7 Complements 211
It was conjectured by Frobenius (1913) that a Markov triple is uniquely determined
by its greatest element. This has been verified whenever the greatest element does not
exceed 10^140. It has also been proved when the greatest element is a prime (and in
some other cases) by Baragar (1996), using the theory of quadratic fields.
7 Complements...............................................
There is an important analogue of the continued fraction algorithm for infinite series.
LetKbe an arbitrary field and letFdenote the set of all formal Laurent series
f=
∑
n∈Z
αntn
with coefficientsαn∈Ksuch thatαn=0 for at most finitely manyn>0. If
g=
∑
n∈Z
βntn
is also an element ofF, and if we define addition and multiplication by
f+g=
∑
n∈Z
(αn+βn)tn, fg=
∑
n∈Z
γntn,
whereγn=
∑
j+k=nαjβk,thenFacquires the structure of a commutative ring. In
fact,Fis a field. For, iff=
∑
n≤vαnt
n,whereαv=0, we obtaing=∑
n≤−vβnt
n
such thatfg=1 by solving successively the equations
αvβ−v= 1
αvβ−v− 1 +αv− 1 β−v= 0
αvβ−v− 2 +αv− 1 β−v− 1 +αv− 2 β−v= 0
·····
Define the absolute value of an elementf=
∑
n∈Zαnt
nofFby putting
|O|= 0 , |f|= 2 v(f) iff=O,
wherev(f)is the greatest integernsuch thatαn=0. It is easily verified that
|fg|=|f||g|,|f+g|≤max(|f|,|g|),
and|f+g|=max(|f|,|g|)if|f| =|g|.
For any f=
∑
n∈Zαnt
n∈F,let
f=
∑
n≥ 0
αntn, {f}=
∑
n< 0
αntn
denote respectively its polynomial and strictly proper parts. Then|{f}|< 1, and
|f| = |f|if|f|≥1, i.e. iff =O.