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
FATIHú ERDUVAN and REFIKú KESKINú ∗
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|}. We can take B := 2n. Thus, taking into account the inequality (12) and using Lemma 1, we obtain
−2m 6 4.5 2 8 · α > |1| > exp(−1.4 · 30 · 3 · 2 (1 + log 2)(1 + log 2n)(0.5)(4.7)(12.1)).
By a simple computation, it follows that
2m log α<2.8 · 1013 · (1 + log 2n) + 2.08. (13)
Rearranging equation (3)as
αn d · 10k βn d √ − = √ − (14) 5 9Fm 5 9Fm and taking absolute values on both sides of (14), we get n k n α d · 10 |β| d √ − ≤ √ + . (15) 5 9Fm 5 9Fm αn Dividing both sides of (15)by√ , we obtain 5 √ √ √ 5 · d · 10k |β|n d 5 1 5 − ≤ + < + < · α−n. 1 n n n 2n n 4 9 · α · Fm α 9 · α · Fm α α
From this, it follows that √ 5 · d · 10k − < · α−n. 1 n 4 (16) 9 · α · Fm √ Taking γ1 := α, γ2 := 10,γ3 := 9Fm/ 5d , and b1 := −n, b2 := k, b3 := −1, we can apply Theorem 1√. The numbers γ1,γ2 and γ3 are positive real numbers and elements of the field K = Q( 5) and so D = 2. Now, we show that √ · · k := − 5 d 10 2 1 n 9 · α · Fm 28 Page 6 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 √ √ n k is nonzero. For,√ if 2 = 0, then α = 5 · d · 10 /9Fm. Conjugating in Q( 5) gives n k n n us β =− 5 · d · 10 /9Fm and so, Ln = α + β = 0. This is a contradiction. By using (5) and the properties of the logarithmic height (see (6), (7) and (8)), we get log α 0.4812 ... h(γ ) = ≈ , h(γ ) = log 10, and 1 2 2 2
√ √ 1 2 9Fm h(γ3) ≤ log 5d + 2log √ = log( 5d) + log(9Fm) − log( 5d) 2 5d ≤ 1.8 + m log α.
So we can take A1 := 0.5, A2 := 4.7 and A3 := 4 + 2m log α. Since k ≤ n and B ≥ max{| − n|, |k|, |−1|}, we can take B := n. Thus, taking into account the inequality (16) and using Lemma 1, we obtain
−n 6 4.5 2 4 · α >|2| > exp(−1.4 · 30 · 3 · 2 (1+ log 2)(1+ log n)(0.5)(4.7)(4+2m log α)). or
n log α<2.3 · 1012 · (1 + log n)(4 + 2m log α) + log 4. (17)
Inserting the equality (13) into the inequality (17), we get n < 5.4 · 1029. This inequality can be quickly verified using the Mathematica command
∧ N[Reduce[n ∗ Log[GoldenRatio]ÐLog[4] < 2.3 ∗ 10 12 ∗ (1 + Log[n]) ∧ ∗(2.3 ∗ 10 13 ∗ (1 + Log[2n]) + 6.08), n]].
Now, let us try to reduce the upper bound on n by applying Lemma 2 .Let
z1 := k log 10 − (n + m) log α + log(5d/9).
If z1 > 0, then by (12), we have the inequality
z1 z1 −2m |z1|=z1 < e − 1 =|1 − e | < 8 · α
x since x < e − 1forx > 0. If z1 < 0, then
− 1 1 − ez1 =|1 − ez1 | < 8 · α 2m < . 2
z1 > 1 From this, we get e 2 and therefore | | − e z1 = e z1 < 2.
Consequently, we get
|z1| |z1| z1 −2m |z1| < e − 1 = e |1 − e | < 16 · α .
In both cases, the inequality
−2m 0 < |z1| < 16 · α Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 Page 7 of 14 28 is true. That is,
− 0 < |k log 10 − (n + m) log α + log(5d/9)| < 16 · α 2m.
