On a mixed Littlewood conjecture in Diophantine approximation

On a mixed Littlewood conjecture in Diohantine aroximation Yann BUGEAUD*, Michael DRMOTA & Bernard de MATHAN Abstract. In a recent aer, de Mathan Teulié asked whether lim inf q + q qα q = 0 holds for every badly aroximable real number α every rime number. We establish that if the sequence of artial quotients of a real number α satisfies a simle, combinatorial condition, then their conjecture is true for the air (α, ) with an arbitrary rime. 1. Introduction A famous oen roblem in simultaneous Diohantine aroximation is the Littlewood conjecture [8]. It claims that, for every given air (α, β) of real numbers, we have lim inf q + q qα qβ = 0, (1) where denotes the distance to the nearest integer. The first significant contribution on this question goes back to Cassels & Swinnerton-Dyer [3] who showed that (1) holds when α β belong to the same cubic field. Further exlicit examles of airs (α, β) of real numbers satisfying (1) have been given in [9,1]. Desite some recent remarkable rogress [12,5], the Littlewood conjecture remains an oen roblem. Recently, de Mathan Teulié [11] roosed a mixed Littlewood conjecture, that can be stated as follows. Let D = (d k ) k Z be a sequence of integers greater than or equal to 2. Set e 0 = 1, for any n 1, e n = 0<k n d k. For q Z, set w D (q) = su{n N : q e n Z} * Suorted by the Austrian Science Fundation FWF, grant M822-N12. 1

q D = 1/e wd (q) = inf{1/e n : q e n Z}. When D is the constant sequence equal to, where is a rime number, then D is the usual -adic value, normalized by = 1. In analogy with the Littlewood conjecture, de Mathan Teulié asked whether lim inf q qα q D = 0 (2) q + holds for every real number α. They roved that (2) holds for every quadratic irrationality α when the sequence D is bounded. In the resent aer, we are focusing on the articular case when D is the constant sequence equal to a rime number. Thus, we investigate the following conjecture. Mixed Littlewood Conjecture. For every real number α every rime number, we have lim inf q qα q = 0. (3) q + Obviously, the above conjecture holds if α is rational or has unbounded artial quotients. Thus, we only consider the case when α is an element of the set Bad of badly aroximable numbers, where Bad = {α R : inf q qα > 0}. q 1 We are concerned with the following question: Problem 1. Is there any α in Bad, which is irrational not quadratic, such that, for any rime number, the air (α, ) satisfies (3)? As briefly outlined on age 231 of [11], the answer to Problem 1 is ositive when α lies in a subset of Bad with Hausdorff dimension 1 (see also [6] for a stronger result). Nevertheless, these aroaches do not rovide any new exlicit examles of airs (α, ) satisfying (3) with α in Bad. The urose of the resent note is recisely to construct exlicitly uncountably many real numbers α in Bad such that the air (α, ) satisfies (3) for any rime number. We further extend the roblem osed by de Mathan Teulié, by considering an inhomogeneous mixed Littlewood conjecture. We ask whether lim inf q qα q y = 0 (4) q + holds for any real number α, any rime number any y in Z. It turns out that our methods allow us to establish (4) for a wide class of real numbers α in Bad. Our roofs heavily deend on -adic analysis, our key tool is the -adic logarithm function. It is not clear to us whether our Theorem 1 has a real analogue, or an analogue for formal ower series over a finite field. 2

