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\topmatter
\title
Complete solution of a family of quartic thue equations
\endtitle

\author
G\"unter Lettl \& Attila Peth\H{o} \\ \\
\endauthor

\address
G\"unter Lettl \newline
Institut f\"ur Mathematik \newline
Karl-Franzens-Universit\"at \newline
Heinrichstra{\ss}e 36 \newline
A-8010 Graz, AUSTRIA
\endaddress

\address
Attila Peth\H{o} \newline
Laboratory of Informatics \newline
University of Medicine \newline
Nagyerdei krt. 98 \newline
H-4032 Debrecen, HUNGARY
\endaddress

\thanks
This paper evolved from a visit of the first author to the university of
Debrecen, Hungary, which was supported by the Austrian Science Foundation
project Nr. 10-U-3.
\endthanks
\endtopmatter


\document
\head
1. Introduction
\endhead
\medskip

Let \ $F \in \Z [X,Y]$ \ be an irreducible binary form of degree \ $\ge 3$
\ and \ $0 \ne m \in \Z $~. The diophantine equation
$$ F=m \tag{1} $$
is called {\it Thue equation } in honour of the Norwegian mathematician Axel
Thue, who proved in 1909 (see \cite{Thu}) that (1) has only finitely many
integral solutions (i.e. \ $(x,y) \in \Z ^2$ \ with \ $F(x,y)=m$~).
In 1968, A. Baker \cite{B} showed that Thue equations can be solved
effectively. Since then, much work was done to lower the effective upper bound
for the solutions of (1) to some reasonable size, and nowadays it is no problem
to calculate all solutions of a Thue equation of low degree with a computer
(see e.g. \cite{Pet1}, \cite{Pet-S}, \cite{G}).
\smallskip

On the other hand, it was quite recently that parametrized families of Thue
equations were completely solved. For example, one knows all solutions of the
Thue equations
$$
X^3 - (n-1)X^2 Y - (n+2)X Y^2 - Y^3 = \pm 1 \qquad (n \in \Z ) \tag{2.a}
$$
and
$$
X^4 - aX^3 Y - X^2 Y^2 + aX Y^3 + Y^4 = 1 \qquad (a \in \Z ) \tag{2.b}
$$
(see \cite{Tho2}, \cite{Mi1}, \cite{Pet2}, \cite{MPR}). \newline
For the following families the solutions are known for large parameters:
$$
X^3 - (n-1)X^2 Y + n X Y^2 - Y^3 = 1 \qquad (n \ge 3.33 \cdot 10^{23})
\tag{2.c} $$
and
$$
X^4 - aX^3 Y - 3 X^2 Y^2 + aX Y^3 + Y^4 = \pm 1 \qquad (a > 9.9 \cdot 10^{27})
\tag{2.d} $$
(see \cite{Mi-T}, \cite{Pet2} ).

All these families of equations are stably solvable in the sense of
\cite{Tho3}, i.e. up to finitely many exceptions, all solutions are given by
finitely many pairs of rational functions of the parameter.
\smallskip

In this paper we consider the family of Thue equations
\smallskip

$$
\matrix F_a := X^4 - a X^3 Y - 6 X^2 Y^2 + a X Y^3 + Y^4 = c \\
\text{with }\ a \in \Z \ \text{ and } \ c \in \{ \pm 1, \pm 4 \}\ .
\endmatrix  \tag{3}
$$
\medskip

\noindent
If \ $a \in \{ 0, \pm 3 \}$~, \ $F_a$ \ factors into two quadratic forms of
positive discriminant (5, 8, or 20, resp.). Each of these forms represents \
$-1$~, thus there are infinitely many integral solutions of (3) for each \ $c
\in \{ \pm 1, \pm 4 \}$~. \newline
For all other values of \ $a$~, \ $F_a$ \ is irreducible, and we have the
following result:

\proclaim{Theorem}
Let \ $a \in \Z \setminus \{ 0, \pm 3 \}$~. Then the only solutions of (3)
are
$$
\matrix\format \r & \ \c\ & \c & \ \c\ & \l \\
F_a(\pm 1,0) & = & F_a(0, \pm 1)  & = & \ \ 1 \quad , \\
F_a(\pm 1,1) & = & F_a(\pm 1, -1) & = & -4 \quad , \endmatrix
$$
(these solutions we call \,{\rm `trivial solutions' }), and
$$
\matrix\format \r & \ \c\ & \c & \ \c\ & \c & \ \c\ & \c & \ \c\ & \l \\
F_1(\pm 1, \pm 2) & = & F_1(\mp 2, \pm 1) & = & F_{-1}(\pm 2, \pm 1) & = &
F_{-1}(\mp 1, \pm 2) & = & -1 \quad , \\
F_1(\pm 3, \pm 1) & = & F_1(\pm 1, \mp 3) & = & F_{-1}(\pm 1, \pm 3) & = &
F_{-1}(\mp 3, \pm 1) & = & \ \ 4 \quad , \\ 
F_4(\pm 3, \pm 2) & = & F_4(\pm 2, \mp 3) & = & F_{-4}(\pm 2, \pm 3) & = &
F_{-4}(\pm 3, \mp 2) & = & 1 \quad , \\
F_4(\pm 5, \pm 1) & = & F_4(\pm 1, \mp 5) & = & F_{-4}(\pm 1, \pm 5) & = &
F_{-4}(\mp 5, \pm 1) & = & \ \- 4 \quad . \endmatrix
$$
\endproclaim
\bigskip

At first, we will state some elementary transformation properties of the form
\ $F_a$~.
\proclaim{Lemma 1}
Let \ $a, c, x, y \in \Z$~.
\smallskip
\itemitem{\bf (i)} If \ $F_a(x,y)=c$ \  then \ $F_a( x-y, x+y ) = -4c$~.
\smallskip
\itemitem{\bf (ii)} If \ $F_a(x,y)=-4c$ \  then \ $F_a( \frac{x-y}2, \frac{x+y}
2 ) = c$~. \smallskip
\itemitem{\bf (iii)} $F_a( x,y ) = F_a( -x,-y ) = F_a( y,-x ) = F_a( -y,x )$
\smallskip
\itemitem{\bf (iv)} $F_{-a}( x,y ) = F_a( y,x )$
\endproclaim

