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\begin{document}

\title{On the system of diophantine equations $x^2-6y^2=-5$ and $x=2z^2-1$.}
\author{Maurice Mignotte \\
D\'epartement de Math\'ematique et Informatique\\
Universit\'e Louis Pasteaur\\
Rue Ren\'e Descartes 7\\
Strasbourg, France\\[1.5ex]
Attila Peth\H{o}\thanks{Research partially supported by Hungarian National
Foundation for Scientific Research Grant No 1641/90.} \\
Laboratory for Computer Science\\
University Medical School Debrecen\\
Nagyerdei krt. 98\\
H-4028 Debrecen, Hungary}
\date{\today}
\maketitle

\section{Introduction}
The aim of this paper is to prove the following
\begin{theorem}
The system of diophantine equations
\begin{equation}
\label{1} x^2-6y^2=-5 \qquad \mbox{and} \qquad x=2z^2-1
\end{equation}
has only the solutions $(x,y,z)=(16561,\pm6761,\pm91);(71,\pm29,\pm6);(17,\pm7,\pm3);$
$(7,\pm3,\pm2);$ $(1,\pm1,\pm1) $ and $(-1,\pm1,0). $
\end{theorem}
Our system of equations is a quartic model of an elliptic curve. It has only 
finitely many integer solutions by a well known result of Siegel
\cite{Si}, moreover they are effectively computable by Baker \cite{Ba}. It is
still interesting to solve it, because the elementary method of J.H.E. Cohn
\cite{Co}, which was further developed by McDaniel and Ribenboim \cite{DR} failed. The Siegel-Baker method, which is the combination of algebraic and transcendental number theoretical tools is complicated. It requires detailed 
knowledge of certain quartic number fields and the solution of several quartic
Thue equations. \\[1ex]
There are two crucial points in our proof:
\begin{enumerate}
\item We prove in section 2. that under general conditions a diophantine
equation $x^2-dy^2=m $ with the side condition $x=az^2-b$ can be 'homogenized',
i.e. can be transformed to finitely many equations $x^2-dy^2=m_i $ with
$x=a_i z^2.$ In this step we use an idea of Mordell \cite{Mo}.
\item After the 'homogenization' we get mixed exponential-polynomial equations
in $n,z \in \Z$ of type
\begin{equation}
\label{2a} a\alpha^n-b\beta^n =c z^2.
\end{equation}
This can be solved theoretically by using results of transcendental number
theory, see Peth\H{o} \cite{P} and Shorey and Stewart \cite{ShSt}. Unfortunately
none of these methods is applicable in practice. Generalizing the argument of Mignotte
\cite{M1} and Mignotte and Peth\H{o} \cite{MP1}, \cite{MP2} we are able to
reformulate (\ref{2a}) directly enough into linear forms in logarithms of
suitable algebraic numbers to use efficiently the known reduction techniques.
\end{enumerate}

\section{Homogenization of the problem}
In the first step toward the proof of our main theorem we use an idea of
Mordell \cite{Mo} to translate (\ref{1}) into finitely many 'homogeneous' equations.
\begin{theorem}
Let $a,b,d,m {\in} \Z$, $d$ square--free. Assume that $(x_0=-b,y_0,z_0){\in} \Z^3$
satisfies
\begin{equation}
\label{2} x^2-dy^2=m \qquad \mbox{and} \qquad x=az^2+b.
\end{equation}
Then there exist for all solutions $(x,y,z){\in} \Z^3$ of (\ref{2}) integers
$e,f,{\Delta}$ with $(e,f)=1$, $f^2-de^2={\Delta}$, where $\Delta$ divides $2dm$,
if $d$ is odd, and $dm$, if $d$ is even,
\begin{eqnarray*}
x&=&2ed\frac{fy_0-ex_0}{\Delta}-x_0,\\
y&=&2e\frac{de y_0-fx_0 }{\Delta }+y_0;\\
az^2&=&2ed\frac{fy_0-ex_0}{\Delta}.