2. Results Our main result shows that (3) holds for any real number α whose sequence of artial quotients is, in some sense, quasi-eriodic. Theorem 1. Let α be in Bad write α = [a 0 ; a 1, a 2,...]. Let T 1 be an integer b 1,..., b T be ositive integers. If there exist two sequences (m k ) k 1 (h k ) k 1 of ositive integers with (h k ) k 1 being unbounded then we have a mk +j+nt = b j, for every j = 1,..., T every n = 0,..., h k 1, lim inf q qα q y = 0. (5) q + for every rime number every y Z. If, furthermore, there exists a constant C such that m k Ch k, for k 1, (6) then we have for every rime number every y Z. lim inf q log q qα q y < +, (7) q + It is worth rehrasing the assumtion of Theorem 1 by using the terminology from combinatorics on words. Let α = [a 0 ; a 1, a 2,...] be in Bad view its sequence of artial quotients as the infinite word a = a 0 a 1 a 2... on the alhabet A = {1,..., M}, where M is an uer bound for the a i s. Theorem 1 asserts that, if there exists a finite, non-emty word B on the alhabet A such that, for any k 1, the concatenation of k coies of B occurs in the word a, then (5) holds for every rime number every y in Z. Consequently, Theorem 1 rovides an uncountable, exlicit class of badly aroximable real numbers for which the mixed Littlewood conjecture, even the inhomogeneous mixed Littlewood conjecture, is true. We further mention that, if the real number α satisfies the assumtion of Theorem 1 with sequences (m k ) k 1 (h k ) k 1 such that (6) holds, then α is either a quadratic irrationality, or a transcendental number. This follows from Theorem 3.2 of [2]. We dislay an immediate consequence of the roof of Theorem 1. Theorem 2. Let α be a quadratic real number. Let be a rime number. For any y in Z, there exist a ositive constant c(α,, y), deending only on α, y, arbitrarily large ositive integers q with qα c(α,, y)/q q y c(α,, y)/ log q. (8) 3

In articular, we have lim inf q log q qα q y < +. (7) q + The case y = 0 was already established in Théorème 2.1 from [11]. Notice that their result actually covers the more general case of a bounded sequence D of ositive integers. In view of [10], we cannot relace log q in (8) by (log q) λ with λ very large. Theorem 2 also holds for an arbitrary bounded sequence D when y is an integer. Using some of the ideas occurring in the roof of Theorem 1, we get another uncountable, exlicit class of badly aroximable real numbers for which the mixed Littlewood conjecture, even the inhomogeneous mixed Littlewood conjecture, is true. Theorem 3. Let α = [a 0 ; a 1, a 2,...] be in Bad. If for every integer h 1 there exists an integer T such that a j+nt = a j, for every j = 1,..., T every n = 0,..., h, then, for any rime number any integer y we have lim inf q qα q y < +. q + More generally, Theorem 3 holds for an arbitrary (bounded) sequence D. Its analogue in the function fields case can also easily be established. Using arguments from [12], we get that (4) holds for many real numbers α with bounded artial quotients. Theorem 4. The set of real numbers α with bounded artial quotients for which lim inf q qα q y = 0, q + for every rime number for every y in Z, has Hausdorff dimension 1. A stronger result holds when y = 0. Namely, Einsiedler & Kleinbock [6] established that the set of α for which (3) does not hold has Hausdorff dimension 0. Presumably, their aroach could be modified to get an analogous result for (4), but this is not clear to us. In view of Theorem 1, we would like to address the following roblem. Problem 2. Let ε be real with 0 < ε 1. Find a real number α in Bad, a rime number, a rational integer y such that lim inf q + q1+ε qα q y < +. (9) It follows from the -adic analogue of the Schmidt Subsace Theorem, established by Schlickewei [13], that any real number α satisfying (9) must be transcendental. Aarently, our construction does not allow us to give a ositive answer to Problem 2. 4