The proof of lemma 1 is done by direct calculation. Note that \ $F_a(x,y)= -4c$
\ implies \ $x \equiv y \ \text{ mod } (2)$~, so \ $\frac{x \pm y} 2$ \ are
indeed integers.
\smallskip

Parts (i), (ii) and (iv) of lemma 1 show that there is a bijection between the
solutions of \ $F_a=c$~, \ $F_a=-4c$ \ and \ $F_{-a}=c$~. If \ $(x,y) \in \Bbb
Z^2$ \ is a solution of (3) with \ $x=0$ \ (or \ $y=0$~), this implies \ $y=
\pm 1$ \ (or \ $x= \pm 1$~) and \ $c=1$~. By (iii), we can also restrict to
positive \ $y$~. \newline
Therefore, to prove our theorem we have only to show the following:
\medskip
{\sl
\centerline{For \ $a \in \N \setminus \{ 3 \}$ \ the only solutions \
$(x,y) \in \Z ^2$ \ of the equation}
\smallskip
\centerline{$F_a = \pm 1$}
\smallskip
\centerline{with \ $x \ne 0$ \ and \ $y \ge 1$ \ are \ $F_1(1,2) = F_1
(-2,1)=-1$~.} }
\bigskip
\pagebreak

\head
2. Simplest quartic number fields
\endhead
\medskip

We consider the polynomials
$$
f_a := F_a(X,1) = X^4 - a X^3 - 6 X^2 + a X + 1 \in \Z [X]\ , \tag{4}
$$
which are irreducible for each \ $a \in \N \setminus \{ 3\}$~.
If \ $f_a (\alpha) =0$~, one can calculate that also \ $f_a ( \frac{\alpha -1}
{\alpha +1} ) =0$~. Thus the rational map \ $x \mapsto \frac{x-1}{x+1}$ \
permutes the roots of \ $f_a$ \ and, for \ $a \ne 3$~, \ $K:= \Q (\alpha)$
\ is a real quartic number field with cyclic Galois group \ $G = \langle \tau
\rangle$~, generated by the automorphism \ $\tau: \alpha \mapsto \frac{\alpha
-1} {\alpha +1}$~.

These socalled ``simplest'' quartic number fields were investigated by M.N.
Gras in \cite{Gras},I.5) (see also \cite{Laz}). They have very similar
properties as the ``simplest'' cubic number fields, which were called so by D.
Shanks \cite{Sh} and are generated by the polynomials coming from the Thue
equation (2.a).
\medskip

We order the four real roots of \ $f_a$ \ by their size and denote them by
$$ \alpha_4 < \alpha_3 < \alpha_2 < \alpha_1 \ .$$
For the size of the roots we have the following estimations, where the first
inequality holds only for \ $a > 20$ \ and the others either for all \ $n \in
\N$ \ or for the indicated values of \ $a$~:
$$
\alignat 3
a + \frac 5a - \frac 1{a^2} \, \underset (a>20) \to < & & \alpha_1 = & \,\alpha
& & \ \ < \ a + \frac 5a \tag{5.a} \\
1 - \frac 2a \ \ < \ \ & \tau(\alpha)\ = & \alpha_2 = & \,\frac{\alpha -1}
{\alpha +1} & & \ \ < \ 1 - \frac 2a + \frac 3{a^2} \tag{5.b} \\
- \frac 1a + \frac 4{a^3} \underset (a>5) \to < & \tau^2(\alpha) =\, &
\alpha_3 = & - \frac 1 \alpha & & \ \ <\ - \frac 1a + \frac 5{a^3} \tag{5.c}\\
- 1 - \frac 2a - \frac 2 {a^2} \ \ < \ \ & \tau^3 (\alpha) =\, & \alpha_4 = & -
\frac {\alpha +1} {\alpha -1} & & \underset (a>9) \to < - 1 - \frac 2a - \frac
1 {a^2} \tag{5.d} \endalignat
$$

\noindent These inequalities can be obtained by considering the sign of \ $f_a$
\ at the points \ $a + \frac 5a$~, \ $a + \frac 5a - \frac 1{a^2}, \dots $~.
\medskip

Now let \ $a \in \N \setminus \{ 3 \}$~, \ $\alpha = \alpha_1 > 1$ \ be the
largest root of \ $f_a$~, \ $K=\Q (\alpha)$ \ and \ $\beta := \frac
{\alpha +1} {\alpha -1} = - \tau ^3 (\alpha) > 1$~. We will repeatedly use that
$$
\tau(\alpha) = \frac 1 \beta , \quad \tau(\beta) = - \alpha , \quad
\tau^2 (\alpha) = - \frac 1 \alpha , \quad \tau^2 (\beta) = - \frac 1 \beta
\ . \tag{6} $$
For the algebraic part of the proof of our theorem we will work with the order
\ $\frak O := \Z [\alpha, \beta]$ \ and its group of units, \ $\frak O
^\times$~.

\proclaim{Lemma 2}
The order \ $\frak O$ \ is invariant under \ $G = \text{Gal } (K/\Q)$ \ and
has discriminant \ $(16+a^2)^3$~.
\endproclaim

\demo{Proof}
By (6), the conjugates of \ $\alpha$ \ are \ $\frac 1 \beta,\, - \frac 1
\alpha,\, - \beta$~. Thus \ $- \frac 1 \alpha = \alpha^3 - a \alpha^2 - 6
\alpha +a$ \ and \ $\frac 1 \beta = -\beta^3 - a \beta^2 + 6 \beta + a$ \
show the invariance of \ $\frak O$ \ under \ $G$~.

Expressing \ $\beta,\beta^2,\beta^3$ \ with respect to the basis \ $ \{ 1,
\alpha,\alpha^2,\alpha^3 \}$ \ shows that \ $\{ \tilde \alpha, \alpha,
\alpha^2, \alpha^3 \}$ \ is a \ $\Z$~- basis for \ $\frak O$~, where
$$
\tilde \alpha = \cases \frac{1+\alpha^3} 2 & \qquad \text{if }\ a \ \text{ is
odd} \\
\frac{1+\alpha+\alpha^2+\alpha^3} 2 & \qquad \text{if }\ a \ \text{ is even}
\endcases \ . $$
Therefore \ $(\frak O : \Z [\alpha] ) = 2$ \ and \ $\text{ disc } (\frak O)
= \frac 14 \text{ disc }(\Z [\alpha]) = \frac 14 \text{ disc } (f_a) =
(16+a^2)^3$~.
\enddemo
\medskip

Obviously, \ $\alpha$ \ and \ $\beta$ \ are units of \ $\frak O$~. The
following proposition is the quartic analogue of theorem 3.10 in \cite{Tho1},
which deals with simplest cubic fields. Furthermore, our proof shows that \
$\{ \alpha, \beta \}$ \ is a basis for the relative units of \ $\frak O$~.