\end{eqnarray*}
\end{theorem}
\begin{proof}
Let $(x,y,z){\in} \Z^3$ be a solution of (\ref{2}). If $x=x_0$ then the choice
${\Delta}=-d,f=0,e=1$ satisfies the assertion. In the sequel we may assume
$x \not= x_0$. Let $e$ and $f$ be coprime integers with
\[ y-y_0=\frac{ e}{f}(x-x_0).\]\\[1ex]
Inserting this formula for $y$ into (\ref{2}) and using $x_0^2-dy_0^2=m$
we get
\[ x+x_0-d\displaystyle\frac{e^2}{f^2}(x-x_0)=2\displaystyle\frac{e}{f}dy_0,\]
which proves the stated parametrized form of $x$ and $y$.\\[1ex]
As $(e,f)=1$ and $x$ and $y$ are integers, the numbers $2d\displaystyle
\frac{fy_0-ex_0}{\Delta}$ and $2\displaystyle \frac{dey_0-fx_0}{\Delta}$
are integers too and $\Delta$ is coprime with $e$. We have further
\[ 2dx_0\displaystyle \frac{fy_0-ex_0}{\Delta}+2dy_0\displaystyle
\frac{dey_0-fx_0}{\Delta}=-\displaystyle \frac{2dem}{\Delta}. \]
Thus $\Delta$ divides $2dm$. If $d$ is even then 4 does not divide $f^2-de^2$
because $d$ is square--free and $(e,f)=1.$ Thus ${\Delta} |dm $ in this
case.\\[1ex]
Inserting the paremetrized formula for $x$ into the second equation of (\ref{2})
we get the equation for $z$ and the proof of the theorem is completed.
\end{proof}
\begin{cor}
All integer solutions $x,y,z$ of (\ref{1})
have the form $x=12e\displaystyle\frac{f-e}{\Delta}-1$, $y=2e
\displaystyle\frac{6e-f}{\Delta}+1$,
$z^2=6e\displaystyle \frac{f-e}{\Delta}$, where ${\Delta}=1,-2,3,-6$ and
\begin{equation}
\label{4} f^2-6e^2={\Delta}.
\end{equation}
\end{cor}
\begin{proof}
We apply Theorem 1. with $d=6$, $m=-5$, $a=2$, $b=-1$,
$(x_0,y_0,z_0)=(1,1,1).$ Then there exist $e,f{\in} \Z$, wich satisfy (\ref{4})
with a ${\Delta}|30$. The only values of ${\Delta}$ with these conditions are 
$1,-2,3,-6,-5,10,-15$ and $30.$ \\
Assume that $5|{\Delta}$ and there exist $e,f{\in}\Z$ with (\ref{4}) and
\begin{equation}
\label{5} 6e(f-e)={\Delta}z^2.
\end{equation}
As by (\ref{4}) $5$ does not divide $e$, we have $5|(f-e)$ by (\ref{5}). We can
rewrite (\ref{4}) and (\ref{5}) as follows
\[
-\frac{\Delta}{5}=\left(\frac{6e-f}{5}\right)^2-6\left(\frac{e-f}{5}\right)^2
\quad \mbox{and} \quad -6e\frac{f-e}{5}=-\frac{\Delta}{5}z^2.
\]
Put $E_1=\displaystyle\frac{e-f}{5}$ and $F_1=\displaystyle\frac{6e-f}{5}$,
then $E_1,F_1 {\in}\Z$ and they satisfy $F_1^2-6E_1^2=-\displaystyle\frac{\Delta}{5}$
and $6E_1(F_1-E_1)=-\displaystyle\frac{\Delta}{5}z^2$.
Thus it is enough to solve (\ref{4}) and (\ref{5}) for those values of $\Delta$
which are not divisible by $5$.
\end{proof}
\begin{lemma}
Let ${\Delta}=1,-2,3$ or $6 $ and $e,f{\in}\Z$ $e \not \doteq 0$ be a solution of (\ref{4}) and
(\ref{5}). Put ${\alpha}=5+2\sqrt{6}$ and ${\beta}=5-2\sqrt{6}.$
Then there exist $n,w{\in}\Z$ such that\\[1ex]
\begin{tabular}{@{\hspace{2cm}}lcl}
$\displaystyle\frac{e}{2}=\displaystyle\frac{\alpha^{2n+1}-\beta^{2n+1}}{4\sqrt6}=w^2,$ &if& $\Delta=1$,\\
$e=\displaystyle\frac{(2+\sqrt{6})\alpha^{2n}-(2-\sqrt{6})\beta^{2n}}{2\sqrt6}=w^2
,$ &if& $\Delta=-2,$ \\
$e=\displaystyle\frac{(3+\sqrt6)\alpha^{2n}-(3-\sqrt6)\beta^{2n}}{2\sqrt6}=w^2
,$ &if& $\Delta=3,$ and \\
$e=\displaystyle\frac{\alpha^{2n}+\beta^{2n}}{2}=w^2,$ &if&  $\Delta=-6.$
\end{tabular}
\end{lemma}
\begin{proof}
We way assume $e\geq0$ without loss of generality. In the sequel $\Box$ denotes
an unspecified square.\\
(i) {\bf Let $\Delta=1$.}
We have $e$ even and $f$ odd, hence $f-e$ odd by (\ref{4}). Hence
$e=2\Box$ or $e=6\Box$. By the theory of Pellian equations there exist
$m{\in}\Z$ and $\varepsilon{\in}\{-1,1\}$ such that
\[
e={\varepsilon}\frac{\alpha^m-\beta^m}{2\sqrt6}.
