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%         On the family of Thue equations...
%                       by
%   M. Mignotte, A. Peth\H{o} and F. Lemmermeyer
%   Reference No.:ACTARITH 2811
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\centerline{ On the family of Thue equations}

\vskip 4mm

\centerline{$x^3-(n-1)x^2 y-(n+2)xy ^2 -y^3=k$}

\vskip 8mm

\centerline{by  M. Mignotte, A. Peth\H{o} \footnote{$^*$}{Research supported 
in part by Hungarian National Foundation for Scientific Research Grant 
No. 16791/95.} and F. Lemmermeyer}

\vskip 12 mm
{\bf Abstract :} { We study the double family of
Thue equations ${\scriptstyle x^3-(n-1)x^2 y-(n+2)xy ^2
-y^3=k}$.}

\vglue 25 mm

\beginsection
1. Introduction

The family of cubic Thue equations
$x^3-(n-1)x^2 y-(n+2)xy ^2 -y^3=\pm 1$, with $n\ge 0$, was
studied by E.~Thomas. He proved in [T1] that this family of
equations has only ``trivial solutions'' except for a finite
number of values of the parameter~$n$, explicitly for $n<10^8$.
Then, M. Mignotte [M] could solve completely this family of
equations and proved that non-trivial solutions occur only for
$n\le 3$.

\medskip

Then, other infinite families of Thue equations $F(x,y)=\pm 1$
were studied. In the cubic case there are works of Thomas [T2],
Mignotte and Tzanakis [MT]; in these cases the family was completely
solved except for an explicit finite range for the parameter. In the
quartic case, there are works of Peth\H{o} [P2], Lettl and Peth\H{o} [LeP] and
Mignotte, Peth\H{o} and Roth [MPR], where the last two studies were completely
finished.

\medskip

Here, we come back to the family of cubics investigated by E.~Thomas,
the case where $F_n(x,y)=x^3-(n-1)x^2 y-(n+2)xy ^2 -y^3$, but
we add a new parameter: the number $k$ on the right hand side. In
other words, we study a family of cubic diophantine equations which
depend on two parameters.

Our results are explicit bounds for the integer solutions in terms of
the two parameters $n$ and~$k$; and we give a special study to the
case $x^3-(n-1)x^2 y-(n+2)xy ^2 -y^3 = \pm (2n+1)$. This case has
some interest since we prove that the diophantine equation
$F_n(x,y)=k$ has no solution in the range $1<\vert k\vert <2n+1$,
except when $k$ is a cube (the existence of a solution in this case
is trivial).

We formulate now the three theorems, which are the main results of the
present paper.

\proclaim Theorem 1.
Let $n \ge 1650, k$ be integers. If
$$|x^3-(n-1)x^2y-(n+2)xy^2-y^3|=k$$
holds for some $x,y \in {\bf Z}$, then
$$\log|y|<c_1 \log^2(n+2)+c_2 \log \; n \log \; k,$$
where
$$
c_1=700+476.4 \left( 1- {1432.1 \over n} \right)^{-1} \left(1.501 -
{1902 \over n}\right) < 1956.4
$$
and
$$
c_2=29.82+ \left(1- {1432.1 \over n}\right)^{-1} {1432 \over n \; \log
\; n} < 30.71.
$$
\rm

\bigskip
It is well known, see [GyP] and the references therein, that the general
estimates of the solutions of  a Thue equation depend polynomially in the
constant
term, but exponentially with respect to the height of the form itself, i.e.
the
maximum of the absolute value of the  coefficients of the form
corresponding to the
Thue equation. In the present case, the dependence is polynomial in $k$ and
``nearly'' polynomial in the height of $F_n(x,y)$ (which is $n+2$).
Moreover the
constants appearing here are much better than in other similar results, see
e.g. [BSt].

\medskip

To illustrate how sharp is the estimate of Theorem 1 let us take $n=1650$
and $k=10^9$, then we get

$$|y|<10^{48698}.$$
This is certainly a bound which is reachable with the present numerical
techniques, and in any case such a bound is much smaller than previous
bounds
obtained for cubic Thue equations.

\medskip

From Theorem 1, we are able to derive an effective improvement of Liouville's
inequality for the rational approximations  of the zeros of $F_n(x,1)$.

\proclaim Theorem 2.
Let $\lambda$ be one of the zeros of $F_n(x,1)$. If $n \ge 1650$ and
$\lambda \in
]-1,0[$ or $\lambda \in ]-2,-1[$ then
$$
\left|\lambda - {x \over y}\right|
> (n+2)^{-c_3} y^{- 3+1/(c_2 \; \log \; n)}
$$
and if $\lambda \in (n,n+1)$ then
$$
\left|\lambda - {x \over y}\right|>
(n+2)^{-c_3-1} y^{-3+1/(c_2 \;\log \; n)},
$$
holds for all $(x,y)\in {\bf Z}^2, y \not = 0,$ where
$$c_3= c_1 \cdot {\log (n+2) \over \log \; n} + 1 < 64.72.
$$
\rm

\bigskip

We are dealing in the second part of the paper with the inequality

$$|x^3-(n-1)x^2y-(n+2)xy^2 -y^3| \le 2n+1 \eqno(1).$$
Combining the method of the proof of Theorem 1 with a result of Lemmermeyer
and Peth\H{o} [LP] we are able to solve the above inequality completely and
prove the following.

\proclaim Theorem 3.
Let $n$ be a nonnegative integer. If $(x,y) \in {\bf Z}^2$ is a solution of
equation $(1)$, then either
$(x,y)=t(u,v)$ with an integer $0 \le |t| \le \root 3\of{2n +1}$ and $\pm
(u,v) \in \{ (1,0), (0,1), (-1,1) \}$
or $\pm (x,y) \in \{ (-1,-1), (-1,2), (-1,n+1),
(-n,-1), (n+1,-n), (2,-1) \}$,
except when $n=2$, in which case (1) has the
extra solutions: $\pm (x,y) \in \{ (-4,3), (8,3),$ $(1,-4), (3,1), (3,-11)
\}.$
\rm

\vfill \eject

The plan of this paper is the following. Section~2 contains two general
lemmas.
The next section is devoted to a pure numerical study of the roots of the
polynomials $x^3-(n-1)x^2-(n+2)x-1$; these estimates are used in the linear
forms
of logs which are studied in the sequel. Section~4, which is short and
easy, contains the proof of Theorem~2 assuming Theorem~1.

The next section is an
instance of the classical Siegel-Baker's reduction of a Thue equation to a
linear form
in logs. This study is detailed in order to lead to sharp estimates; here
we get
linear forms in three logs. Then, the linear form in logs obtained in
Section~5 is
studied a first time in Section~6. In this study, this linear form is
considered as
a linear form in two logs (we just group two terms). To get sharp
estimates, we
apply the general result of Laurent-Mignotte-Nesterenko. Of course, this
implies a
tedious reconstruction of the proof of the lower bound of this linear form;
for example we have to choose the parameters of the ``auxiliary function''
(more exactly, of the interpolation determinant): this is the price to pay.
The conclusion of this section is an upper bound on~$y$, in terms of the
heights of the terms of the linear form. The very long Section~6 contains 
a second study of this linear form in ``two'' logs; first, we prove an 
estimate of the coefficients of this linear form in terms of~$y$, then we 
bound these coefficients and (after some
computation) at the end we get an explicit upper bound of~$y$, in terms of
the initial parameters~$n$ and~$k$. This ends the proof of Theorem~1.

Section~8,
which is short and easy, give a simple upper bound of~$y$ depending only
on~$n$, when $k< n^4$; this result is used in the next section.

The last two sections deal with the proof of Theorem~3. Here, a result of 
Lemmermeyer and Peth\H{o} [LP] plays an essential role. Section~9
contains a study of the special linear form in ``two'' logs which occur in
this
case; this leads to the proof of Theorem~3 for~$n>1700$. To cover the range
$n \le 1700$, in Section~10 we consider --- for the first time --- the linear
form in
logs as a linear form in three logs. This study is classical and we can
easily finish the proof of Theorem~3.

\vfill \eject

\beginsection
2. Preliminary lemmas

This section contains two lemmas on cubic fields and cubic forms
which have a general interest.

\medskip

\proclaim Lemma~1. [P1, Theorem 2.]
Let $f(X,Y)= X^3 +bX^2Y+cXY^2+dY^3$ be a cubic form with positive
 discriminant $ D_f$, and
suppose that $f(x,y)=k$, where $x$, $y$ are rational integers, $y\ne
0$. Let $\alpha_i$, $i=1$, 2, 3, be the roots of the polynomial
$f(X,1)$.
 Put $L_i = \alpha_i - {x \over y}$, $i=1$, 2, 3, and suppose
that $|L_1| \le |L_2| \le |L_3|$; then
$$
|L_1|^4 |L_2|^2 \le {4 |k|^4 \over D_f \, |y|^{12}}.
$$
\rm

{\bf Proof}
Notice the formula
$$
D_f = \prod_{i<j} (\alpha_i-\alpha_j)^2
= L_3^4 L_2^2
\left(1-{L_2\over L_3}\right)^2
\left(1-{L_1\over L_3}\right)^2
\left(1-{L_1\over L_3}\right)^2 \!.
$$
Since the function
$$
g(u,v)=(1-u)\,(1-v)\left( 1 - {u\over v}\right) $$
satisfies $0\le g(u,v) \le 2$ on the domain $-1\le u, v \le 1$, $|v|\le
|u|$, we have
$$
D_f \le 4 \, L_3^4 L_2^2.
$$
Multiplying both sides of this inequality by $L_2^2 L_1^4$
and using the relation
$L_1 L_2 L_3=k/y^3$, we get the result.
\medskip

\proclaim Corollary 1.
Using $|L_1|\le |L_2|$, we get
$$
|L_1| \le \left({ 4 k^4 \over D_f} \right)^{1/6}\times {1\over |y|^2}.
$$
\rm

