On
amixed Littlewood conjecture for quadratic
numbers
par BERNARD DE MATHAN
RÉSUMÉ. Nous étudions un
problème diophantien
simultané relié à la conjecture de Littlewood. En utilisant des minorations con- nues de formes linéaires delogarithmes p-adiques,
nous montronsqu’un
résultat que nous avonsprécédemment
obtenu, concernantles nombres
quadratiques,
est presqueoptimal.
ABSTRACT. We
study
a simultaneousdiophantine problem
re-lated to Littlewood’s conjecture.
Using
known estimates for linear forms inp-adic logarithms,
we prove that a previous result, con- cerning theparticular
case ofquadratic
numbers, is close to bethe best
possible.
1. Introduction
In a
joint
paper, with O. Teuli6[5],
we have considered thefollowing problem.
Let ,~ be a sequence ofintegers greater
than 1. Consider the sequence(rn)n>o,
where ro = 1 and rn = for n > 0. For q EZ,
set-
and
Notice that
I.IB
is notnecessarily
an absolutevalue,
but when 13 is theconstant sequence p, where p is a
prime number,
then1. 1 L3
is the usualp-adic
value.For x E
R,
we denoteby fxl
the number in(-1/2, 1~2~
such thatx - E Z. As
usual,
we putllxll
=Let a be a real number. Given a
positive integer M,
Dirichlet’s Theo-rem asserts that for any n, there exists an
integer
q, with 0q Mrn, satisfying simultaneously
theapproximation
conditionllqall 1/M
andthe
divisibility
condition i. e.I/rn- Indeed,
it isenough
toapply
Dirichlet’s Theorem to the number rna. We thus findpositive
inte-gers q with
By analogy
with Littlewood’sconjecture,
we ask whetherholds. The
problem
is trivial for arational,
and for an irrational number a, one caneasily
see[5]
that condition(1)
isequivalent
to thefollowing:
for each n E
N,
consider the continued fractionexpansion
We have
(1)
if andonly
ifHowever,
we shall not use this characterization here.We do not know whether
(1)
is satisfied for any real number a. In[5],
wehave
proved
that if we assume that the sequence Zi = isbounded, (1)
is true for everyquadratic
number a. Moreprecisely:
Theorem 1.1.
(de
Mathan and Teulié[5]) Suppose
that the sequence B is bounded. Let a be aquadratic
real number. Then there exists aninfinite
set
of integers
q > 1 withand
In
particular,
we haveAs
msual,
forpositive
functions x and y, the notation x C y means that there exists apositive
constant C such that x CCy.
In our lecture at
Graz,
for the "Journ6esArithm6tiques 2003" ,
it wasdiscussed whether the factor
Inq
in(3)
is bestpossible.
We do not knowthe answer to this
question,
but we shall prove:Theorem 1.2. Assume that the sequence L3 is bounded. Let a be a real
quadratic number,
and let S be a setof integers
q > 1 withThen there exists a constant A =
A(S)
such thatfor
any q E S.One may expect that
(4)
holds forany A
>1,
but we are not able toprove this. We do not even know whether there exists a real number A for which
(4)
holds for any set S ofintegers q
> 1satisfying (2). Indeed,
Theorem 1.2 does not ensure that sups
A(S)
+oo.There is some
analogy
between thisproblem,
and the classical simultane-ous
Diophantine approximation.
Forinstance,
let us recall Peck’s Theorem.Let n be an
integer greater
than1,
and let al,..., I an, i be n numbers in areal
algebraic
number field ofdegree
n -I- 1 overQ.
Then it wasproved by
Peck
[7]
that there exists an infinite set ofintegers
q > 1 withfor
Assume that
1, aI,
..., an arelinearly independent
overQ,
and let S be aninfinite set of
integers q
>1,
withfor each 1 k n. Then we have
proved
in[3]
that there exists a constantK = such that
Theorem 1.2 can be
regarded
as ananalogue
of this result with n =1,
and its
proof
is similar.2. Proof of the result 2.1. Some rational
approximations
of a.In the
quadratic
field~(a),
there exists a unit w of infinite order. Re-placing,
if necessary, cvby w2
or1/ w2,
we may suppose w > 1. In hisoriginal work,
Peck uses units which are"large"
and whose otherconjugates
are "small" and close to be
equal. Here,
Peck’s units arejust
thewm,s,
with rn e N. We shall use these units in order to describe the rational
approximations
of a whichsatisfy (2).