\]
Let $a_m=\displaystyle\frac{\alpha^m-\beta^m}{4\sqrt6}$ for $m{\in}\Z$.
As $a_{-m}=-a_m$ we may
assume $\varepsilon=1$ and $m\geq0$. Thus we get the equations
\begin{equation}
\label{10} a_m=\frac{\alpha^m-\beta^m}{4\sqrt6}=\delta_1\Box
\end{equation}
with $\delta_1=1$ or $3$. It is easy to see that $3|a_m$ holds iff $3|m$.
Let $m=3k$ and $a'_k=\displaystyle \frac{a_{3k}}{3}$. We have $a_0'=0, a_1'=33$
and $a'_{k+2}=970a'_{k+1}-a'_k$ for $k\geq0$.\\[1ex]
The sequence ${\{a'_k \ \mbox{mod} \ 5\}}^{\infty}_{k=0}$ is periodic, and its period is
$(0,3,0,2).$ As $(\frac{2}{5})=(\frac{3}{5})=-1$  we see that if (\ref{10}) holds
with $\delta_1=3$, then $6|m$. Let $m=6k$, and for $n{\in}\Z$ put $b_n=\alpha^n+\beta^n$. Then
\begin{equation}
\label{11} a'_{2k}=\frac{a_{6k}}{3}=\frac{a_{3k}}{3}b_{3k}.
\end{equation}
We have also
\[
\left(\frac{b_k}{2}\right)^2-\left(\frac{\alpha-\beta}{2}\right)^2a_k^2=1,
\]
hence $(a_k,b_k)=\cases{2& \ \mbox{for}\ k \ \mbox{even} \cr 1& \
\mbox{for} \ k \ \mbox{odd} }$.\\  
Assume that $m>0$ is the smallest even integer with $a'_{m}=x^2.$ Then, by (\ref{11}) $a'_mb_{3m}=x^2$,
hence $m$ must be even, say $m=2m_1$  and $a'_{2m_1}=2x^2_1$, and $b_{6m_1}=2x^2_2$.
Continuing this process, assume that $2m_k$ is the smallest even divisor of $m$ such
that $a'_{2m_k}=2x^2_k$ with an integer $x_k$. Then
$a'_{2m_k}b_{3m_k}=2x^2_k$. Let $m_k$ odd. Then $a'_{m_k}$ is odd, and a
square, which is a contradiction. Hence $m_k$ is even and either
a square or $2\Box$ in contradiction with the choice of $m$ and $m_k$.\\
This means, that in (\ref{10}) $\delta_1=3$ is not possible. \\[1ex]
{\em Now we claim, that if a positive even integer $m$ satisfies (\ref{10}),
then there exists an odd divisor of $m$, which satisfies (\ref{10})
too.}\\[1ex]
Let $m=2m_1>0$ be the smallest even solution of (\ref{10}). Then as
$a_{2m_1}=a_{m_1}b_{m_1}=\Box$ and $(a_{m_1},b_{m_1})=\cases{1,& \ \mbox{if}\ $m_1$ \ \mbox{odd}
\cr 2,& \ \mbox{if}\ $m_1$ \ even \cr}$, either $m_1$ is
odd and $a_{m_1}=\Box$ or $m_1$ is even and $a_{m_1}=2\Box$. Continuing this argument we
get the proof of the claim and the lemma in the present case.\\[1ex]
(ii) {\bf Let ${\Delta}=-2$.} We have $e=\Box$ or $3\Box$ by (\ref{5}) and
\[
e=\varepsilon\frac{(2+\sqrt6)\alpha^m-(2-\sqrt6)\beta^m}{2\sqrt6}=a_m
\]
with a suitable ${\varepsilon}{\in}\{-1,1\}$ and $m{\in}\Z$ by (\ref{4}). It is easy to see that
$a_{-m}=a_{m-1}$, hence we may assume again $\varepsilon =1$ and $m\geq0$. We have
$3|a_m$ if and only if $m\equiv1$ (mod $3$).\\
Let $b_m=\displaystyle \frac{a_{3m+1}}{3}$ for $m\geq0$.
