each solution $(x,y)$
  listed in Table~\ref{tab:small-solutions-all-t}, $\bigl(y, -(x+y)\bigr)$ and
  $\bigl(-(x+y),x)$ are also contained in the table.
\end{proof}


\section{Computation of the Galois Group}\label{sec:galois}

For the remainder of this paper, we fix a positive squarefree integer $D$ and
assume that $t\in\Z_k\setminus \Z$, where $\Z_k$ is the
ring of integers in the algebraic number field $k:=\Q(\sqrt{-D})$.

\begin{lemma}\label{le:galois:irreducible}
  For $t\neq (-1\pm 3\sqrt{-3})/2$, $f_t$ is irreducible over $k$.
\end{lemma}
\begin{proof}
  Assume first $\betrag{t}\ge 6$.
  From \eqref{eq:elem:alpha-2-3-estimate} and $\alpha\conj3\notin\R$ we see that $0<\betrag[normal]{\alpha\conj 3}\le
  1/2$ which implies that $\alpha\conj3\notin\Z_k$. However, $\alpha\conj3$ is
  an algebraic integer. This yields $\alpha\conj3\notin k$. From
  \eqref{eq:elem:galois-action} we conclude that $\alpha\conj
  j\notin k$ for $j=1,\dots,3$.

  For $\betrag{t}<6$, the assertion has been checked using Pari~\cite{Pari:2000}.
\end{proof}

Let $K:=\Q(\alpha)$. From \eqref{eq:elem:galois-action} and $t-1=\alpha\conj
1+\alpha\conj 2+\alpha\conj 3$ we see that
$K=\Q(\alpha\conj1,\alpha\conj2,\alpha\conj3,\sqrt{-D})$. 
For $t\neq \exceptt$,  we have $[K:\Q]=6$ and
the minimal polynomial of $\alpha$ over $\Q$
is $f_t\cc{f_t}$.

\begin{theorem}\label{th:galois:galois}
  Let $t\in\Z_k\setminus\Z$, 
  $L:=\Q(\alpha,\cc{\alpha})$ 
  and let $G:=\Gal(L/\Q)$. Let $\tau$ denote
  the complex conjugation.
  \begin{enumerate}
  \item If $t\in T_1:=\{\exceptt\}$, then $L=\Q(\sqrt{-3})$ and $G=\group{\tau}\simeq C_2$.
  \item If $t\in T_2:=\{\pm\sqrt{-7}, -1\pm\sqrt{-7}\}$, then  $L=K=\Q(\alpha)$ and
    $G=\group{\sigma}\simeq C_6$ with $\sigma=(\alpha\conj1 \cc{\alpha\conj 3} \alpha\conj2
  \cc{\alpha\conj1} \alpha\conj3 \cc{\alpha\conj 2})$.
%  $\Q(\sqrt{-D})$ and $\Q(\alpha+\cc{\alpha})\subset \R$ are the only fields between $\Q$ and $K$.
  \item If $(\myRe t =-1/2$ and $t\notin T_1)$ or if 
    \begin{multline*}
      t\in T_3:=\{
    (- 5\pm \sqrt{-11})/2,
    (3 \pm \sqrt{-11})/2,
    (- 13\pm\sqrt{-23})/2,
    (- 3\pm \sqrt{-23})/2,\\
    (1\pm \sqrt{-23})/2,
    (11\pm \sqrt{-23})/2,
    (- 7 \pm \sqrt{-31})/2,
    (5 \pm \sqrt{-31})/2\},
    \end{multline*}
    then $L=K=\Q(\alpha)$ and
    $G=\group{\sigma,\tau} \simeq S_3$ with
    $\sigma=(\alpha\conj1\alpha\conj2\alpha\conj3)(\cc{\alpha\conj1}\cc{\alpha\conj3}\cc{\alpha\conj2})$.
    There are four fields between $\Q$ and $K$, namely the quadratic field
    $k=\Q(\sqrt{-D})$ and three cubic fields. One of the cubic fields is real.
% the cubic fields $\Q(\alpha\conj2+\cc{\alpha\conj2})\Q(\alpha(\alpha+1)) \subset \R$,
%     $\Q(\alpha+1/\alpha)$, and $\Q(\alpha+1/(\alpha+1))$.
  \item If $\myRe t\neq -1/2$ and $t\notin T_2\union T_3$, then $[L:K]=3$ and
    $G=\group{\sigma,\sigma',\tau}$ with $\betrag{G}=18$, $\sigma=(\alpha\conj1
    \alpha\conj2\alpha\conj3)$ and
    $\sigma'=(\cc{\alpha\conj1}\cc{\alpha\conj2}\cc{\alpha\conj3})$.
    Furthermore, the only field between $\Q$ and $K$ is $k$.
  \end{enumerate}
\end{theorem}
\begin{proof}
  We assume $\betrag{t}\ge 21$, leaving the remaining cases to Pari~\cite{Pari:2000}.
  We note that the complex conjugation $\tau$ is an element of $G$. 
  
