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Repdigits As Products of Two Fibonacci Or Lucas Numbers

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Repdigits As Products of Two Fibonacci Or Lucas Numbers Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 https://doi.org/10.1007/s12044-020-0554-0 Repdigits as products of two Fibonacci or Lucas numbers FATIH˙ ERDUVAN and REFIK˙ KESKIN˙ ∗ Mathematics Department, Sakarya University, Sakarya, Turkey ∗ Corresponding author. E-mail: [email protected]; [email protected] MS received 12 October 2018; revised 19 November 2019; accepted 2 December 2019 Abstract. In this study, we show that if 2 ≤ m ≤ n and Fm Fn represents a repdigit, then (m, n) belongs to the set {(2, 2), (2, 3), (3, 3), (2, 4), (3, 4), (4, 4), (2, 5), (2, 6), (2, 10)}. Also, we show that if 0 ≤ m ≤ n and Lm Ln represents a repdigit, then (m, n) belongs to the set (0, 0), (0, 1), (1, 1), (0, 2), (1, 2), (2, 2), (0, 3), . (1, 3), (1, 4), (1, 5), (2, 5), (3, 5), (4, 5) Keywords. Fibonacci number; Lucas number; repdigit; Diophantine equations; linear forms in logarithms. Mathematics Subject Classification. 11B39, 11J86, 11D61. 1. Introduction Let (Fn) and (Ln) be the sequences of Fibonacci and Lucas numbers given by F0 = 0, F1 = 1, L0 = 2, L1 = 1, Fn = Fn−1 + Fn−2 and Ln = Ln−1 + Ln−2 for n ≥ 2, respectively. Binet formulas for Fibonacci and Lucas numbers are αn − βn n n Fn = √ and Ln = α + β , √ 5 √ 1 + 5 1 − 5 where α = and β = , which are the roots of the characteristic equations 2 2 x2 −x −1 = 0. It can be seen that 1 <α<2,−1 <β<0 and αβ =−1. For more about Fibonacci and Lucas sequences with their applications, see [7]. The relation between n-th Fibonacci number Fn and α is given by n−2 n−1 α ≤ Fn ≤ α (1) for n ≥ 1. Also, the relation between n-th Lucas number Ln and α is given by n−1 n α ≤ Ln ≤ 2α . (2) The inequality (1) and (2) can be proved by induction. A repdigit is a non-negative inte- ger whose digits are all equal. Investigation of the repdigits in the second-order linear © Indian Academy of Sciences 0123456789().: V,-vol 28 Page 2 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 recurrence sequences has been of interest to mathematicians. In [8], Luca has found all Fibonacci and Lucas numbers which are repdigits. The largest repdigits in Fibonacci and Lucas sequences are F5 = 55 and L5 = 11. Later, in [1], it was shown that the largest Fibonacci number which is a sum of two repdigits is F20 = 6765 = 6666 + 99. Further- more, in [9], Luca found all repdigits which are sums of three Fibonacci numbers. In this paper, we study the Diophantine equations d · (10k − 1) Fm Fn = (3) 9 and d · (10k − 1) Lm Ln = , (4) 9 where d and k are non-negative integers with 1 ≤ d ≤ 9. Assuming that 2 ≤ m ≤ n,itis shown that if Fm Fn is a repdigit, then m ≤ 4 and n ≤ 10. Also, it is shown that if Lm Ln is a repdigit, then m ≤ 4 and n ≤ 5for0≤ m ≤ n. In Section 2, we introduce necessary lemmas and theorems. Then we prove our main theorems in Section 3. 2. Auxiliary results As in [5], we use Baker’s theory of linear forms to solve equations (3) and (4)inm and n for non-negative integers d and k. Since such bounds are of crucial importance in effectively solving the Diophantine equations of similar form, we start by recalling some basic notions from algebraic number theory. Let η be an algebraic number of degree d with minimal polynomial d d d−1 (i) a0x + a1x +···+ad = a0 (X − η ) ∈ Z[x], i=1 (i) where the ai ’s are relatively prime integers with a0 > 0 and η ’s are conjugates of η. Then d 1 ( ) h(η) = log a + log(max{|η i |, 1}) (5) d 0 i=1 is called the logarithmic height of η. In particular, if η = a/b is a rational number with gcd(a, b) = 1 and b > 1, then h(η) = log (max {|a|, b}). The following properties of logarithmic height can be found in the book [4]: h(η ± γ) ≤ h(η) + h(γ ) + log 2, (6) ± h(ηγ 1) ≤ h(η) + h(γ ), (7) h(ηm) =|m|h(η). (8) Now we give a lemma which is deduced from Corollary 2.3 of Matveev [10] and provide a large upper bound for the subscript n in equations (3) and (4) (also see, Theorem 9.4 in [3]). Lemma 1. Assume that γ1,γ2,...,γt are positive real algebraic numbers in a real alge- braic number field K of degree D, b1, b2,...,bt are rational integers, and := γ b1 ...