Then $b_0=3, b_1=2907$ and $b_{m+2}=970b_{m+1}-b_m$ for
$m\geq0$. The period of the sequence $\{b_m \ \mbox{mod}\ 5\}^{\infty}_{m=0}$
is $(3,2,2,3)$ which means that $e=3\Box$ is not possible. \\
If $e=\Box$, then $f-e=-3\Box$ this implies that $m$ has to be even.\\[1ex]
(iii) {\bf Let $\Delta=3$.} In this case $e$ is odd, hence a square by (\ref{5}).
Let
\[
a_m=\frac{(3+\sqrt6)\alpha^m-(3-\sqrt6)\beta^m}{2\sqrt{6}}
\]
for $m{\in}\Z$. Then $e=a_m$ for a suitable $m$. Considering $a_m$ modulo 4 we see
that $a_m=\Box$ is only possible if $m$ is even.\\[1ex]
(iv) {\bf Let $\Delta=-6$.} Now $e=\Box$ by (\ref{5}) and
\[
e=\sqrt6\frac{\alpha^m+\beta^m}{2\sqrt6}=\frac{\alpha^m+\beta^m}{2}=a_m
\]
for a suitable $m{\in}\Z.$ Considering $a_m$ modulo 3 we see that $a_m=\Box$ is
only possible if $m$ is even.
\end{proof}
\section{Application of linear forms}
Let the algebraic number $\beta$ be a zero of the irreducible polynomial
$p(x)=a_nx^n+...+a_0 \in \Z[x]$, where $(a_n,...,a_0)=1$. Denote $\beta_1=
\beta,\dots,\beta_n$ the zeros of $p(x)$. The absolute logarithmic height of
$\beta$ is defined by
\[
h(\beta)=\frac{1}{n}\log\left(\prod_{i=1}^n \max \{1,|\beta_i| \} \right).
\]
In this section we will use the following theorem of Waldschmidt \cite{W}.
\begin{theorem}
Let $\alpha_1,...,\alpha_n$ be non-zero algebraic numbers; for $i=1,...,n$, let
$\log \alpha_i$ be a determination of the logarithm of $\alpha_i$. Suppose that
the numbers $\log \alpha_1,...,\log \alpha_n$ are $\Q$-linearly independent. Put 

\[
D=[\Q(\alpha_1,...,\alpha_n):\Q]  \quad \mbox{and} \quad 
g=[\R (\log \alpha_1,...,\log \alpha_n):\R].
\]
Let $A_1,...,A_n,A,E$ and $f$ be positive real numbers such that 

\[
\log A_i \geq h(\alpha_i), \quad (1\leq i \leq n), \quad A=\max\{A_1,...,A_n\}
\]
and
\[
e\leq E \leq \min\left\{ A^D_1,...,A^D_n,\frac{nD}{f} \left(\sum^n_{i=1} \frac{|\log \alpha_i|}{\log A_i}\right)^{-1} \right\}.
\]
Let $b_1,...,b_n$ be rational integers with $b_n\not= 0.$ Put
\[
M=\max_{1 \leq j \leq n} \left\{ \frac{|b_n|}{\log A_j} + \frac{|b_j|}{\log A_n} \right\},
\]
\[
Z_0= \max \left\{ 7+3 \log n, \frac{g}{D} \log E, \log \left(\frac{D}{\log E}\right) \right\},
\quad G_0=\max\{ 4nZ_0; \log M \}
\]
and
\[
U_0= \max \{ D^2 \log A,D^{n+2} G_0 Z_0 \log A_1 \cdots \log A_n (\log E)^{-n-1} \}.
\]
Then the linear form
\[
\Lambda =b_1 \log \alpha_1 + \dots + b_n \log \alpha_n.
\]
satisfies
\[
|\Lambda \geq exp \left\{ -1500g^{-n-2} 2^{2n} n^{3n+5} \left(1 + \frac{g}{f}\right)^n U_0 \right\}.