  Assume first that $G$ is cyclic, generated by some $\sigma$. Without loss of
  generality, we may assume that $\sigma^2$ is the automorphism which induces
  \eqref{eq:elem:galois-action}.  It is clear that $\tau=\sigma^3$, which
  implies $\sigma=(\alpha\conj1 \cc{\alpha\conj 3} \alpha\conj2
  \cc{\alpha\conj1} \alpha\conj3 \cc{\alpha\conj 2})$.  Let
  \begin{equation*}
    u_1:=\left(4-\prod_{j=1}^3 (\alpha\conj j+\cc{\alpha\conj j})\right)^2
    -2\prod_{j=1}^3 \bigl((\alpha\conj j)^2+(\cc{\alpha\conj j})^2\bigr).
  \end{equation*}
  Since $u_1$ is invariant under $\sigma$ and $u_1\in\Z_L$, we have $u_1\in\Z$.
  Using \eqref{eq:elem:root-estimate-alpha-1-order-8}, we get
  $\betrag{u_1}<1$ for $\betrag{t}\ge 21$. This implies $u_1=0$.
  Eliminating $\alpha$ and $\cc{\alpha}$ from the set of equations $u_1=0$,
  $f(\alpha)=0$ and $\cc{f(\alpha)}=0$, we get $D=0$ or $\myIm{t}=0$, which has
  been excluded. 

  Next, we assume that $L=K$ but $G$ is not cyclic. Let $\sigma\in G$ be the
  automorphism which maps $\alpha\conj1$ to $\alpha\conj2$. Since
  $\sigma(\cc{\alpha\conj1})\neq \cc{\alpha\conj1}$ because $\betrag{G}=6$,
  the only possibility for $\sigma$ is
  $\sigma=(\alpha\conj1\alpha\conj2\alpha\conj3)(\cc{\alpha\conj1}\cc{\alpha\conj3}\cc{\alpha\conj2})$. 
  We note that $\tau\sigma=\sigma^2\tau$ and get $G=\group{\sigma,\tau}$. We
  consider $\theta_j:=\sigma^j(\alpha+\cc{\alpha})$, $j=1,2,3$, and
  \begin{equation*}
    u_2:=1-\theta_1\theta_2\theta_3-\theta_1\theta_2-\theta_2\theta_3-\theta_3\theta_1-t-\cc{t}.
  \end{equation*}
  As above, we obtain $u_2=0$ for $\betrag{t}\ge 20$. This yields
  $\myRe t=-1/2$.

  Conversely, if $\myRe t=-1/2$, we get $\cc{f_t}(-1-\alpha)=0$ and we obtain
  the claimed Galois group. 

  Finally, $\cc{\alpha}\notin K$ yields $[L:K]=18$ and the asserted Galois
  group.
  
  The assertions on the fields between $\Q$ and $K$ follow from Galois Theory.
\end{proof}

For the case  $\myRe t=-1/2$, we will need the following relation.

\begin{lemma}\label{le:galois:re-t--1-2-relation}
  For $\myRe t=-1/2$, $\betrag{t}\ge 6$, and $\alpha=\alpha\conj 1$ as
  calculated in Lemma~\ref{le:elem:root-estimate-alpha-1}, we have
  \begin{equation*}
    \cc{\alpha\conj 1}=-1-\alpha\conj 1.
  \end{equation*}
\end{lemma}
\begin{proof}
  Using~\eqref{eq:le:elem:root-estimate-alpha-1}, we get
  \begin{equation*}
    \betrag{\cc{\alpha\conj 1}-(-1-\alpha\conj 1)}=\betrag{\frac{3 + 10t +
    10t^2}{( t + t^2 )^3}+L_6\left(\frac{10}{\betrag{t}^{7/2}}\right)}\le 0.86.
  \end{equation*}
  In the proof of Theorem~\ref{th:galois:galois}, we noticed that
  $-1-\alpha\conj 1$ is a root of $\cc{f_t}$. From
  \eqref{eq:elem:alpha-2-3-estimate} we see that $\betrag[normal]{\cc{\alpha\conj
  i}-\cc{\alpha\conj 1}} >1$ for $i=2$, $3$. This yields the assertion.
\end{proof}

\section{Unit group}\label{sec:unit-group}
Let $(x,y)\in\Z_k$ be a solution to \eqref{eq:thue}. From 
\begin{equation}\label{eq:thue-norm}
  \prod_{l=1}^3(x-\alpha\conj l y)=F_t(x,y)=\mu
\end{equation}
we conclude that $\beta\conj l:=x-\alpha\conj l y$, $l=1$, $2$, $3$,
are units in $\order:=\Z_k[\alpha]$. 

