Fibonacci and Lucas Numbers As Products of Two Repdigits

Home , Fibonacci number, Lucas number

Turkish Journal of Mathematics Turk J Math (2019) 43: 2142 – 2153 http://journals.tubitak.gov.tr/math/ © TÜBİTAK Research Article doi:10.3906/mat-1905-24

Fibonacci and Lucas numbers as products of two repdigits

Fatih ERDUVAN∗, Refik KESKİN Department of Mathematics, Faculty of Science, Sakarya University, Sakarya, Turkey

Received: 08.05.2019 • Accepted/Published Online: 24.06.2019 • Final Version: 28.09.2019

Abstract: In this study, it is shown that the largest Fibonacci number that is the product of two repdigits is F10 =

55 = 5 · 11 = 55 · 1 and the largest Lucas number that is the product of two repdigits is L6 = 18 = 2 · 9 = 3 · 6.

Key words: Fibonacci number, Lucas number, repdigit, Diophantine equations, linear forms in logarithms

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 n − n α β n n Fn = √ and Ln = α + β , 5 √ √ 1 + 5 1 − 5 where α = and β = , which are the roots of the characteristic equation x2 − x − 1 = 0. It can 2 2 be seen that 1 < α < 2 , −1 < β < 0, and αβ = −1. For more about Fibonacci and Lucas sequences with their history, one can see [7]. The relation between the nth Fibonacci number Fn and α is given by

n−2 n−1 α ≤ Fn ≤ α (1) for n ≥ 1. Also, the relation between the nth Lucas number Ln and α is given by

n−1 n α ≤ Ln ≤ 2α . (2)

Inequalities (1) and (2) can be proved by induction. A repdigit is a positive integer whose digits are all equal. Investigation of the repdigits in the second-order linear recurrence sequences has been of interest to mathematicians. In [10], Luca found all Fibonacci and Lucas numbers that are repdigits. The largest repdigits in Fibonacci and Lucas sequences are F5 = 55 and L5 = 11. After that, in [2], the authors showed that the largest Fibonacci number that is a sum of two repdigits is F20 = 6765 = 6666 + 99. In [9], the authors found all repdigits in the Pell and Pell–Lucas sequences. The largest repdigits in Pell and Pell–Lucas sequences are

P3 = 5 and Q2 = 6. Later, the authors [1] found all Pell and Pell–Lucas numbers that are sums of two repdigits.

It was shown that the largest Pell number that is a sum of two repdigits is P6 = 70 = 4 + 66 and the largest

∗Correspondence: [email protected] 2010 AMS Mathematics Subject Classification: 11B39, 11J86, 11D61

2142 This work is licensed under a Creative Commons Attribution 4.0 International License. ERDUVAN and KESKİN/Turk J Math

Pell–Lucas number that is a sum of two repdigits is Q6 = 198 = 99+99. In this paper, we investigate Fibonacci and Lucas numbers that are the product of two repdigits. In other word, we solve the Diophantine equations

d (10m − 1) d (10n − 1) F = 1 · 2 (3) k 9 9 and d (10m − 1) d (10n − 1) L = 1 · 2 . (4) k 9 9

It is shown that the largest Fibonacci number that is a product of two repdigits is F10 = 55 = 5 · 11 = 55 · 1 and the largest Lucas number that is a product of two repdigits is L6 = 18 = 2 · 9 = 3 · 6. In Section 2, we introduce necessary lemmas and theorems. Then we prove our main theorems in Section 3.

2. Auxiliary results In [5], in order to solve Diophantine equations of a similar form, the authors used Baker’s theory of lower bounds for a nonzero linear form in logarithms of algebraic numbers. Since such bounds are of crucial importance in effectively solving Diophantine equations (3) and (4), we start by recalling some basic notions from algebraic number theory. Let η be an algebraic number of degree d with minimal polynomial

Yd 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 the η s are conjugates of η. Then ! d n o 1 X 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}). Some known properties of logarithmic height are found in many works stated in the references:

h(η ± γ) ≤ h(η) + h(γ) + log 2, (6)

h(ηγ±1) ≤ h(η) + h(γ), (7)

h(ηm) = |m|h(η). (8)

The following lemma, Lemma 1, is deduced from Corollary 2.3 of Matveev [11] and provides a large upper bound for the subscript n in equations (3) and (4)(also see Theorem 9.4 in [4]).