We end this Section by another question. Problem 3. Given α in Bad, is there a rime number such that (3) holds for the air (α, )? Aarently, there are no contribution towards Problem 3, which seems to be quite difficult. 3. Proof of Theorem 1 Preliminaries to the roof. Let (q n ) n 1 be the sequence of the denominators of the convergents of α. These numbers satisfy the recurrence relation with q 0 = 1 q 1 = 0. For n 0, set q n = a n q n 1 + q n 2 Q n = ( qn q n 1 ) We can write Q n = P n Q n 1 where P n is the matrix For j = 1,..., T, set P n = M j = ( ) an 1 1 0 ( ) bj 1 1 0 M = M T...M 1. Relacing if necessary b 1,..., b T by b 1,..., b T, b 1,..., b T, we may relace T by 2T, thus we can suose detm = 1. The matrix M is diagonalizable its eigenvalues are quadratic units ω > 1 1/ω. We have N Q(ω)/Q ω = 1. Set a = T r Q(ω)/Q. Then a > 2 is a ositive integer, ω is a root of the olynomial X 2 ax +1. We have Q mk +nt = MQ mk +(n 1)T for every 1 n h k, hence Q mk +nt = M n Q mk. The Cayley Hamilton theorem imlies that, for 2 n h k, we have Q mk +nt = M 2 Q mk +(n 2)T = (am I)Q mk +(n 2)T = aq mk +(n 1)T Q mk +(n 2)T. Hence the sequence (q mk +nt ) n 0 satisfies when 2 n h k. q mk +nt = aq mk +(n 1)T q mk +(n 2)T (10) 5

Note that we have Q mk +1+nT = M Q mk +1+(n 1)T for 0 < n < h k, where M = M 1 M T... M 2 = M 1 MM1 1. Hence, M has the same characteristic olynomial as M, the sequence (q mk +1+nT ) n 0 also satisfies the recurrence q mk +1+nT = aq mk +1+(n 1)T q mk +1+(n 2)T (10) when 2 n < h k. Consequently, we may relace m k by m k + 1. Since q mk q mk +1 are corime, we can, without any loss of generality, suose that does not divide q mk. We have q mk +nt = A k ω n + B k ω n, (11) for 0 n h k, where A k B k are given by (11) with n = 0 n = 1: A k = ωq m k +T q mk ω 2 1 (12) B k = ω q m k +T ωq mk ω 2 1. (13) We also denote by ω a zero of the olynomial X 2 ax + 1 in the algebraic closure of Q. We still denote by the -adic value extended to this field. As ω is a unit, we have ω = 1. The formulæ(12) (13) hold in Q. Since ω ω 1 are conjugated numbers in Q(ω), as well as A k B k, we can change A k into B k by changing the embedding of ω in Q (ω) it is convenient that this embedding deend uon k. We thus can suose that A k B k (i.e., q mk +T ωq mk ωq mk +T q mk ). Since we suose that A k + B k = q mk = 1, we then have 1 A k 1/ ω 2 1. The field Q (ω) is comlete. The ball G = {x Q (ω) : x 1 < 1/( 1) } is a subgrou of finite index in the multilicative grou {x Q (ω) : x = 1}. Hence, relacing again T by lt, therefore ω by ω l, with a suitable ositive integer l, we may also suose that ω 1 < 1/( 1). In the sequel, we shall make use of the -adic logarithm function, which is defined on the multilicative grou {x Q (ω) : x 1 < 1} in Q (ω) by log x = + n=1 ( 1) n 1 (x 1) n /n. We have log xy = log x + log y, for x, y in {x Q (ω) : x 1 < 1},, for every x, y in G, log x = x 1, 6

log x log y = log x y = x y 1 = x y. The constants imlied in the symbols occurring below will only deend uon b 1,..., b T on. The roofs in the cases y = 0 y 0 are rather different. First, we deal with the case y = 0, which is slightly more difficult. An auxiliary result for the case y = 0. Kee the above notations set either ε k = ω 2 1 q mk +T ωq mk or ε k = 1/2 ω 2 1 q mk +T ωq mk, in such a way that ε k < 1 is an integral ower of. Lemma 1. There exist integers x k y k, with y k = 1, such that x k q mk + y k q mk +T ε k (14) Furthermore, we have max{ x k, y k } ε 1/2 k. (15) x k q mk + y k q mk +T ω 2 1 x k + y k ω (max{ x k, y k }) 2. (16) In course of the roof of Lemma 1, we need the following version of Liouville s Lemma. Lemma 2. For any integers X Y, not both zero, we have X + Y ω (max{ X, Y }) 2. Proof. We have N Q(ω)/Q (X + Y ω) max{ X, Y } 2 N Q(ω)/Q (X + Y ω) = (X + Y ω)(x + Y/ω) X + Y ω. As N Q(ω)/Q (X + Y ω) is a non zero integer, we get N Q(ω)/Q (X + Y ω) (N Q(ω)/Q (X + Y ω) 1. It then follows that X + Y ω max{ X, Y } 2 1, as claimed. 7