\]
\end{theorem}
Let $\alpha$ be a real quadratic unit and $\K=\Q(\alpha)$. Let $\gamma'$ denotes
the conjugate of $\gamma{\in}\K$. Take $\beta={\alpha}'$ and assume that $\alpha>{|}{\beta}{|}$.
Let $a,b$ and $c{\in}\Z_{\K}.$ Assume that the integers $m,x{\geq}0$ satisfy
the equation
\[
a\alpha^{2m}-b^2\beta^{2m}=cx^2.
\]
{\em Our aim in this section is to prove an upper bound for $m$.} \\
Let $\L=\K(\sqrt{-c})$ and assume that $\L$ is a quadratic extension of
$\K$, i.e. $[\L:\Q]=4.$ Then our equation implies
\begin{equation}
\label{15}
N_{\L/\Q}(b\beta^m+\sqrt{-c}x)=N_{\L/\Q}(a)=A,
\end{equation}
with an integer $A$. \\
Choose in $\Z_{\L}$ units $\eta_2,...,\eta_r;$ $r=1,2$ or $3$ such that the
group $\cal{U}$ generated by $\eta_1=\alpha,\eta_2,...,\eta_r$ has finite index in
the group of units of $\Z_{\L}$. There exists in $\Z_{\L}$ a maximal finite set
of, with respect to $\cal{U}$, non--associated elements of norm $A$. This set
will be denoted by $\cal A$. Then there exist for all
$m,x{\in}\Z$ with (\ref{15}) a $\gamma{\in}\cal A$ and $\varepsilon{\in}\cal U$
such that
\begin{equation}
\label{16}
b\beta^m+\sqrt{-c}x=\gamma \varepsilon.
\end{equation}
Let order the conjugates $\L^{(i)},i=1,2,3,4$ of $\L$ according the following
ordering of the conjugates of
$\sqrt{-c}:\sqrt{-c},-\sqrt{-c},\sqrt{-c'},-\sqrt{-c'}.$ \\
It is easy to see that if $m>m_0$ then
\begin{equation}
\label{17a}
\frac{1}{2} \frac{\sqrt{| a |}}{| \gamma^{(i)} |} \alpha^m <
| \varepsilon^{(i)} | < \frac{2\sqrt{| a |}}{| \gamma^{(i)}|} \alpha^m
\end{equation}
for $i=1,2$; and if $b'>0$, which we may assume without loss of generality, then
\begin{equation}
\label{17b}
\frac{b'}{2|\gamma^{(3)}|} \alpha^m < | \varepsilon^{(3)} | < 2
\frac{b'}{| \gamma^{(3)} |} \alpha^m
\end{equation}
and
\begin{equation}
\label{17c}
\frac{|a'|}{2b'|\gamma^{(4)}|}  \alpha^{-3m} < | \varepsilon^{(4)} | <
\frac{2|a'|}{b'|\gamma^{(4)}|} \alpha^{-3m}
\end{equation}
hold. We remark that if $b'<0$ than only the role of $\varepsilon^{(3)}$
and $\varepsilon^{(4)}$ changes. \\
The last inequalities imply that if $c'>0$ then (\ref{15}) has only finitely
many solutions and they are very easy to compute. In fact $\varepsilon^{(3)}$ and
$\varepsilon^{(4)}$ are in this case conjugate complex numbers, hence
\[
\frac{b'}{2| \gamma^{(3)} |} \alpha^m < | \varepsilon^{(3)} | =
| \varepsilon^{(4)} | < \frac{2|a'|}{b'|\gamma^{(4)}|} \alpha^{-3m},
\]
i.e $m<\displaystyle\frac{1}{4} \log \left|\displaystyle\frac{4a'\gamma^{(3)}}
{b'^2\gamma^{(4)}} \right|.$ \\[1ex]
The situation is more interesting when $c'<0.$ Then $\varepsilon^{(3)}$ and
$\varepsilon^{(4)}$ are real numbers and we will use estimations on linear forms
in logarithms of algebraic numbers to establish an upper bound for $m$. \\
Let first $c>0$ (and $c'<0$). Then $\L$ has two nonreal and two real
conjugates, and there exist $u_1,u_2 \in \Z$ with $\varepsilon= \eta^{u_1}_1
\eta^{u_2}_2.$ The estimations (\ref{17a}) with $i=2$ and (\ref{17b}) yield
\[
| u_1 | < \frac{2m \log\alpha | \log | \eta^{(3)}_2 |
-\log | \eta^{(2)}_2 ||}{R} +c_1
\]
and
\[ | u_2 | < \frac{4m\log^2\alpha}{R} +c_1, \]
where $R$ denotes the regulator of $\cal U$ and
\[
c_1=2\log\left(3\overline{|a|} \overline{|b^2|} \overline{\left| \frac{1}{\gamma}
\right|} \right)\max \{\log\alpha,\log\overline{|\eta_2|} \}/ R.