\bigskip

\proclaim Lemma 2.
Let $\alpha \in K$, where $K$ is a cubic Galois field.
Denote by $\alpha$, $\alpha'$ and $\alpha''$ the conjugates of
$\alpha$. Choose such an order that these conjugates satisfy
$|\alpha_1|\ge |\alpha_2| \ge |\alpha_3|$.  Then the measure of
the quotient $\alpha/\alpha'$ satisfies
$$
M (\alpha/\alpha') \le |\alpha_1|^2 |\alpha_2| .
$$
\rm
\medskip
{\bf Proof}
Let $k$ be the norm of $\alpha$. Consider the polynomial
$$
k(X-\alpha/\alpha')\,(X-\alpha'/\alpha'')\,(X-\alpha''/\alpha)
= (\alpha' X-\alpha)\,(\alpha'' X-\alpha') \,(\alpha X-\alpha''),
$$
clearly, this polynomial has rational integer coefficients,
and $\alpha/\alpha'$  is a root of it. Thus,
$$
\eqalign{
M(\alpha/\alpha') &\le |k|\,\max \{1, |\alpha/\alpha'|\}
\, \max \{1,|\alpha'/\alpha''|\}\, \max \{1,|\alpha''/\alpha|\}
\cr
&= k \,{|\alpha_1|\over |\alpha_3|} = |\alpha_1|^2 |\alpha_2|.}
$$
\medskip


{\bf Remark}.--- It is easy to verify that the previous lemma can be
generalized in the following way:
Let $\alpha \in K$, where $K$ is a cyclic Galois field of degree $d$.
Let $\alpha'$ be some conjugate  of
$\alpha$, with $\alpha '\ne \alpha $. Choose an order on the set
$\alpha _1$, \dots,~$\alpha _d$ of the conjugates of~$\alpha $ for
which  $|\alpha_1|\ge |\alpha_2| \ge \cdots \ge  |\alpha_d|$.  Then
the measure of the quotient $\alpha/\alpha'$ satisfies
$$
M (\alpha/\alpha') \le
|\alpha_1|^{d-1} |\alpha_2|^{d-2}\cdots |\alpha _{d-1}| .
$$


\vfill \eject


\beginsection 3.
Numerical study of the roots of $x^3-(n-1)x^2-(n+2)x-1$

In this section, we gather several sharp estimates for the roots of
the polynomial~$f$ associated to the cubic form
$$
F(x,y)= F_n(x,y)=x^3-(n-1)x^2 y-(n+2)xy ^2 -y^3,
$$
that is
$$
f(x)=f_n(x)=x^3-(n-1)x^2-(n+2)x-1.
$$
From now on, we shall keep these definitions for $f$, $f_n$, $F_n$
and~$F$.

\medskip
Since $f(-2)<0$, $f(-1)=1$, $f(0)=-1$, $f(n)=-2n-1$ and
$f(n+1)=n^2+n+1$, this polynomial has three real roots, say
$\lambda=\lambda^{(1)}$, $\lambda^{(2)}$, $\lambda^{(3)}$,
and we may suppose  that
$$
\lambda \in \left]-1,0 \right[, \qquad
\lambda^{(2)} \in \left] -2,-1\right[,
\qquad \lambda^{(3)} \in \left] n, n+1\right[ .
$$
Moreover, the polynomial $f$ is invariant under the change of variable
$\sigma  : x \mapsto -1/(1+x)$, which satisfies
$\sigma ^2 (x)= -(1+x)/x$ and
$\sigma ^3(x)=x$. This proves that the field ${\bf Q} (\lambda )$
is a Galois field
and that
$$
\lambda^{(2)} = - {1\over \lambda+1},
\quad {\rm and} \quad \lambda^{(3)} = - \left( 1+{1\over
\lambda}\right).
$$
More generally,
$$
\sigma ^j \bigl(\lambda ^{(i)}\bigr) = \lambda ^{(k)},\qquad
\hbox{where \ $k \equiv i+j$  mod 3}.
$$

\medskip

Since
$$
f_n \!\left( n+{\textstyle {2\over n}}\right) = 1 + {4 \over n} + {4
\over n^2} + {8 \over n^3}, $$ we see that
$$
\lambda^{(3)} < n+ {2 \over n}.
$$
A second computation gives
$$
f_n \left( n+{\textstyle {2\over n+1}} \right) = -1 + {4\over n+1} -{4
\over (n+1)^2 } + { 8\over (n+1)^3} , $$
which shows that
$$
\lambda^{(3)} > n+ {2\over n+1}
$$
for $n>2$. In the sequel, we suppose $n\ge 3$, and the two cases $n=1$ 
and $n=2$ will be considered separately.

In terms of $\lambda^{(3)}$, we have
$$
\lambda^{(3)} = - \left( 1+{1\over \lambda} \right)
\ \Longrightarrow \ \lambda = - {1\over \lambda^{(3)}+1},
$$
and
$$
\lambda^{(2)} = - {1\over 1+\lambda}
= -{1\over 1 - {\displaystyle {1\over \lambda^{(3)} +1}}}
= - {\lambda^{(3)} +1 \over \lambda^{(3)}}.
$$
These estimates imply (for $n\ge 3$)
$$
- { n+1 \over n^2+2n+3 } = - {1 \over 1+n+{2 \over n+1}}
< \lambda < - {1 \over 1+n+ {2 \over n}} = - {n \over n^2+n+2},
$$
and
$$
-1 - {n+1\over n^2+n+2} <
\lambda^{(2)} = - \left( 1 +{1\over \lambda ^{(3)}} \right)
< -1 - {n\over n^2+2}.
$$
These estimates lead to estimates for logarithms. Indeed,
$$
\eqalign{
\log n +{2\over n(n+1)} - {2\over \bigl(n(n+1)\bigr)^2}
< \log \left( n+{2\over n+1} \right) &<
\ell_3 := \log \lambda^{(3)} < \log \left(n+{2\over n}\right)
< \log n + {2\over n^2},
\cr
{n-{1\over 2}\over n^2+2}<
{n\over n^2+2} - { n^2\over 2\bigl(n^2+2\bigr)^2}
&< \ell_2 := \log \left| \lambda^{(2)} \right| < {n+1 \over n^2+n+2},
}
$$
and
$$
- \log (n+1) - {2\over n(n+1)}<
\ell_1 := \log |\lambda| < - \log (n+1) - {2 \over (n+1)^2 }
+{2\over (n+1)^4}.
$$

\vfill
\eject

\beginsection 4.
Proof of Theorem 2.

Assuming that Theorem 1 is true we now give the proof of Theorem 2
because it is short and depends only on the estimates of section 3. 
We consider only the case $\lambda \in ]1,0[.$ The proofs of 
the other cases are similar.
\def\Z{{\bf Z}}
Assume that there exists $(x,y) \in \Z^2$, $y \not=0$ such that
$$
\left|\lambda - {x \over y}\right|
\le (n+2)^{-c_3} y^{- 3+1/(c_2 \;\log \; n)}.
$$
Let $\lambda_2$ and $\lambda_3$ denote the other zeros
of $F_n(x,1)$ such that $\lambda_2 \in ]-2,-1[$ and $\lambda_3 \in
]n,n+1[.$ Then
$$
\left|\lambda_i - {x \over y}\right|
\le |\lambda- \lambda_i| + (n+2)^{-c_3}
$$ 
for $i=2,3.$ Using these inequalities and the
estimates of Section 3, one can easily prove that
$$
|\lambda_2y - x||\lambda_3y - x| < (n+2) |y|,
$$
hence
$$
|F_n(x,y)| < (n+2)^{-(c_3-1)} |y|^{1/(c_2 \; \log \; n)} .
$$
Putting $k=(n+2)^{-(c_3-1)} |y|^{1/(c_2 \; \log \; n)}$,
Theorem 1 implies
$$
\eqalign{
\log |y| & <  c_1 \log^2(n+2) + c_2 \log n \log k
\cr
& <   c_1 \log^2(n+2) + c_2 \log n \left( -c_1 {\log^2 (n+2) \over
c_2 \log n} + { \log|y| \over c_2 \log n} \right)
\cr
& =  \log |y|}
$$
a contradiction, and the Theorem is proved.

\beginsection
5. Reduction of the Thue equation $F_n (x,y)=k$

Let $x$, $y$ be rational integers, $y>0$, such that
$$
F_n(x,y) = x^3-(n-1)x^2y -(n+2)xy^2 -y^3
= \pm k, \qquad k>0.\eqno (2)
$$
The condition $k>0$ is not restrictive since $F(-x,-y)=-F(x,y)$.

We know (see [T1]) that $\lambda$, $\lambda^{(2)}$
constitute a fundamental system of units for the order ${\bf Z}
[\lambda]$.

\medskip
The following lemma is useful for the reduction of the Thue equation
in the case $k>1$.
\medskip

\def\bR{{\bf R}}
\proclaim Lemma 3.
Let $0\ne \beta  \in {\bf Z} [\lambda ]$ and let $0<c_1$, $c_2\in \bR$. Then,
there exist rational integers $a_1$ and $a_2$ and some element $\gamma
\in {\bf Z} [\lambda]$ such that
$$
\beta = \gamma \lambda^{a_1}
(\lambda^{(2)})^{a_2}
$$
with, for $n\ge 3$,
$$
c_i \le \bigl| \gamma  ^{(i)}| < (n+3) c_i,
\qquad
\hbox{$i=1$, 2,}
$$
and
$$
{|N (\beta )| \over (n+3)^2 c_1c_2}
<|\gamma  ^{(3)}| \le  {|N (\beta )| \over  c_1c_2} .
$$
\rm

