1. Introduction
Let
and
be two positive integers with
and
be matrices of the group 
Denote
the group, respectively
the semigroup, generated by the matrices 
The following problem
has been studied in several papers:
Instance: 
Question:
or
are they free with
as generators?
Recall that in 1991 D. Klarner, J.-C. Birget and W. Satterfield in [3] proved that if
then the problem
is not decidable. Moreover in 1999 J. Cassaigne, T. Harju and J. Karhümaki in [4] proved that the same result is true if we suppose that all the matrices
are lower triangular.
The case
is open and seems difficult. In [5] and [6] results concerning the freeness of the semigroups and groups generated by two matrices are established. In this paper we are studying this problem restricted to the case of Möbius groups.
Let
and
. The Möbius group
is the subgroup of
generated by
and
.
The problem of characterization of the set of complex values of
or
for which the group
is free, was studied in several papers. Thus in [1] it is proved that if
is transcendental or
then
is free.
R.C. Lyndon and J.L. Ullman in [1] remarked that
is not free if and only if there exists a word
whose
letters are non zero integers so that the product of the powers of matrices
is a lower triangular matrix. The element in the right upper corner of the matrix
is of the form
where
is a polynomial in
of degree
with coefficients
Moreover
are polynomials with integers coefficients in the variables 
Results concerning the set of algebraic values of
or
for which the group
is not free were obtained in [1] [2] [7] -[11] .
Deciding if for
the group
is not free seems very difficult. Let us recall some important results in this direction.
The group
is not free if
belongs to one of the following sets:
(see [1] [2] [7] [8] [10] [11] ).
In this paper we check if for a given
there exists a non trivial word of non zero integers
such that 
The main results of our paper concern the freeness of Möbius groups:
We prove that if the length of
is small then the problem is decidable (cases
and
) (see Theorems 1, 2 and 3).
We give algorithms which solve the problem for
(see Corollary 1 and the proof of Theorem 3). Moreover, we give an arithmetical criteria for this problem when
(see 2 of Theorem 1).
We give a lower bound numerical function
defined from
to
, increasing and unbounded, such that for each
if
is a lower triangular matrix then the length of
is bigger than
(see Theorem 4 and Corollary 3).
As proved by A.F. Beardon ([2] ) in these two cases
we have to find solutions for the equations (B1) and (B2). In fact in our paper we consider and study two more general equations:
(B'1) 
(B'2) 
2. Sequences of Polynomials Associated to Matrices
In this section, we study the properties of some sequences of polynomials in a fixed
associated to matrices of the group
.
We consider
the free monoid of words on non zero integers with the concatenation operation. We denote by
the empty word of the free monoid
and a non empty word
by
, where
are non zero integers. Then
is called the length of
and is denoted by
. The reversal of a word
is
and the opposite of
is 
For every word
of
of length
we consider the matrix product

For instance, for
non zero integers we have:
and 
We use the notation:

We remark that
and
are polynomials in
with coefficients in
We also have
and 
If
then
and if
then
Also
and if
then 
We use the notation
to indicate that
is a lower triangular matrix or that
.
From now on, in order to simplify the notation we write:

For instance,
is an abbreviation for the polynomial in
with parameters
defined by:

Using the fact that
we have:
(1)
The sequences of polynomials in
,
,
,
and
verify the following relations:
(2)
(3)
(4)
(5)
The relations (4) and (5) follow from the equality

In the following sections, we also use the following two relations:
(6)
(7)
Using the previous relations we obtain Proposition 1 The sequences
and
of polynomials in
verify the following identities:
(8)
(9)
Proof. From (5) we have

These identities and the equation
give the equation (8). The equation (9) can be similarly obtained .
Let us suppose that
where
and
are non zero integers and
.
If
the group
is not free because in this case
(see [1] ).
In the following we consider that
. Then
, and
. Indeed, if
then using the fact that
we deduce
which is in contradiction with the fact that
.
This remark allows us to define a new sequence
by
This sequence satisfies the following relation:
(10)
Thus we obtain