\]
We have $$\gamma^{(1)} \varepsilon^{(1)}+\gamma^{(2)}\varepsilon^{(2)}=2b\beta^m,$$
hence
\[
\left|1+\frac{\gamma^{(2)}}{\gamma^{(1)}}\left(\frac{\eta^{(2)}_2}{\eta^{(1)}_2}
\right)^{u_2} \right| < \frac{4b}{\sqrt{|a|}} \alpha^{-2m}.
\]
If $m>m_0$, then $\displaystyle\frac{4b}{\sqrt{|a|}}\alpha^{-2m} <
\displaystyle \frac{1}{2}$ and so
\[
| \Lambda_1 | = \left| \arg\left(-\frac{\gamma^{(2)}}{\gamma^{(1)}}\right)+u_2
\arg\left(\frac{\eta^{(2)}_2}{\eta^{(1)}_2}\right)+u_0\pi \right| < \frac{4.1
| b |}{\sqrt{| a |}}\alpha^{-2m},
\]
with $u_0{\in}\Z$ and $-\pi \leq \arg(z) \leq \pi$ for every $z \in \C.$ The
last inequality yields $| u_0 | < | u_2 | +2.$ \\[1ex]
We can set in Theorem 3 \\
$n=3, D=4, g=1$ \\
$\log A_1=h\left(\displaystyle\frac{\gamma^{(2)}}{\gamma^{(1)}}\right),
\log A_2=h\left(\displaystyle\frac{\eta^{(2)}_2}{\eta^{(1)}_2}\right),
\log A_3=\displaystyle\frac{1}{2}$ \\
$E=e, M=4(| u_2 | +1)$ \\
$Z_0=7+3\log 3, G_0=\log M$ \\
$U_0=4^5(7+3\log 3)\displaystyle\frac{1}{2}h\left(\displaystyle
\frac{\gamma^{(2)}}{\gamma^{(1)}}\right)
h\left(\displaystyle\frac{\eta^{(2)}_2}{\eta^{(1)}_2}\right)\log M, $ \\
and get
\[
| \Lambda_1 | > \exp\left\{-2\cdot10^{16}h\left(\frac{\gamma^{(2)}}{\gamma^{(1)}}\right)
h\left(\frac{\eta^{(2)}_2}{\eta^{(1)}_2}\right) \log M \right \}.
\]
Comparing the lower and upper bounds for $| \Lambda_1 |$ we conclude
\begin{equation}
\label{20}
2m\log\alpha -\log\frac{4.1|b|}{\sqrt{|a|}}< 2\cdot10^{16}
h\left(\frac{\gamma^{(2)}}{\gamma^{(1)}}\right)h\left(\frac{\eta^{(2)}_2}
{\eta^{(1)}_2}\right) \log\left(\frac{16m\log^2\alpha}{R} +4c_1 +4\right).
\end{equation}
This inequality yields an upper bound for $m$, which we shall only compute
knowing the actual values of the accouring parameters. \\[1ex]
Let now $c<0$ (and $c'<0$). Then all conjugates of $\L$ are real and there
exist $u_1,u_2,u_3 \in \Z$ with $\varepsilon = \eta^{u_1}_1
\eta^{u_2}_2\eta^{u_3}_3.$ We recall $\eta_1=\alpha.$ The estimations
(\ref{17a}) with $i=1,2$ and (\ref{17b}) yield
\[
| u_i | < \frac{4m\log^2\alpha \log h}{R}+c_2, \ i=2,3
\]
with $c_2=3\sqrt3 \log(3\overline{|a|} \overline{|b^2|} \overline{|\frac{1}{\gamma}|})
\log\alpha \log\overline{|\eta_2|} \log \overline{|\eta_3|}/R$ and
$h=max\{\overline{|\eta_2|},\overline{|\eta_3|}\}.$ \\
Similarly to the above case, but working with real instead of complex
logarithms we get
\[
| \Lambda_2 | =\left| \log \left| \frac{\gamma^{(2)}}{\gamma^{(1)}} \right|
+u_2 \log \left| \frac{\eta^{(2)}_2}{\eta^{(1)}_2} \right| + u_3 \log \left|
\frac{\eta^{(2)}_3}{\eta^{(1)}_3}\right|\right| < \frac{5.6|b|}
{\sqrt{| a |}} \alpha^{-2m}.