Therefore, we investigate the unit structure of $\order$. It can easily be
checked that 
\begin{equation*}
  1=\alpha(\alpha^2-(t-1)\alpha-(t+2))=(\alpha+1)(2+t\alpha-\alpha^2),
\end{equation*}
which implies that $\alpha$ and $(\alpha+1)$ are units in $\order$.



\begin{lemma}\label{le:unit:unit-group}
  Let $\betrag{t}\ge 20$ and
  \begin{equation*}
    \zeta=
    \begin{cases}
      (1+\sqrt{-3})/2&\text{if $D=3$},\\
      i&\text{if $D=1$},\\
      -1&\text{otherwise.}
    \end{cases}
  \end{equation*}
  Then the index $I$ of 
  $\group{\zeta, \alpha,\alpha+1}$ in the unit group $\order^\times$ can be
  bounded by 
  \begin{equation}\label{eq:unit:index-bound}
    I\le 5.03 \log^2\betrag{t}.
  \end{equation}
\end{lemma}
\begin{proof}
  Assume that $\rho\in\order$ is a root of unity. Since $\Q(\rho)$ is Galois
  and since by Theorem~\ref{th:galois:galois},
  the only Galois subfield between $\Q$ and $K$ is $k$,
  we see that $\rho\in k$. It is clear that the torsion group of $\Z_k^\times$
  (and therefore of $\order^\times$) is generated by $\zeta$.

  Using \eqref{eq:elem:root-estimate-alpha-1-order-8}, we calculate
  \begin{equation}\label{eq:unit:regulator-estimate}
    0.97\log^2\betrag t\le 
    -\det
    \begin{pmatrix}
      \log\betrag[normal]{\alpha\conj 1}&\log\betrag[normal]{\alpha\conj 1+1}\\
      \log\betrag[normal]{\alpha\conj 2}&\log\betrag[normal]{\alpha\conj 2+1}
    \end{pmatrix}\le 1.03\log^2\betrag t.
  \end{equation}
  This implies that $\alpha$ and $\alpha+1$ are independent units of
  $\order$.

  By Friedman~\cite[Theorem~B]{Friedman:1989}, the regulator of $K$
  can be bounded by $\Reg \Z_K\ge 0.2052$. From Pohst and
  Zassenhaus~\cite[p.~361]{Pohst-Zassenhaus:1989}, we conclude that 
  \begin{equation*}
    I=[\order^\times : \group{\zeta,\alpha,\alpha+1}]\le 
    [\Z_K^\times : \group{\zeta,\alpha,\alpha+1}]=
    \frac{\Reg \group{\zeta,\alpha,\alpha+1}}{\Reg \Z_K}\le
    \frac{1.03 \log^2\betrag t}{0.2052}\le 5.03\log^2\betrag t.
  \end{equation*}
\end{proof}

\section{Approximation Properties}\label{sec:Approximation-Properties}
We call a  solution  $(x,y)$ to \eqref{eq:thue} a solution of
\emph{type $j$}, if 
\begin{equation*}
  \betrag[normal]{\beta\conj j}=\min_{l=1,2,3}\betrag[normal]{\beta\conj l}.
\end{equation*}

>From 
\begin{equation*}
  \betrag[normal]{y}\betrag[normal]{\alpha\conj l-\alpha\conj j}\le 
  \betrag[normal]{x-\alpha\conj  l y}+\betrag[normal]{x-\alpha\conj j y}\le 
  2\betrag[normal]{\beta\conj l},
\end{equation*}
and \eqref{eq:thue-norm} we conclude that
\begin{equation}\label{eq:approx:beta-j-estimate}
  \betrag[normal]{\beta\conj j}=
  \frac1{\prod_{l\neq j}  \betrag[normal]{\beta\conj l}}\le 
  \frac{4}{\prod_{l\neq j}\betrag{\alpha\conj
  l-\alpha\conj j}}\cdot\frac{1}{\betrag{y}^2}=\frac{4}{\betrag[normal]{f_t'(\alpha\conj j)}}\cdot\frac1{\betrag{y}^2}.
\end{equation}

\begin{lemma}\label{le:approx:solution-galois-type}
  Let $\betrag{t}\ge 20$ and $(x,y)$ be a solution to \eqref{eq:thue} of type
  $j$. 
  Then $\bigl(-(x+y),x\bigr)$ is a solution to
  \eqref{eq:thue} of type $j':=(j+1)\bmod 3$.
\end{lemma}
\begin{proof}
  We may assume $\betrag{y}\ge\sqrt{5}$ by Lemma \ref{le:small-solutions}.
  By \eqref{eq:elem:polynomial-identities}, $\bigl(-(x+y), x\bigr)$ is also a
  solution to \eqref{eq:thue}. From \eqref{eq:elem:galois-action} and
  \eqref{eq:approx:beta-j-estimate} we see that
  \begin{equation*}
    \betrag{\frac{-(x+y)}x-\alpha\conj{j'}}=
    \betrag{\frac{\alpha\conj j-x/y}{\alpha\conj j \cdot x/y}}\le
    \frac{4}{\betrag{y}^3\cdot\betrag[normal]{f_t'(\alpha\conj j)}}\cdot
    \frac{1}{\betrag[normal]{\alpha\conj j}}\cdot
    \frac{1}{\betrag[normal]{\alpha\conj j}-4/(\betrag{y}^3\cdot\betrag[normal]{f_t'(\alpha\conj j)}) }.
  \end{equation*}