Lemma 1 Assume that γ1, γ2, ..., γt are positive real algebraic numbers in a real algebraic number field K of degree D, b1, b2, ..., bt are rational integers, and

b1 bt − Λ := γ1 ...γt 1

2143 ERDUVAN and KESKİN/Turk J Math is not zero. Then t+3 4.5 2 |Λ| > exp −1.4 · 30 · t · D (1 + log D)(1 + log B)A1A2...At , where

B ≥ max {|b1|, ..., |bt|} , and Ai ≥ max {Dh(γi), | log γi|, 0.16} for all i = 1, ..., t.

Now we give a lemma that was proved by Dujella and Pethő [8] and is a variation of a lemma of Baker and Davenport [3]. 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. That is, ||x|| = min {|x − n| : n ∈ Z} for any real number x. Then we have:

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, µ be some real numbers with A > 0 and B > 1. Let ϵ := ||µq|| − M||γq||. If ϵ > 0, then there exists no solution to the inequality

0 < |uγ − v + µ| < AB−w, in positive integers u, v, and w with log(Aq/ϵ) u ≤ M and w ≥ . log B

The following lemma was given in [6].

Lemma 3 Let a, x ∈ R and 0 < a < 1. If |x| < a, then

− log(1 − a) |log(1 + x)| < · |x| a and a |x| < · |ex − 1| . 1 − e−a

3. Main Theorems

Theorem 4 Let m, n, d1, d2 be positive integers with m ≤ n and d1, d2 ≤ 9. Let

d (10m − 1) d (10n − 1) M = 1 and N = 2 . 9 9

If Fk = M · N , then

(k, Fk,M,N) = (1, 1, 1, 1), (2, 1, 1, 1), (3, 2, 1, 2), (4, 3, 1, 3), (5, 5, 1, 5), (6, 8, 2, 4) and

(k, Fk,M,N) = (6, 8, 1, 8), (8, 21, 3, 7), (10, 55, 5, 11), (10, 55, 1, 55).

2144 ERDUVAN and KESKİN/Turk J Math

Proof Assume that Fk = M · N. Now assume that 1 ≤ m ≤ n ≤ 22. Then by using the Mathematica program, we see that k ≤ 212. In this case, we obtain only the solutions Fk ∈ {1, 2, 3, 5, 8, 21, 55}. From now on, assume that n ≥ 23. Then the inequality

d (10n − 1) d (10m − 1) d (10n − 1) α4n−4 < 10n−1 ≤ 2 ≤ 1 · 2 = F ≤ αk−1 9 9 9 k implies that k > 4n − 3. That is, k > 89 for n ≥ 23. Combining the left side of (1) with (3), we obtain

d (10m − 1) d (10n − 1) αk−2 ≤ F = 1 · 2 ≤ (10n − 1)2 < 102n ≤ α10n. k 9 9

From this, we get k < 10n + 2. On the other hand, we rewrite equation (3) as

αk − βk d (10m − 1) d (10n − 1) √ = 1 · 2 5 9 9 to obtain d d 10m+n αk βk d d 10m d d 10n d d 1 2 − √ = −√ + 1 2 + 1 2 − 1 2 . (9) 81 5 5 81 81 81 Taking absolute values of both sides of (9), we get

m+n k | |k m n d1d210 α β d1d210 d1d210 d1d2 − √ ≤ √ + + + . (10) 81 5 5 81 81 81

d d 10m+n Dividing both sides of (10) by 1 2 gives us the following inequality: 81

81 · αk · 10−(m+n) 81 |β|k 1 1 1 1 − √ ≤ √ + + + m+n n m m+n d1d2 5 5d1d210 10 10 10 < 4 · 10−m.

From this, it follows that

· k · −(m+n) 81 α 10 −m 1 − √ < 4 · 10 . (11) d1d2 5 √ Now, let us apply Lemma 1 with γ := α, γ := 10, γ := 81/d d 5 and b := k, b := −(m + n), b := 1. 1 2 3 1 2 1 2 √ 3 Note that the numbers γ1, γ2, and γ3 are positive real numbers and elements of the field K = Q( 5). The degree of the field K is 2, so D = 2. Now we show that