{\bf Proof}
Put
$$
\tilde c_i = c_i \exp \biggl\{
{1\over 2}\bigl| \log |\lambda ^{(i)}| \bigr|
+ {1\over 2}\bigl| \log |\lambda ^{(i+1)}| \bigr|
\biggr\}, \qquad \hbox{$i=1$, 2,}
$$
and consider the system of equations
$$
\log \bigl| \beta ^{(i)}/\tilde c_i \bigr|
= r_1 \log |\lambda ^{(i)}| + r_2 \log |\lambda ^{(i+1)}|,
\qquad \hbox{$i=1$, 2.}
$$
As the vectors
$(\log |\lambda ^{(1)}|, \log |\lambda ^{(2)}|, \log |\lambda ^{(3)}|)$
and
$(\log |\lambda ^{(2)}|, \log |\lambda ^{(3)}|, \log |\lambda ^{(1)}|)$
are linearly independent in $\bR ^3$, this system has a unique real
solution, $r_1$ and $r_2$.
Put $r_1=a_1+\theta_1$
and $r_2=a_2+\theta_2$, where $a _1$, $a_2 \in {\bf Z}$, and
$|\theta_1|$, $|\theta_2|\le 1/2$.
Then consider the algebraic
number $\gamma = \beta \lambda^{-a_1} (\lambda ^{(2)})^{-a_2}$.
Then  $|N \gamma |= |N \beta |$ and $\gamma  \in {\bf Z} [\lambda ]$,
and we get
$$
\bigl| \log |\gamma  ^{(i)}/\tilde c_i|\bigr|
\le  {\textstyle {1\over 2}}
\bigl( \bigl|\log |\lambda ^{(i)}|\bigr|+\bigl|\log |\lambda
^{(i+1)}|\bigr|\bigr),
\qquad \hbox{$i=1$, 2,}
$$
which implies
$$
\log c_i
< \log |\gamma  ^{(i)}|
\le
\log c_i +
\bigl( \bigl|\log |\lambda ^{(i)}|\bigr|+\bigl|\log |\lambda
^{(i+1)}|\bigr|\bigr),
\qquad \hbox{$i=1$, 2.}
$$
By the estimates of the previous section, for $n\ge 3$,
$$
\eqalign{
\bigl| \log |\lambda ^{(1)}|\bigr|+\bigl|\log |\lambda ^{(2)}|\bigr|
&= \log (\lambda ^{(3)}+1)+\log \left(1+{1\over \lambda ^{(3)}}\right)
= \log \left(\lambda ^{(3)}+2+{1\over \lambda ^{(3)}}\right)
\cr
&< \log \left(n+{2\over n}+2+{n\over n^2+2}\right)
< \log (n+3),
}
$$
and
$$
\bigl| \log |\lambda ^{(2)}|\bigr|+\bigl|\log |\lambda ^{(3)}|\bigr|
=\log (\lambda ^{(3)})+\log \left(1+{1\over \lambda ^{(3)}}\right)
= \log (\lambda ^{(3)}+1) < \log (n+2).
$$
This proves the two first inequalities. The third follows immediately.
\medskip

\proclaim Corollary 2.
Let $0\ne  \beta  \in {\bf Z} [\lambda ]$ with norm $k$.
Then, there exist rational integers $a_1$ and $a_2$ and some element
$\gamma \in {\bf Z} [\lambda]$ such that
$$
\beta = \gamma \lambda^{a_1} (\lambda^{(2)}) ^{a_2}
$$
with
$$
M \bigl( \gamma  ^{(2)}/\gamma  ^{(3)}\bigr) \le  k (n+3)^{5/2},
$$
and
$$
\Bigl| \log \bigl| \gamma  ^{(2)}/\gamma  ^{(3)}\bigr| \Bigr|
\le  {\textstyle {3\over 2}} \log (n+3) .
$$
\rm

{\bf Proof.}
Put $c_1=c_2 = \root 3 \of {k}/\sqrt{n+3}$. The application of Lemma~3
implies the existence of $\gamma  \in {\bf Z} [\lambda ]$, which is 
associated to $\beta $ and which satisfies
$$
\root 3\of k /(n+3) < \bigl|\gamma  ^{(3)}\bigr|
< \root 3\of k \cdot (n+3) .
$$
From Lemma 2, we get
$$
M \bigl( \gamma  ^{(2)}/\gamma  ^{(3)}\bigr) \le  k (n+3)^{5/2}.
$$
Finally, as
$$
\log \bigl|\gamma  ^{(2)}\bigr| - \log \bigl|\gamma  ^{(3)}\bigr|
\le  {1\over 3} \log k + \log (n+3)
- {1\over 3} \log k + {1\over 2} \log (n+3)= {3\over 2} \log (n+3)
$$
and
$$
\log \bigl|\gamma  ^{(2)}\bigr| - \log \bigl|\gamma  ^{(3)}\bigr|
\ge  {1\over 3} \log k - \log (n+3)
- {1\over 3} \log k - {1\over 2} \log (n+3)= - {3\over 2}\log (n+3),
$$
the statement is proved.

\bigskip


If $(x,y)$ is a solution of (2), then there exist $u_1$,
$u_2 \in {\bf Z}$,
$$
x-\lambda y = \gamma   \lambda ^{u_1}(\lambda ^{(2)})^{u_2}, \eqno (3)
$$
where $\gamma  $ satisfies the conditions of the preceding corollary.

We may assume (see Thomas [T1]) that
$$
\left| {x\over y}-\lambda \right|
\le
\left| {x\over y}-\lambda ^{(2)}\right|,\ \;
\left| {x\over y}-\lambda ^{(3)}\right|.
$$
Applying Lemma~1 a first time, we get
$$
|L_1| \le \left( 4 k^4 \over (n^2+n+7)^2 \right)^{1/6} {1 \over y^2}, $$
using the fact that the discriminant of the form $F_n$ is equal to
$(n^2+n+7)^2$.

\vfill \eject

We suppose that $n >0$ and
$$
y \ge \max\{3,k,n\}.
$$
Then the previous inequality implies $|L_1| \le {1 \over 7}$, thus
$$
|L_2|\ge \vert \lambda ^{(2)}-\lambda \vert - \vert L_1\vert > {3\over
4}, $$
and a second application of  Lemma~1 gives
$$
|x- \lambda y| \le \left({64 k^4 \over 9(n^2+n+7)^2}\right)^{\!1/4} \!
{1\over | y|^2}
\le {0.55 \over y}\le 0.19 . \eqno (4)
$$
Notice also that, for $n\ge 3$,
$$
\eqalign{y &\le \bigl| x - \lambda^{(2)} y \bigr| \le y \, 
\left(1+{2\over n^2}\right), \cr
y \left( n+{2 \over n} \right) &\le \bigl|x-\lambda^{(3)} y \bigr| \le
y \left(n +{4\over n}\right).} \eqno (5)
$$

\bigskip
Let us give the proofs of these two pairs of inequalities.
By the estimates of the preceding section, we have
$$
\lambda^{(2)}-\lambda = - {\lambda^{(3)} +1 \over \lambda^{(3)}}
+ {1\over \lambda^{(3)} +1}
=-1-{1\over \lambda^{(3)} (\lambda^{(3)}+1)} ,
$$
hence
$$
1+{1\over (n+2)^2} < \bigl| \lambda^{(2)}-\lambda\bigr|
< 1+{1\over n^2}.
$$
This leads to the estimate
$$
\eqalign{
y &\le \left(1+{1\over (n+2)^2}\right) y -{0.55\over y}
\le \vert \lambda -\lambda ^{(2)}\vert y -\vert x-\lambda y\vert \le
\bigl| x - \lambda^{(2)} y \bigr|
\cr
&\le \vert \lambda -\lambda ^{(2)}\vert y +\vert x-\lambda y\vert
\le {0.55\over y}+\left(1+{1\over n^2}\right) y
\le y \left( 1+{2\over n^2}\right).
}
$$
In the same way,
$$
n+{3\over n+1}<n+{2\over n+1}+{n+1\over n^2+n+2}<
\lambda^{(3)}- \lambda = \lambda^{(3)} + {1\over \lambda^{(3)}+1}
< n+{2\over n}+{n\over n^2+2}<n+{3\over n},
$$
which implies
$$
\left(n+{2\over n}\right) y <
\bigl| x-\lambda^{(3)} y \bigr| < \left( n+{4\over n}\right) y.
$$
Now, using (5), we can refine (4),
$$
\vert x-\lambda y\vert
= {k\over \vert x-\lambda ^{(2)}y\vert \cdot \vert x-\lambda ^{(3)}y\vert}
< {k\over ny^2}. \eqno (6)
$$
\vfill \eject

Considering the conjugates of (3), we get Siegel's identity
$$
\bigl( \lambda^{(3)} - \lambda \bigr) \gamma^{(2)}
(\lambda^{(2)})^{u_1} (\lambda^{(3)})^{u_2}
- \bigl( \lambda^{(2)} - \lambda\bigr) \gamma^{(3)}
(\lambda^{(3)})^{u_1} \lambda^{u_2}
= \bigl( \lambda^{(3)}-\lambda^{(2)} \bigr)
\gamma \lambda^{u_1} (\lambda^{(2)}) ^{u_2}.
$$
Dividing by $\bigl(\lambda^{(2)} - \lambda\bigr) \gamma^{(3)}
(\lambda^{(3)})^{u _1} \lambda^{u_2}$ gives
$$
{\bigl(\lambda^{(3)} - \lambda \bigr) \gamma^{(2)}
\over  \bigl(\lambda^{(2)} - \lambda \bigr) \gamma^{(3)}}
\left({\lambda^{(2)} \over \lambda^{(3)}}\right)^{u_1}
\left({\lambda^{(3)} \over \lambda}\right)^{u_2}
- 1
= { \lambda^{(3)}-\lambda^{(2)} \over \lambda^{(2)} - \lambda}
{\gamma \lambda^{u_1} (\lambda^{(2)}) ^{u_2}
\over \gamma^{(3)} (\lambda^{(3)})^{u _1} \lambda^{u_2}}.
$$
The right-hand side is nonzero and is also equal to
$$
{ \lambda^{(3)}-\lambda^{(2)} \over \lambda^{(2)} - \lambda}
{x-\lambda y \over x-\lambda ^{(3)}y},
$$
thus, by $(6)$ and $(5)$, its absolute value is
$$
\le
\left|{ \lambda^{(3)}-\lambda^{(2)}\over\lambda^{(2)}-\lambda}\right|
{k\over (n^2+2)y^3}.
$$
We notice that
$$
{ \lambda^{(3)}-\lambda^{(2)}\over\lambda^{(2)}-\lambda}
= {\lambda^{(3)}+{1\over \lambda^{(3)}+1}
\over -{\lambda^{(3)}+1\over \lambda^{(3)}}+{1\over \lambda^{(3)}+1}}
= -\lambda^{(3)} \cdot
{(\lambda^{(3)})^2+\lambda^{(3)}+1
\over (\lambda^{(3)})^2+\lambda^{(3)}+1}
= - \lambda^{(3)},
$$
hence, the considered right-hand side is
$$
\le {k\over ny^3}.
$$
Now put
$$
\Lambda =
u_1 \log \Bigl|{\lambda^{(2)} \over \lambda^{(3)}}\Bigr|
+ u_2 \log \Bigl| {\lambda^{(3)} \over \lambda} \Bigr|
+ \log \Bigl| {\gamma^{(2)} \over \gamma^{(3)}} \Bigr|
+ \log \Bigl |
{\lambda^{(3)}-\lambda \over \lambda^{(2)}-\lambda}\Bigr|.
$$
Using the inequality $\vert \log z \vert <1.5 \times \vert z-1\vert$,
which is true for $\vert z -1\vert <1/2$, we get
$$
0< |\Lambda|  <{3k \over 2ny^3}.
$$
The relation
$$
{\lambda^{(3)} - \lambda \over \lambda^{(2)} - \lambda}
= {-1-{1 \over \lambda} -\lambda \over -{1\over \lambda +1} -\lambda}
= \lambda+1 = -{1\over \lambda^{(2)}}
$$
combined with
$\lambda^{(3)} /\lambda = (\lambda^{(3)})^2 \lambda^{(2)}$
gives
$$
\Lambda = (u_1+u_2-1) \log \bigl| \lambda^{(2)}\bigr|
+ (2u_2-u_1) \log \bigl| \lambda^{(3)}\bigr|
+ \log \left| {\gamma^{(2)} \over \gamma^{(3)}} \right|.
$$