Let \ $k$ \ be the quadratic subfield of \ $K$ \ and \ $\eps > 1$ \ be the
fundamental unit of the quadratic order \ $\frak o := \frak O \cap k$~.
\medskip

\proclaim{Proposition 1}
If \ $a \ge 20$ \ then \ $\frak O ^\times = \langle -1, \alpha, \beta, \eta
\rangle$ \ with \ $\eta \in \{ \eps, \sqrt{\eps}, \sqrt{\alpha\beta\eps} \,
\}$~. \endproclaim

For the proof we will need the following

\proclaim{Lemma 3}
Let \ $\gamma$ \ be an algebraic integer of degree \ $d$ \ and \ $M(\gamma) :=
\prod\limits_{k=1}^d \text{ \rm max } \{1, |\gamma^{(k)}| \}$~, where
\ $\gamma^{(k)} \quad (1 \le k \le d)$ \ are the conjugates of \ $\gamma$~.
Then
$$
|\text{\rm disc } (\Z [\gamma]) | \,\le\, d^d\, M(\gamma)^{2(d-1)} \ .$$
\endproclaim

\demo{Proof of Proposition 1}
Let us first show that
\smallskip
\centerline{\sl $\{ \alpha, \beta \}$ \ are multiplicatively independent .}
\medskip
Suppose that there are \ $m, n \in \Z$ \ with \ $(m,n) \ne (0,0)$ \ and \
$\alpha^m \beta^n =1$~. Let us choose such a pair \ $(m,n)$ \ with minimal \
$|m|+|n| >0$~.
Using (6), we have
$$\alignat 2
1 & = \tau (\alpha^m \beta^n) & & = \beta^{-m} (-1)^n \alpha^n \quad \text{
and} \\
1 & = \tau^{-1} (\alpha^m \beta^n) & & = (-1)^m \beta^m \alpha^{-n} \qquad .
\endalignat $$

Since \ $\alpha, \beta > 0$~, we obtain \ $m \equiv n \equiv 0 \ \text{ mod }
(2)$ \ and \ $\alpha^{\frac m2} \beta^{\frac n2} =1$~, which is a contradiction
to the above choice of \ $(m,n)$ \ with minimal \ $|m|+|n|$~.
\medskip

Next we will show that
\smallskip
\centerline{\sl $\{ \alpha, \beta \}$ \ can be extended to a basis of \ $\frak
O^\times$~, if \ $a \ge 20$~.}
\medskip

Let us assume on the contrary that there exists a \ $\gamma\in \frak O^\times$
\ and \ $k,l,n\in \Z$ \ with \ $n\ge 2,\quad (k,l,n)=1$ \ and \
$\gamma^n=\alpha^k\beta^l$\,. Without restriction, we suppose that \ $|k|,|l|
\le \frac{n}{2}$ \ and \ $\gamma > 0$\,.

From \ $\tau(\gamma)^n = (-1)^l\alpha^l\beta^{-k} $ \ and \
$\tau^{-1}(\gamma)^n = \alpha^{-l}(-1)^k\beta^k $ \ we see that an even \ $n$ \
also forces \ $l$ \ and \ $k$ \ to be even, which contradicts \ $(k,l,n) =
1$\,. Thus \ $n$ \ must be odd, and we have \ $n\ge 3$ \ and \ $|k|,|l| <
\frac{n}{2}$\,.

If \ $\frac 23 n< |k|+|l|$\,, we consider \ $\gamma' :=
\alpha^{\text{ sgn}(k)}\, \beta^{\text{ sgn}(l)}\, \gamma^{-2}$ \ instead of \
$\gamma$\,, which yields \ $(\gamma')^n = \alpha^{k'} \beta^{l'}$ \ with \ $k'=
n \text{ sgn }(k) -2k,\quad l' = n \text{ sgn }(l)-2l$ \ and \ $0 <
|k'|+|l'| \le \frac{2}{3}n$\,.
Therefore we can assume that we have a unit \ $\gamma \in \frak O^\times$ \
with \ $\gamma = \alpha^{\frac kn} \, \beta^{\frac ln}$ \ and \ $0 < |\frac
kn|+|\frac ln| \le \frac{2}{3}$\,.
\smallskip

Let \ $M(\gamma)$ \ be defined as in lemma 3. From (5.a-d) we obtain \
$M(\alpha) = M(\beta) = \alpha\cdot\beta \le (a+\frac 5a)(1+\frac 2a +
\frac{2}{a^2})$\,. Observe that \ $M(\gamma_1\gamma _2) \le M(\gamma_1)\cdot
M(\gamma_2)$\,, thus \ $M(\gamma) \le M(\alpha)^{|\frac kn|} \,
M(\beta)^{|\frac ln|} \le M(\alpha)^\frac 23$\,. Now lemma 3 yields
$$(16+a^2)^3 = \text{disc }(\frak O) \le \text{disc }(\Z[\gamma])\le
4^4\, M(\gamma)^6 \le 4^4\, M(\alpha)^4 \ .$$
But \ $(16+a^2)^3\le 4^4 \, \bigl( (a+\frac 5a) (1+\frac 2a + \frac{2}{a^2})
\bigr)^4$ \ does not hold for \ $a \ge 20$~, which contradicts the existence of
\ $\gamma$ \  and proves our assertion above.
\bigskip

Now let \ $\eta \in \frak O^\times$ \ with \ $\frak O^\times=\langle -1,
\alpha,\beta,\eta \rangle$\,, and consider the absolute value of the relative
norm map from\ $K$ \ to \ $k$
$$\align
\Cal N: \ \frak O^\times & \ \to\  \langle \varepsilon\rangle \\
\gamma & \ \mapsto\ |N_{K/k} (\gamma)| \endalign $$
Since \ $N_{K/k}(\alpha) = N_{K/k}(\beta)=-1$\,, we have \
$\langle-1,\alpha,\beta \rangle \subset \ker\ (\Cal N)$\,. On the other hand, \
$\Cal N(\eps) = \eps^2$ \ shows that \ $\bigl(\langle \eps \rangle : \Cal
N(\frak O^\times)\bigr) \le 2$~, \ $\Cal N(\eta)\not=1$ \ and \ $\ker\ (\Cal N)
= \langle -1,\alpha,\beta \rangle$~.