Dividing this inequality by log α, we get ( / ) log 10 log 5d 9 − m 0 < k − (n + m) + < 33.3 · α 2 . (18) log α log α
log 10 Now, we show that is irrational. On the contrary, assume that log α
log 10 = p log α q for some positive integers p and q. This shows that 10q = α p , which is impossible since log 10 α p is irrational. Take γ = and M = 5.4·1029. Then we found that the denominator log α of the 63rd convergent
p63 = 123084391290586534132720655418148 q63 25723116487424714265759180025093 of γ exceeds 6M. Now take ( / ) μ := log 5d 9 . log α
In this case, considering the fact that 1 ≤ d ≤ 9, a quick computation with Mathematica gives us the inequality 0 <= (μ) := ||μq63|| − M||γ q63|| ≤ 0.162533. To calculate this value of , we use the following command in the Mathematica program:
For[d = 1, d ≤ 9, d ++, z = Max[Denominator[Convergents[γ,k]]]; x = Min[N[Abs[FractionalPart[μ ∗ z]], 20], 1 −N[Abs[FractionalPart[μ ∗ z]], 20]] −M ∗ Min[N[Abs[FractionalPart[γ ∗ z]], 20], 1 −N[Abs[FractionalPart[γ ∗ z]], 20]]; If[z > 6 ∗ M, Print[{d, x}];];]. (19)
Let A := 33.3, B := α and w := 2m in Lemma 2. Thus, we can say that the inequality (18) has no solution for
log(Aq /(μ)) log(Aq /0.162533) 2m = w ≥ 63 ≥ 63 ≥ 162.9. log B log B
So
m ≤ 81. 28 Page 8 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28
Here and later on, for evaluating this task, we use the following command in the Mathe- matica program:
Log[A ∗ q /] N k . (20) Log[B]
Substituting this upper bound for m into (17), we obtain n < 1.5 · 1016.Now,let √ z2 := k log 10 − n log α + log( 5d/9Fm).
If z2 > 0, then by (16), we get the inequality
z2 z2 −n |z2|=z2 < e − 1 =|1 − e | < 4 · α .
− 1 If z < 0, then 1 − ez2 =|1 − ez2 | < 4 · α n < as n ≥ 100. Therefore, we get ez2 > 3 2 4 4 | | − z2 = z2 < 4 . and so e e 3 Thus it follows that
|z2| |z2| z2 −n |z2| < e − 1 = e |1 − e | < 5.4 · α .
This means that the inequality
−n 0 < |z2| < 5.4 · α is always true. That is, √ −n 0 < |k log 10 − n log α + log( 5d/9Fm)| < 5.4 · α .
Dividing both sides of the above inequality by log α, we obtain √ log 10 log( 5d/9F ) − 0 < k − n + m < 11.3 · α n. (21) log α log α
log 10 Putting γ := and taking M := 1.5 · 1016, we found that the denominator of the log α 40-th convergent
p40 = 41924177609269798247 q40 8761634947982290092 of γ exceeds 6M. Taking √ ( / ) μ := log 5d 9Fm log α and considering the fact that m ≤ 81 and 1 ≤ d ≤ 9, a quick computation with Math- ematica gives us the inequality 0 <= (μ) =||μq40|| − M||γ q40|| ≤ 0.497798. Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 Page 9 of 14 28
Let A := 11.3, B := α and w := n in Lemma 2. Thus, with the help of Mathematica command given in (20), we can say that the inequality (21) has no solution for
log(Aq /) log(Aq /0.497798) n = w ≥ 40 ≥ 40 ≥ 97.1283. log B log B
Therefore n ≤ 97. This contradicts our assumption that n ≥ 100. Thus, the proof is completed.
Theorem 6. Let 0 ≤ m ≤ n. A product of the form Lm Ln represents a repdigit if and only if (m, n) belongs to the set (0, 0), (0, 1), (1, 1), (0, 2), (1, 2), (2, 2), (0, 3), . (0, 5), (1, 3), (1, 4), (1, 5), (2, 5), (3, 5), (4, 5)
Proof. Assume that Lm Ln represents a repdigit for 0 ≤ m ≤ n. If m = 0, then 2Ln = d·(10k −1) . 9 for some non-negative k and d Therefore d must be even. That is, there exists ·( k − ) ∈{, , , } = . = d1 10 1 . ∈ d1 1 2 3 4 such that d 2d1 Thus Ln 9 By Lemma 3, we get Ln {1, 2, 3, 4, 11} and this implies that
(m, n) = (0, 0), (0, 1), (0, 2), (0, 3), (0, 5).
If m = 1, then Ln is a repdigit and so by Lemma 3, we get Ln ∈{1, 3, 4, 7, 11}.This shows that
(m, n) = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5).
Now assume that 2 ≤ m ≤ n ≤ 99. Thus, a simple computation shows that k ≤ 41. Then by using the Mathematica command given for Fibonacci numbers in the proof of Theorem 5, we obtain the solutions
(m, n) = (2, 2), (2, 5), (3, 5), (4, 5).