\beginsection 6.
A first study of the linear form in logs

The previous linear form $\Lambda$ can be written as
$$
\Lambda = V \ell_2 + v \ell_3
+ \log \left| \gamma^{(2)}\over \gamma^{(3)} \right|,
$$
where we have put
$$
V = u_1+u_2-1, \qquad v=-u_1+2 u_2.
$$
It satisfies
$$
|\Lambda | < {3k \over 2n y^3}.
$$
Grouping the last two terms we get a linear form of the type
(we want to use the
results of [LMN], so that we shall use the notations of this paper),
$$
\Lambda = b_1 \log \alpha_1 + b_2 \log \alpha_2,
$$
where
$$
b_1=V, \quad b_2 = 1, \quad \alpha_1=\lambda^{(2)},
\quad \alpha_2 = \left|(\lambda ^{(3)})^v \cdot
\gamma^{(2)}/\gamma^{(3)}\right|.
$$
Concerning $\alpha_1$, we have
$$
\bigl|\log |\lambda^{(2)}|\bigr| \le {1\over n}, \qquad
M(\alpha_1)= \bigl|\lambda^{(2)} \cdot \lambda^{(3)} \bigr| ,
$$
which implies
$$
 h(\alpha_1)= h(\lambda)\le {\textstyle {1\over 3} }\,\log(n+2) .
$$
By Corollary~2,
$$
\log \Bigl(M \bigl( \gamma^{(2)}/\gamma^{(3)} \bigr)\Bigr)
 \le \log k+ {5\over 2} \log (n+3).
$$
Thus,
$$
h(\alpha_2) \le {|v| \over 3} \log (n+2) + {1\over 3}
\,\Bigl( \log k + {\textstyle {5\over 2}} \log (n+3) \Bigr).
$$

\bigskip

For the convenience of the reader, we quote the main result
of [LMN].

\medskip
\proclaim
Theorem A.
Let $K$ and $L$ be integers $\ge 3$, $R_1$, $S_1$, $R_2$, $S_2$
positive integers. Let $\rho >1$ be a real number. Put $R=R_1+R_2-1$,
$S=S_1+S_2-1$,
$$
g = {1\over 4} - {KL\over 12RS}, \qquad
b = \bigl((R-1)b_2 + (S-1)b_1\bigr)
\left( \prod_{k=1}^{K-1} k! \right)^{-2/(K^ 2-K)} .
$$
Let $a_1$, $a_2$ be real numbers such that
$$
a_i \ge \rho |\log \alpha_i| - \log|\alpha_i|
+2D h(\alpha_i), \quad \hbox{$i=1 $, 2,}
$$
where $D=[{\bf Q}(\alpha_1,\alpha_2) \,:\, {\bf Q}]$. Suppose that
$$
\eqalign{
\# \bigl\{ \alpha_1 ^r \alpha_2^s \,;
\, 0\le r <R_1,\, 0\le s\le S_1 \bigr\}
&\ge L, \cr \# \bigl\{ b_1 r + s b_2 \,;\, 0\le r <R_2,\
0\le s <S_2 \bigr\} &> (K-1)L } \eqno \rm (7) $$
and that
$$
K(L-1) \log \rho +(K-1)\log 2- (D+1)\log (KL)
- D(K-1) \log b -gL(Ra_1+Sa_2) >0.\eqno \rm (8)
$$
Then,
$$
|\Lambda'| \ge \rho^{-KL+0.5}, \qquad {\rm where\ }\quad
\Lambda' = \Lambda \cdot
\max \left\{ {LRe^{LR|\Lambda|/(2b_1)}\over
2b_1}, {L Se^{LS|\Lambda|/(2b_2)}\over 2b_2} \right\}.
$$
\rm

\bigskip

In our case, we have $D=3$. The study above shows that one can take
$$
a_1 \ge  2 \log (n+2)+(\rho -1)/n
$$
and
$$
a_2 \ge \Bigl(2|v|+ 5 \Bigr) \log (n+3) + 2 \log k + (\rho -1)\ell_0,
$$
where $\ell_0 = |\log \alpha _2|$. For further reference, we notice that
(by Corollary~2)
$$
\ell_0 \le \left(|v|+ {\textstyle {3\over 2}} \right) \,\log (n+3) .
$$
\smallskip

The algebraic numbers $\alpha_1$ and $\alpha_2$ are multiplicatively
independent. In fact, assume that they are not. Then there exist non-zero
integers $r$ and $s$ such that

$$1=\alpha_1^r \alpha_2^s = \lambda^{(2)r} \lambda^{(3)sv}(\gamma^{(2)}/
\gamma^{(3)})^s.$$
Thus $\gamma^{(2)}/\gamma^{(3)}$ is a unit in ${\bf Z}_{\bf K}$ and can be
written as

$$\gamma^{(2)}/\gamma^{(3)}= \lambda^{(2)t_2} \lambda^{(3)t_3}$$
with suitable $t_2, t_3 \in {\bf Z}$. Combining the last two equations and
using
that $\lambda^{(2)}$ and $\lambda^{(3)}$ are multiplicatively independent
we get
$s(v+t_3)=0$ or $t_3=-v$. This implies

$$\Lambda = (V+t_2) \log|\lambda^{(2)}|,$$
and so either $\Lambda=0$ or $|\Lambda|> {1 \over 2n}.$
We already noticed that $\Lambda$ is nonzero, and the second
alternative contradicts
$$|\Lambda|<{3k \over 2ny^3}<{1 \over 2ny}.$$
The claim is proved.

\vfill
\eject

Thus, in any case, the first one of conditions $(7)$ holds if we suppose
$$
R_1 S_1 \ge  L,
$$
and the choices of these two parameters will satisfy this condition.
\smallskip

To satisfy the second of conditions $(7)$, we take
$$
R_2 = \left[\sqrt {(K-1) La_2/a_1}\right]+1, \quad
S_2 = \left[\sqrt {(K-1) La_1/a_2}\right]+1,
$$
so that $R_2S_2 > (K-1)L$.
\smallskip

Now, there are two possibilities: either the numbers
$rb_1+sb_2$, $0\le r<R_2$ and $0\le s<S_2$ are pairwise different
and then the second of conditions $(7)$ holds, or (since
$b_2=1$) we have $|V|\le S_2$.
\medskip

We also take
$$
K = [\mu ^2 L a_1 a_2 ] +1,
$$
where $0<\mu \le 0.5$ and $L$ will be chosen later.
Now we study the term $g(Ra_1+Sa_2)$.
Since $R=R_1+R_2-1$ and $S=S_1+S_2-1$,
$$
\eqalign{
g(Ra_1+Sa_2) &= \left({1\over 4}-{KL\over 12 RS} \right)
(Ra_1+Sa_2) \cr
&\le
{1\over 4} \bigl( (R_2-1) a_1 +(S_2-1) a_2\bigr)
+ {1\over 4} ( R_1 a_1 +S_1 a_2)
 -{KL\over 12}\left( {a_1\over (S_2-1)+S_1}
+ {a_2\over (R_2-1)+R_1} \right). }
$$
Then,  the definitions of $R_2$ and $S_2$ imply
$$
g(Ra_1+Sa_2) = {1\over 2}\sqrt {KLa_1a_2}
+{1\over 4}(R_1a_1+S_1a_2) -{KL\over 12}\left(
{a_1\over \sqrt {KLa_1/a_2}+S_1}
+ {a_2\over \sqrt {KLa_2/a_1}+R_1} \right).
$$
Now we use the inequality
$$
{a_1\over \sqrt {KLa_1/a_2}+S_1} \le
\left({a_1\over \sqrt {KLa_1/a_2}}\right)
\left(  1 - {S_1\over \sqrt {KLa_1/a_2}}\right)
= {\sqrt{a_1 a_2} \over \sqrt{KL}} - {a_2S_1 \over KL},
$$
and the similar one with $R_1$, and get
$$
g(Ra_1+Sa_2)
\le   {1\over 3}  \sqrt {KLa_1 a_2} +{1\over 3} (a_1R_1+a_2S_1).
$$
Choosing
$$
R_1 = \left[ \sqrt {La_2/a_1}\right] +1, \qquad
S_1 = \left[ \sqrt {La_1/a_2}\right] +1
$$
gives
$$
g(Ra_1+Sa_2) \le  {1\over 3}  \sqrt {KLa_1 a_2} +
{2\sqrt{La_1a_2} \over 3} + {a_1 +a_2\over 3}.
$$