Therefore either \ $\Cal N(\eta) = \eps^2$~, which yields \ $\eta\eps^{-1}\in
\ker\ (\Cal N)$ \ and \ $\frak O^\times=\langle -1,\alpha,\beta,\eps \rangle$~,
or \ $\Cal N(\eta) = \eps$~, which yields \ $\eta^2\eps^{-1} \in \ker\ (\Cal
N)$~. In the latter case, we can choose \ $\eta$ \ such that \ $\eta^2 =
\eps\,\alpha^k\,\beta^l$ \ with \ $k,l\in \{0,1\}$~. Considering the signs of
the conjugates, we can see that either \ $\frak O^\times = \langle -1, \alpha,
\beta, \sqrt\eps\, \rangle$ \ (and \ $N_{k/\Q}(\eps) = 1$~) or \ $\frak O
^\times = \langle -1, \alpha, \beta, \sqrt{\alpha\beta\eps}\, \rangle$ \ (and \
$N_{k/\Q}(\eps) =-1$~).
\enddemo
\bigskip

\head
3. Proof of the Theorem for \ $a \ge 488\,050 $
\endhead
\medskip

We will use the notations introduced in the last chapter.
Let \ $a \in \N$ \ with \ $a > 20$ \ and suppose that there exist
\ $x,y \in \Z$ \ with \ $x \ne 0,\ y \ge 1$ \ and
$$ F_a(x,y) = \pm 1 \ . $$
Since \ $F_a(x,y) = \prod_{i=1}^4 (x- \alpha_i y)$ \ is just the norm from \
$K= \Q (\alpha)$ \ to \ $\Q$ \ of
$$ \gamma_1 := x - \alpha_1 y \, \in \frak O \ ,$$
proposition 1 yields that there exist \ $u_1, u_2, u_3 \in \frac 12 \Z$ \
with
$$ |\gamma_1| = \eps^{u_1} \alpha^{u_2} \beta^{u_3} \tag{7.a} $$
Using (6), we obtain for the conjugates
$$\gamma_{i+1}:= \tau(\gamma_i)= x- \alpha_{i+1}y \qquad (1\le i \le 3)\ :$$
$$ \align
|\gamma_2| = & \,\eps^{-u_1} \beta^{-u_2} \alpha^{u_3} \tag{7.b} \\
|\gamma_3| = & \,\eps^{u_1} \alpha^{-u_2} \beta^{-u_3} \tag{7.c} \\
|\gamma_4| = & \,\eps^{-u_1} \beta^{u_2} \alpha^{-u_3} \tag{7.d}
\endalign $$
\medskip

Now from \ $\prod_{i=1}^4 |\frac xy - \alpha_i| = \frac 1{y^4} \le 1$ \ one
sees that the rational \ $\frac xy$ \ is close to one of the zeroes of \
$f_a$~. Let \ $j \in \{ 1, \dots , 4 \}$ \ with \ $|\frac xy - \alpha_j| =
\text{min } \{ |\frac xy - \alpha_i| \mid 1 \le i \le 4 \}$~.
At first we will show that
\smallskip
\centerline{\sl it suffices to consider the cases \ $j=1$ \ and \ $j=4$~.}
\medskip

Suppose that \ $j \in \{2,3,4\}$~. Since obviously \ $|\frac xy - \alpha_j| <
1$~, (5.a) and (5.b) yield \ $\frac xy < \alpha_2 +1 < 2$ \ and \ $|\frac xy -
\alpha_1| = \alpha_1 -\frac xy > a-2 \ge 19$ \ for \ $a > 20$~. \newline
Thus \ $\prod_{i=2}^4 |\frac xy - \alpha_i| = \frac 1 {y^4 |\frac xy -
\alpha_1|} < \frac 1{19}$ \ and
$$\Bigl| \frac xy - \alpha_j \Bigr| \,<\, \frac 1{\root 3 \of {19}} < 0.38
\ . \tag{8}$$

If \ $j=2$~, (5.b) and (8) yield \ $0.52 < \frac xy < 1.38$~.
Since \ $F_a(x,y)=F_a(-y,x)$ \ and \ $-1.93 < - \frac yx < -0.72$ \ we see that
\ $(-y,x)$ \ is a solution of \ $F_a= \pm 1$ \ with \ $x \ge 1$ \ and \ $
|- \frac yx - \alpha_4| = \text{min } \{ |- \frac yx - \alpha_i| \mid 1 \le i
\le 4 \}$~.

Similarly, if \ $j=3$~, we have \ $-0.43 < \frac xy < 0.38$ \ and \ $(y,-x)$ \
is a solution satisfying all our assumptions and \ $|- \frac yx - \alpha_1| =
\text{min } \{ |- \frac yx - \alpha_i| \mid 1\le i \le 4 \}$~.
\medskip

We call \ $(x,y)$ \ a solution of type I (resp. type II) if \ $j=1$ \ (resp. \
$j=4$~). In the following we will show that in both cases we obtain non
vanishing integers \ $N_1, N_2 \in \Z$ \ with
$$ \text{log } |N_1 \text{ log }\alpha + N_2 \text{ log }\beta | < - c \text
{ max } \{ |N_1|,|N_2| \}  \text{ log }\alpha\ .$$
Using a lower bound for this linear form in two logarithms, we will arrive at a
contradiction to the existence of a solution \ $(x,y)$ \ for sufficiently large
\ $a$~.
\bigskip

\subhead
Case 1: Solutions of type I
\endsubhead
\medskip

We suppose that the solution \ $(x,y)$ \ is of type I, i.e.
$$\Bigl| \frac xy - \alpha_1 \Bigr| = \text{min } \left \{ \Bigl| \frac xy
- \alpha_i \Bigr| \ \ \biggl|\ 1 \le i \le 4 \biggr. \right \} \ .$$
Then obviously \ $0 < \gamma_2 < \gamma_3 < \gamma_4$ \ and \ $\text{sgn }
(\gamma_1) = \text{sgn } \bigl( F_a(x,y) \bigr)$~.