Denote
by
ao = id and ai = a theautomorphisms
of~(a).
Asusual,
we denote
by
Tr the trace form ao + cri. The basis(1, a)
ofadmits a dual basis
(/30, (3l)
for thenon-degenerate Q-bilinear
form(x, y)
~---~ onQ(a).
That meansthat,
if we set ao = 1 and al = a,we have =
6ki ,
=0,1
and 1 =0, l,
where611
=1,
and6ki
= 0l. Here it is easy to calculate
(3o = 2013 ~’~~ and (3¡
= ~_)~~~ . Hence,
if we
put
, ,
where q
andq’
are rationalnumbers,
we haveAlso notice that
(5)
and(6) imply
thatLet D be a
positive integer
such thatDa, D ,
I arealge-
braic
integers.
The notation A ~
B,
where A and B arepositive quantities,
means thatLemma 2.1. Let 1 be a
positive
number inQ(a).
Let A be apositive integer
such that~1
is analgebraic integer.
For eachm E N, define
therational number
Then
Aq
is a rationalinteger,
one has q > 0 when m islarge,
and theintegers DAq satisfy (2).
Proof.
Also define- -
As
A-yw’
and arealgebraic integers, Aq
andDAq’
are rationalintegers.
As = wehave q
= +o-(~y)w-’~’2,
hence q > 0 as soon asw2m
>-,7 (7)/7,
and thenFrom
(7),
we getqa - q’
=(a -
henceAs
DAq
andDAq’
areintegers,
it follows from(10)
that forlarge
m wehave
~~DOqa~~
=DD~qa-q’~,
andby (9)
and(10),
theintegers DAq satisfy
(2).
Conversely:
Lemma 2.2. Let S be a set
of positive integers
qsatisfying (2~ .
Thenthere exists a
finite
set rof numbers 7
E~ 0,
such thatfor
any q ES,
there E r and m E N such thatProof.
For q Es,
letm(q)
= m be thepositive integer
such thatWe thus have cv’~’2 ~ q.
Let q’
be the rationalinteger
such that
~qa ~
=q’ .
SetFirst,
notice thatD,
is analgebraic integer.
From(5),
weget (8). Writing
we see
that q
> 0when q
islarge,
andycv’’2
q. As we have c,’n q, wethus
get q
1. We also havehence, by (2), Io-()’) I
Wwmlq,
andthus, ]
W 1.Then,
asD-y
is analgebraic integer
inQ(a),
and~Q(7)~)
C1,
the set of theq’s
isfinite.
2.2. End of
proof.
Denote
by
P the set of allprime
numbersdividing
one of thebk.
Sincewe assume that the sequence
(bk)
isbounded,
this set is finite. Forp E P,
we extend the
p-adic
absolute value to~(a).
Thecompletion
of this field isQp(a).
Asabove,
let w be a unit inQ(a)
with w > 1. Note that = 1.The ball
{~
E is asubgroup
of finite index in themultiplicative
groupIx
EQp(a); Ixlp
=If. Hence, replacing
by w",
where n is a suitablepositive integer,
we may also suppose thatIw -
for everyp E P.
We shall use the
p-adic logarithm function,
which is defined on the mul-tiplicative
group{j-
ECp; I x - 1}
CCp by
This function satisfies
We prove:
Lemma 2.3.
Let p
be a numberof P.
be apositive
numberof Q(a) .
For
m E N,
setThen, if
we have
for large
rrt;if
then
Proof.
Recall that q > 0 when m islarge (Lemma 2.1).
From thedefinition,
if 16 - I lp then, by (11),
we writeIw2m-8lp
and we obtain
(12).
Accordingly,
in order to achieve theproof
of theresult,
we shall useknown lower bounds for linear forms in
p-adic logarithms.