By Lemma 9 of [LMN],
$$
b\le {(R-1) b_2 +(S-1) b_1\over K-1} \times
\exp\left\{ {3\over 2} - {\log (2\pi (K-1)/\sqrt e) \over K-1}
+ {\log K \over 6K(K-1)} \right\},
$$
where
$$
(R-1)b_2 +(S-1)b_1 \le
\Bigl(\sqrt{(K-1) La_2/a_1} +\sqrt{La_2/a_1} +1\Bigr)
+ \Bigl(\sqrt{(K-1) La_1/a_2} +\sqrt{La_1/a_2} +1\Bigr) V.
$$
These remarks show that condition $(8)$ holds if
$$
\eqalign{
K(L-1) \log \rho &+ (K-1) \log 2 - 4 \log (KL)
-3(K-1) H_0 - {3\over 2} (K-1) \cr
&+ \log \bigl( \pi (K-1) /\sqrt e\,\bigr) - {\log K \over 6K}
-{L \over 3} \sqrt{KLa_1a_2} - {2L^{3/2} a_1a_2 \over 3}
-{L(a_1+a_2)\over 3} \ge 0, }
$$
where we have put
$$
H_0 = \log \left( { \sqrt{La_2/a_1} \bigl( \sqrt{K-1} +1\bigr) 
+1 + \Bigl( \sqrt{La_1/a_2} \bigl( \sqrt{K-1} +1\bigr)+1\Bigr) 
V \over K-1}  \right).
$$
We have
$$
\eqalign{
&{ \sqrt{La_2/a_1} \bigl( \sqrt{K-1} +1\bigr) +1
+ \Bigl( \sqrt{La_1/a_2} \bigl( \sqrt{K-1} +1\bigr)+1\Bigr)
 V \over K-1}
\cr
&\qquad \le {K\over K-1}\times \left(\bigl(1+1/\sqrt{K}\bigr)
{\sqrt{La_2/a_1}  +V\sqrt{La_1/a_2} \over\sqrt{K}}
+ {V+1\over K}\right)
\cr
&\qquad \le {K\over K-1}\times \left(\Bigl(1+{1\over
\sqrt{K}}\Bigr) {1\over \mu}\Bigl({1\over a_1} +{V\over
a_2}\Bigr) + {V+1\over \mu^2La_1a_2}\right)
\cr
\cr
&\qquad \le {K\over K-1}\times {1\over \mu}\times
\left({1\over a_1} +{V\over a_2}\right)\left(1+{1\over
\sqrt{K}}
 + {1\over \mu L}\Bigl({1\over a_1} +{1\over a_2}\Bigr)\right).
\cr }
$$
We put $\theta  = \log \rho$. Choosing $\mu = 2/(3\theta  )$,
and using the definition $K=[\mu^2 L a_1 a_2]+1$, we get
$$
H_0 \le H := \log \left({K\over K-1}\right)
+ \log \left(1+ {1\over \sqrt{K}}
+{3\theta \over 2L} \Bigl({1\over a_1}
+{1\over a_2}\Bigr)\right)
+ \log ^+ \left({1\over\mu a_1} +
{V\over \mu a_2} \right),
$$
where, as usual, $\,\log ^+ x = \max \{0, \log x\}$.
Then, we see that condition~$(8)$ is satisfied when
$$
\Phi := {2 \over 3 \theta } \left(
{2(L-1) \over 3} - {2H\over \theta } - {L\over 3}
+ {2\log 2 \over 3 \theta } \right) %\cr
- {7 \log L \over a_1a_2 L }
- {3 \log \bigl(4a_1a_2 /(9 \theta ^2)\bigr) \over a_1a_2 L}
-{2 \sqrt{L}\over 3 \sqrt{ a_1a_2}}
- {1\over 3} \left( {1\over a_1} + {1\over a_2} \right) \ge 0.
$$
[We divided by $La_1a_2$.] Now, we choose
$$
L = \max \bigl\{16, [\varepsilon +6H/\theta ]+3 \bigr\},
$$
where $\varepsilon >0$ is to be chosen later.
This choice implies
$$
\eqalign{
\Phi &\ge {4\log 2\over 9 \theta ^2}
- {7 \log L \over a_1a_2 L }
- {3 \log \bigl(4a_1a_2 /(9 \theta ^2)\bigr) \over a_1a_2 L}
- {1\over 3} \left( {1\over a_1} + {1\over a_2} \right)
+ {4\varepsilon L \over 9 \theta }
-{2 \sqrt{L}\over 3 \sqrt{ a_1a_2}}
\cr
&\ge {4\log 2\over 9 \theta ^2}
- {1.22\over a_1a_2 }
- {3 \log \bigl(4a_1a_2 /(9 \theta ^2)\bigr) \over 16 a_1a_2}
- {1\over 3} \left( {1\over a_1} + {1\over a_2} \right)
+ \sqrt{L}\left({4\varepsilon \sqrt{L}\over 9 \theta }
-{2 \over 3 \sqrt{ a_1a_2}}\right).
\cr
}
$$
Thus, $\,\Phi \ge 0\,$ when both
$$
\Theta := {4\log 2\over 9 \theta ^2}
- {1.22\over a_1a_2 }
- {3 \log \bigl(4a_1a_2 /(9 \theta ^2)\bigr) \over 16 a_1a_2}
- {1\over 3} \left( {1\over a_1} + {1\over a_2} \right) \ge 0
$$
and
$$
\varepsilon \ge {3\theta \over 8 \sqrt{a_1a_2}}.
$$
We take $\rho =15$, and in this case $\Theta \ge 0$ for
$$
a_2\ge a_1\ge 15, \qquad
a_2\ge 24, \qquad \varepsilon =0.054.
\eqno (9)
$$
Then $(8)$ holds, and we get
$$
\log |\Lambda'| \ge - (\mu^2 L^2 a_1 a_2 - L)\theta
= - {4\over 9\theta } L^2 a_1 a_2 - L \theta .
$$
The choices of $R$ and $S$ imply $\ \max \{R, S\} \le L
\max\{a_1,a_2\}$, and the definition of $\Lambda'$ leads to
$$
\log |\Lambda'| \le \log |\Lambda| + 2 \log L
+ \log \bigl(\max\{a_1,a_2\}\bigr),
$$
except maybe when
$$
|\Lambda |\ge {1\over L^2 \cdot \max\{a_1,a_2\}}.
$$
Hence, we always have
$$
\log |\Lambda| \ge - {4\over 9\theta } L^2 a_1 a_2 - L \theta
- 2 \log L - \log \bigl(\max\{a_1,a_2\}\bigr)
\ge - \left({4\over 9\theta } L^2 +{0.01\over a_1}\right) a_1 a_2 -
(\theta +0.35) L.
$$
Recall that
$$
\log |\Lambda| \le \log {3k\over 2n} - 3 \log y .
$$
Thus, under condition $(9)$, we get
$$
\log y \le {1\over 3}\left({4\over 9\theta }+{0.01\over a_1}\right) L^2
a_1 a_2 +{\theta+0.35\over 3} L   + {1\over 3} \log {3k\over 2n}.
$$



\vfill \eject

\beginsection 7.
A second study of linear forms and the proof of Theorem 1

Applying conjugations to relation $(3)$ and taking logs, we get the 
system of relations
$$
\cases{ u_1 \ell_2 + u_2\ell_3 = \log y + \eta_2 &\cr
u_1 \ell_3 + u_2\ell_1 = \log y  + \eta_3,&\cr }
$$
where
$$
\eta _2 = \log \left| {x\over y}-\lambda ^{(2)}\right|
- \log |\gamma^{(2)}|
$$
and
$$
\eta _3 = \log \left| {x\over y}-\lambda ^{(3)}\right|
- \log |\gamma  ^{(3)}|.
$$
By (5),
$$
0 \le  \log \left| {x\over y}-\lambda ^{(2)}\right| \le {2\over n^2},
$$
and
$$
\log n \le  \log \left| {x\over y}-\lambda ^{(3)}\right| \le\log (n+1).
$$
Moreover, the choice of $\gamma  $ implies
$$
{k^{1/3}\over \sqrt{n+3}}\le |\gamma  ^{(2)}| \le k^{1/3}\,\sqrt{n+3},
\qquad
{k^{1/3}\over (n+3)}\le |\gamma  ^{(3)}| \le k^{1/3}\, (n+3).
$$
These estimates lead to
$$
-{1\over 2}\log (n+3)- {1\over 3} \log k \le \eta _2 \le  {2\over n^2}-
{1\over 3} \log k   +{1\over 2}\log (n+3)
$$
and
$$
-\log (n+3)+\log n- {1\over 3} \log k \le \eta _3 \le
- {1\over 3} \log k  +2\log (n+3).
$$

The above system implies
$$
\eqalign{
u_1 (\ell_3^2-\ell_1 \ell_2)
&= (\ell_3-\ell_1) \log y -\ell_1 \eta_2 +\ell_3 \eta_3, \cr
u_2 (\ell_3^2-\ell_1 \ell_2) &= (\ell_3-\ell_2) \log y +\ell_3 \eta_2
-\ell_2 \eta_3. \cr
 }
$$
Hence,
$$
V = u_1+u_2 -1 = {1\over \ell_3^2-\ell_1 \ell_2}
\Bigl( (2\ell_3-\ell_1-\ell_2) \log y
+(\ell_3-\ell_1 )\eta_2+(\ell_3 -\ell_2)\eta_3
\Bigr)-1.
$$

Put
$$
\Delta  =\ell_3^2-\ell_1\ell_2,
$$
then
$$
\Delta   (V+1) - (2\ell_3-\ell_1-\ell_2) \log y =
(\ell_3-\ell_1 )\eta_2+(\ell_3 -\ell_2)\eta_3  .
$$