81 · αk · 10−(m+n) Λ1 := 1 − √ d1d2 5 √ √ is nonzero. On the contrary, assume that Λ = 0. Then αk = 10(m+n)d d 5/81. Conjugating in Q( 5) gives √ 1 1 2 k (m+n) k k us β = −10 d1d2 5/81. This implies that Lk = α + β = 0, which is a contradiction. Since

log α 0.4812... h(γ ) = h(α) = = , h(γ ) = h(10) = log 10 1 2 2 2

2145 ERDUVAN and KESKİN/Turk J Math and √ √ h( γ3) = h(81/d1d2 5) ≤ h(81) + h( 5) + h(d1) + h(d2) < 9.6 by (7), we can take A1 := 0.5,A2 := 4.7, and A3 = 19.2. Also, since k < 10n + 2, m ≤ n, and B ≥ max {|k|, | − (n + m)|, |1|}, we can take B := 10n + 2. Thus, taking into account inequality (11) and using Lemma 1, we obtain −m 6 4.5 2 4 · 10 > |Λ1| > exp −1.4 · 30 · 3 · 2 (1 + log 2)(1 + log(10n + 2) · C , where C = (0.5) (4.7) (19.2). By a simple computation, it follows that

m log 10 < 4.38 · 1013 · (1 + log(10n + 2)) + log 4. (12)

Rearranging equation (3) as

d 10n 9 · αk 9βk d 2 − √ = − √ + 2 (13) m m 9 d1(10 − 1) 5 d1(10 − 1) 5 9 and taking absolute values of both sides of (13), we get

d 10n 9 · αk 9 · |β|k d 2 − √ ≤ √ + 2 . (14) m m 9 d1(10 − 1) 5 d1(10 − 1) 5 9

n Dividing both sides of (14) by d210 /9 gives us the following inequality:

81 · αk · 10−n 81 · |β|k 1 1 − √ ≤ √ + < 2 · 10−n. m n m n d1d2(10 − 1) 5 d1d210 (10 − 1) 5 10

From this, it follows that

81 · αk · 10−n 1 − √ < 2 · 10−n. (15) m d1d2(10 − 1) 5 √ Taking γ := α, γ := 10, γ := 81/d d (10m − 1) 5, and b := k, b := −n, b := 1, we can apply Lemma 1. 1 2 3 1 2 1 2 3 √ The numbers γ1, γ2, and γ3 are positive real numbers and elements of the field K = Q( 5) and so D = 2. As before, it can be shown that 81 · αk · 10−n Λ2 := 1 − √ m d1d2(10 − 1) 5 log α 0.4812... is nonzero. By using (5) and the properties of the logarithmic height, we get h(γ ) = = , h( γ ) = 1 2 2 2 log 10, and √ √ m m h(γ3) = h(81/d1d2(10 − 1) 5) ≤ h(81) + h( 5) + h(d1) + h(d2) + h(10 − 1) < 10.3 + m log 10.

We can thus take A1 := 0.5, A2 = 4.7, and A3 := 20.6 + 2m log 10. Since k < 10n + 2 and B ≥ max {|k|, | − n|, |1|}, we can take B := 10n + 2. Thus, taking into account inequality (15) and using Lemma 1, we obtain −n 2 · 10 > |Λ2| > exp (C · (1 + log(10n + 2)) (0.5) (4.7) (20.6 + 2m log 10)) ,

2146 ERDUVAN and KESKİN/Turk J Math where C = −1.4 · 306 · 34.5 · 22 · (1 + log 2). Thus,

n log 10 − log(2) < 2.28 · 1012 · (1 + log(10n + 2)) (20.6 + 2m log 10) . (16)

Using inequalities (12) and (16), a computer search with Mathematica gives us that n < 4.45 · 1029 . Now let us try to reduce the upper bound on n by applying Lemma 2. Let √ z1 z1 := k log α − (n + m) log 10 + log(81/d1d2 5) and x = e − 1.

From (11), we have 4 1 |x| = |ez1 − 1| < < 10m 2 ≥ 1 for m 1. Choosing a := 2 , we get the inequality

log 2 4 |z | = |log(x + 1)| < · < 5.55 · 10−m 1 (1/2) 10m by Lemma 3. Thus, it follows that √ −m 0 < k log α − (n + m) log 10 + log(81/d1d2 5) < 5.55 · 10 .