These relations are similar with formulas for continued fractions. The properties of these sequences will be used in the next sections of our paper.
Let us also consider the sequence
defined by:

We remark that
if and only if

The following lemma is the key element of Section 5.
Lemma 1 Let
be
non zero integers and suppose that
. If
then 
Proof. If
we have

Let
such that
is not free. We define the following numerical function:
The number
will be called the calibre of the group
.
Hence
if and only if there are non zero integers
such that
. Also we have
if and only if there are non zero integers
such that 
3. The Diophantine Equation (B1)
In the next three sections, we consider the following problem
, where
,
:
Instance: Two non zero integers
with 
Question: Is there a word of length
of non zero integers
such that
, where
?
So we check solutions in non zero integers
for the diophantine equation
(11)
The set of
for which the Möbius group
is not free coincides with the set of
for which there exists
such that the Equation (11) admits solutions.
In this section, we consider the case
and in the next section the case
. The relation
is equivalent to the Equation (B'1) and the relation
is equivalent to the equation (B'2). If
and
are perfect squares we obtain the equations (B1) and (B2).
We will prove that the problem
is decidable. The decidability of the problem
has already been established by A.F. Beardon (Theorem 2, [2] ) for the case when
and
are perfect squares. Our algorithm is simpler and allows us to give an arithmetical criteria for integers
and
for which the problem
has solutions (see Theorem 1 below).
First, we prove a result concerning the equation (B'1).
Proposition 2 Let
and
be two integers with
and
. Denote

Then:
1. If
and
we have


2. The set
is finite.
Proof.
1) Let
and for
put
. Then
,
and
.
As
we deduce
. Because
and
we have

2) results from 
Using the previous proposition we can obtain the decidability of the problem
.
Theorem 1 Let
and
be two integers with
. The following sentences are equivalent:
1. The equation
has solutions in non zero integers.
2. There exists a divisor
of
,
such that
.
3.
.
Proof. The equivalence between (1) and (2) results from the Proposition 2. It is enough to consider
in that proposition. The equivalence between (1) and (3) is obvious.
Remark 1 Let
be the set of all divisors of the integer
. If
is like in (2) of the previous Theorem 1 then a solution
to the equation (B'1) can be obtained by taking
•
and
•
for 
where
and
Moreover any solution
of the equation (B'1) can be obtained by this method. We can write
as in (3) of the Theorem 1

The results of A.F. Beardon ([2] , theorem 2) concerning the problem
for the case when
and
are perfect squares (or equivalently when
) result immediately from the next corollary.
Corollary 1 Let
and
be two non zero integers with
and
The group
is not free with the calibre
if and only if there exists a divisor
of
,
such that
.
From the previous theorem it also follows:
1) The equation (B'1) has no solution if
.
2)
in the following cases: a)
; b)
; c)
and
with 
Below we present another form of the Theorem 1 in which we use the decomposition of
as a product of prime numbers.
Theorem 2 Let
and
be two integers with
and
. Let us suppose that the decomposition of
as a product of powers of distinct prime numbers
is 
Then
if and only if there exist:
• two disjoint subsets
and
of
with 
• a set of integers
with
for every 
• 
such that 
Proof. Let
be a divisor of
We can drop the case
because
Hence
and
for every
We put
and
Then
and
Let:
.
We have
for every
The condition
is equivalent to

Corollary 2 Let
and
be two non zero integers and
be a prime number. Suppose that
Then
if and only if there exists an integer
with
such that
where 
Proof. We take
in the previous theorem.
Example: Using the previous results and an example from ([7] ) we have
and 
4. The Beardon Diophantine Equation (B2)
Now we consider the problem
. We mention that the equation
has been considered in several papers (see [2] [8] [10] ) for the case when
and
are perfect squares.
From now on, we suppose that
for every
i.e. following Theorem 1,
does not belong to
Hence we can define a function
by
We remark that
and a)
if 
b)
if
where 
Using the relations (8) for the sequence of polynomials
we prove that the problem
is decidable.
Theorem 3 Let
such that
does not belong to the set
. Then the equation

has a finite number of solutions 
Proof. Using the relations (8) we deduce that

Hence
Using the function
we have:

