A Diophantine equation in k{generalized Fibonacci numbers and repdigits Jhon J. Bravo Departamento de Matem´aticas Universidad del Cauca Calle 5 No 4{70, Popay´an,Colombia E-mail: [email protected] Carlos Alexis G´omezRuiz Departamento de Matem´aticas Universidad del Valle 25360 Calle 13 No 100-00, Cali, Colombia E-mail: [email protected] Florian Luca School of Mathematics University of the Witwatersrand Private Bag X3 Wits 2050 Johannesburg, South Africa Max Planck Institute for Mathematics Vivatsgasse 7 53111 Bonn, Germany Department of Mathematics Faculty of Sciences University of Ostrava 30 Dubna 22 701 03 Ostrava 1, Czech Republic E-mail: [email protected] Abstract 2010 Mathematics Subject Classification: 11B39; 11J86. Key words and phrases: Generalized Fibonacci numbers, lower bounds for nonzero linear forms in logarithms of algebraic numbers, repdigits. 1 2 J. J. Bravo, C. A. G´omezand F. Luca (k) The k−generalized Fibonacci sequence fFn gn starts with the value 0;:::; 0; 1 (a total of k terms) and each term afterwards is the sum of the k preceding terms. In the present paper, we study on members of k{generalized Fibonacci sequence which are sum of two repdigts, extending a result of D´ıazand Luca [5] regarding Fibonacci numbers with the above property. 1 Introduction Given an integer k ≥ 2, we consider the k{generalized Fibonacci sequence (k) (k) or, for simplicity, the k{Fibonacci sequence F := fFn gn≥2−k given by the recurrence (k) (k) (k) (k) (1.1) Fn = Fn−1 + Fn−2 + ··· + Fn−k for all n ≥ 2; (k) (k) (k) (k) with the initial conditions F−(k−2) = F−(k−3) = ··· = F0 = 0 and F1 = 1. (k) We shall refer to Fn as the nth k−Fibonacci number. We note that in fact each choice of k produces a distinct sequence which is a generalization of the usual Fibonacci sequence fFngn≥0, obtained for k = 2. Cooper and Howard [4] proved the following formula. For k ≥ 2 and n ≥ k + 2, n+k b k+1 c−1 (k) n−2 X n−(k+1)j−2 (1.2) Fn = 2 + Cn;j 2 ; j=1 where n − jk n − jk − 2 C = (−1)j − : n;j j j − 2 a In the above, we used the convention that b = 0 if either a < b or if one of a or b is negative and we denoted by bxc the greatest integer less than or equal (k) n−2 to x. We have Fn < 2 for all n ≥ k + 2 (see [2]). Furthermore, the first (k) (k) minf0;n−2g k + 1 non{zero terms in F are powers of two, namely Fn = 2 for all 1 ≤ n ≤ k + 1. We next recall some facts and properties of the k−Fibonacci sequence which will be used later. First, it is known that the characteristic polynomial of F (k), namely k k−1 Ψk(x) = x − x − · · · − x − 1; is irreducible over Q[x] and has just one zero outside the unit circle. Through- out this paper, α := α(k) denotes that single zero, which is a Pisot number of degree k since the other zeros of the characteristic polynomial Ψk(x) are k{Fibonacci numbers as sum of two repdigits 3 strictly inside the unit circle (see, for example, [14], [15] and [16]). Moreover, it is known from Lemma 2.3 in [11] that (1.3) 2(1 − 2−k) < α(k) < 2; holds for all k ≥ 2; a fact rediscovered by Wolfram [16]. To simplify notation, we will omit the dependence on k of α. It was proved in [2] that n−2 (k) n−1 (1.4) α ≤ Fn ≤ α holds for all n ≥ 1: We now consider the function fk(z) := (z−1)= (2 + (k + 1)(z − 2)) for k ≥ 2. Dresden and Du remarked in [6] that k (k) X (i) (i)n−1 (k) n−1 1 (1.5) F = fk(α )α ; F − fk(α)α < ; n n 2 i=1 (1) (k) where α := α ; : : : ; α are the zeros of Ψk(x). The expression on the left (k) hand side is known as Binet{like formula for Fn . Furthermore, the inequality on the right{hand side in (1.5), shows that the contribution of the zeros which (k) are inside the unit circle to Fn is very small. −k From the formula of fk(z), is easy see that if z 2 (2(1 − 2 ); 2), then @zfk(z) < 0. Thus, from inequality (1.3), we conclude that −k 1=2 = fk(2) ≤ fk(α) ≤ fk 2(1 − 2 ) ≤ 3=4; for all k ≥ 3: p (i) Also, f2((1 + 5)=2) = 0:72360 : : : < 3=4. Further, if z = α with i = (i) 