γbt − 1 t 1 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 Page 3 of 14 28 is not zero. Then t+3 4.5 2 || > exp(−1.4 · 30 · t · D (1 + log D)(1 + log B)A1 A2 ...At ), where B ≥ max{|b1|,...,|bt |}, and Ai ≥ max {Dh(γi ), | log γi |, 0.16} for all i = 1,...,t. The following lemma was proved by Dujella and Peth˝o[6] and is a variation of a lemma of Baker and Davenport [2]. This lemma will be used to reduce the upper bound for the subscript n in equations (3) and (4). Let the function || · || denote the distance from x to the nearest integer, i.e., ||x|| = min {|x − n|:n ∈ Z} for any real number x. Then we have as follows. Lemma 2. Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational number γ such that q > 6M, and let A, B,μbe some real numbers with A > 0 and B > 1. Let := ||μq|| − M||γ q||. If >0, then there exists no solution to the inequality −w 0 < |uγ − v + μ| < AB , in positive integers u,vand w with log(Aq/) u ≤ M and w ≥ . log B The following lemmas are given in [8]. Lemma 3. The only repdigits in the Lucas sequence are 1, 2, 3, 7, 11. Lemma 4. The only repdigits in the Fibonacci sequence are 0, 1, 2, 3, 5, 8, 55. The identity mn L2mn+k ≡ (−1) Lk(mod Fm) (9) is well known, which will be needed in the proof of Theorem 6. 3. Main theorems Theorem 5. Let 2 ≤ m ≤ n. A product of the form Fm Fn represents a repdigit if and only if (m, n) belongs to the set {(2, 2), (2, 3), (3, 3), (2, 4), (3, 4), (4, 4), (2, 5), (2, 6), (2, 10)}. Proof. Assume that Fm Fn represents a repdigit for 2 ≤ m ≤ n. If m = 2, then we have Fn ∈ {1, 2, 3, 5, 8, 55}, by Lemma 4. This implies that n = 2, 3, 4, 5, 6, 10. Assume that 28 Page 4 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 3 ≤ m ≤ n ≤ 99. Thus, a simple computation shows that k ≤ 40. Then, with the help of the Mathematica command, For[n = 3, n ≤ 99, n ++, For[m = 3, m ≤ n, m ++, x = Fibonacci[m]∗Fibonacci[n]; For[d = 1, d ≤ 9, d ++, For[k = 1, k ≤ 40, ∧ k ++, y = (d ∗ (10 k − 1))/(9); If[x ==y, Print[{m, n, k, d, x}]];];];];], we obtain only the solutions Fm Fn = 4, 6, 9. This gives us that (m, n) = (3, 3), (3, 4), (4, 4). From now on, assume that n ≥ 100. Combining the right side of (1) with (3), we obtain · ( k − ) k−1 d 10 1 n+m−2 n+m 2n 10 ≤ = Fm Fn ≤ α <α ≤ α . 9 From this, we get k ≤ n. On the other hand, we rewrite equation (3)as αm − βm αn − βn d · ( k − ) √ · √ = 10 1 5 5 9 to obtain αm+n d · 10k d αmβn + βmαn − βm+n − =− + . (10) 5 9 9 5 Taking absolute values on both sides of (10), we get m+n k m n m n m+n α d · 10 d α |β| |β| α |β| − ≤ + + + . (11) 5 9 9 5 5 5 αn+m Dividing both sides of (11)by , we obtain 5 k n m m+n 5 · d · 10 5d |β| |β| |β| 1 − ≤ + + + 9 · αn+m 9 · αn+m αn αm αn+m 5 1 1 1 − < + + + < 8 · α 2m. α2m α2m α2m α4m From this, it follows that · · k 5 d 10 − m 1 − < 8 · α 2 . (12) 9 · αn+m Now, let us apply Lemma 1 with γ1 := α, γ2 := 10,γ3 := 5d/9 and b1 := −(n + m), b2 := k, b3 := 1. Note that√ the numbers γ1,γ2 and γ3 are positive real numbers and elements of the field K = Q( 5). It is obvious that the degree of the field K is 2. So D = 2. Now, we show that 5 · d · 10k := 1 − 1 9 · αn+m Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 Page 5 of 14 28 n+m k n+m is nonzero. For, if 1 = 0, then α = 5 · d · 10 /9, which is impossible since α is irrational. Moreover, since log α 0.4812 ... h(γ ) = h(α) = ≈ , h(γ ) = h(10) = log 10 1 2 2 2 and 5 · d h(γ ) = h ≤ h(5) + h(d) + h(9)<6.01, 3 9 by (7), we can take A1 := 0.5, A2 := 4.7 and A3 = 12.1. Also, since k ≤ n, m ≤ n and B ≥ max {|−(n + m)|, |k|, |1|}.
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