\]
The parameters in the application of Waldschmidt's theorem are the same as
earlier except that
$\log A_3= h\left(\displaystyle\frac{\eta^{(2)}_3}{\eta^{(1)}_3}\right),$
$M=2 | u_2 |$ and\\ $U_0=4^5(7+3\log3) h\left(\displaystyle\frac{\gamma^{(2)}}
{\gamma^{(1)}}\right)h\left(\displaystyle\frac{\eta^{(2)}_2}{\eta^{(1)}_2}\right)
h\left(\displaystyle\frac{\eta^{(2)}_3}{\eta^{(1)}_3}\right) \log M.$
Hence Theorem 3 implies
\begin{equation}
\label{21}
2m\log\alpha-\log\frac{5.6|b|}{\sqrt{| a |}} < 4.10^{16} h\left(\displaystyle
\frac{\gamma^{(2)}}{\gamma^{(1)}}\right)h\left(\displaystyle\frac{\eta^{(2)}_2}
{\eta^{(1)}_2}\right)h\left(\displaystyle\frac{\eta^{(2)}_3}
{\eta^{(1)}_3}\right) \log\left(\frac{8m\log^2 \alpha \log h}{R} +
2c_2\right).
\end{equation}
\begin{rem}
The argument of this section can be easily generalized to the case when $\alpha$
is not a unit. Then we have to apply lower bounds for linear forms in p-adic
logarithms. 
\end{rem}
\section{Proof of Theorem 1}
In table 1. we are listing the data necessary for the application of the method
of section 3 to the equations given in Lemma 1. We are using the notations
$\alpha = 5+2\sqrt6,$ $\beta=5-2\sqrt6$ and $\vartheta=\sqrt{-c}.$\\ [1ex]
\begin{tabular}{rccccccc}
$\Delta$&a&b&c&r&$\eta_2$&$\eta_3$&$\gamma$\\[1ex]
1&1&$\beta^2$&$-12+5\sqrt{6}$&2&$5-2\vartheta-2\sqrt{6}$&&1\\
-2&-$\alpha$&1&$-(6+2\sqrt{6})$&3&$1+\vartheta$&$2+\vartheta-\frac{\vartheta}
{2}$&1\\
3&$\alpha$&1&$4+2\sqrt{6}$&2&$1+ \vartheta$&&1\\
-6&-1&1&-2&3&$\sqrt{2+\sqrt{3}}$&$\frac{(1+\sqrt{3})(2+\sqrt{2})}{2}$&$\sqrt{2}+\sqrt{3}$\\
\multicolumn{8}{c}{Table 1.}
\end{tabular}\\[2ex]
We are giving the details of the proof only for the case $\Delta = 1,$ when our 
equation has by Lemma 1 the form 
$$\alpha^{2m}-\beta^2\beta^{2m} = 4\beta\sqrt{6}w^2=4(5\sqrt{6}-12)w^2.$$
It is easy to see that it has only one solution $(m,w)=(0,1)$ in the range 
$ 0 \le m \le 10 .$ If $m>10$ then (10) is obviously true. Thus we may 
assume in the sequel $m>10.$
The algebraic number field $\L=\Q(\sqrt{6},\sqrt{12-5\sqrt{6}}),$ has two real 
and two non-real conjugates. Its regulator is $R = 6.83836$ and we get 
$$ |u_2| < 3.07398 m + 8.95847. $$

As $\gamma=1$ there are only two summands in $\Lambda_1,$ actually it has the 
form
$$\Lambda_1=\left| u_2 \arg \left( \frac {5-2\sqrt{6}+2\sqrt{12-5\sqrt{6}}} 
{5-2\sqrt{6}-2\sqrt{12-5\sqrt{6}}} \right) +u_0 \pi \right| < 
0.042 \alpha^{-2m}.$$
As we proved, there are generaly three logarithms in $\Lambda_1,$ but in the 
actual example we have only two, therefore in the, to (14) analogous inequality
we get a much better constant. More precisely we have
$$4.58486 m + 3.17387 < 6.81595 \cdot 10^{11} \log(12.4 m + 40), $$
which implies $ m < 5 \cdot 10^{12} $ and $|u_2| < 1.55 \cdot 10^{13}.$
Dividing the inequality for $\Lambda_1$ by $u_2 \pi$ we see that, as $m>10$, 
$u_0/u_2$ is a convergent of 
$$\arg\left(\frac{\eta_2^{(2)}}{\eta_2^{(1)}}\right)/\pi= \delta=
 .93557845273700309088141600367180617252445255312155. $$
The denominator of the 26-th convergent of $\delta$, 
$\displaystyle\frac{ 51706546491839}{ 55266927472061},$ is larger then $10^{14},$ hence 
$$|u_0-u_2\gamma| \ge | 51706546491839 -  55266927472061\gamma|>
 .16132\cdot10^{-13}, $$ which implies $m \le 5.$ Thus our equation has only
the trivial solution. 