Dividing this inequality by log 10, we get ! √ log α log(81/d d 5) 0 < k − (n + m) + 1 2 < 2.5 · 10−m. (17) log 10 log 10

log α Take γ := ∈/ Q and M := 4.45 · 1030. Then we find that q , the denominator of the 63rd convergent of log 10 63 γ , exceeds 6M. Now take √ log(81/d d 5) µ := 1 2 . log 10

In this case, considering the fact that 1 ≤ d1, d2 ≤ 9, a quick computation with Mathematica gives us the inequality 0 < ϵ = ϵ(µ) := ||µq63|| − M||γq63|| < 0.485826. Let A := 2.5,B := 10, and w := m in Lemma 2. Thus, with the help of Mathematica, we can say that inequality (17) has no solution for

log(Aq /ϵ(µ)) m = w ≥ 63 ≥ 33.0598. log B

Thus, m ≤ 33.

Substituting this upper bound for m into (16), we obtain n < 6.8 · 1015 . Now, let √ m z2 z2 := k log α − n log 10 + log 81/(10 − 1)d1d2 5 and x = e − 1.

2147 ERDUVAN and KESKİN/Turk J Math

From (15), we have 2 1 |x| = |ez2 − 1| < < 10n 10 ≥ 1 for n 23. Choosing a := 10 , we get the inequality

log(10/9) 2 |z | = |log(x + 1)| < · < 2.2 · 10−n 2 (1/10) 10n by Lemma 3. Thus, it follows that √ m −n 0 < k log α − n log 10 + log 81/(10 − 1)d1d2 5 < 2.2 · 10 .

Dividing both sides of the above inequality by log 10, we obtain

√ log α log 81/(10m − 1)d d 5 0 < k − n + 1 2 < 0.96 · 10−n. (18) log 10 log 10

log α Put γ := and take M := 6.8 · 1016 . Then we find that q , the denominator of the 42nd convergent of log 10 42 γ , exceeds 6M. Taking √ log 81/(10m − 1)d d 5 µ := 1 2 log 10 and considering the fact that m ≤ 33 and 1 ≤ d1, d2 ≤ 9, a quick computation with Mathematica gives us the inequality 0 < ϵ = ϵ(µ) = ||µq42|| − M||γq42|| < 0.499112. Let A := 0.96,B := 10, and w := n in Lemma 2. Thus, with the help of Mathematica, we can say that inequality (18) has no solution for

log(Aq /ϵ(µ)) n = w ≥ 42 ≥ 22.6951. log B

Therefore, n ≤ 22, which contradicts our assumption that n ≥ 23. This completes the proof. 2

Theorem 5 Let m, n, d1, d2 be positive integers with m ≤ n and d1, d2 ≤ 9. Let

d (10m − 1) d (10n − 1) M = 1 and N = 2 . 9 9

If Lk = M · N , then

(k, Lk,M,N) = (0, 2, 1, 2), (1, 1, 1, 1), (2, 3, 1, 3), (3, 4, 1, 4), (3, 4, 2, 2) and

(k, Lk,M,N) = (4, 7, 1, 7), (5, 11, 1, 11), (6, 18, 2, 9), (6, 18, 3, 6).

Proof Assume that Lk = M · N. Now assume that 1 ≤ m ≤ n ≤ 40. Then by using the Mathematica program, we see that k ≤ 385. In this case, we obtain only the solutions Lk ∈ {1, 2, 3, 4, 7, 11, 18}. From now on, assume that n > 40. Then we get the inequality

d (10n − 1) d (10m − 1) d (10n − 1) α4n−4 < 10n−1 ≤ 2 ≤ 1 · 2 = L ≤ 2αk < αk+2 9 9 9 k

2148 ERDUVAN and KESKİN/Turk J Math and therefore k > 4n − 6. That is, k > 158 for n > 40. Combining the left side of (2) with (4), we obtain

d (10m − 1) d (10n − 1) αk−1 ≤ L = 1 · 2 ≤ (10n − 1)2 < 102n ≤ α10n. k 9 9

This implies that k < 10n + 1. Now we rewrite equation (4) as

d (10m − 1) d (10n − 1) αk + βk = 1 · 2 9 9 to obtain d d 10m+n d d 10m d d 10n d d 1 2 − αk = −βk + 1 2 + 1 2 − 1 2 . (19) 81 81 81 81 Taking absolute values of both sides of equation (19), we get

m+n m n d1d210 k k d1d210 d1d210 d1d2 − α ≤ |β| + + + . (20) 81 81 81 81

m+n Dividing both sides of (20) by d1d210 /81 gives us the following inequality:

· k · −(m+n) · | |k − 81 α 10 ≤ 81 β 1 1 1 1 m+n + n + m + m+n d1d2 d1d210 10 10 10 < 4 · 10−m.