Our next aim is to obtain estimations for the \ $\gamma_i$~. Put \ $P:= \prod
_{i=2}^4 \bigl| \frac xy - \alpha_i \bigr|$~. We will repeatedly use the
estimations (5.a-d). From \ $|\frac xy - \alpha_1| < 1$ \ we deduce \ $A:= a-1
< \alpha_1 -1 < \frac xy$~. Inserting this in
$$P > \Bigl( A-1+ \frac 2a - \frac 3{a^2} \Bigr) \Bigl( A + \frac 1a - \frac
5{a^3} \Bigr) \bigl( A+1 \bigr) \tag{9} $$
yields \ $P> \frac{a^3}2$ \ for \ $a>4$ \ and \ $|\frac xy - \alpha_1| \le |
\gamma_1| = \frac{1}{y^3 P} < \frac{2}{a^3}$~.
Thus we can sharpen our lower bound for \ $\frac xy$ \ to \ $A:= a + \frac 5a -
\frac 1{a^2} - \frac 2{a^3} < \alpha_1 - \frac 2{a^3} < \frac xy$~, and
inserting this in (9) yields \ $P > a^3$ \ and
$$ |\gamma_1| < \frac 1{a^3} \ . \tag{10.a} $$

Combining \ $a + \frac 5a - \frac 1{a^2} - \frac 1{a^3} < \frac xy < a + \frac
5a + \frac 1{a^3}$ \ with (5.b,c) yields
$$\align
y\,\bigl( a-1+ \frac 7a - \frac 5{a^2} \bigr) < & \,\gamma_2 \tag{10.b} \\
y\, a < & \,\gamma_3 < y\, \bigl( a+ \frac 6a \bigr) \tag{10.c} \endalign $$
\bigskip

Eliminating \ $x$ \ and \ $y$ \ from the equations
$$\gamma_i = x - \alpha_i y \qquad (i=1,2,4) $$
yields via the socalled {\it Siegel's identity } the following:
$$ 0< \frac {\gamma_4 (\alpha_1 - \alpha_2)} {\gamma_2 (\alpha_1 - \alpha_4)}
=1-\frac {\gamma_1(\alpha_2 - \alpha_4)} {\gamma_2(\alpha_1 - \alpha_4)}\ .$$
Using \ $\frac {\alpha_1 - \alpha_2} {\alpha_1 - \alpha_4} = \beta^{-1}$~,
\quad\ $\frac {\alpha_2 - \alpha_4} {\alpha_1 - \alpha_4} = \frac 2{\alpha+1}$
\ and (7.b,d) we obtain
$$ \alpha^{-2u_3} \beta^{2u_2 -1} = 1 - \frac{\gamma_1}{\gamma_2} \,\frac 2
{\alpha +1} \ . \tag{11} $$
Since we deal with solutions of type I, \ $|\gamma_1|$ \ is very small and the
above expression very near to \ $1$~.
\medskip

\proclaim{Lemma 4} 
For $a > 0, a\not= 3 $ we have
{\rm $$ \bigl| 2u_3 \text{ log }\alpha - (2u_2 -1) \text{ log }\beta \bigr| <
\frac{|\gamma_1|}{\gamma_2} \,\frac 2 \alpha $$ }
\endproclaim

\demo{Proof}
If \ $\gamma_1 >0$~, (11) yields
$$ \bigl| 2u_3 \text{ log }\alpha - (2u_2 -1) \text{ log }\beta \bigr| = -
\text{ log } \Bigl(1 - \frac{\gamma_1}{\gamma_2} \,\frac 2 {\alpha +1} \Bigr)
\ . $$
Using \ $- \text{ log } (1-x) < x+x^2$~, which holds for \ $x < 0.68$~, and \
$\frac{\gamma_1}{\gamma_2} < \frac{1}{a^3 (a-1)}$~, one easily shows the
assertion.

\noindent
If \ $\gamma_1 < 0$~, the assertion follows with \ $\text{log } (1+x) \le x$~.
\enddemo
\medskip

Since \ $\text{log } \alpha > \text{log }a$~, \ $\text{log } \beta \sim \frac
2a$ \ and \ $ \frac{|\gamma_1|}{\gamma_2} \frac 2{\alpha} < \frac 2{a^5-a^4}$ \
we can deduce that \ $2u_3 \ne 0$~, \ $2u_2 -1 \ne 0$~, \ $ \text{sgn } (2u_3)
= \text{sgn } (2u_2 -1)$ \ and \ $M:= |2u_2 -1| = \text{max } \{|2u_3|, |2u_2
-1| \}$~.
Furthermore, \ $\bigl| \frac{\gamma_1}{\gamma_3} \bigr| = \alpha^{2u_2}
\beta^{2u_3} < \frac{1}{a^4}$ \ shows that \ $u_2 <0, \quad u_3 <0$~.
\smallskip

Next we will derive bounds for \ $|2u_2-1|$ \ and \ $|2u_3|$~:

\proclaim{Lemma 5}
If $a>1$ then 
$$ M:= |2u_2 -1| > \frac a2 \text{ \rm log } a$$
and
$$ |2u_2-1| + |2u_3| < M \Bigl( 1 + \frac{2}{a \text{ \rm log }a} \Bigr)\ .$$
\endproclaim

\demo{Proof}
From lemma 4 we obtain
$$|2u_2 -1| \text{ log }\beta \,>\, |2u_3| \text{ log } \alpha -
\frac {|\gamma_1|}{\gamma_2} \,\frac 2\alpha \,\ge \text{ log } \alpha -
\frac {|\gamma_1|}{\gamma_2} \,\frac 2\alpha \ ,$$
so it suffices to prove
$$\text{log } \alpha - \frac{|\gamma_1|} {\gamma_2} \,\frac 2\alpha \ge \frac
a2 \text{ log }a \text{ log }\beta \ . $$
We have the estimation
$$\frac{\alpha\,\gamma_2} {|\gamma_1|} > a (a-1) a^3 = a^5 - a^4 \tag{12} $$
and use
$$\text{log }(1+x) \le x - \frac{x^2}2 + \frac{x^3}3 $$
to obtain
$$\text{log }\beta < \text{log } \Bigl( 1+ \frac 2a + \frac 2{a^2} \Bigr) <
\Bigl( \frac 2a + \frac 2{a^2} \Bigr) - \frac 12 \Bigl( \frac 2a + \frac
2{a^2} \Bigr)^2 + \frac 13 \Bigl( \frac 2a + \frac 2{a^2} \Bigr)^3 <
\frac 2a - \frac{1}{a^3} \ , \tag{13} $$
where the last inequality holds for \ $a \ge 20$~, but $\log \beta < \frac 2a -
\frac{1}{a^3} $ holds for $a \geq 1$ too. \newline
To prove the first inequality of our lemma, it suffices to show that
$$\text{log }a - \frac{2}{a^5-a^4} \,\ge\, \frac a2 \Bigl( \frac 2a - \frac
1{a^3} \Bigr) \text{ log }a \ ,$$
but this is true for \ $a \ge 3$~. A direct computation shows that the first
inequality of this lemma is true for $a=2$ too.