2; : : : ; k, then jfk(α )j < 1 for all k ≥ 2. Indeed, this follows for k ≥ 3 from the fact that jα(i)j < 1, so jα(i) − 1j < 2, and j2 + (k + 1)(α(i) − 2)j > k − 1. p Also, f2((1 − 5)=2) = 0:2763 :::. Thus, we conclude that for all k ≥ 2 we have both (1.6) 1=2 ≤ fk(α) ≤ 3=4 and (i) (1.7) fk(α ) < 1 for all i = 2; : : : ; k: In this paper, we study a problem regarding the representation of k{ Fibonacci numbers as sums of repdigits. Recall that a positive integer is called a repdigit if it has only one distinct digit in its decimal expansion. In particular, such number has the form a(10m − 1)=9 for some m ≥ 1 and 1 ≤ a ≤ 9. In 2000, Luca [10] showed that 55 is the largest repdigit Fibonacci number. Marques [12] proved in 2013 that 44 is the largest repdigit in the 4 J. J. Bravo, C. A. G´omezand F. Luca Tribonacci sequence (k = 3). The same year, Bravo and Luca [5] showed that there are no repdigits having at least 2 digits in any k{generalized Fibonacci sequence for any k > 3, confirming a conjecture of Marques. In 2011, D´ıaz y Luca [5] found all Fibonacci numbers as sum of two repdigts. Here, we study an analogue of the problem of D´ıazand Luca when the sequence of Fibonacci numbers is replaced by the sequence of k{generalized Fibonacci numbers. More precisely, we have the following result. Main Theorem. For k ≥ 3 and n ≥ k + 2, the Diophantine equation 10m − 1 10l − 1 (1.8) F (k) = a + b ; 1 ≤ a; b ≤ 9; n 9 9 has only 17 positive integer solutions (n; k; m; l; a; b) with m ≥ maxfl; 2g: (3) (3) (3) F6 = 13 = 11 + 2 F7 = 24 = 22 + 2 F9 = 81 = 77 + 4 (4) (4) (4) F6 = 15 = 11 + 4 F7 = 29 = 22 + 7 F8 = 56 = 55 + 1 (4) (5) (5) F9 = 108 = 99 + 9 F7 = 31 = 22 + 9 F8 = 61 = 55 + 6 (5) (6) (7) F9 = 120 = 111 + 9 F8 = 63 = 55 + 8 F12 = 1004 = 999 + 5 (3) (3) (6) F8 = 44 = 11 + 33 F8 = 44 = 22 + 22 F12 = 976 = 888 + 88 (8) (9) F10 = 255 = 222 + 33 F12 = 1021 = 999 + 22 (k) n−2 On the other hand, for n < k + 2, Fn is the power of two 2 and the only solutions of (1.8) with m ≥ maxfl; 2g, are (k) (k) F6 = 16 = 11 + 5 (k ≥ 5) and F8 = 64 = 55 + 9 (k ≥ 7): We clarify that the condition m ≥ maxfl; 2g ≥ 2, in the above theorem, (k) is only meant to insure that Fn has at least 2 digits, and so to avoid small numbers which are the sums of two one-digit numbers. 2 Upper bounds for the solutions of (1.8) We begin our work with the case n ≥ k+2. Assume throughout that equation (1.8) holds. Combining the fact that 10m − 1 10l − 1 10m−1 < a + b = F (k) 9 9 n with inequality (1.4), we have log α (2.1) l ≤ m < (n − 1) + 1: log 10 We need to bound k and n. k{Fibonacci numbers as sum of two repdigits 5 2.1 A polynomial upper bound on n in terms of k Using (1.5), we obtain from (1.8) that l−m n−1 a + b10 m 1 a + b 5 (2.2) fk(α)α − 10 < + ≤ : 9 2 9 2 n−1 Dividing both sides of the above inequality by fk(α)α , we obtain l−m −1 a + b10 −(n−1) m 5 (2.3) fk(α) α 10 − 1 < ; 9 αn−1 where we used the fact that fk(α) > 1=2. We put −1 l−m γ1 := fk(α) (a + b10 )=9; γ2 := α; γ3 := 10; (2.4) b1 := 1; b2 := −(n − 1); b3 := m; b1 b2 b3 Λ1 := γ1 · γ2 · γ3 − 1: So, by (2.3), 5 (2.5) jΛ j < : 1 αn−1 Our next step will be to find a lower bound for jΛ1j. For this purpose we use the following result of Matveev (see [13] or Theorem 9.4 in [3]). Lemma 1. Let K be a number field of degree D over Q; γ1; : : : ; γt be positive real numbers of K, and b1; : : : ; bt rational integers. Put b1 bt Λ := γ1 ··· γt − 1 and B ≥ maxfjb1j;:::; jbtjg: Let Ai ≥ maxfDh(γi); j log γij; 0:16g be real numbers, for i = 1; : : : ; t: Then, assuming that Λ 6= 0, we have t+3 4:5 2 jΛj > exp(−1:4 × 30 × t × D (1 + log D)(1 + log B)A1 ··· At): In the above and in what follows, for an algebraic number η of degree d over Q and minimal primitive polynomial over the integers d Y (i) f(X) := a0 (X − η ) 2 Z[X] i=1 with positive leading coefficient a0, we write h(η) for its logarithmic height, given by d ! 1 X h(η) := log a + log maxfjη(i)j; 1g : d 0 i=1 6 J.