  From \eqref{eq:elem:root-estimate-alpha-1-order-8}, the assumption
  $\betrag{y}\ge \sqrt{5}$, and \eqref{eq:approx:beta-j-estimate} we conclude that
  \begin{equation*}
    \min_l\betrag{\frac{-(x+y)}x-\alpha\conj
    l}=\betrag{\frac{-(x+y)}x-\alpha\conj {j'}}.
  \end{equation*}
\end{proof}

Lemma~\ref{le:approx:solution-galois-type} shows that if there is a solution
to \eqref{eq:thue} which is not listed in
Table~\ref{tab:small-solutions-all-t}, then there is also a solution $(x,y)$ of type
$1$ which is not contained in Table~\ref{tab:small-solutions-all-t}.


\section{Stable Growth}\label{sec:stable-growth}
Let $(x,y)$ be a solution to \eqref{eq:thue} of type $1$ 
which is not listed in Table~\ref{tab:small-solutions-all-t} and $\betrag{t}\ge
20$. By Lemma~\ref{le:small-solutions}, we may assume
$\betrag{y}\ge
\sqrt 5$. 
For $l=2$, $3$, we have
\begin{equation}\label{eq:stable:beta-conj-l-exact}
  \betrag[normal]{\beta\conj l}=\betrag{y}\cdot\betrag[normal]{\alpha\conj 1-\alpha\conj l}
  \cdot\betrag{1+\frac{x/y-\alpha\conj 1}{\alpha\conj 1-\alpha\conj l}}.
\end{equation}
Taking logarithms and using \eqref{eq:approx:beta-j-estimate} and
\eqref{eq:elem:root-estimate-alpha-1-order-8}, 
we get 
\begin{equation}\label{eq:stable:log-beta-l-asymptotic}
  \log\betrag[normal]{\beta\conj l}=\log\betrag{y}+\log\betrag[normal]{\alpha\conj
  1-\alpha\conj l}+L_{20}\left(\frac{c_l}{\betrag{t}^3\betrag{y}^3}\right)
\end{equation}
for $c_2=4.6$ and $c_3=4.37$.
Taking this for $l=3$ and $l=2$ and eliminating 
$\log\betrag{y}$, we get
\begin{equation}\label{eq:stable:linear-form-proto-1}
  \betrag{\Lambda}:=\betrag{\log\betrag{\frac{\beta\conj 2}{\beta\conj 3}}-\log\betrag{\frac{\alpha\conj
  1-\alpha\conj 2}{\alpha\conj1-\alpha\conj 3}}}\le \frac{8.97}{\betrag{t}^3\betrag{y}^3}.
\end{equation}

From \eqref{eq:stable:beta-conj-l-exact} we see that $\Lambda=0$ would imply
\begin{equation*}
  \frac{x/y-\alpha\conj 1}{\alpha\conj 1- \alpha\conj 2}=\frac{x/y-\alpha\conj
  1}{\alpha\conj1-\alpha\conj3}.
\end{equation*}
This would lead to $x/y=\alpha\conj 1$ or $\alpha\conj3=\alpha\conj2$, which are
both contradictions. Therefore we have $\Lambda\neq 0$.

Lemma~\ref{le:unit:unit-group} implies that $\bigl(\beta\conj
l\bigr)^I=\zeta^{u_0}\bigl(\alpha\conj l\bigr)^{u_1}\bigl(\alpha\conj l+1\bigr)^{u_2}$ for
some integers $u_0$, $u_1$, and $u_2$. Taking logarithms yields
\begin{equation}\label{eq:stable:log-beta-l-basis-representation}
  \log\betrag[normal]{\beta\conj
    l}=\frac{u_1}{I}\log\betrag[normal]{\alpha\conj l}
  +\frac{u_2}{I}\log\betrag[normal]{\alpha\conj l +1}, \qquad l=1,2,3.
\end{equation}

From \eqref{eq:stable:linear-form-proto-1}, we obtain
\begin{equation*}
  \betrag{\Lambda}=\betrag{\frac{u_1}{I}\log\betrag{\frac{\alpha\conj2}{\alpha\conj3}}+
    \frac{u_2}{I}\log\betrag{\frac{\alpha\conj2+1}{\alpha\conj3+1}}-
    \log\betrag{\frac{\alpha\conj
  1-\alpha\conj 2}{\alpha\conj1-\alpha\conj 3}}}\le \frac{8.97}{\betrag{t}^3\betrag{y}^3}.
\end{equation*}