For the proof of the second inequality we start with
$$|2u_3| \text{ log }\alpha < |2u_2 -1| \text{ log }\beta + \frac{|\gamma_1|}
{\gamma_2}\, \frac 2\alpha\ .$$
Using again (12) and (13), we arrive at
$$|2u_3| \, < \, M \Bigl( \frac 2a - \frac 1{a^3} \Bigr) \frac 1{\text{log }a}
+ \frac 2{(a^5-a^4) \text{ log }a} \, < \, M \, \frac 2{a \text{ log }a}\ ,
\tag{13a} $$
which immediately yields our assertion.
\enddemo

\proclaim{Lemma 6}
If \ $a \ge 1034$ \ then
{\rm $$ \text{log } \bigl| 2u_3 \text{ log } \alpha - (2u_2 -1) \text{ log }
\beta \bigr| < -0.98 \ |2u_2 -1|  \text{ log } \alpha \ ,$$ }
and if \ $a \ge 4$ \ then
{\rm $$ \text{log } \bigl| 2u_3 \text{ log } \alpha - (2u_2 -1) \text{ log }
\beta \bigr| < -0.8 \ |2u_2 -1|  \text{ log } \alpha \ .$$ }
\endproclaim

\demo{Proof}
Taking logarithms of the equations (7.b-d) we get a regular system of linear
equations for the \ $u_i$~. Using Cramer's rule and \ $\prod_{i=1}^4 |\gamma
_i| = 1$~, we obtain
$$ u_2 = - \,\frac {\text{log }\alpha \text{ log } \frac{\gamma_3}{|\gamma_1|}
- \text{log } \beta \text{ log } \frac{\gamma_4}{\gamma_2} }
{2( \text{log}^2\,\alpha + \text{log}^2\, \beta)} $$
and
$$|2u_2 -1| = \frac {\text{log }\alpha \text{ log } \frac{\gamma_3}{|\gamma_1|}
- \text{log } \beta \text{ log } \frac{\gamma_4}{\gamma_2} }
{ \text{log}^2\,\alpha + \text{log}^2\, \beta } +1 <
\frac{1}{\text{log }\alpha} \text{ log } \frac{\gamma_3}{|\gamma_1|}\, +1\ .
\tag{14} $$
From lemma 4 we get
$$\text{log }\bigl| 2u_3\text{ log }\alpha - (2u_2 -1)\text{ log }\beta \bigr|
< - \,\text{log } \Bigl( \frac{\gamma_2}{|\gamma_1|} \,\frac{\alpha}2 \Bigr)
\ ,$$
thus we have to show that
$$ 0.98 \ |2u_2 -1| \text{ log }\alpha < \text{log } \Bigl( \frac{\gamma_2}
{|\gamma_1|} \, \frac{\alpha}2 \Bigr)\ .$$
Using (14), it remains to show
$$ 0.98 \Bigl( \text{log } \frac{\gamma_2}{|\gamma_1|} + \text{log } \frac
{\gamma_3}{\gamma_2} + \text{log }\alpha \Bigr) < \text{log } \frac
{\gamma_2}{|\gamma_1|} + \text{log }\alpha - \text{log }2\ ,$$
or equivalently,
$$ 0.98 \text{ log } \frac {\gamma_3}{\gamma_2} + \text{log } 2 < 0.02
\text{ log } \frac {\gamma_2 \,\alpha}{|\gamma_1|} \ .$$
With (10.b,c) we get
$$\frac{\gamma_3}{\gamma_2} < \frac {a+\frac 6a} {a-1+\frac 7a -\frac 5{a^2}}
< 1 + \frac 1a \qquad \text{and} \qquad
\text{log } \frac {\gamma_3}{\gamma_2} < \frac 1a \ .$$
Using this and (12), lemma 6 follows from
$$ \frac{0.98}a + \text{log }2 < 0.02 \text{ log } (a^5-a^4) \ , $$
which holds for \ $a \ge 1034$~.

The proof of the second inequality is analogous.
\enddemo
\bigskip

\subhead
Case 2: Solutions of type II
\endsubhead
\medskip

Now we suppose that the solution \ $(x,y)$ \ is of type II, i.e.
$$\Bigl| \frac xy - \alpha_4 \Bigr| = \text{min } \left \{ \Bigl| \frac xy
- \alpha_i \Bigr| \ \ \biggl|\ 1 \le i \le 4 \biggr. \right \} \ .$$
This case differs from case 1 just by applying the Galois automorphism \
$\tau^{-1}$~. Thus the algebraic relations can be transformed directly, whereas
another archimedian valuation makes the metrical results of this case similar,
but not identical to those of case I.