We obtain a finite number of possibilities for
and
So
and
remain to be studied. From the equation

it follows that

Hence there exists
such that






Thus there exists a finite number of possibilities for
and 
If
from the inequality
we obtain a) If
then
has no solution.
b) If
then 
We also remark that the equation (B'2) is equivalent to the following equation
(12)
This enables us to obtain some explicit expressions for the rationals
such that equation (B'2) has solutions in 
Proposition 3 Let
be two non zero integers and
be two divisors of
If 
then the equation (B'2) has solutions in 
Proof. Let
and
Then (10) is equivalent to 
Note that if in equation (B'2) we have
then
is exactly given by the above expression.
Using once again (10) we obtain Proposition 4 Let
and
be in
with
If

then the equation (B'2) has solutions in 
Proof. Consider (10) for
and
Then
and
where
It follows that if we take
then (10) is verified.
In the next proposition we give another method to obtain solutions of Equation (B'2). It is similar to those presented in [8] and [10] .
Proposition 5 Let
and
be two integers with
Suppose that there exist
and 
in
such that
If
then the equation (B'2) has solutions in 
Proof. Let
and
Then
Hence the equation (B'2) has solutions.
We end this section with the following open questions:
wang#title3_4:spQuestions:
1) Find all the solutions of (B2).
2) Find arithmetical characterizations (similar to those given in Theorem 1 for the positive integers
and
for which the problem
has solutions.
5. Increasing Unbounded Lower Bound Function for (
In this section, we prove that in order to show that the group
is not free for a rational
with
and
close to 4, we have to consider longer and longer words in
and
. Similar remarks (without any proof) have been made by A.F. Beardon in [2] and S.P. Farbman in [7] .
Everywhere in this section, we consider that
is a rational number in the open interval
.
From the Lemma 1, Section 2, if
then
For this reason we consider the sequence
of rational functions in the variable
,
defined by:

For example
and 
We also define the function
by the formula:

Thus one has
if and only if
,
if and only if
and
if and only if 
Now we will calculate
.
Note that
For this reason we find the matrix
, where
. We suppose now that
with
, so
. As
the matrix
verifies the equation:

Using this relation we find that

Hence 
Lemma 2 Let
,
be such that
. Then for every
we have
.
Proof. Since
we obtain that
. Hence 
The previous expression for
and Lemma 2 show that
is well defined and
, for every
in the open interval
. So
is a lower bound numerical function for the function
restricted to 
Theorem 4 For any
and
one has
if and only if there exists
such that 
Proof. Let
, where
From the definition of the function
this previous equality holds if and only if
and
, for all
. But
if and only if 
Thus we obtain the system of two inequalities
and
.
Finally,
if and only if we have
and
for all
. These inequalities give 
Corollary 3 The function
is increasing and unbounded.
Therefore

Example: We consider the sequence
, for
.
• For
we have
. So
, hence
. As
, it follows that
.
• For
we have
and
,
,
Hence
and since

we have 
• For
we have
and
,
,
,
,
Hence
and
. From [7] we have
.
wang#title3_4:spQuestions:
1) Is it true that for every
and
, the problem
is decidable?
2) Is it true that for every
there exists
such that the problem
is decidable?
3) Is it true that for every
there exists
, such that the problem
has solutions?
4) Find
, for
.
Acknowledgements
I thank Elias Tahhan (University S. Bolivar, Caracas) and Jerzy Tomasik (Universite d’Auvergne, ClermontFerrand) for discussion concerning some logical aspects of my paper.