\noindent
Using the estimates of the $\ell_i$'s (end of Section~3) and those of
$\eta _2$ and $\eta  _3$, we get
$$
\eqalign{
&\Bigl| \Delta   (V+1) -(2\ell_3-\ell_1-\ell_2) \log \bigl(y
\bigm/ k^{1/3}\bigr) - (\ell_3-\ell_2) \log (n+3) \Bigr|
\cr
&\qquad \qquad \le
\Bigl( (\ell_3-\ell_2)\log \bigl((n+3)^2/n\bigr)+
(\ell_3-\ell_1)\log (n+3)\Bigr)
\cr
&\qquad \qquad \le
\left( \Bigl(\log n + {2\over n^2}\Bigr)\log \bigl((n+3)^2/n\bigr)+
2\Bigl(\log (n+1) + {2\over n^2}\Bigr)\log (n+3)\right)
\cr
&\qquad \qquad \le
3 \,\log  (n+2) \times \log (n+3)
}
$$
for $n\ge 5$.
Notice  that
$$
\Delta   \ge  (\log n)^2 + {\log (n+1)\over n+1} > \log^2 n
$$
and
$$
2\ell_3-\ell_1-\ell_2 < 2 \log n + {4\over n^2}+
\log (n+1)+{2\over n(n+1)} -{n-{1\over 2}\over n^2+2}
\le 3\,\log (n+1),
$$
and also
$$
2\ell_3-\ell_1-\ell_2 > 2 \log n + {4\over (n+1)^2}+
\log (n+1)+{2\over (n+1)^2}-{1\over n} > 3\,\log n ,
$$
for $n\ge 3$.
\medskip
We notice that this implies $V\ge 0$ for $n\ge 5$ and,
$$
\eqalign{
V+1&\le {1\over \Delta  } \left( 3 \log (n+1) \log {y\over k^{1/3}}
+4 \,\log  (n+2) \times \log (n+3)\right)
\cr
&\le {\log (n+2)\over \log ^2 n}
\Bigl( 3 \log \bigl(y/k^{1/3}\bigr) +4 \log (n+3)\Bigr),
}
$$
for $n\ge 5$.
\vfill \eject

\medskip

After this list of estimates, let us bound $V$. Recall that
$$
\log y \le {1\over 3}\left({4\over 9\theta }+0.01\right) L^2
a_1 a_2 +{\theta+0.35\over 3} L   + {1\over 3} \log {3k\over 2n}.
$$
By the conditions on $a_1$ and $a_2$, and since $\rho =15$ and $L\ge 16$
this implies
$$
\log y \le {1\over 3} 0.17423 \times  L^2
a_1 a_2 + {1\over 3} \log {3k\over 2n}.
$$
Since $a_1\ge 2\log (n+2)$, we get (for $n\ge 5$)
$$
V+1 \le
{\log (n+2)\over \log ^2 n}
\bigl( 0.17423 \times  L^2 a_1 a_2+4\log (n+3)\bigr)
\le
{\log (n+2)\over \log ^2 n}
\times  0.17456 \cdot  L^2 a_1 a_2.
$$
We consider the two cases
$$
L = \cases {16, \cr
\noalign{\vskip 2pt}
[6H/\theta +\varepsilon ]+3\ge 17.}
$$
In the first case,
$$
V +1\le  {\log (n+2)\over \log ^2 n}
\times  44.688 \cdot a_1 a_2.
$$
Now suppose $L\ge 17$. The definition of $H$ implies
$$
H \le  1.522 + \log^+ \left({|V|\over a_2}\right),
$$
and the definition of $L$, combined with $L\ge 17$, implies
$$
\log (V/a_2)\ge 4.77.
$$
Then, the inequality
$$
{V\over a_2} \le  {\log (n+2)\over \log ^2 n}
\times  0.17456 \cdot  L^2 a_1
\le  0.17456 \times {a_1\log (n+2)\over \log ^2 n}
\left[ 3.054+{6\bigl(1.522 + \log (V/a_2)\bigr)\over \theta }\right]^2
$$
leads to a contradiction. Thus we have proved that
$$
V+1 \le  {\log (n+2)\over \log ^2 n}
\times  44.688 \cdot a_1 a_2,
$$
and that
$$
L=16.
$$
\vfill \eject
We have just obtained an upper bound for $|V|$, using a non trivial
lower bound of~$\Lambda$. Using again $\Lambda$, we directly get a
lower bound for~$|V|$. Indeed,
$$
\Lambda = V \ell_2 \pm  \ell_0,
$$
where $|\ell_2|<(n+1)/(n^2+n+2)$; since $|\Lambda|$ is very small,
this implies
$$
V \ge n \ell_0.
$$
Suppose that $n\ge 1650$, then (in this case $V\ge 0$)
$$
V+1 \le  {\log (n+2)\over \log ^2 n}
\times  44.688 \cdot a_1 a_2 \le {44.696 \over \log n}\cdot a_1 a_2,
$$
where
$$
a_1 = \max\bigl\{ 2 \log (n+2)+14/n, 15\bigr\} <2.03 \log n,
$$
and
$$
a_2 = \max\left\{\Bigl(2|v|+ {5\over 2}\Bigr) \log (n+3) + \log k
+ 14\ell_0, 24\right\}.
$$
Combining these estimates gives
$$
V <89.5 \,a_2\qquad \hbox{for \ $n\ge 1650$.}
$$
If $v\ne 0$ then
$\,a_2 = \Bigl(2|v|+ {5\over 2}\Bigr) \log (n+3)+\log k+14\ell_0\,$
and
$$
n\ell_0 <89.5 \left(\Bigl(2|v|+ {5\over 2}\Bigr) \log (n+3) + 
\log k+ 14\ell_0 \right).
$$
Hence, if $v\ne 0$,
$$
(n-1253) \ell_0 < \bigl(179 |v| + 223.75 \bigr) \log (n+3) + 89.5 \log k
$$
for $n\ge 1650$. Since
$$
\ell_0 \ge  |v| \log n -{3\over 2} \log (n+3)
> (|v|-1.501) \log n,
$$
for $n\ge 1650$, we get also
$$
(n-1253-179.1) |v|
< \bigl(1.501\,(n-1253) + 223.75 \bigr)  + 89.5 {\log k \over \log n},
$$
or,
$$
|v| < \left(1-{1432.1\over n}\right)^{-1}
\left\{ 1.501 - {1657\over n} +89.5 \,{\log k \over n\,\log n}\right\}.
$$

\vfill \eject

Now we consider the $y$ term and we suppose $n\ge 1650$. We have
$$
\log y \le {1\over 3} 0.17423 \times  L^2
a_1 a_2 + {1\over 3} \log {3k\over 2n},
$$
with $a_1 <2.03 \,\log n$ and $L=16$, hence
$$
\log y \le  29.77 \, a_2 \log n+ {1\over 3} \log k.
$$
Using the inequality $\ell_0\le |v|\log (n+1)+1.5\,\log (n+3)$ and the
definition of $a_2$, we get
$$
\log y < 29.77 \, (16 |v|+23.5)\,\log^2 (n+2)
+29.82\,\log n\times \log k.
$$
And the upper bound on $|v|$ gives
$$
\eqalign{
\log y &< \left(700 + 476.4
\left(1-{1432.1\over n}\right)^{-1} \left\{ 1.501 - {1902\over n}\right\}
\right) \log^2 (n+2)
\cr
&\qquad + \left(29.82 + \left(1-{1432.1\over n}\right)^{-1}
{1432 \over n\,\log n}\right) \log n\times \log k.
}
$$
Theorem 1 is proved.


\vfill \eject

\beginsection 8.
An improved bound, if $k$ is small

In this section we add some hypothesis on $k$ and then get an upper
bound for $\ell_0$, which leads to upper bounds on $|v|$ and $y$.
\medskip

We suppose $n\ge 1650$ and
$$
k \le  n^4.
$$
Then, the upper bound on $v$ becomes
$$
|v| < \left(1-{1432.1\over n}\right)^{-1}
\left\{ 1.501 - {1657\over n} + {358 \over n}\right\},
$$
which implies
$$
|v|\le 1\qquad {\rm for}\quad n\ge 2750.
$$

Using the definition of $a_2$, we also obtain
$$
\ell_0 < \max\left\{{761 \,\log n\over n-1253}, {2148\over n}\right\}
= {761 \,\log n\over n-1253}
\qquad {\rm for}\quad n\ge 3150.
$$
This implies
$$
\ell_0 < {\log n \over n^{1/3}} \qquad {\rm for}\quad n\ge 23500.
$$
In this case, we choose
$$
\rho = n^{1/3};
$$
then $\mu =2/\log n$. Since $a_1>2\,\log n$ and $a_2>2.5\,\log n$, taking
again $L=16$, we have $K\ge 360$. It is also easy to verify that
$$
H <4.67+\log \log n
$$
and that we can take $\varepsilon =0.17$
(the number $\varepsilon $ occurs in the definition of the
integer~$L$). These estimates prove that the choice $L=16$ is
legitimate. Then applying again Theorem~A, for $n\ge 23500$, we get
$$
\eqalign{
\log y
&\le {1\over 3}\left({4\over 9\theta }+{0.01\over a_1}\right) L^2
a_1 a_2 +{\theta+0.35\over 3} L   + {1\over 3} \log {3k\over 2n}
\cr
&
\le {1\over 3}\left({4\over 3 }+0.005\right) 256
\times 2.02 \times 9.5 \times \log n +{\theta+0.35\over 3} L   + {1\over
3} \log {3k\over 2n} .}
$$
Thus, we have proved that
$$
\log y < 2180 \,\log n \qquad {\rm if} \quad
n\ge 23500 \quad {\rm and} \quad k\le n^4.
$$

\beginsection 9.
Proof of Theorem 3.

