1 2 Journal of Integer Sequences, Vol. 17 (2014), 3 Article 14.8.3 47 6 23 11 Powers of Two as Sums of Two Lucas Numbers Jhon J. Bravo Mathematics Department University of Cauca Street 5 No. 4-70 Popay´an, Cauca Colombia [email protected] Florian Luca School of Mathematics University of the Witwatersrand P. O. Box Wits 2050 South Africa and Mathematical Institute UNAM Juriquilla Santiago de Quer´etaro 76230 Quer´etaro de Arteaga Mexico [email protected] Abstract Let (Ln)n≥0 be the Lucas sequence given by L0 = 2, L1 = 1 and Ln+2 = Ln+1 + Ln for n 0. In this paper, we are interested in finding all powers of two which are ≥ a sums of two Lucas numbers, i.e., we study the Diophantine equation Ln + Lm = 2 in 1 nonnegative integers n, m, and a. The proof of our main theorem uses lower bounds for linear forms in logarithms, properties of continued fractions, and a version of the Baker- Davenport reduction method in diophantine approximation. This paper continues our previous work where we obtained a similar result for the Fibonacci numbers. 1 Introduction Let (Fn)n≥0 be the Fibonacci sequence given by F0 = 0, F1 = 1 and Fn+2 = Fn+1 + Fn for all n 0. The Fibonacci numbers are famous for possessing wonderful and amazing properties.≥ They are accompanied by the sequence of Lucas numbers, which is as important as the Fibonacci sequence. The Lucas sequence (Ln)n≥0 follows the same recursive pattern as the Fibonacci numbers, but with initial conditions L0 = 2 and L1 = 1. The study of properties of the terms of such sequences, or more generally, linear recurrence sequences, has a very long history and has generated a huge literature. For the beauty and rich applications of these numbers and their relatives, one can see Koshy’s book [9]. For example, a remarkable property of the Fibonacci sequence is that 1, 2 and 8 are the only Fibonacci numbers which are powers of 2. One proof of this fact follows from Carmichael’s primitive divisor theorem [6], which states that for n greater than 12, the nth Fibonacci number Fn has at least one prime factor that is not a factor of any previous Fibonacci number (see the paper of Bilu, Hanrot, and Voutier [2] for the most general version of the above statement). Similarly, it is well known that 1, 2 and 4 are the only powers of 2 that appear in the Lucas sequence. The problem of finding all perfect powers in the Fibonacci sequence and the Lucas se- quence was a famous open problem finally solved in 2006 in a paper in Annals of Mathematics by Bugeaud, Mignotte, and Siksek [3]. In their work, they applied a combination of Baker’s method, the modular approach and some classical techniques to show that the only per- fect powers in the Fibonacci sequence are 0, 1, 8 and 144, and the only perfect powers in the Lucas sequence are 1 and 4. A detailed account of this problem can be found in [3, Section 10]. In our recent paper [5], we found all powers of 2 which are the sums of at most two Fibonacci numbers. Specifically, we proved the following. a Theorem 1. The only solutions of the Diophantine equation Fn + Fm = 2 in positive integers n,m and a with n m are given by ≥ 2F1 =2, 2F2 =2, 2F3 =4, 2F6 = 16, and F2 + F1 =2, F4 + F1 =4, F4 + F2 =4, F5 + F4 =8, F7 + F4 = 16. In this paper, we prove an analogue of Theorem 1 when the sequence of Fibonacci numbers is replaced by the sequence of the Lucas numbers, i.e., we extend our previous work [5] and 2 determine all the solutions of the Diophantine equation a Ln + Lm =2 (1) in nonnegative integers n m and a. Similar problems have≥ recently been investigated. For example, repdigits which are sums of at most three Fibonacci numbers were found by Luca [12]; Fibonacci numbers which are sums of two repdigits were obtained by D´ıaz and Luca [7], while factorials which are sums of at most three Fibonacci numbers were found by Luca and Siksek [11]. We prove the following result. Theorem 2. All solutions of the Diophantine equation (1) in nonnegative integers n m and a, are ≥ 2L0 =4, 2L1 =2, 2L3 =8, L2 + L1 =4, L4 + L1 =8 and L7 + L2 = 32. Let us give a brief overview of our strategy for proving Theorem 2. First, we rewrite equation (1) in suitable ways in order to obtain two different linear forms in logarithms which are both nonzero and small. Next, we use a lower bound on such nonzero linear forms in two logarithms due to Laurent, Mignotte, and Nesterenko as well as a general lower bound due to Matveev to find an absolute upper bound for n; hence, an absolute upper bound for m and a, which we then reduce using standard facts concerning continued fractions. In this paper, we follow the approach and the presentation described in [5]. 