Repdigits as Sums of Four Fibonacci or Lucas Numbers

Benedict Vasco Normenyo

Institut de Mathématiques et de Sciences Physiques Dangbo Bénin

bvnormenyo@imsp-uac.org

Florian Luca School of Mathematics

University of the Witwatersrand

Private Bag X3 Wits 2050 South Africa and

Department of Mathematics Faculty of Sciences University of Ostrava

30 Dubna 22 701 03 Ostrava 1 Czech Republic

florian.luca@wits.ac.za

Alain Togbé

Department of Mathematics, Statistics and Computer Science

Purdue University Northwest 1401 S, U.S. 421 Westville, IN 46391 USA

    atogbe@pnw.edu

Abstract

In this paper, we determine all base-10 repdigits expressible as sums of four Fibonacci or Lucas numbers.

1 Introduction

The Fibonacci sequence (F_{n})_{n} and the Lucas sequences (L_{n})_{n} are given, respectively, by

F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n} for n ≥ 0

and

L_{0} = 2, L_{1} = 1, L_{n+2} = L_{n+1} + L_{n} for n ≥ 0.

Luca [2] answered the question of which repdigits can be written as sums of three Fibonacci numbers by following a general method (see [3]). Luca [2] showed that all nonnegative integer solutions (m_{1}, m_{2}, m_{3}, n) of the equation

N = F_{m_{1}} + F_{m_{2}} + F_{m_{3}} = d (\frac{10^{n} − 1}{9}) with d ∈ \{1, …, 9\}

have

N ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 44, 55, 66, 77, 99, 111, 555, 666, 11111}.

Luca, Normenyo, and Togbe [6, 7] obtained analogous results for Pell numbers and Lucas numbers.

Luca [2] conjectured that the method he employed could be used to compute all solutions of the equation

d (\frac{10^{n} − 1}{9}) = F_{m_{1}} + F_{m_{2}} + F_{m_{3}} + F_{m_{4}}

with d ∈ {1, …, 9} and m_{1} ≥ m_{2} ≥ m_{3} ≥ m_{4}. Luca et al. [8] investigated this idea for Pell numbers. Luca et al. [8] showed that, all nonnegative integer solutions (m_{1}, m_{2}, m_{3}, n) of the equation

N = P_{m_{1}} + P_{m_{2}} + P_{m_{3}} + P_{m_{4}} = d (\frac{10^{n} − 1}{9}) with d ∈ \{1, …, 9\}

have

N ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 77, 88, 99, 111, 222, 444, 888, 999}.

In this paper, we compute all repdigits that can be expressed as sums of four Fibonacci or Lucas numbers. We prove Theorem 1 and Theorem 2 below.

Theorem 1. All nonnegative integer solutions (m_{1}, m_{2}, m_{3}, m_{4}, n) of the equation

N = F_{m_{1}} + F_{m_{2}} + F_{m_{3}} + F_{m_{4}} = d (\frac{10^{n} − 1}{9}) with d ∈ \{1, …, 9\}  (1)

have

N ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 99, 111, 222, 333, 555, 666, 777, 999, 1111, 2222, 11111, 66666}.

Theorem 2. All nonnegative integer solutions (m_{1}, m_{2}, m_{3}, m_{4}, n) of the equation

N = L_{m_{1}} + L_{m_{2}} + L_{m_{3}} + L_{m_{4}} = d (\frac{10^{n} − 1}{9}) with d ∈ \{1, …, 9\}  (2)

have

N ∈ {4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, 555, 666, 999, 2222, 4444, 11111, 88888}.

Here is the organization of this paper. In the next section, we recall the useful results to prove our two main results. We use them in Section 3 to prove Theorem 1. In Section 4, for the sake of completeness, we apply the same method for the entire proof of Theorem 2.

2 Preliminaries

In this section, we recall some results that are useful for the proof of Theorem 1 and Theorem 2. Let 𝕂 be an algebraic number field of degree D over ℚ, let α_{1}, …, α_{n} ∈ 𝕂 \ {0} and let b_{1}, …, b_{n} ∈ ℤ. Set

B = max{|b_{1}|, …, |b_{n}|}

and

Λ = α_{1}^{b_{1}} ⋯ α_{n}^{b_{n}} − 1.

Let A_{1}, …, A_{n} be real numbers with

max{Dh(α_{i}), | log α_{i}|, 0.16} ≤ A_{i}, 1 ≤ i ≤ n,

where h(η) is the logarithmic height of an algebraic number η which is given by the formula

h(η) = \frac{1}{d(η)} (log |a_{0}| + ∑_{i=1}^{d(η)} log (max\{|η^{(i)}|, 1\})) ,

where d(η) is the degree of η over ℚ and

f(X) = a_{0} ∏_{i=1}^{d(η)} (X − η^{(i)}) ∈ ℤ[X]

the minimal polynomial of η of degree d(η) over ℤ.