Proof of Lemma 1. Let s be a ositive integer. By the igeonhole rincile, there exist integers x k,s y k,s, not both zero, such that x k,s q mk + y k,s q mk +T ε k s (17) max{ x k,s, y k,s } ε 1/2 k s/2. (18) First, we rove that there exists a non-negative integer S, deending only on ω, such that, if s > S, then y k,s cannot be divisible by s. Indeed, let σ be a ositive integer, with 0 σ s, such that y k,s is divisible by σ. Then, x k,s as well is divisible by σ. Indeed, we have y k,s q mk +T σ x k,s q mk + y k,s q mk +T σ, hence x k,s q mk σ. Since we have assumed that q mk = 1, we get x k,s σ. Setting x k,s = σ x k,s y k,s = σ y k,s, we deduce from (17) that we have from (18) that Hence, writing x k,s + q y mk +T k,s q mk ε k s+σ (19) max{ x k,s, y k,s } ε 1/2 k s/2 σ. (20) x k,s + y k,s ω = x k,s + q y mk +T k,s q mk noticing that (19) imlies that x k,s + q y mk +T k,s q mk + y k,s ( ω q m k +T q mk < q mk +T ωq mk, ), (21) we get Then by (20) If σ = s, we thus get x k,s + y k,s ω q mk +T ωq mk ε k. x k,s + y k,sω max{ x k,s, y k,s } 2 s 2σ. x k,s + y k,sω max{ x k,s, y k,s } 2 s it follows from Lemma 2 that we must have s 1, i.e., s S, where S 0 is an integral constant, deending only uon ω. Thus, if s > S, then y k,s cannot be divisible by s. Take s = S + 1. If we define σ by y k,s = σ, we then have σ < s. We define x k = σ x k,s, y k = σ y k,s, so that y k = 1. (22) 8

By (19) (20), as s = S + 1, we have x k q mk + y k q mk +T ε k (14) max{ x k, y k } ε 1/2 k. (15) To conclude, let us show that the air of integers (x k, y k ) which we have constructed satisfies (16). Indeed, as ε k < q mk +T ωq mk, we deduce from (22), (14) (21) that x k + y k ω = q mk +T ωq mk. Accordingly, conditions (14) (15) lead to Lemma 1. Comletion of the roof for the case y = 0. For any k 1, let x k y k be the integers given by Lemma 1. For any n 0, set Q k,n = x k q mk +nt + y k q mk +(n+1)t. (23) Our aim is to rove that there exists an integer n(k) at most equal to h k 1 such that Q k,n(k) 0, inf k 1 Q k,n(k) Q k,n(k) Q k,n(k) α = 0. First, we note that, for every 0 n < h k, we have Q k,n max{ x k, y k } q mk +nt We thus get Q k,n α max( x k, y k )/q mk +nt. Q k,n Q k,n α max{ x k, y k } 2. (24) Further, with A k B k being defined as in (12) (13), we can write Q k,n = C k ω n + D k ω n where C k = A k (x k + y k ω) D k = B k (x k + y k ω 1 ). Since A k 1, we deduce from Lemma 1 that C k + D k = x k q mk + y k q mk +T ω 2 1 x k + y k ω ω 2 1 C k. 9