2 Auxiliary results Before proceeding further, we recall that the Binet formula n n Ln = α + β holds for all n 0, ≥ where 1+ √5 1 √5 α := and β := − 2 2 2 are the roots of the characteristic equation x x 1=0of(Ln)n≥0. This will be an important ingredient in what follows. In particular,− the− inequality n−1 n α Ln 2α (2) ≤ ≤ holds for all n 0. In order to≥ prove Theorem 2, we need a result of Laurent, Mignotte, and Nesterenko [10] about linear forms in two logarithms. But first, some notation. 3 Let η be an algebraic number of degree d with minimal polynomial d d d−1 (i) a x + a x + + ad = a (X η ), 0 1 ··· 0 − i=1 Y (i) where the ai’s are relatively prime integers with a0 > 0 and the η ’s are conjugates of η. Then d 1 h(η)= log a + log max η(i) , 1 d 0 {| | } i=1 ! X is called the logarithmic height of η. In particular, if η = p/q is a rational number with gcd(p,q)=1 and q > 0, then h(η) = logmax p ,q . The following properties of the logarithmic{| height,| } which will be used in the next section without special reference, are also known: h(η γ) h(η)+ h(γ)+log2. • ± ≤ h(ηγ±1) h(η)+ h(γ). • ≤ h(ηs)= s h(η). • | | With the above notation, Laurent, Mignotte, and Nesterenko [10, Corollary 1] proved the following theorem. Theorem 3. Let γ1,γ2 be two non-zero algebraic numbers, and let log γ1 and log γ2 be any determinations of their logarithms. Put D =[Q(γ1,γ2): Q]/[R(γ1,γ2): R], and Γ := b log γ b log γ , 2 2 − 1 1 where b1 and b2 are positive integers. Further, let A1,A2 be real numbers > 1 such that log γi 1 log Ai max h(γi), | |, , i =1, 2. ≥ D D Then, assuming that γ1 and γ2 are multiplicatively independent, we have 21 1 2 log Γ > 30.9 D4 max log b′, , log A log A , | | − · D 2 1 · 2 where b b b′ = 1 + 2 . D log A2 D log A1 We shall also need the following general lower bound for linear forms in logarithms due to Matveev [13] (see also the paper of Bugeaud, Mignotte, and Siksek [3, Theorem 9.4]). 4 Theorem 4 (Matveev’s theorem). Assume that γ1,...,γt are positive real algebraic numbers in a real algebraic number field K of degree D, b1,...,bt are rational integers, and b1 bt Λ := γ γt 1, 1 ··· − is not zero. Then t+3 4.5 2 Λ > exp 1.4 30 t D (1 + log D)(1 + log B)A At , | | − · · · 1 ··· where B max b ,..., bt , ≥ {| 1| | |} and Ai max Dh(γi), log γi , 0.16 , for all i =1,...,t. ≥ { | | } In 1998, Dujella and Peth˝oin [8, Lemma 5 (a)] gave a version of the reduction method based on the Baker-Davenport lemma [1]. To conclude this section of auxiliary results, we present the following lemma from [4], which is an immediate variation of the result due to Dujella and Peth˝ofrom [8], and will be one of the key tools used in this paper to reduce the upper bounds on the variables of the equation (1). Lemma 5. Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational γ such that q > 6M, and let A, B, µ be some real numbers with A > 0 and B > 1. Let ǫ := µq M γq , where denotes the distance from the nearest integer. || || − || || ||·|| If ǫ> 0, then there is no solution to the inequality 0 < uγ v + µ < AB−w, − in positive integers u,v and w with log(Aq/ǫ) u M and w . ≤ ≥ log B 3 The Proof of Theorem 2 Assume throughout that equation (1) holds. First of all, observe that if n = m, then the a−1 original equation (1) becomes Ln = 2 . But the only solutions of this latter equation are (n,a) (0, 2), (1, 1), (3, 3) and this fact has already been mentioned in the Introduction. So, from∈ { now on, we assume} that n>m. If n 200, then a brute force search with Mathematica in the range 0 m<n 200 gives the≤ solutions (n,m,a) (2, 1, 2), (4, 1, 3), (7, 2, 5) .