Obviously we now have \ $\gamma_1 < \gamma_2 < \gamma_3 < 0$ \ and \ $\text{sgn
} (\gamma_4) = - \text{ sgn } \bigl( F_a(x,y) \bigr)$~.
Again we derive estimations for the \ $\gamma_i$~.
From (8) we deduce \ $\frac xy < \alpha_4 + 0.38 < -0.62 - \frac 2a =: A$~, and
$$
P := \prod_{i=1}^3 \ \Bigl| \frac xy - \alpha_i \Bigr| >
\Bigl(a + \frac 5a - \frac 1{a^2} - A \Bigr) \Bigl( 1 - \frac 2a - A\Bigr)
\Bigl( - \frac 1a - A \Bigr) \tag{9'}$$
yields \ $P>a$ \ and \ $|\gamma_4| < \frac 1a$~.
This sharpens the above inequality to \ $\frac xy < \alpha_4 + \frac 1a <
-1 - \frac 1a$~, and inserting this in (9') yields \ $P > 2a$ \ and
$$ |\gamma_4| < \frac 1{2a} \ . \tag{10.d'} $$

Combining \ $-1 - \frac {2.5}a - \frac 2{a^2} < \frac xy < -1 - \frac {1.5}a -
\frac 1{a^2}$ \ with (5.a-c) yields
$$\alignat 2
y\,\Bigl( a+1+ \frac{6.5}a \Bigr) & < & \,|\gamma_1| < & \,y\, \Bigl( a+1+
\frac {7.5}a +\frac 2{a^2} \Bigr) \tag{10.a'} \\
y\,\Bigl( 2- \frac {0.5}a \Bigr) & < & \,|\gamma_2| < & \,y\, \Bigl( 2+ \frac
{0.5}a + \frac  5{a^2} \Bigr) \tag{10.b'} \\
y\, \Bigl( 1+ \frac {0.5}a \Bigr) & < & \,|\gamma_3| < & \,y\, \Bigl( 1+
\frac{1.5}a + \frac 3{a^2} \Bigr) \tag{10.c'} \endalignat $$
\bigskip

Applying \ $\tau^{-1}$ \ to (11) yields
$$ \alpha^{-2u_2+1} \beta^{-2u_3} = 1 + \frac{\gamma_4}{\gamma_1} \, (\alpha
- 1) \ . \tag{11'} $$
Since we deal with solutions of type II, \ $|\gamma_4|$ \ is small and the
above expression near to \ $1$~.
\medskip

\proclaim{Lemma 4'}
For $a \ge 1, a \not = 3$ we have
{\rm $$ \bigl| (2u_2-1) \text{ log }\alpha + 2u_3 \text{ log }\beta \bigr|
< \Bigl| \frac{\gamma_4}{\gamma_1} \Bigr| \, \alpha $$ }
\endproclaim

\demo{Proof}
The proof is analogue to that of lemma 4.
\enddemo
\medskip

Since \ $\text{log } \alpha > \text{log }a$~, \ $\text{log } \beta > \frac 2a -
\frac 1{a^2}$ \ and \ $\bigl| \frac{\gamma_4}{\gamma_1} \bigr| \, \alpha <
\frac 1{2a}$~, we can deduce that \ $2u_3 \ne 0$~, \ $2u_2 -1 \ne 0$~, \ $
\text{sgn } (2u_3) = - \text{sgn } (2u_2 -1)$ \ and \ $M:= |2u_3| =
\text{max } \{|2u_3|, |2u_2 -1| \}$~.
Furthermore, \ $\alpha^{-2u_3} \beta^{2u_2} = \bigl| \frac{\gamma_4}{\gamma_2}
\bigr| < \frac{1}{4a-1}$ \ shows that \ $u_2 <0$ \ and \ $u_3 >0$~.
\smallskip

For \ $|2u_2-1|$ \ and \ $|2u_3|$ \ we obtain nearly the same bounds as
in case I:

\proclaim{Lemma 5'}
For $a \ge 2$ we have
$$ M:= |2u_3| > \frac a2 \text{ \rm log } a - \frac 14$$
and
{\rm $$
|2u_2 -1| + |2u_3| < M \Bigl( 1 + \frac 2{a \text{ log }a} \Bigr) + \frac 1
{2a \text{ log }a}\ .$$}
\endproclaim

\demo{Proof}
From lemma 4' we obtain
$$|2u_3| \text{ log }\beta \,>\, |2u_2-1| \text{ log } \alpha - \Bigl|
\frac {\gamma_4}{\gamma_1} \Bigr| \,\alpha \,\ge \text{ log } \alpha -
\Bigl| \frac{\gamma_4}{\gamma_1} \Bigr| \,\alpha \ ,$$
so using (13), it suffices to prove
$$\text{log } \alpha - \Bigl| \frac{\gamma_4}{\gamma_1} \Bigr| \,\alpha  \ge
\Bigl( \frac a2 \text{ log }a - \frac 14 \Bigr) \Bigl( \frac 2a - \frac 1{a^3}
\Bigr) \ .$$
Estimating
$$\Bigl| \frac{\gamma_4}{\gamma_1} \Bigr| \,\alpha < \frac {a + \frac 5a}
{2a(a+1)} \ , \tag{12'}$$
it suffices to show that
$$\text{log }a - \frac{a+ \frac 5a} {2a(a+1)} \,>\, \Bigl( \frac a2 \text{ log
}a - \frac 14 \Bigr) \ \Bigl( \frac 2a - \frac 1{a^3} \Bigr) \ ,$$
which is easily shown to be true for \ $a \ge 3$~.

With (12') and (13) we obtain for the second inequality:
$$\align
|2u_2 -1| & < \, M \frac {\text{log }\beta} {\text{log }\alpha} + \Bigl| \frac
{\gamma_4}{\gamma_1} \Bigr| \, \frac \alpha {\text{log }\alpha} <
M \Bigl(\frac 2a - \frac 1{a^3} \Bigr) \frac 1{ \text{log }a} +
\frac{a+ \frac 5a} {2a(a+1)\text{ log }a} < \\
& < M \frac 2{a \text{ log }a} - \frac{\frac a2 \text{ log }a -\frac 14} {a^3
\text{ log }a} + \frac{a+ \frac 5a} {2a(a+1)\text{ log }a} < M \frac 2
{a \text{ log }a} + \frac 1 {2a \text{ log }a}\ .\endalign
\tag{12a'}$$
\enddemo
\medskip

\proclaim{Lemma 6'}
If \ $a \ge 262183$ \ then
{\rm $$ \text{log } \bigl| (2u_2-1) \text{ log } \alpha + 2u_3 \text{ log }
\beta \bigr| < -0.95 \ |2u_3| \text{ log } \alpha \ ,$$ }
and if \ $a \ge 2$ \ then
{\rm $$ \text{log } \bigl| (2u_2-1) \text{ log } \alpha + 2u_3 \text{ log }
\beta \bigr| < -0.8 \ |2u_3| \text{ log } \alpha \ .$$ }
\endproclaim