Taking advantage of \eqref{eq:elem:galois-action}, this linear form can be
rewritten as a linear form in two logarithms:
\begin{equation}\label{eq:stable:linear-form-upper}
  \betrag{I\cdot \Lambda}=\betrag{v_1\log\betrag{\alpha}+v_2\log\betrag{1+\frac1\alpha}}
  \le \frac{8.97\cdot I}{\betrag{t}^3\betrag{y}^3},
\end{equation}
where
\begin{equation*}
  v_1=u_1-u_2\qquad\text{and}\qquad v_2=2u_1+u_2-I
\end{equation*}
are integers. 

If $\myRe t=-1/2$, we have 
\begin{equation*}
 \betrag{1+\frac{1}{\alpha}}^2=1+\frac1{\alpha}+\frac1{\cc{\alpha}}
 +\frac{1}{\alpha\cc{\alpha}}
 = 1+\frac{\alpha+\cc{\alpha}+1}{\alpha\cc{\alpha}}=1
\end{equation*}
by Lemma~\ref{le:galois:re-t--1-2-relation}. This implies
$\log\betrag{1+\alpha^{-1}}=0$. Therefore, \eqref{eq:stable:linear-form-upper}, \eqref{eq:unit:index-bound},
and \eqref{eq:elem:root-estimate-alpha-1-order-8} yield
$\betrag{v_1}<1$ for $\betrag{t}\ge 20$. This is a contradiction to $\Lambda\neq
0$. Hence, there are no solutions which are not listed in
Table~\ref{tab:small-solutions-all-t} for $\myRe t=-1/2$ and $\betrag t\ge
20$. This proves Theorem~\ref{th:result:re-t--1-2} for $\betrag{t}\ge 20$.

It is now crucial to observe that $\log\betrag{\alpha}\sim \log\betrag t$,
whereas $\log\betrag{1+\alpha^{-1}}=O(\betrag{t}^{-1})$. This means that $v_2$
has to be much bigger than $v_1$.

Assume that $v_1=0$. Using~\eqref{eq:elem:root-estimate-alpha-1-order-8}, we
get 
\begin{equation*}
  \log\betrag{1+\alpha^{-1}}=\frac12\left(\frac1t+\frac1{\cc{t}}\right)-
  \frac14\left(\frac1{t^2}+\frac1{\cc
  t^2}\right)+L_{20}\left(\frac{93}{1000\betrag{t}^2}\right).
\end{equation*}
Simple estimates show that this implies
\begin{equation}\label{eq:stable:log-abs-alpha-2-lower}
  \betrag{\log\betrag{1+\alpha^{-1}}}\ge \frac{0.4}{\betrag{t}^2}
\end{equation}
for $\myRe t\neq -1/2$. Combining this with \eqref{eq:stable:linear-form-upper},
\eqref{eq:unit:index-bound}, and the assumption $v_1=0$ yields
\begin{equation}\label{eq:stable:v2-min}
  \betrag{v_2}\le \frac{113\log^2\betrag{t}}{\betrag{y}^3\betrag{t}}.
\end{equation}
For $\betrag{t}\ge 351$, this bound is smaller than $1$. This yields a
contradiction to $\Lambda\neq 0$.

Therefore, we get $v_1\neq 0$. This implies $\betrag{v_1}\ge 1$ and 
from \eqref{eq:stable:linear-form-upper},
\eqref{eq:stable:log-abs-alpha-2-lower}, and
\eqref{eq:elem:root-estimate-alpha-1-order-8}, we obtain
\begin{equation*}
  \betrag{v_2}\ge
  \frac1{\log\betrag{1+\alpha^{-1}}}
  \left(\betrag{v_1\log\betrag[normal]{\alpha}}-\betrag{\Lambda}\right)\ge
  \frac{\log\betrag{\alpha}}{\log\betrag{1+\alpha^{-1}}}-\frac{8.97\cdot
    I}{0.4\betrag{y}^3 \betrag{t}}
  \ge 0.89 \betrag{t}\log\betrag{t}.
\end{equation*}

We solve the system \eqref{eq:stable:log-beta-l-basis-representation} of linear equations in $u_1/I$ and $u_2/I$
by Cramer's rule and using
\eqref{eq:stable:log-beta-l-asymptotic}. We obtain
\begin{equation*}
  -\frac{v_2}{I}=\left(\frac{3}{\log\betrag{t}}+L_{20}\left(\frac{2.93}{\betrag{t}\log\betrag{t}}\right)\right)
  \log\betrag{y}+L_{20}(4.3).
\end{equation*}