\bigskip

The proof Theorem 1 is similair in the other cases. We may 
always set $m_0=10$ and the upper bound for $m$ computed from (\ref{20})
or (\ref{21}) depending on the value of $r$ is in all cases less then
$10^{20}$. To fill the gap between 10 and $10^{20}$ we can use the above 
reduction procedure, originally due to Baker and Davenport \cite{BD}.\\
The solution $(-1,\pm 1,0)$ of (\ref{1}) comes from the equation $\displaystyle \frac{\alpha^m-\beta^m}{4\sqrt6}=w^2$ for even
exponents, and is $n=w=e=0.$ The other solutions given in Theorem 1 follow from
the solutions of the equations in Lemma 1, which are
$(\Delta,n,e)=(1,0,2);(-2,0,1);(3,0,1);(-6,0,1)$ and $(-6,1,49).$


\begin{thebibliography}{99}
\bibitem{Ba} A. Baker, {\em Bounds for the solutions of the hyperelliptic
equation}, Proc. Cambridge Philos. Soc. {\bf 65} (1969), 439--444.
\bibitem{BD} A. Baker and H. Davenport, {\em The equations $3x^2-2=y^2$ and
$8x^2-7=z^2$,} Quart. J. Math. Oxford, {\bf 20} (1969), 129--137.
\bibitem{Co} J.H.E. Cohn {\em Squares in some recurrent sequences,} Pacific
J. Math. {\bf 41} (1972), 631--646.
\bibitem{DR} W.L. McDaniel and P. Ribenboim, {\em Squares and double-squares in
Lucas sequences,} C.R. Math. Rep. Acad. Sci. Canada {\bf 14} (1992), 104--108.
\bibitem{M1} M. Mignotte, {\em Su una classe di equazioni del tipo
$a^n + b^n = z^2$}, Rend. del Sem. Univ. Cagliari, {\bf 62} (1992) fasc. 1.
\bibitem{MP1} M. Mignotte et A. Peth\H{o}, {\em Sur les carr\'es dans certaines
suites de Lucas}, J. Th\'eorie des Nombres de Bordeaux, {\bf 5} (1993), 333--341.
\bibitem{MP2} M. Mignotte \'es A. Peth\H{o}, {\em Az $a^n+b^n=z^3$ diofantoszi
egyenletr\H{o}l}, Matematikai Lapok, to appear.
\bibitem{Mo} L.J. Mordell, {\em Diophantine equations}, Academic Press, 1969.
\bibitem{P} A. Peth\H{o}, {\em Perfect powers in second order linear
recurrences,} J. Number Theory, {\bf 15} (1983), 117--127.
\bibitem{ShSt} T.N. Shorey and C.L. Stewart, {\em On the Diophantine
equation $ax^{2t} + bx^ty + cy^2 = d$ and pure powers in recurrences,}
Math. Scand., {\bf 52} (1983), 24--36.
\bibitem{Si} C.L. Siegel, {\em The integral solution of the equation $y^2=ax^n+
bx^{n-1}+...+k$}, J. London Math. Soc. {\bf 1} (1926), 66--68.
\bibitem{W} M. Waldschmidt, {\em Minorations de combinaisons lin\'eaires de
logarithmes de nombres alg\'ebriques,} Journal Canadien de Math\'ematiques,
{\bf 45} (1993), 176--224.
\end{thebibliography}
\end{document}