Lemma 3. ([1, Theorem 9.4]) Assume that Λ ≠ 0. We then have

log |Λ| > −3 × 30^{n+4} × (n + 1)^{5.5}D^{2}(1 + log D)(1 + log nB)A_{1} ⋯ A_{n}.  (3)

Furthermore, if 𝕂 is real, we have

log |Λ| > −1.4 × 30^{n+3} × n^{4.5}D^{2}(1 + log D)(1 + log B)A_{1} ⋯ A_{n}.  (4)

We now discuss a computational method for reducing upper bounds for solutions of Diophantine equations.

Let ϑ_{1}, ϑ_{2}, β ∈ ℝ be given, and let x_{1}, x_{2} ∈ ℤ be unknowns. Let

Λ = β + x_{1}ϑ_{1} + x_{2}ϑ_{2}.  (5)

Let c, δ be positive constants. Set X = max{|x_{1}|, |x_{2}|}. Let X_{0} be a (large) positive constant. Assume that

|Λ| < c · exp(−δ · Y),  (6)
X ≤ X_{0}.  (7)

When β = 0 in (5), we get

Λ = x_{1}ϑ_{1} + x_{2}ϑ_{2}.

Put ϑ = −ϑ_{1}/ϑ_{2}. Let the continued fraction expansion of ϑ be given by

[a_{0}, a_{1}, a_{2}, …],

and let the kth convergent of ϑ be p_{k}/q_{k} for k = 0, 1, 2, . . .. We may assume without loss of generality that |ϑ_{1}| < |ϑ_{2}| and that x_{1} > 0. We have the following results.

Lemma 4. ([9, Lemma 3.1]) (i) If (6) and (7) hold for x_{1}, x_{2} with X ≥ X^{∗}, then (−x_{2}, x_{1}) = (p_{k}, q_{k}) for an index k that satisfies

k ≤ −1 + \begin{matrix}log & (1 + X_{0}\sqrt{5}) \\ log & (\frac{1+\sqrt{5}}{2}) \end{matrix} := Y_{0}.  (8)

Moreover, the partial quotient a_{k+1} satisfies

a_{k+1} > −2 + \frac{|ϑ_{2}| exp(δq_{k})}{cq_{k}}.  (9)

(ii) If for some k with q_{k} ≥ X^{∗}, we have

a_{k+1} > \frac{|ϑ_{2}| exp(δq_{k})}{cq_{k}},  (10)

then (6) holds for (−x_{2}, x_{1}) = (p_{k}, q_{k}).

Lemma 5. ([9, Lemma 3.2]) Let

A = max_{0≤k≤Y_{0}} a_{k+1}.

If (6) and (7) hold for x_{1}, x_{2} and β = 0, then

Y < \frac{1}{δ} log (\frac{c(A + 2)X_{0}}{|ϑ_{2}|} ) .  (11)

When βϑ_{1}ϑ_{2} ≠ 0 in (5), put ϑ = −ϑ_{1}/ϑ_{2} and ψ = β/ϑ_{2}. Then we have

\frac{Λ}{ϑ_{2}} = ψ − x_{1}ϑ + x_{2}.

Let p/q be a convergent of ϑ with q > X_{0}. For a real number x we define ∥x∥ = min{|x − n|, n ∈ ℤ} be the distance from x to the nearest integer. We have the following result.

Lemma 6. ([9, Lemma 3.3]) Suppose that

∥ qψ ∥> \frac{2X_{0}}{q} .

Then, the solutions of (6) and (7) satisfy

Y < \frac{1}{δ} log (\frac{q^{2}c}{|ϑ_{2}|X_{0}} ) .

3 Proof of Theorem 1

It is well known that the Fibonacci numbers are given by

F_{m} = \frac{1}{\sqrt{5}} (α^{m} − β^{m}) for m ≥ 0, where (α, β) = (\frac{1 + \sqrt{5}}{2}, \frac{1 − \sqrt{5}}{2}) .

In equation (1) we suppose that m_{1} ≥ m_{2} ≥ m_{3} ≥ m_{4}. A search with Maple in the range 0 ≤ m_{1} ≤ 599 yielded only the solutions shown in the statement of Theorem 1.

Let us suppose that solutions of equation (1) exist for m_{1} ≥ 600. For m_{1} ≥ 600, we have that

F_{600} ≤ F_{m_{1}} ≤ F_{m_{1}} + F_{m_{2}} + F_{m_{3}} + F_{m_{4}} = d (\frac{10^{n} − 1}{9}) ≤ 10^{n} − 1,

and so

125 ≤ \frac{log(1 + F_{600})}{log 10} ≤ n.