Here, Mathematica command for Lucas numbers is “LucasL”. From now on, assume that n ≥ 100. Then
· ( k − ) d 10 1 k L ≤ Lm Ln ≤ ≤ 10 − 1, 100 9 which gives us
log(1 + L ) 20 ≤ 100 ≤ k. log 10
That is, k ≥ 20. Combining the right side of inequality of (2) with (4), we obtain
· ( k − ) k−1 d 10 1 n+m 2n 10 ≤ = Lm Ln ≤ 4α ≤ 4α . 9 28 Page 10 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28
From this, we get k < n. On the other hand, we can rewrite equation (4)as
· ( k − ) (αm + βm) · (αn + βn) = d 10 1 9 to obtain
k + d · 10 d + αm n − =− − αmβn − βmαn − βm n. (22) 9 9 Taking absolute values on both sides of equation (22), we get · k m+n d 10 d m n m n m+n α − ≤ + α |β| +|β| α +|β| . (23) 9 9
Dividing both sides of (23)byαn+m, we obtain k n m m+n d · 10 d |β| |β| |β| 1 − ≤ + + + 9 · αn+m 9 · αn+m αn αm αn+m 1 1 1 1 − < + + + < 4 · α 2m. α2m α2m α2m α4m
From this, it follows that · k d 10 − m 1 − < 4 · α 2 . (24) 9 · αn+m
Now, let us apply Lemma 1 with γ1 := α, γ2 := 10,γ3 := d/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. It can be k n+m seen that 3 := 1 − d · 10 /9 · α is nonzero. Moreover, since
log α 0.4812 ... h(γ ) = h(α) = ≈ , h(γ ) = h(10) = log 10 1 2 2 2 and
d h(γ ) = h ≤ h(d) + h(9)<4.4, 3 9 by (7), we can take A1 := 0.5, A2 := 4.7 and A3 := 8.8. On the other hand, as k < n, m ≤ n and B ≥ max{| − (n + m)|, |k|, |1|}, we can take B := 2n. Thus, taking into account the inequality (24) and using Theorem 1, we obtain
−2m 6 4.5 2 4 · α > |3| > exp(−1.4 · 30 · 3 · 2 (1 + log 2)(1 + log 2n)(0.5)(4.7)(8.8)).
By a simple computation, it follows that
2m log α<2.1 · 1013 · (1 + log 2n) + 1.4. (25) Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 Page 11 of 14 28
Rearranging equation (4)as
d · 10k d αn − =−βn − (26) 9Lm 9Lm and taking absolute values on both sides of equation (26), we get · k n d 10 n d α − ≤|β| + . (27) 9Lm 9Lm
Dividing both sides of (27)byαn, we obtain · k |β|n − d 10 ≤ + d < 1 + 1 < · α−n. 1 n n n 2n n 2 9 · α · Lm α 9 · α · Lm α α From this, it follows that · k − d 10 < · α−n. 1 n 2 (28) 9 · α · Lm
Taking γ1 := α, γ2 := 10,γ3 := 9Lm/d and b1 := −n, b2 := k, b3 := −1, we can apply Lemma 1√. The numbers γ1,γ2 and γ3 are positive real numbers and elements of the field k n K = Q( 5) and so, D = 2. One can verify that 4 := 1 − d · 10 /9 · α · Lm = 0. By log α 0.4812 ... using (5) and the properties of the logarithmic height, we get h(γ ) = ≈ , 1 2 2 h(γ2) = log 10 and
9L h(γ ) ≤ log d + log m = log(9L )<2.9 + m log α. 3 d m
So, we can take A1 := 0.5, A2 := 4.7 and A3 := 6 + 2m log α.Ask < n and B ≥ max{| − n|, |k|, |−1|}, we can take B := n. Thus, taking into account the inequality (28) and using Lemma 1, we obtain
−n 6 4.5 2 2 · α > |4| > exp(−1.4 · 30 · 3 · 2 (1 + log 2)(1 + log n)(0.5)(4.7)(6 + 2m log α)) or
n log α<2.3 · 1012 · (1 + log n)(6 + 2m log α) + log 2. (29)
Using the inequalities (25) and (29), a computer search with Mathematica command
∧ N[Reduce[n ∗ Log[GoldenRatio]-Log[2] < 2.3 ∗ 10 12 ∗ ∧ (1 + Log[n]) ∗ (2.1 ∗ 10 13 ∗ (1 + Log[2n]) + 6), n]] gives us that n < 4.9 · 1029. Now, let us try to reduce the upper bound on n by applying Lemma 2.Let
z3 := k log 10 − (n + m) log α + log(d/9). 28 Page 12 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28
If z3 > 0, then we have the inequality
z3 z3 −2m |z3|=z3 < e − 1 =|1 − e | < 4 · α .