Let us consider the inequality
$$
|x^3-(n-1)x^2y-(n+2)xy^2-y^3| \le k
$$
with $k=2n+1$. The family of equations $|F_n(x,y)| = 1 $ was completely
solved by Thomas [T1] and Mignotte [M]. Moreover, it was proved by
Lemmermeyer and Peth\H{o} [LP] that if $\gamma \in {\bf Z}[\lambda]$ has
norm, in absolute value, less then $2n+1$, then $\gamma$ is a rational
integer multiple of a binomial unit from ${\bf Z}[\lambda]$. Thus to prove
Theorem 3 it remains to study the equation

$$
|x^3-(n-1)x^2y-(n+2)xy^2-y^3| = 2n+1. \eqno(10)
$$
It is easy to see that
$$
F_n(1,1) =
F_n(1,-2)=F_n(1,-n-1)=F_n(n,1)=F_n(-n-1,n)=F_n(-2,1)=-(2n+1),
$$
and
$$
F_n(-1,-1)
= F_n(-1,2) = F_n(-1,n+1) = F_n(-n,-1)=F_n(n+1,-n)= F_n(2,-1) = 2n+1.
$$

The continued fraction expansion of $\lambda$ is
$$
\lambda=\left[ -1,1,n,\left[ {n\over 2} \right] ,\dots \right],
$$
if $n>0$, and hence its first convergents are
$$
-1,\quad 0,\quad - {1\over n+1},
\quad {-\left[{n\over 2}\right] \over (n+1)\left[{n\over 2}\right]+1}.
$$

Let $(x,y) \in {\bf Z}^2$ be a solution of $(10)$, such that $y \ge 1$ and
$$
|\lambda y-x|<|\lambda^{(2)}y-x|, \ |\lambda^{(3)}y-x|.
$$
By Corollary 1, and by $D_f=(n^2+n+7)^2$ we have
$$
|L_1| \le \left( {4(2n+1)^4\over (n^2+n+7)^2}\right)^{1/6}
{1\over y^2} < {2\over y^2},
$$
thus
$$
\left| {x\over y}-\lambda \right| < {2\over 25},
$$
if $|y| \ge 5$. Using the estimations for $\lambda$, $\lambda^{(2)}$ 
and $\lambda^{(3)}$ we get
$$
|x-\lambda^{(2)}y|> \left(|\lambda^{(2)}-\lambda|-
{2\over 25}\right) |y| > \left( 1-{2\over 25} \right) |y|=
{23\over 25}|y|
$$
and
$$
|x-\lambda^{(3)}y|>\left(|\lambda^{(3)}-\lambda|-
{2\over 25}\right) |y|>\left( n-{2\over 25}\right)|y|.
$$
Therefore
$$
\left|{x\over y}-\lambda
\right|<{2n+1\over {23\over 25}\left(n-{2\over 25}\right)|y|^3} <
{1\over 2y^2}
$$
if $\,|y| \ge 5\,$ and $\,n \ge 3$, which means that in those
cases $\,{x\over y}\,$ is a convergent of $\lambda$. We conclude the
same for $n=1$ when~$|y| \ge 8$ and for $n=2$ when $|y| \ge 6$.
\medskip
If $n=0$, then $(10)$ is already solved, see Thomas [T1] and Mignotte [M]. For
$n=1$ with $|y| \le 7$ and $n=2$ with $|y| \le 5$ we get by a direct
computation the following solutions.
\bigskip

{\settabs 10\columns
\+&$n$ & 1 & 2 & 2 & 2 & 2 & 2 & 2 \cr
\+&$x$ & -1 & -1 & -3 & -4 & -1 & 8 & -1 \cr
\+&$y$ & 2 & 2 & 2 & 3 & 3 & 3 & 4 \cr
\+&$f_n(x,y)$ & 3 & 5 & -5 & 5 & 5 & 5 & -5 \cr
}
\bigskip

\noindent
We remark that exactly the values given in the table serve the non-trivial
solutions for $n=2.$

If $n \ge 3$ and $4 \ge |y| \ge 2$ then $|\lambda| < {2\over 9}$
and by using
$$
|x-\lambda y| < {2\over |y|}
$$
we get
$$
-2<- {2\over |y|}-|\lambda||y|<|x|< {2\over |y|}+|\lambda||y|<2,
$$
but the solutions with $|x|\le 1$ are listed at the beginning. We did the
same for $|y| \le 1$. Thus apart from the cases listed in the table, the
solutions of $(10)$ are coming from convergents of $\lambda$, hence 
$|y| \ge n+1$. If $y=n+1$, then
$$
F_n(x,n+1)-(2n+1)=(x+1)\bigl(x^2-n^2x-(n^3+3n^2+5n+2)\bigr).
$$

If the second factor $q(x)$ is reducible over ${\bf Q} [x]$, then it
has an integer zero. But a simple calculation shows, that $q(-n-1)<0$,
 $q(-n-3)>0$ and $q(-n-2)=0$ iff $n=2$. Thus $q(x)$ is reducible
over ${\bf Q} [x]$ iff $n=2$ and these cases are listed in table 1.
\medskip
Let $\,q_1(x)=F_n(x,n+1)+2n+1$.
Then by $\,q_1(0)<0$, $\,q_1(-1)>0$, $\,q_1(-n-2)=n^2+3n>0\,$
and
$$
\eqalign{
q_1(-n-3)&=-n^3-3n^2-5n- 12<0,
\cr
q_1(n^2+n+1)&=-(n^4+4n^3+6n^2+5n+1),
\cr
q_1(n^2+n+2)&=-n^3+n^2+n+7,
\cr
q_1(n^2+n+3) &=
n^4+2n^3+12n^2+13n+29
}
$$
we see that $q_1(x)$ has no integer zero if
$n \ge0$. Thus we may assume $y>n+1$, whence the continued fraction
expansion of $\lambda$ implies $y \ge (n+1) \lbrack {n\over 2}
\rbrack +1 > {n^2\over 2}.$
\vfill
\eject

By Theorem 3.1 of Lemmermeyer and Peth\H{o} [LP], all elements of
${\bf Z} [\lambda]$ of norm $2n+1$ are associated to one of the  
conjugates of $\lambda-1$. It is easy to see that $\lambda^{(2)}-1=
-{\lambda+2 \over \lambda+1}$ and $\lambda^{(3)}-1=-{2\lambda+
1 \over \lambda}$ and that $\lambda-1,\lambda+2$ and $2\lambda+1$ 
satisfy the conclusion of Corollary 2. Hence we may apply
the results of sections 6 and 7.

\medskip

First we prove that if $n \ge 1700$ then $(10)$ has only the trivial solutions.
We distiguish three cases and adopt the notations of section 6. We remember
that if $n \ge 5$ then $V \ge 0$ and so $v \le 1.$

\medskip

{\bf Case I.} $\gamma = \gamma_1 = \lambda -1.$ Now we have 
$$\alpha_2=\lambda^{(3)v} \left| \lambda^{(2)}-1 \over \lambda^{(3)}-1 \right|,
\quad 
h(\alpha_2) \le {1 \over 3} (|v| \log(n+2) + \log(n(2n+1))).$$ 
By the choice 
$\rho = 15$ we have also $ a_2= 14 \ell_0 +2 (|v|+2) \log(n+2) + 2\log2,$ 
where
$$
|v-1| \log\lambda^{(3)} + \log 2 + {1 \over 2(n-1)} \le \ell_0 \le
 |v-1| \log\lambda^{(3)} + \log 2 + {3 \over 2(n-1)} .
 $$
As $V\ge 0$ and $v\le 1$ we have $V>n \ell_0 > n(|v-1|\log \; n + \log 2).$
On the other hand 
$$
V < 89.5 a_2 \le 1253 \ell_0 +179(|v|+2)\log(n+2) +179 \log 2.
$$
Combining these inequalities we get $n<8450$ for $v=1$ and $n<1620$ for
$v<1 .$

\medskip

{\bf Case II.} $\gamma = \gamma_2 = \lambda +2.$ We now transform
$$\alpha_2=\lambda^{(3)v} \left|
\lambda^{(2)}+2 \over \lambda^{(3)}+2 \right|$$ 
to a more appropriate algebraic
number to prove a sharp bound for $n.$ Indeed, using $\lambda^{(2)}+2 =
(\lambda -n)\lambda$ and $\lambda^{(3)}+2 = (\lambda^{(2)} -n)\lambda^{(2)}$
we get
$$
{\lambda^{(2)}+2 \over \lambda^{(3)}+2} =
{(\lambda -n)\lambda (\lambda+1)^2 \over n (\lambda+1)+1} =
{n - \lambda \over n (\lambda+1)+1} 
{1 \over \lambda^{(3)} \lambda^{(2)3} }.
$$
Thus our linearform becomes
$$
\Lambda=(V-3)\log|\lambda^{(2)}|+(v-1)\log
|\lambda^{(3)}|+ \log \left|
{n-\lambda \over n(\lambda+1) + 1 } \right|.
$$
By the estimates of $\lambda $, we also have 
$$
1-{2\over n^3}<
 {n-\lambda\over n+n\lambda+1}
=1-  {(n+1)\lambda+1\over n(\lambda+1)+1}<1 - {1 \over n^3} .
$$

We proved $|\Lambda|< {3k \over 2ny^3}$ in section 5 for all $n \ge 3$
and $k \in {\bf N}$. With our special value $k=2n+1$ using the estimate
$y>{n^2 \over 2}$ we get

$$|\Lambda|< {12(2n+1) \over n^7}< {1 \over n^3}$$
if $n \ge 4$ immediately. This inequality remains true for $n=3$ too, which
one can verify by a direct computation.
Hence $v=1$ is impossible and if $v<1$ then 
$$|(V-3) \log |\lambda^{(2)}||>|v-1| \log |\lambda^{(3)}|-|\Lambda|
- \left| \log{n-\lambda \over n+n\lambda+1}\right| >|v-1| \log n,
$$
which implies
$$V-3 > |v-1| n \log n.$$
On the other hand 
$$V<89.5 a_2 $$
with
$$
a_2 = (16|v| + 18) \log(n+2) + 2\log2.
$$
These inequalities imply again
$$
n<1620.$$

\medskip

{\bf Case III.} $\gamma = \gamma_3 = 2\lambda +1.$ This case is similar to 
case I, and therefore we have given only the most important data: 
$$\alpha_2=\lambda^{(3)v} \left|
2\lambda^{(2)}+1 \over 2 \lambda^{(3)}+1 \right|, \quad
h(\alpha_2) \le {1 \over 3} ((|v|+2) \log(n+2) + 2\log2)$$
and
$$
|v-1|\log \; n -\log2 < \ell_0 < |v-1| \log(n+2) - \log 2.
$$
Thus 
$$
a_2 < (16|v|+18)\log(n+2) -10 \log 2.
$$
If $v=1,$ then $ n< 8640$ and if $v<1,$ then $n<1700.$

\bigskip

 We have proved that equation $(10)$ has just the
``trivial solutions'' for $n>8640$; moreover $v=1$ if $1700 \le n \le 8640$ 
and this can occur only in cases I and III. We shall prove now the 
impossibility of $v=1.$ 

\smallskip

From $|\Lambda|< {12(2n+1) \over n^7}$ we get

$$\left| V + v {\log|\lambda^{(3)}| \over \log |\lambda^{(2)}|}+
{ \log | \gamma_i^{(2)}|- \log| \gamma_i^{(3)}|  \over 
\log |\lambda^{(2)}|} \right| < {1 \over n^5}$$
for $n \ge 30$ and $i = 1,2,3.$ Putting
$$\delta_1= {\log |\lambda^{(3)}| \over \log |\lambda^
{(2)}|}  \quad {\rm \  and \ } \quad
\delta_{2i}={\log | \gamma_i^{(2)}| - \log|\gamma_i^{(3)}| \over
\log |\lambda^{(2)}|}$$
 this can be reformulated to


$$ \hfil ||v \delta_1+\delta_{2i} ||< {1 \over n^5}, \eqno (11)$$
$i=1,2$ or $3$, where $||\; \; ||$ denotes the distance to the nearest integer.