Combining this with~\eqref{eq:stable:v2-min} and \eqref{eq:unit:index-bound}
yields 
\begin{equation}\label{eq:stable:log-y-lower}
  \log{\betrag{y}}\ge \log y_0:= 
  5.6\cdot 10^{-2}\cdot\betrag{t}-1.37\cdot\log\betrag{t}.
\end{equation}

We have proved the property which is called ``stable growth'' by
E.~Thomas~\cite{Thomas:1993}:
\begin{proposition}
  Let $\betrag{t}\ge 351$ and $x$, $y\in\Z_{\Q(t)}$ be a solution to
  \eqref{eq:thue} which is not listed in Theorem~\ref{th:result}. 
  Then
  \begin{equation*}
    \max\{ \betrag{x}, \betrag{y}, \betrag{x+y}\} \ge \frac{1.057^{\betrag{t}}}{\betrag{t}^{1.37}}.
  \end{equation*}
\end{proposition}

\section{Lower Bound for $\betrag{\Lambda}$}\label{sec:lower-bound-lambda}



We use the following refinement, due to Mignotte~\cite{Mignotte:1998:lauren-mignot-nester}, of a theorem of Laurent, Mignotte, and Nesterenko~\cite{Laurent-Mignotte-Nesterenko:1995} on linear forms in two logarithms. For any non-zero algebraic number $\gamma$ of degree $d$ over $\Q$, whose minimal polynomial over $\Z$ is $c\prod_{j=1}^d(X-\gamma\conj j)$, we denote by 
\begin{equation*}
  h(\gamma)=\frac{1}{d}\left(\log\betrag{c}+\sum_{j=1}^d \log\max(1,\betrag[normal]{\gamma\conj j})\right)
\end{equation*}
its absolute  logarithmic height.


\makeatletter
\protected@edef\MignotteTheoremTwo{Mignotte~\cite[Theorem~2]{Mignotte:1998:lauren-mignot-nester}}
\makeatother

\begin{lemma}[\MignotteTheoremTwo]\label{le:linear:mignotte}
  Consider the linear form
  \begin{equation*}
    \Lambda=b_2\log\alpha_2-b_1\log\alpha_1,
  \end{equation*}
  where $b_1$ and $b_2$ are positive integers, $\alpha_1$ and $\alpha_2$ are
  non-zero algebraic numbers of absolute value at least $1$, and $\log\alpha_1$
  and $\log\alpha_2$ are any values of their logarithms.

  Suppose that $\alpha_1$ and $\alpha_2$ are multiplicatively independent. Put
  \begin{equation*}
    D=[\Q(\alpha_1,\alpha_2):\Q]/[\R(\alpha_1,\alpha_2):\R].
  \end{equation*}
  Let $a_1$, $a_2$, $h$, $k$ be real positive numbers, and $\rho$ a real number $>1$. Put $\lambda=\log \rho$, $\chi=h/\lambda$ and suppose that $\chi\ge\chi_0$ for some number $\chi_0\ge 0$ and that
  \begin{align*}
    h&\ge D\left(\log\left(\frac{b_1}{a_2}+\frac{b_2}{a_1}\right)+\log\lambda+f(\ceiling{K_0})\right)+0.023,\\
    a_i&\ge \max\{1,\rho\betrag{\log \alpha_i}-\log\betrag{\alpha_i}+2D h(\alpha_i)\}, \quad (i=1,2)\\
    a_1\cdot a_2&\ge \lambda^2,
  \end{align*}
  where 
  \begin{equation*}
    f(x)=\log\frac{(1+\sqrt{x-1})\sqrt{x}}{x-1}+\frac{\log x}{6x(x-1)}+\frac32 +\log\frac34+\frac{\log\frac{x}{x-1}}{x-1}
  \end{equation*}
  and
  \begin{equation*}
    K_0=\frac{1}{\lambda}\left(\frac{\sqrt{2+2\chi_0}}{3}+\sqrt{\frac{2(1+\chi_0)}{9}+\frac{2\lambda}{3}\left(\frac{1}{a_1}+\frac{1}{a_2}\right)+\frac{4\lambda\sqrt{2+\chi_0}}{3\sqrt{a_1a_2}}}\right)^2a_1a_2.
  \end{equation*}
  Put
  \begin{equation*}
    v=4\chi+4+1/\chi, \qquad m=\max\{2^{5/2}(1+\chi)^{3/2}, (1+2\chi)^{5/2}/\chi\}.
  \end{equation*}
  Then we have the lower bound
  \begin{align*}
    \log\betrag{\Lambda}\ge &-\frac1{\lambda}\left(\frac v6+\frac12
    \sqrt{\frac{v^2}{9}+\frac{4\lambda v}3\left(\frac{1}{a_1}+\frac{1}{a_2}\right)+\frac{8\lambda m}{3\sqrt{a_1a_2}}}\right)^2a_1a_2\\
  &-\max\bigl\{\lambda(1.5+2\chi)
  +\log(((2+2\chi)^{3/2}+(2+2\chi)^2\sqrt{k^*})A+(2+2\chi)), D\log 2\bigr\},
  \end{align*}
  where 
  \begin{equation*}
    A=\max\{a_1,a_2\} \text{ and }
    k^*=\frac{1}{\lambda^2}\left(\frac{1+2\chi}{3\chi}\right)^2+\frac{1}{\lambda}\left(\frac{2}{3\chi}+\frac{2}{3}\cdot \frac{(1+2\chi)^{1/2}}{\chi}\right).
  \end{equation*}
\end{lemma}