Forinstance,
itfollows from
[8]
that:Lemma 2.4.
(K.
Yu[8])
Let xand y
bealgebraic
numbers inCp,
withIx - P-,/(P-1)
andly - lip p-’I(P-1).
Then there exists a realconstant such that
for
anypair (k, ~) of
rationalintegers
withNote that this result is
trivial,
with r, =1,
iflog x
andlog yare
notlinearly independent
overQ,
andlog x =1= 0, Le, x =1=
1.Indeed,
ifa log x
=b log y,
where a and b are rationalintegers
withb # 0,
then we writeHence we »
1,
when klog x y =1=
0.We can then achieve the
proof
of Theorem 1.2.Applying
Lemma2.2,
wecan suppose that the set r contains a
unique element 7
>0, i.e.,
for any q ES,
there exists Tn e N such that we have(8).
It follows from Lemma2.3 and 2.4 that there exists a constant such that
Iqlp
W m-K(one
maytake K = 0 if
10";1’)
+lip
>p-l~~p-1> ).
As q ~I.A)m,
hence m ~Inq,
weget
Iqlp
» Now set K _ ~~,(the
constant ~~, maydepend
uponp E P).
Note thatTIpEP lqlp. Indeed, putting
=l/rn,
we haveq E hence and
I1PEP lqlp
Cflpep 1/rl"
We thusget
(4)
with A =Epep
Kp, and Theorem 12 isproved.
2.3. A remark.
Note that one may also use Lemma 2.3 for
solving
theopposite problem.
For
simplicity,
consider the case where is thep-adic
value for aprime
number p. If we take a
positive number -y
EQ(a)
such that~(7) _
2013~, forinstance, q
= a -a(a) (one
mayreplace
aby
-a, and so, we can suppose a - >0),
then we have =0,
andby (12),
we
get imlp. By
Lemma2.1,
there exists apositive integer
A such that for every
large
m, the numbers q =q(m)
= arepositive integers satisfying (2).
For m =ps
with s EN,
weget Imlp
=hence
lqlp
~ Since m ~lnq,
we have thusproved
that there existsan infinite set of
integers q
> 1satisfying (2)
and(3) (which
is Theorem1.1).
In this way we obtainintegers q
> 1satisfying (2)
and such thatIqlp
One can ask whether there exists an infinite set of
integers q
> 1 satis-fying (2),
withGiven a
positive decreasing
sequence(Em)
+oo, ap-adic
version[4]
of Khintchine’s Theorem ensures that for almost all x EZp,
there existinfinitely
manypositive integers
m suchthat I x -
em .One often considers as reasonable the
hypothesis
that agiven "special"
irrational number x E
Zp
satisfies thiscondition,
with Em =1/(mlnrrc)
for m > 1
(which
is false if x EZp
nQ,
since in this case, we haveIx -
W for mlarge).
Let us prove that we canchoose -y
> 0 inQ(a),
, withd
l’ +11
PIw - 11
p, such that log w is an irrationalnumber in
Zp.
In order to make thisobvious,
we prove:Lemma 2.5. There
exists ç
EQ(a)
suchthat ~
is not aunit, NQ(a):QÇ
=1,
and
lçlp
= 1.Proof.
The number is a root of theequation
+ 1 =0,
where S isa rational
integer,
S = Tr w. Thenumber ~
must be a root of anequation
~2-t~-f 1 = 0,
where t is a rational number for which there exists apositive
rational number p such
that t2 -
4 =p2(,S’2 - 4).
Suchpairs (t, p)
can beexpressed by using
a rational parameter 0:Let us show that we can choose 0 E
Q*
suchthat t (j. Z
and 1.It is
enough
to take 0 = p. As we haveS2
>4,
henceS2 >
9 and(S2 - 4)p2 -
1 >4, t
cannot be aninteger
for this choice of 0. But we have1,
since~(S’2 - 4)p2 _ lip
= 1. Then there exists anumber ~
EQ(a)
such that
~2 - t~ + 1 = 0, and ~
is neither a rationalnumber,
since p >0,
nor an
algebraic integer, since t g
Z. Then we have =1,
and
lçlp
= 1 because either conditionlçlp
1 orlçlp
> 1 wouldimply Replacing ~ by Çn,
where rc is a suitablepositive integer,
we thus may finda satisfying
Lemma2.5,
withmoreover
-lip Iw - lip.