Hence, we have C k + D k < C k, thus C k = D k. We then get 1 + D k /C k ω 2 1 < 1/( 1). (25) Write Q k,n = C k ω n (D k /C k + ω 2n ). Since we have C k x k + y k ω (max{ x k, y k }) 2, we thus get Q k,n (max{ x k, y k }) 2 D k /C k + ω 2n, by (24), Q k,n Q k,n Q k,n α D k /C k + ω 2n. (26) Now, inequality (25) enables us to use the -adic logarithm in the domain G where the equality log x log y = x y holds. As D k /C k +ω 2n = log ( D k /C k ) 2n log ω, we may also write (26) as Q k,n Q k,n Q k,n α log( D k /C k ) 2 log ω n. Now let us show that (log( D k /C k ))/(2 log ω) lies in Q. This is trivial if ω Q. In the case where ω Q, there exists a unique Q -automorhism σ of Q (ω), different from the identity. We have σ(ω) = 1/ω, σ is isometrical. We have σ(log ω) = log(σ(ω)) = log ω, σ( D k /C k ) = C k /D k, since σ(c k ) = D k. We thus have Hence, σ ( log( D k /C k ) ) = log( C k /D k ) = log( D k /C k ). ( ) log( Dk /C k ) σ 2 log ω = log( D k/c k ), 2 log ω the number log( D k /C k )/(2 log ω) of Q (ω) lies in Q. Further, this number lies in Z, since log( D k /C k ) = D k /C k + 1 ω 2 1 = 2 log ω, by (25). If t k is a ositive integer with 1 2 h k < t k 1 2 h k, then there exists an integer n, with 0 n < t k, such that log( D k /C k ) 2 log ω 10 n t k.

Relacing if necessary n by n + t k, we may ensure that log( D k /C k ) 2n log ω, hence, D k /C k + ω 2n 0. Taking either n(k) = n or n(k) = n + t k, we have thus constructed an integer n(k) satisfying 0 n(k) < h k, Q k,n(k) 0 Q k,n(k) Q k,n(k) Q k,n(k) α t k 1 h k. (27) Since the sequence (h k ) k 1 is unbounded, we have inf Q k,n(k) Q k,n(k) Q k,n(k) α = 0. k 1 It remains for us to rove that the estimate (7) holds when (6) is satisfied. Note that as the artial quotients of α are bounded, we have by (23): hence, Now, by (15) Lemma 2, we have Thus, we get log Q k,n(k) log max{ x k, y k } + m k + n(k), log Q k,n(k) log max{ x k, y k } + m k + h k. max{ x k, y k } q mk +T ωq mk 1/2 q mk. log max{ x k, y k } m k log Q k,n(k) m k + h k. Accordingly, if (6) is satisfied, then we have log Q k,n(k) h k, (7) follows from (27). The case y 0. We shall rove that for any non-zero y in Z, there exists an infinite set Q of ositive integers Q satisfying Q Qα 1 (28) lim inf Q y = 0. (29) Q Q + Note that if (29) is true for y, then it is true for My, where M is any ositive integer. Indeed, we may relace Q by MQ, while reserving (28). Then it is enough to rove (29) when y = 1. Actually we shall construct a set of ositive integers Q, with (28), such that the -adic toological closure of this set contains the unit circumference y 1 = 1. For this urose, it is enough to construct a set of Q, with (28), whose -adic closure contains the ball y 1 λ, for some ositive integer λ. Indeed, relacing then the set of Q by the µq, where µ runs among the integers 0 < µ < λ with µ = 1, the -adic closure of this set will contain the unit circumference y 1 = 1. 11