\smallskip

For $1700 \le n \le 8640$ and $i=1$ and $3$ we tested $(11)$ with $v=1,$ 
but did not found any $n$ satisfying $(11)$ even with the much larger 
right hand side $10^{-5}$. We performed the same test with $v=1$ for 
the values $1 \le i \le 3$ and $0 \le n < 1700$ with the same result.
Thus it remains to prove Theorem 3 only in the range  $0 \le n \le 1700$ 
and with $v \not= 1.$

\vfill \eject

\beginsection 10.
The case $0 \le n \le 1700$ and finish the proof of Theorem 3.

\bigskip

In the range $0 \le n \le 1700$ application of linear forms in two 
logarithms fails to work
because we are not able to prove an upper bound for $|v|$ and $V$.
Therefore we
apply the following more general theorem of Baker and W\"ustholz [BW]:

\proclaim Theorem B.
For a linear form $\Lambda \not= 0$ in logarithms of $m$ real algebraic
numbers $\alpha_1,\dots,\alpha_m$ with rational integer coefficients
$b_1,\dots,b_m$ we have
$$
\log|\Lambda| \ge -18(m+1)! m^{m+1}(32D)^{m+2}h(\alpha_1)\dots
h(\alpha_m) \log B,
$$
where $B=\max \{ |b_1|,\dots|b_m| \}$, and where $D$ is the degree of the
number field generated by $\alpha_1,\dots,\alpha_m$.

\rm

\bigskip

In the actual case we have $m=3, D=3,$
$$
h(\lambda^{(2)})=h(\lambda^{(3)}) \le {1 \over3} \log(n+2),
\quad
h\left({\gamma_i^{(2)} \over \gamma_i^{(3)}}\right)
\le {1 \over 3} \log \bigl(n(2n+1)\bigr) 
\quad i=1,2,3, 
\quad
B=V,
$$
and we get
$$
\eqalign{
\log|\Lambda| &\ge -18 \cdot 4 ! \cdot 3^6 \cdot 32^5 \cdot \log^2(n+2)
\log \bigl(n(2n+1)\bigr) \log V
\cr
&>-1.06 \cdot 10^{13} \log^2(n+2) \log(n(2n+1)) \log V.
}
$$
On the other hand, if $|y| \ge 2$ and $n \ge 5$ then
$$
\log|\Lambda|< -(V+1){\log^2n \over \log(n+2)} +4 \log(n+3)-\log n +\log
{3 \over 2}.
$$
A comparison of these estimates leads to

$$V<10^{17}.$$

This bound is much worse than that we got for $n >1650$ and demonstrates
that the estimates for linear forms in two logarithms are much more 
suitable for solving
diophantine problems completely than the general estimates. Unfortunately,
estimates for two logs are not always applicable.

\smallskip

Despite the large bound for $V$ we can prove that our problem has no
non-trivial solution in the actual range too. For this purpose we use a
variant of the Baker-Davenport lemma  [BD], which was helpful in
[MPR] too. We remember that

$$ |\Lambda_i|=|V+ v \delta_1+\delta_{2i}|<{3(2n+1) \over 2ny^3
\log|\lambda^{(2)}|}
<\exp
\left\{-(V+1){\log^2n \over \log(n+2)}+4\log(n+3)+\log 3\right\}
\eqno (12) $$  
$(i= 1,2,3),$ where the numbers $\delta_1$ and $\delta_{2i}$ were
defined in the previous section.

\bigskip

\proclaim Lemma 4.
Suppose that $n \ge 5$, $1 \le i \le 3$, $V<10^{17}$, $v \not= 1$ and
$$|\Lambda_i|< 10^{-50}$$
and that ${\tilde \delta}_1$ and ${\tilde \delta}_{2i}$ are rational numbers
such that
$$
|\delta_1 - {\tilde \delta}_1|<10^{-50}, |\delta_{2i} - {\tilde
\delta}_{2i}|<10^{-50}.
$$
If there exists a convergent $p/q$ in the continued fraction expansion of
${\tilde \delta}_1$ such that $q \le 10^{25}$ and
$$q||q{\tilde \delta}_{2i}||>{5 \cdot 10^{17} \over n \log n}$$
then $(12)$ cannot hold for $v,V \in {\bf Z}$.
\rm

\medskip

{\bf Proof}
Assume that there exist $v,V \in {\bf Z}$ which satisfy $(12)$ and 
$|\Lambda_i|< 10^{-50}$. We remember that $(12)$ implies 
$V>|v-1| n (\log n - \log 2)$. Let us fix $i.$

Let $p/q$ be a convergent of ${\tilde \delta}_1$ with the properties given in
the lemma. Multiplying $(12)$ by $q$ and inserting $\delta_1$ and $\delta_{2i}$
we get
$$
q|\Lambda_i|=|q{\tilde \delta}_2+q(\delta_{2i}-{\tilde
\delta}_{2i})+ v (q{\tilde
\delta}_1-p)+ v q(\delta_1-{\tilde \delta}_1)+
v p + V q| < 10^{-25}.
$$
Thus
$$
||q{\tilde \delta}_{2i}||< 2 \cdot 10^{-25}+ |v| |q{\tilde \delta}_1-p|+ 
|v|10^{-25}
$$
and
$$
q||q{\tilde \delta}_{2i}||<2+|v|q|q{\tilde \delta}_1-p|+|v|
<{5V \over n \log n}<{5 \cdot 10^{17} \over n \log n},
$$
which is a contradiction. The lemma is proved. \medskip

\bigskip

As $|v-1| \not = 0$ we have $V>n \log (n/2)$, which implies
$|\Lambda_i|<10^{-50}$ for $n \ge 25$ immediately. The same is 
true for $n \ge 8$ and $|v|>10$. In
these cases we computed $\delta_1$ and $\delta_{2i}$ with 50 decimal digit
precision and found for all $n$, where $8 \le n \le 1700$, a convergent
$p/q$ of ${\tilde \delta}_1$ with the properties given in the lemma. For
those values $(12)$ can not hold. We tested $(11)$ for $8 \le n \le 25$
and $|v| \le 10$ separately.

In the remaining cases $0 < n \le 7$ we used the method from Peth\H{o} und
Schulenberg [PSch], but did not found any more solutions. Thus, Theorem 3 is
proved.

\medskip

For the computation we used PARI-GP on a notebook. The total computation time
took some hour.

\bigskip

\beginsection
References

\let\sc=\rm
\let\em=\sl

[BD] \
{\sc A. Baker} and  {\sc 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.

\smallskip 
[BSt] \
{\sc A. Baker} and {\sc C.L. Stewart}, {\em On effective
approximations to cubic irrationals}, in New Advances in Transcendence
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[BW] \
{\sc A. Baker} and {\sc G. W\"ustholz}, {\em Linear forms and
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[GyP] \
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\smallskip 
[LMN] \
 {\sc M. Laurent, M. Mignotte} et {Y. Nesterenko},
{\em Formes lin\'eaires en deux logarithmes et d\'eterminants
d'interpolation},
J. Number Theory, (to appear).

\smallskip 
[LP] \
{\sc F. Lemmermeyer} and {\sc A. Peth\H{o}}, {\em  Simplest
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\smallskip 
[LeP] \
{\sc G. Lettl} and {\sc A. Peth\H{o}}, {\em Complete solution of
a family of quartic Thue equations}, Abh. Math. Sem. Univ.
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[M] \
{\sc M. Mignotte}, {\em Verification of a conjecture of
E. Thomas}, J. Number Theory, {\bf 44} (1993), 172--177.

\smallskip 
[MPR] \
{\sc M. Mignotte, A. Peth\H{o} } and {\sc R. Roth,}
{\em Complete solutions of quartic Thue and index form equations},
Math. Comp. (to appear).

\smallskip
[MT] \
{\sc M. Mignotte} and {\sc N. Tzanakis}, {\em On a family of cubics}, 
J. Number Theory, {\bf 39} (1991), 41--49.

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[P1] \
{\sc A. Peth\H o}, {\em On the representation of 1 by binary
cubic forms with positive discriminant.} In Number Theory, Ulm 1987,
ed. H.P. Schlickewei and E.~Wirsing, Springer-Verlag Lect. Notes
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\smallskip 
[P2] \
 {\sc A. Peth\H o}, {\em Complete solutions to a family of
quartic diophantine equations}, Math. Comp. {\bf 57} (1991), 777-798.

\smallskip 
[PSch] \
 {\sc A. Peth\H{o}} und {\sc R. Schulenberg}, {\em
Effektives L\"osen von Thue Gleichungen},
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\smallskip
[T1] \
{\sc E. Thomas,} {\em Complete solutions to a family of cubic
diophantine equations}, J. Number Theory {\bf 34} (1990), 235-250.

\smallskip
[T2] \
{\sc E. Thomas,} {\em Solutions to certain families of Thue equations}, J. Number
Theory, {\bf 43} (1993), 319-369. 

\bigskip

{\obeylines
Maurice Mignotte 
D\'epartement de Math\'ematique et Informatique
Universit\'e Louis Pasteur
7, rue Ren\'e Descartes
67084 Strasbourg Cedex
France 
\medskip
Attila Peth\H{o}  
Laboratory of Informatics  
University of Medicine  
Nagyerdei Krt. 98  
H--4032 Debrecen  
Hungary  
\medskip
Franz Lemmermeyer    
Erwin-Rohde-Str. 19  
D--69120 Heidelberg,  
Germany  }

\end{document}

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