Then wehave
] log g]p
Further let us prove thatlog wE Qp.
Indeed that is trivial if a EQp,
since in thiscase ~
and lie inQp,
hence so dolog ~
and If
Qp (a)
hasdegree
2 overQp,
thenlog~
andlogw
lie inQp (a).
But Q can be extended into a continuous
Qp-automorphism
ofQp(a) ,
andwe
get
since = = 1.That proves that E
Qp,
and since] logg]p
we concludethat
Zp. Lastly, log wis
not a rationalnumber, since ~
is not aunit.
Now, by
Hilbert’sTheorem,
thereexists -y
EQ(a),
with ~y >0,
such
that g = ~cr(7)/~’
We thus have found ~y > 0 inQ(ca),
such thatand is an irrational element
ofZp.
Underi p 2 log w p
the above
hypothesis,
it would existinfinitely
manyintegers
m > 1 with1/(m log m), and, by (12),
we could obtain an infinite 2 log wset of
integers q
>1, q
= where A is apositive integer, satisfying (2)
and such thatIqlp
q n 9* Inparticular, (3’)
would be satisfied.3. Conclusion
For a sequence ,13
bounded,
the Roth-Ridout Theorem[6]
allows us tosee that for any irrational
algebraic
real number a, thus inparticular
for aquadratic,
we have:(see [5]).
Of course, our method is far fromenabling
us to prove that there exists a real constant A such thatWe can
only study
theapproximations
withqllqall
« 1. It seems difficultto
study approximations
in the"orthogonal
direction"1,
with forinstance, q
=p’,
for aprime
number p. For suchapproximations,
it is notknown whether
infnEN IIpnaII
= 0holds,
neither if there exists A such that0. It is very difficult to obtain more
precise
results than the Roth-Ridout Theorem(see [1]).
Even for rational
approximations satisfying (2),
we are not able to provethat the constants
A(S)
are bounded. This is related to Lemma 2.4. It would be necessary to prove that there exists a real constant K for which this Lemma holds for x = cv and for any y EQ(a) with Iy -lip p-1/(p-i)
and= 1. There exist many effective estimates of
I k log
x + tlog y lp (see
for instance[2]
and[8]),
butthey
do notprovide
the needed result. Itseems difficult to take the
particular
conditionsrequired
into account.Acknowledgements.
The author thankswarmly
Michel Waldschmidt for theirlighting
conversations about linear forms inlogarithms,
at Graz. Theauthor also thanks the referee.
References
[1] M. BAUER, M. BENNETT, Applications of the hypergeometric method to the generalized Ramanujan-Nagell equation. Ramanujan J. 6 (2002), 209-270.
[2] Y. BUGEAUD, M. LAURENT, Minoration effective de la distance p-adique entre puissances de nombres algébriques. J. Number Theory 61 (1996), 311-342.
[3] B. DE MATHAN, Linear forms in logarithms and simultaneous Diophantine approximation.
(To appear).
[4] B. DE MATHAN, Approximations diophantiennes dans un corps local. Bull. Soc. math.
France, Mémoire 21 (1970).
[5] B. DE MATHAN, O. TEULIÉ, Problèmes diophantiens simultanés. Monatshefte Math. 143
(2004), 229-245.
[6] D. RIDOUT, Rational approximations to algebraic numbers. Mathematika 4 (1957), 125-
131.
[7] L. G. PECK, Simultaneous rational approximations to algebraic numbers. Bull. Amer.
Math. Soc. 67 (1961), 197-201.
[8] K. Yu, p-adic logarithmic forms and group varieties II. Acta Arith. 89 (1999), 337-378.
Bernard DE MATHAN Universite Bordeaux I
UFR Math-Info. Laboratoire A2X 351 cours de la Lib6ration 33405 Talence, France
E-mail : Bernard . de-Mathan@math . u-bordeaux l . fr