First, suose that 2. We take the sequences (m k ) k 1 (h k ) k 1 as in the statement of the theorem. The numbers A k B k are defined as in (12) (13). Recall that we suose that q mk = 1 A k B k, hence A k 1. We note that we may also suose that A k B k ω 1. Indeed, we may relace q mk either by q mk +T or q mk +2T, while reserving the condition q mk = 1, since (10) ensures that if q mk is not divisible by, then either q mk +T or q mk +2T is not divisible by. Now, if we relace q mk by q mk +T (res. by q mk +2T ), then A k is relaced by A k ω (res. by A k ω 2 ), B k is relaced by B k ω 1 (res. by B k ω 2 ). If A k B k < ω 1, then we have A k ω B k ω 1 = A k (ω ω 1 ) + (A k B k )ω 1 = A k (ω ω 1 ) ω 2 1 = ω 1 in virtue of the roerties of the logarithm function, since ω 1 < 1/( 1). In the same way, if A k B k < ω 1, then we have A k ω 2 B k ω 2 ω 1. Accordingly, by these changes, we may suose that q mk = 1 Let λ be the ositive integer such that A k B k ω 1. (30) ω 1 2 = λ+1. As q mk = 1, there exists an integer L k with 0 < L k < λ L k = 1, such that L k (A k + B k ) 1 mod λ. (31) We shall rove that the -adic closure of the set comosed by the integers L k q mk +nt, for k 1 0 n h k, contains the ball y 1 λ. Note that, L k being bounded, the integers L k q mk +nt satisfy (28), thus the result will be roved. Set L k A k = A k L kb k = B k. For 0 n h k, we have L k q mk +nt = A k ωn + B k ω n. Consider the ma ϕ k from the ball {x : x 1 < 1} of Q (ω) into Q (ω) such that Let y be a number of Q with ϕ k (x) = A kx + B kx 1. y 1 λ. (32) We shall find x Q (ω) such that ϕ k (x) = y. We must have A k x2 yx + B k = 0. Accordingly we take x = y + y 2 4A k B k 2A (33) k However we must recise the meaning of the symbol. We define the function over the ball {z : z 1 1 } in Q, with values in the same ball, by the fact that log z = 1 2 log z (that is to say z = ex log z 2, where ex u = + n=0 un n! for u 1 ). Therefore we have z 1 = z 1. (34) 12

Now, by (31) (32), we have y 2 4A kb k (A k B k) 2 = y 2 (A k + B k) 2 λ (35), by (30) A k B k 2 λ+1, we get y 2 4A k B k (A k 1 B k )2 1. (36) Moreover the number (y 2 4A k B k )/(A k B k )2 lies in Q, since in the case where ω Q, it is invariant under the above automorhism σ. Inequality (36) allows us to define the number (y 2 4A k B k )/(A k B k )2 in Q, we ut in (33) y 2 4A k B k = (A k B k) y 2 4A k B k (A k. B k )2 Then the number x is well defined in Q (ω). Further let us rove that Indeed, writing y 2 4A k B k (A k B k ) = A k B k we deduce from (34) that from (35), that x 1 ω 1. (37) y 2 4A k B k (A k 1 B k )2, y 2 4A k B k (A k B k) = y2 4A k B k (A k B k )2 A k B k, y 2 4A k B k (A k B k ) Since A k B k λ/2, it follows that λ A k B k. y 2 4A k B k (A k B k ) λ/2. As y (A k + B k ) λ, we thus get which leads to (37) since A k 1. y + y 2 4A k B k 2A k λ/2 13

Lastly we rove that log x log ω belongs to Q. Indeed, in the case where ω Q, we have σ( y 2 4A ) k B k (A k = B k )2 y 2 4A k B k (A k B k )2 since this number lies in Q. Hence, as σ(a k B k ) = B k A k, we get σ( y 2 4A k B k ) = y 2 4A k B k. We thus have It follows immediatly that ( y + y2 4A ) k σ B k 2A k = y y 2 4A k B k 2B k. σ(x) = 1/x (38) as ω also satisfies (38), we conclude as above that log x log ω belongs to Q. Moreover, as log x log ω = x 1 / ω 1, we conclude from (37) that log x log ω belongs to Z. Accordingly, given a ositive integer N, there exists an integer n, with 0 n < N, such that log x log ω n N, i.e., log x log ω n N ω 1, thus x ω n N ω 1. As B k A k 1/ ω 1, we thus get ϕ k (x) ϕ k (ω n ) N. Now ϕ k (x) = y, if 0 n h k, then ϕ k (ω n ) = L k q mk +nt. Let us select k such that h k N, we thus have found an integer n, with 0 n h k, such that L k q mk +nt y N. (39) Accordingly, the -adic closure of the set of L k q mk +nt, with k 1 0 n h k contains the ball y 1 λ. There are some minor changes when = 2. First note that, using the logarithm function over Q 2 for z 1 1/4, we get z 2 1 2 = 1 2 z 1 2. The function is then defined over the ball z 1 2 1/8 in Q 2, satisfies z 1 2 = 2 z 1 2. Also note that if A k B k 2 < 1, then A k 2 = B k 2 = 2, since A k + B k 2 = 1. Reasoning as above, we thus may suose that The number λ is then determined by A k B k 2 1 2 ω 1 2. L k is determined by 2 λ = 1 16 ω 1 2 2 L k q mk 1 mod 2 λ 14