From this, it follows that

· k · −(m+n) 81 α 10 −m 1 − < 4 · 10 . (21) d1d2

Now let us apply Lemma 1 with γ := α, γ := 10, γ := 81/d d and b := k, b := −(n + m), b := 1. Note 1 2 3 1 2 1 2 √ 3 that the numbers γ1, γ2, and γ3 are positive real numbers and elements of the field K = Q( 5). The degree of k −(m+n) the field K is 2, so D = 2. It can be seen that Λ3 := 1− 81 · α · 10 /d1d2 is nonzero. Moreover, since

log α 0.4812... h(γ ) = h(α) = = , h(γ ) = h(10) = log 10 1 2 2 2 and

h( γ3) = h(81/d1d2) ≤ h(81) + h(d1) + h(d2) < 8.8 by (7), we can take A1 := 0.5,A2 := 4.7, and A3 := 17.6. On the other hand, as k < 10n + 1, m ≤ n, and B ≥ max {|k|, | − (n + m)|, |1|}, we can take B := 10n + 1. Thus, taking into account the inequality (21) and using Lemma 2, we obtain

−m 4 · 10 > |Λ3| > exp (C · (1 + log 2)(1 + log(10n + 1)) (0.5) (4.7) (17.6)) , where C = −1.4 · 306 · 34.5 · 22 . By a simple computation, it follows that

m log 10 − log 4 < 4.1 · 1013 · (1 + log(10n + 1)). (22)

2149 ERDUVAN and KESKİN/Turk J Math

Rearranging equation (4) as n · k · k d210 − 9 α 9 β d2 m = m + (23) 9 d1(10 − 1) d1(10 − 1) 9 and taking absolute values of both sides of equation (23), we get

n · k · | |k d210 − 9 α ≤ 9 β d2 m m + . (24) 9 d1(10 − 1) d1(10 − 1) 9

n Dividing both sides of (24) by d210 /9 gives us the following inequality:

· k · −n · | |k − 81 α 10 ≤ 81 β 1 1 m n m + n d1d2(10 − 1) d1d210 (10 − 1) 10

81 · |β|k 1 < + < 2 · 10−n. 10n 10n

From this, it follows that

· k · −n − 81 α 10 · −n 1 m < 2 10 . (25) d1d2(10 − 1) Taking γ := α, γ := 10, γ := 81/d d (10m − 1), and b := k, b := −n, b := 1, we can apply Lemma 1. The 1 2 3 1 2 1 2 3 √ numbers γ1, γ2, and γ3 are positive real numbers and elements of the field K = Q( 5) and so D = 2. One k −n m can verify that Λ4 := 1 − 81 · α · 10 /d1d2(10 − 1) ≠ 0. By using (5) and the properties of the logarithmic log α 0.4812... height, we get h(γ ) = = , h( γ ) = log 10, and 1 2 2 2

m m h(γ3) = h(81/d1d2(10 − 1)) ≤ h(81) + h(d1) + h(d2) + h(10 − 1) < 9.5 + m log 10.

Thus, we can take A1 := 0.5, A2 = 4.7, and A3 := 19+2m log 10. As k < 10n+1 and B ≥ max {|k|, | − n|, 1}, we can take B := 10n + 1. Thus, taking into account inequality (25) and using Lemma 1, we obtain

−n 2 · 10 > |Λ4| > exp (C · (1 + log(10n + 1)) (0.5) (4.7) (19 + 2m log 10)) , where C = −1.4 · 306 · 34.5 · 22 · (1 + log 2). Thus, it follows that

n log 10 − log 2 < 2.27 · 1012 · (1 + log(10n + 1)) (19 + 2m log 10) . (26)

Using inequalities (22) and (26), a computer search with Mathematica gives us that n < 4.15 · 1029 . Now let us try to reduce the upper bound on n by applying Lemma 2. Let

z3 z3 := k log α − (n + m) log 10 + log(81/d1d2) and x = e − 1.

From (21), we have 4 1 |x| = |ez3 − 1| < < 10m 2 ≥ 1 for m 1. Choosing a := 2 , we get the inequality

log 2 4 |z | = |log(x + 1)| < · < 5.55 · 10−m 3 (1/2) 10m

2150 ERDUVAN and KESKİN/Turk J Math by Lemma 3. Thus, it follows that

−m 0 < |k log α − (n + m) log 10 + log(81/d1d2)| < 5.55 · 10 .