Let $\sigma:=\sign(\betrag[normal]{\alpha\conj 2}-1)$.
We will apply Lemma~\ref{le:linear:mignotte} to the linear form in logarithms
$\sigma2 I \Lambda$, which can be written as
\begin{equation*}
  \sigma 2I\Lambda=\sigma v_1\log(\alpha\conj1\cc{\alpha\conj
  1})-(-v_2)\log(\alpha\conj2\cc{\alpha\conj2})^{\sigma}
\end{equation*}
by \eqref{eq:stable:linear-form-upper} and \eqref{eq:elem:galois-action}.
We note that $\log(\alpha\conj1\cc{\alpha\conj
  1})$ and
$\log(\alpha\conj2\cc{\alpha\conj2})$ are linearly independent over $\Q$ 
by \eqref{eq:unit:regulator-estimate}.

>From now on, we will assume $\betrag{t}\ge 10^9$ in order to obtain sharper
bounds. 


We have
\begin{align*}
  D&=9\\
  h(\alpha\conj1\cc{\alpha\conj1})&=h(\alpha\conj2\cc{\alpha\conj2})\\
  &\le
  \frac23(\betrag[big]{\log\betrag[normal]{\alpha\conj1}}+\betrag[big]{\log\betrag[normal]{\alpha\conj2}})\le
  \frac23\log\betrag{t}+\frac1{3t}+\frac{1}{3\cc{t}}+ 
  L_{10^9}\left(4.87\cdot10^{-2}\cdot\frac{\log\betrag{t}}{\betrag{t}^2}\right).
\end{align*}
We choose $\rho=30.0658$ and obtain
\begin{align*}
  a_1&=12\log\betrag{t}+6\left(\frac1t+\frac1{\cc t}\right)+\frac{62.76}{\betrag{t}}
  +0.877\cdot\frac{\log\betrag{t}}{\betrag{t}^2},\\
  a_2&=74.14\cdot\log\betrag{t}+6\left(\frac{1}t+\frac{1}{\cc t}\right)+
  7.31\cdot\frac{\log\betrag{t}}{\betrag{t}^2},\\
  0<\sigma v_1&\le \frac{14.7}{\betrag{t}}\log\betrag y+19.5\cdot\frac{\log\betrag{t}}{\betrag{t}},\\
  -v_2&\le \left(\frac{2500}{171}\log\betrag{t}+1.46\cdot \frac{\log\betrag{t}}{\betrag{t}}\right)\log\betrag{y}+
  \left(19.50\log^2\betrag{t}+2.42\cdot\frac{\log^2\betrag{t}}{\betrag{t}}\right),\\
  h&=9\log\log\betrag{y}+\left(7.519+177.2\cdot\frac{\log\betrag{t}}{\betrag{t}}+\frac{3.72}{\betrag{t}}\right),\\
  K_0&\ge 4\,862\,274.
\end{align*}
Since the only available asymptotic information on $\betrag{y}$ is 
\eqref{eq:stable:log-y-lower}, we use this bound everywhere but in the asymptotic main term.
After some calculations, we finally get
\newcommand{\linearformlowerfactor}{(3073\log^2\betrag{t}+7505\log\betrag{t}+941.1\sqrt{\log\betrag{t}}+44510)}
\begin{equation}\label{eq:linear:lambda-lower}
  \log\betrag{2 I \Lambda}\ge - \log^2\log\betrag{y}\cdot\linearformlowerfactor.
\end{equation}

We consider the function
\begin{multline*}
  h(y,t):=3\log\betrag{y}
  -\log^2\log\betrag{y}\cdot\linearformlowerfactor\\
  +3\log\betrag{t}-2\log\log{\betrag{t}}-4.384.
\end{multline*}
By~\eqref{eq:linear:lambda-lower}, \eqref{eq:stable:linear-form-upper}, and \eqref{eq:unit:index-bound}, we have
\begin{equation}\label{eq:linear:h-y-t-le-0}
  h(y,t)\le 0
\end{equation}
for $\betrag{t}\ge 10^9$.