0 < L k < 2 λ. Then, the 2-adic closure of the set of integers L k q mk +nt contains the ball {y : y 1 2 2 λ }, (39) holds. In order to obtain (7), it is enough to note that log L k q mk +nt m k + n, thus, if condition (6) is satisfied, then log L k q mk +nt h k. We may choose the integer N above such that h k / < N h k, i.e., from (39), we get (7). N h k, 4. Proof of Theorem 3 Proof of Theorem 3. Let α be as in the statement of Theorem 3. Let be a rime number. First, notice that if ω is any quadratic unit, then the index of the multilicative grou G = {x Q (ω) : x 1 < 1/( 1) } in the unit ball {x Q (ω) : x 1 = 1} is a divisor of 2 ( 2 1). Hence, we have ω 2 ( 2 1) 1 < 1/( 1). Accordingly, in the statement of Theorem 3, if we relace T by 2 ( 2 1)T, we can suose, as in the roof of Theorem 1, that the eigenvalues ω T, 1/ω T of the matrix M = M T... M 1 are quadratic units with ω 2 ( 2 1) T 1 < 1/( 1). Then, in order to rove Theorem 3 for y = 0 it is enough to consider Q n = q nt 1 for 0 n h. This sequence satisfies (10), hence we can write since Q 0 = 0. We have Hence, for 0 < s log h/ log, we have Taking s such that h/ < s h, we get Q n = A(ω n T ω n T ) A 1/ ω 2 T 1. s Q s ω2 T 1 ωt 2 1 Q s 1/h. = s. This roves the result, in virtue of (28). Let y be a non-zero integer. Consider Q n = yq nt = EωT n + F ω n T, where max{ E, F } 1/ ω 2 T 1. 15

Since Q 0 = y, we have as above which, again, roves the result by (28). Remarks on the roofs. s Q 2 y s ω2 T 1 ωt 2 1 = s, Although Theorem 2 for y = 0 is a consequence of Theorem 1, was already roved in [11], we would like to note that in this case, the roof of (7) is very simle, the use of the -adic logarithm function is actually not necessary. Indeed, the assumtions of Theorem 1 are satisfied for some ositive integer m, we have a m+j+nt = b j for every non-negative integer n for any j = 1,..., T. For any k 1, we set m k = m, we choose the integers x k = q m+t y k = q m. Then, as in the roof of Theorem 1, we consider integers Q n = q m+t q m+nt q m q m+(n+1)t. These integers satisfy (28) since m is now a fixed number. In this case, as Q 0 = 0, we have C k = D k, that is to say that we can write Q n = C(ω n ω n ), with C 0 since q m+t ωq m 0. Hence, we have Q n 0 for n > 0, Q n ω 2n 1. Then, we only have to check that ω 2s 1 s ω 2 1. This follows from an elementary induction by writing ω 2s = 1 + u s, using Newton s formula. As log Q n n, we thus see that the integers Q s satisfy (28), with log Q s s Q s s. This rovides the estimation (7). The same holds when y 0 is an integer. Indeed, by the Bezout Theorem, q m q m+1 being corime, we can take integers x y such that x q m + y q m+1 = y (where m 0 is chosen in a such way that a m+j+nt = b j for every non-negative integer n for any j = 1,... T ). Since both the sequences (q m+nt ) n 0 (q m+1+nt ) n 0 satisfy (10), then for every non-negative integer n we can write x q m+nt + y q m+1+nt = Eω n + F ω n where E F are numbers in Q (ω), with E 1/ ω 2 1 F 1/ ω 2 1. The integers Q n = x q m+nt + y q m+1+nt satisfy (28). Since ω 2s 1 s ω 2 1, we see that for n = 2 s, we have Q 2 s Q 0 s, that is to say Q 2 s y s. Further, it is easy to see that Theorem 2 Theorem 3 remain valid when relacing in (8) the -adic absolute value by D, if we take y Z. In the other arts of the roof, the role layed by the -adic logarithm function seems more dee. For instance, it is easy to rove the analogue of Theorem 2 for formal ower series over a finite field, when y is a olynomial. But we do not know whether Theorem 16