If z3 < 0, then by (24), we get the inequality
− 3 1 − ez3 =|1 − ez3 | < 4 · α 2m < . 5
z3 > 2 From this, we get e 5 and therefore,
| | − 5 e z3 = e z3 < . 2 Consequently, we get
|z3| |z3| z3 −2m |z3| < e − 1 = e |1 − e | < 10 · α .
In both cases, the inequality
−2m 0 < |z3| < 10 · α is true. That is,
− 0 < |k log 10 − (n + m) log α + log(d/9)| < 10 · α 2m.
Dividing this inequality by log α, we get ( / ) log 10 log d 9 − m 0 < k − (n + m) + < 21 · α 2 . (30) log α log α
log 10 Take γ := ∈/ Q and M := 4.9 · 1029. Then we found that q , the denominator of log α 61 the 61st convergent
p61 = 123084391290586534132720655418148 q61 25723116487424714265759180025093 of γ exceeds 6M. Take ( / ) μ := log d 9 . log α
Now assume that 1 ≤ d ≤ 8. In this case, a quick computation with Mathematica command given in (19) gives us the inequality 0 <= (μ) := ||μq61|| − M||γ q61|| ≤ 0.201992. Let A := 21, B := α and w := 2m in Lemma 2. Then with the help of the Mathematica command given in (20), we can say that the inequality (30) has no solution for
log(Aq /) log(Aq /0.201992) 2m = w ≥ 61 ≥ 61 ≥ 159.948. log B log B Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28 Page 13 of 14 28
So
m ≤ 79.
Substituting this upper bound for m into (29), we obtain n < 1.5 · 1016.Now,let
z4 := k log 10 − n log α + log(d/9Lm).
If z4 > 0, then by (28), we have the inequality
z4 z4 −n |z4|=z4 < e − 1 =|1 − e | < 2 · α .
− 1 If z < 0, then 1 − ez4 =|1 − ez4 | < 2 · α n < since n ≥ 100. Therefore, we get 4 4 | | − z4 > 3 z4 = z4 < 4 e 4 and so, e e 3 . Then it follows that
|z4| |z4| z4 −n |z4| < e − 1 = e |1 − e | < 2.7 · α .
Therefore, it is always true that
−n 0 < |z4| < 2.7 · α .
That is,
−n 0 < |k log 10 − n log α + log(d/9Lm)| < 2.7 · α .
Dividing both sides of the above inequality by log α, we get ( / ) log 10 log d 9Lm −n 0 < k − n + < 5.7 · α . (31) log α log α
log 10 Putting γ := and taking M := 1.5 · 1016, we found that q , the denominator of log α 39 the 39-th convergent
p39 = 33149137033495047668 q39 6927759924953310265 of γ exceeds 6M. Taking ( / ) μ := log d 9Lm log α and considering the fact that m ≤ 79 and 1 ≤ d ≤ 8, a quick computation with the Mathematica command given in (19) gives us the equality 0 <= (μ) := ||μq39|| − M||γ q39|| ≤ 0.498401. Let A := 5.7, B := α and w := n in Lemma 2. Then with the help of Mathematica command given in (20), we can say that the inequality (31) has no solution for log(Aq /) log(Aq /0.498401) n = w ≥ 39 ≥ 39 ≥ 95.2156. log B log B 28 Page 14 of 14 Proc. Indian Acad. Sci. (Math. Sci.) (2020) 130:28
Therefore n ≤ 95. This contradicts our assumption that n ≥ 100. Now assume that d = 9. Then
k Ln Lm = 10 − 1 and therefore, 9|Ln Lm.Letm = 24k + r, n = 24t + s, 0 ≤ r, s ≤ 23. Thus
Lm = L24k+r ≡ Lr (mod F12) and Ln = L24t+s ≡ Ls(mod F12) by (9). Then it follows that
Lm Ln ≡ Lr Ls(mod F12)
Since k ≥ 3, it follows that
k Ln Lm = 10 − 1 ≡−1 ≡ 7(mod 8) (32)
Moreover, since 9|Ln Lm, we get
Lm Ln ≡ 63(mod 72).
On the other hand, since Lm Ln ≡ Lr Ls(mod F12), we get Lm Ln ≡ Lr Ls(mod 72) and therefore, Lr Ls ≡ 63(mod 72). A simple computation shows that Lr Ls ≡ 63(mod 72) is impossible for 0 ≤ r, s ≤ 23. This completes the proof of the theorem.
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