\demo{Proof}
Like in lemma 6, we use Cramer's rule to obtain
$$|2u_3| = \frac {\text{log }\alpha \text{ log } \bigl| \frac {\gamma_2}
{\gamma_4} \bigr| + \text{log } \beta \text{ log } \bigl| \frac {\gamma_1}
{\gamma_3} \bigr|} {\text{log}^2\,\alpha + \text{log}^2\, \beta} <
\frac{1}{\text{log }\alpha} \text{ log } \Bigl| \frac{\gamma_2}{\gamma_4}
\Bigr| + \frac {\text{log }\beta} {\text{log}^2\,\alpha} \text{ log } \Bigl|
\frac{\gamma_1}{\gamma_3} \Bigr| \ . \tag{14'}$$
Using lemma 4', we have to show that
$$ 0.95\ |2u_3| \text{ log }\alpha < \text{log } \Bigl( \bigl| \frac{\gamma_1}
{\gamma_4} \bigr| \, \frac 1\alpha \Bigr)\ .$$
Using (14'), it remains to show
$$ 0.95 \ \frac {\text{log }\beta} {\text{log }\alpha} \text{ log } \Bigl|
\frac{\gamma_1}{\gamma_3} \Bigr| + \text{log }\alpha <
0.05 \text{ log } \Bigl| \frac {\gamma_2}{\gamma_4} \Bigr| + \text{log } \Bigl|
\frac {\gamma_1}{\gamma_2} \Bigr|\ ,$$
If we estimate \ $\alpha, \beta$ \ and \ $\gamma_i$~, it remains to show
$$ 0.95 \, \Bigl( \frac 2a - \frac{1}{a^3} \Bigr) \frac {\text{log } (a+1)}
{\text{log }a} + \text{log } \bigl( 2+ \frac 5{a^2} \bigr) < 0.05 \text{ log }
(4a-1) \ .$$
This inequality holds for \ $a \ge 262183$~, which proves lemma 6'.
\enddemo
\bigskip

\subhead
End of the Proof of the Theorem
\endsubhead
\medskip

Now we need a lower bound for the absolute value of the linear form
$$\Lambda := N_1 \text{ log }\alpha + N_2 \text{ log }\beta\,, \qquad
N_1, N_2 \in \Z \setminus \{ 0 \}\ .$$
We use the following result of M. Laurent \cite{Wa}, appendix p. A-2, adopted
for our application:

\proclaim{Proposition 2}
Let \ $a_1, a_2 \ge e$ \ with
{\rm $$ h(\alpha) \le \text{ log }a_1\ ,\qquad h(\beta) \le \text{ log }a_2\
,$$}
where \ $h(\cdot)$ \ denotes the logarithmic absolute height of an algebraic
number. Let
{\rm $$b' := \frac{|N_1|}{4 \text{ log }a_2} + \frac{|N_2|}{4 \text{ log }a_1}
\qquad \text{\sl and} \qquad B \ge 0.5 + \text{log }b' \ .$$}
If \ $4B \ge 25$ \ then
{\rm $$\text{log }|\Lambda| \ge -87\ 4^4 B^2 \text{ log }a_1 \text{ log }a_2
\ .$$}
\endproclaim

\remark{Remark}

(explanation, why \ $4B \ge 25$ \ is sufficient.)
\endremark
\bigskip

Since
$$h(\alpha) = h(\beta) = \frac 14 (\text{log }\alpha + \text{ log }\beta )
< \frac 14 \text{ log } \bigl( (a+ \frac 5a) (1 +\frac 2a +\frac 2{a^2}) \bigr)
< \frac 14 \text{ log } (a+2.1) $$
for \ $a \ge 72$~, we can put
$$a_1 = a_2 = \root 4 \of {a+2.1}\ .$$
$a_i \ge e$ \ holds for \ $a \ge 53$~. \newline
With lemma 5 (lemma 5' resp.) we can put
$$\align
B:= &\ 0.5 + \text{log }\biggl( \frac {M(1+ \frac 2{a\text{ log }a})}
{\text{log } (a+2.1)} \biggr) \qquad\qquad\qquad \text{ in case 1} \tag{15} \\
B= B' := &\ 0.5 + \text{log }\biggl( \frac {M(1+ \frac 2{a\text{ log }a}) +
\frac 1 {2a \text{ log }a}} {\text{log } (a+2.1)} \biggr) \qquad \text{in case
2.} \tag{15'} \endalign$$
Using the lower bound for \ $M$~, which is given in lemma 5 (lemma 5', resp.),
one checks, that \ $4B \ge 25$ \ is satisfied for \ $a \ge 629$~.
\smallskip

Combining proposition 2 with lemma 6 (lemma 6' resp.) yields
$$\align
0.98\ M \text{ log }\alpha < &\ 1392\ B^2 \text{ log}^2 (a+2.1) \qquad \text{
 in case 1} \tag{16} \\
0.95\ M \text{ log }\alpha < &\ 1392\ {B'}^2 \text{ log}^2 (a+2.1) \qquad
\text{in case 2.} \tag{16'} \endalign$$
Considered as a function of \ $M$~, the difference of the two sides of these
inequalities are monotonous for \ $M > \frac a2 \text{ log }a$ \ (\ $M >
\frac a2 \text{ log }a - \frac 14$~, resp.), if \ $a > 64000$~.
Inserting the lower bound for \ $M$~, we arrive at
$$\align
0.98\, \frac a2 \text{ log }a \text{ log}\bigl( a + \frac 5a - \frac 1{a^2}
\bigr) < &\, 1392\, \Bigl( 0.5 + \text{log } \frac {\frac a2 \text{ log }a +1}
{\text{log }(a+2.1)} \Bigr)^2 \text{ log}^2(a+2.1) \tag{17} \\
0.95\, \bigl(\frac a2 \text{ log }a - \frac 14 \bigr) \text{ log}\bigl( a +
\frac 5a - \frac 1{a^2} \bigr) < &\, 1392\, \Bigl( 0.5 + \text{log } \frac
{\frac a2 \text{ log }a + \frac 34} {\text{log }(a+2.1)} \Bigr) ^2  \text{ log}
^2(a+2.1)\ . \tag{17'} \endalign$$

But (17) does not hold for \ $a \ge 470415$ \ and (17') does not hold for
\ $a \ge 488050$~, which concludes the proof of the theorem.
\bigskip
\bigskip

\head
4. Proof of the Theorem for \ $1 \le a < 488050$
\endhead
\medskip

\newpage


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\vskip 1in

\endRefs

\enddocument