Dividing this inequality by log 10, we get

log α log(81/d1d2) −m 0 < k − (n + m) + < 2.5 · 10 . (27) log 10 log 10

log α Let γ := ∈/ Q and M := 4.15 · 1030. Then we find that q , the denominator of the 67th convergent of log 10 67 γ , exceeds 6M. Take log(81/d d ) µ := 1 2 . log 10

In this case, a quick computation with Mathematica gives us the inequality 0 < ϵ = ϵ(µ) := ||µq67||−M||γq67|| <

0.4766 for 1 ≤ d1, d2 ≤ 9, except for d1 = d2 = 9. Let A := 2.5,B := 10, and w := m in Lemma 2. Then with the help of Mathematica, we can say that inequality (27) has no solution for

log(Aq /ϵ(µ)) m = w ≥ 67 ≥ 36.1498. log B

Thus, m ≤ 36.

In the case of d1 = d2 = 9, from (27) we have

log α −m 0 < k − (n + m) < 2.5 · 10 . log 10

If we divide this inequality by k, we get

log α n + m 2.5 0 < − < . (28) log 10 k 10m · k

10m Assume that m ≥ 36. Then it can be seen that ≥ 20 · 1034 > 10n + 1 ≥ k, so we have log α − n+m < 5 log 10 k 1 2.5 < . From the known properties of the continued fraction, it is seen that the rational number n+m is 10m·k 2k2 k log α a convergent to γ = log 10 . Now let [a0; a1, a2, ...] = [0; 4, 1, 3, 1, 1, 1, 6, ...] be the continued fraction expansion p p r n+m t · 31 of γ and let be its rth convergent. Assume that k = for some t. Then we have 13 10 > q62 > qr qt 31 12 · 10 > 10n + 1 ≥ k. Thus, t ∈ {0, 1, 2, ..., 61}. Furthermore, aM = max{ai|i = 0, 1, 2, ..., 61} = 106. Again, from the known properties of the continued fraction, we get

− pt 1 1 γ > 2 = 2 . qt (aM + 2)k 108 · k

Thus, from (28), we obtain 2.5 1 > . 10m · k 108 · k2

2151 ERDUVAN and KESKİN/Turk J Math

This implies that 1 ≥ 2.5 1 1 1 34 m > > > 34 , 40 · 10 10 108 · k 108 · q62 1.404 · 10 a contradiction. Therefore, m < 36. Consequently, taking m ≤ 36 and substituting this upper bound for m into (26), we obtain n < 7.26 · 1015 . Now let

m z4 z4 := k log α − n log 10 + log (81/d1d2(10 − 1)) and x = e − 1.

From (25), we have 2 1 |x| = |ez4 − 1| < < 10n 10

1 for n > 40. Choosing a := 10 , we get the inequality

log(10/9) 2 |z | = |log(x + 1)| < · < 2.2 · 10−n 4 (1/10) 10n by Lemma 3. Thus, it follows that

m −n 0 < |k log α − n log 10 + log (81/d1d2(10 − 1))| < 2.2 · 10 .

Dividing both sides of the above inequality by log 10, we get m − log α log (81/d1d2(10 1)) −n 0 < k − n + < 0.96 · 10 . (29) log 10 log 10

log α Let γ := and M := 7.26 · 1016 . Then we find that q , the denominator of the 53rd convergent of γ , log 10 53 exceeds 6M. Taking log (81/d d (10m − 1)) µ := 1 2 log 10 and considering the fact that m ≤ 34 and 1 ≤ d1, d2 ≤ 9, a quick computation with Mathematica gives us the equality 0 < ϵ = ϵ(µ) := ||µq53|| − M||γq53|| < 0.4999 except for the cases d1 · d2 = 9 and m = 1. Let A := 0.96,B := 10, and w := n in Lemma 2. Then with the help of Mathematica, we can say that inequality (29) has no solution for log(Aq /ϵ(µ)) n = w ≥ 53 ≥ 36.4655. log B

Therefore, n ≤ 36. If d1 · d2 = 9 and m = 1, then from (29), we get

log α n 0.96 0 < − < . (30) log 10 k 10n · k

If we follow the way after (28), we find that n ≤ 40. These cases both contradict the assumption that n > 40. This completes the proof of the theorem. 2

n m n m Corollary 6 The equations Fk = (10 − 1)(10 − 1) and Lk = (10 − 1)(10 − 1) have no solution (k, m, n) in positive integers.

2152 ERDUVAN and KESKİN/Turk J Math

References

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