1 has an analogue in this setting. The crucial oint is the use of the -adic logarithm function, that we lose in the formal ower series case. 5. Proof of Theorem 4 Let be a rime number y be an integer. Let µ denote the Kaufman measure, whose existence has been roved in [7]. The measure µ is suorted on the set Bad. For any n 1, set q n := n + y. For any real number α we have q n q n α q n y = q n q n α n (1 + y ) q n α. Thus, (4) holds as soon as α satisfies inf q nα = 0. (40) n 1 Using the exonential growth of the sequence (q n ) n 1 a result of Davenort, Erdős, & LeVeque [4], as exlained in Section 4 of [12], we get that (40) holds for µ-almost all α. Hence, µ-almost all α satisfy (4) for any rime any integer y. Arguing then as in age 294 from [12], we obtain that the Hausdorff dimension of the set of real numbers α in Bad for which (4) holds for any rime any integer y is equal to 1, as claimed. References [1] B. Adamczewski Y. Bugeaud, On the Littlewood conjecture in simultaneous Diohantine aroximation, J. London Math. Soc. To aear. [2] B. Adamczewski Y. Bugeaud, On the Maillet Baker continued fractions. Prerint. [3] J. W. S. Cassels H. P. F. Swinnerton-Dyer, On the roduct of three homogeneous linear forms indefinite ternary quadratic forms, Philos. Trans. Roy. Soc. London, Ser. A, 248 (1955), 73 96. [4] H. Davenort, P. Erdős W. J. LeVeque, On Weyl s criterion for uniform distribution, Michigan Math. J. 10 (1963), 311 314. [5] M. Einsiedler, A. Katok E. Lindenstrauss, Invariant measures the set of excetions to the Littlewood conjecture, Ann. of Math. To aear. [6] M. Einsiedler D. Kleinbock, Measure rigidity -adic Littlewood-tye roblems. Prerint. [7] R. Kaufman, Continued fractions Fourier transforms, Mathematika 27 (1980), 262 267. [8] J. E. Littlewood, Some roblems in real comlex analysis. D. C. Heath Co. Raytheon Education Co., Lexington, Mass., 1968. 17

[9] B. de Mathan, Conjecture de Littlewood et récurrences linéaires, J. Théor. Nombres Bordeaux 13 (2003), 249 266. [10] B. de Mathan, On a mixed Littlewood conjecture for quadratic numbers, J. Théor. Nombres Bordeaux 17 (2005), 207 215. [11] B. de Mathan et O. Teulié, Problèmes diohantiens simultanés, Monatsh. Math. 143 (2004), 229 245. [12] A. D. Pollington S. Velani, On a roblem in simultaneous Diohantine aroximation: Littlewood s conjecture, Acta Math. 185 (2000), 287 306. [13] H. P. Schlickewei, The -adic Thue Siegel Roth Schmidt theorem, Arch. Math. (Basel) 29 (1977), 267 270. Yann Bugeaud Michael Drmota Université Louis Pasteur Institut für Diskrete Mathematik Mathématiques und Geometrie 7, rue René Descartes TU Wien 67084 STRASBOURG Cedex (France) A-1040 WIEN (Austria) bugeaud@math.u-strasbg.fr Bernard de Mathan Institut de Mathématiques Université Bordeaux I 351, cours de la Libération 33405 TALENCE Cedex (France) demathan@math.u-bordeaux1.fr michael.drmota@tuwien.ac.at 18