Since
\begin{equation*}
  \frac{\partial}{\partial \log\betrag{y}} h(y,t)=3-2\linearformlowerfactor\frac{\log\log\betrag{y}}{\log\betrag{y}}>0
\end{equation*}
for $\betrag{y}\ge y_0$ ($y_0$ has been defined in
\eqref{eq:stable:log-y-lower}) 
and $\betrag{t}\ge 2.15\cdot 10^8$, we have $h(y,t)\ge h(y_0,t)$.
But $h(y_0,t)> 0$ for $\betrag{t}\ge \tmin$, which is a contradiction to \eqref{eq:linear:h-y-t-le-0}.
This proves Theorem~\ref{th:result}.

\section{Proof of Theorem \ref{th:result:re-t--1-2}}\label{sec:proof-re-t--1-2}
Theorem~\ref{th:result:re-t--1-2} has already been proved for
$\betrag{t}\ge 20$ in Section~\ref{sec:stable-growth}.
For the remaining small values of $\betrag{t}$, the
asymptotic expansions do not help.  Especially, we have to redefine the order
of the roots of $f_t$, since $\alpha\conj 1$ was defined via an asymptotic
expansion.


We easily calculate
\begin{equation*}
  \hat f_{t_2}(x_2):=f_{-1/2+\complexI t_2}(-1/2+\complexI x_2)
  = -x_2^3+t_2 x_2^2-\frac94 x_2 +\frac{t_2}{4}.
\end{equation*}
Since $\discr \hat f_{t_2}<0$ for $t_2\neq \pm 3\sqrt 3/2$, we conclude
that there is exactly one root of $f_{-1/2+\complexI t_2}$ with real part equal
to $-1/2$. This root will be denoted by $\alpha:=\alpha\conj 1$. Since
$\alpha\conj 1+\cc{\alpha\conj 1}=-1$, we have
\begin{equation*}
  \betrag[normal]{\alpha\conj 1+1}=\betrag[normal]{\alpha\conj 1}.
\end{equation*}
The relations \eqref{eq:elem:galois-action} give us the other roots of $f_t$.

The type of a solution is defined as in
Section~\ref{sec:Approximation-Properties}. However, we cannot use
Lemma~\ref{le:approx:solution-galois-type}. 

For each $1\le \betrag{t}\le 20$ with $\myRe t=-1/2$ and for each pair $(j,l)$
with $1\le l\neq j\le 3$, we calculate $\alpha$ explicitly and set
\begin{equation*}
  \hat c_{tjl}:=\betrag{\frac{4}{f'_t(\alpha\conj j)}\cdot\frac{1}{\alpha\conj j-\alpha\conj
  l}\cdot\frac{1}{11^{3/2}}}, \qquad c_{tjl}:=\max\{\log(1+\hat
  c_{tjl}),-\log(1-\hat c_{tjl})\}.
\end{equation*}
By \eqref{eq:approx:beta-j-estimate} and \eqref{eq:stable:beta-conj-l-exact},
we get 
\begin{equation*}
  \log\betrag[normal]{\beta\conj l}=\log\betrag{y}+\log\betrag[normal]{\alpha\conj
  j-\alpha\conj l}+\varepsilon_{tjl}
\end{equation*}
for some $\betrag[normal]{\varepsilon_{tjl}}\le c_{tjl}$ and
for all solutions $(x,y)$ to \eqref{eq:thue} of type $j$ with $\betrag{y}\ge
\sqrt{11}$.

As in Section~\ref{sec:stable-growth}, we obtain
\begin{equation*}
  \betrag{I\cdot \Lambda}=\betrag{v_1\log\betrag[normal]{\alpha\conj
  j}+v_2\log\betrag{1+\frac1{\alpha\conj j}}} \le 
c_{tj}:=\frac{\log^2\betrag{\alpha}}{0.2052}\cdot \sum_{\substack{l=1\\l\neq j}}^3 c_{tjl}.
\end{equation*}
We note that $\log\betrag[normal]{1+1/\alpha\conj1}=\log\betrag[normal]{\alpha\conj 2}=0$ and
$\log\betrag[normal]{\alpha\conj 1}=-\log\betrag[normal]{1+1/\alpha\conj
  2}=-\log\betrag[normal]{\alpha\conj 3}=\log\betrag[normal]{1+1/\alpha\conj
  3}=\log\betrag[normal]{\alpha}$.
This yields
\begin{equation*}
  \betrag{w_j \log\betrag{\alpha}}\le c_{tj},
\end{equation*}
where $w_1=v_1$, $w_2=-v_2$, and $w_3=v_2-v_1$ are integers.

However, it is checked numerically that ${c_{tj}}/{\log\betrag{\alpha}}<1$
for all relevant $t$ and $j$. Therefore, we get $w_j=0$, which implies
$\Lambda=0$. But this is impossible by the argument in
Section~\ref{sec:stable-growth}. 

Finally, we have to find all solutions with $\betrag{y}< \sqrt{11}$. The
procedure described in Lemma~\ref{le:small-solutions} exactly yields the
solutions listed in Table~\ref{tab:small-solutions-re-t--1-2}.

\bibliographystyle{amsplain}
\bibliography{numth}


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