That is, n ≥ 125. Now,

10^{n−1} ≤ d (\frac{10^{n} − 1}{9}) = F_{m_{1}} + F_{m_{2}} + F_{m_{3}} + F_{m_{4}}

≤ 4F_{m_{1}}

≤ \frac{4}{\sqrt{5}} (α^{m_{1}} + |β|^{m_{1}})

< \frac{8}{\sqrt{5}}α^{m_{1}}

< α^{m_{1}+2.7}, since \frac{8}{\sqrt{5}} < α^{2.7}. This means that 10^{n−1} < α^{m_{1}+2.7}, and thus

4.78(n − 1) < (n − 1)\frac{log 10}{log α} < m_{1} + 2.7.

Consequently,

n < 4.78n − 7.48 < m_{1},

as n ≥ 125. Therefore, 125 ≤ n < m_{1}.

We can put equation (1) in the form

\frac{α^{m_{1}} − β^{m_{1}}}{\sqrt{5}} + \frac{α^{m_{2}} − β^{m_{2}}}{\sqrt{5}} + \frac{α^{m_{3}} − β^{m_{3}}}{\sqrt{5}} + \frac{α^{m_{4}} − β^{m_{4}}}{\sqrt{5}} = \frac{d × 10^{n}}{9} − \frac{d}{9}.  (12)

We examine (12) in four different ways as follows.

Step 1: We express (12) in the form

\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) − \frac{d × 10^{n}}{9} = −\frac{d}{9} + \frac{1}{\sqrt{5}} (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}}) .

(13)

It follows that

|\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) − \frac{d × 10^{n}}{9} |

≤ \frac{d}{9} + \frac{1}{\sqrt{5}} (|β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}}) ,

leading to

|\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) − \frac{d × 10^{n}}{9} | < \frac{α^{4}}{\sqrt{5}}.  (14)

Multiplying both sides of inequality (14) by \frac{\sqrt{5}α^{−m_{1}}}{1+α^{m_{2}−m_{1}}+α^{m_{3}−m_{1}}+α^{m_{4}−m_{1}}} , we obtain

|1 − α^{−m_{4}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)})| < \frac{α^{4−m_{1}}}{1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}} ,

and so

|1 − α^{−m_{4}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)})| < α^{4−m_{1}}.  (15)

Put

Γ_{1} := 1 − α^{−m_{4}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) .  (16)

Suppose that Γ_{1} = 0. Then we have that

α^{m_{1}} + α^{m_{2}} + α^{m_{3}} + α^{m_{4}} = \frac{10^{n} × d\sqrt{5}}{9}.

Conjugating in ℚ (\sqrt{5}) yields

β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}} = −\frac{10^{n} × d\sqrt{5}}{9}.

Thus,

\frac{10^{125} × \sqrt{5}}{9} ≤ \frac{10^{n} × d\sqrt{5}}{9} = |β^{m_{1}} + β^{m_{2}} + β^{m_{3}}β^{m_{4}}|
≤ |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}}
< 4.

This implies that \frac{10^{125}×\sqrt{5}}{9} < 4, which is false. Hence, it follows that Γ_{1} ≠ 0.

In order to apply Lemma 3 to Γ_{1}, we set

α_{1} = α, α_{2} = 10, α_{3} = \frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)},

b_{1} = −m_{4}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. Thus, the degree D of ℚ (\sqrt{5}) is 2 and B = max{m_{4}, n, 1} ≤ m_{1}. The minimal polynomial of α over ℤ is x^{2} − x − 1, and so d(α) = 2 and a_{0}(α) = 1. It follows that

h(α) = \frac{1}{2} log α.

Also, the minimal polynomial of \sqrt{5} over ℤ is x^{2} − 5. Thus,

h (\sqrt{5}) = \frac{1}{2} log 5.

We have

max{2h(α_{1}), | log α_{1}|, 0.16} = log α < 0.49 =: A_{1},
max{2h(α_{2}), | log α_{2}|, 0.16} = 2 log 10 < 4.61 =: A_{2}.

Set

C_{1} = 2.3 × 10^{12} > 1.4 × 30^{6} × 3^{4.5} × D^{2} × (1 + log D) × A_{1} × A_{2}.

Next, we compute A_{3}. We find that,

α_{3} = \frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)} < \sqrt{5},

and

α_{3}^{−1} = \frac{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}{d\sqrt{5}} ≤ \frac{36}{\sqrt{5}}α^{m_{1}−m_{4}}.

Hence, | log α_{3}| < 3 + (m_{1} − m_{4}) log α. Also, we have that

h(α_{3}) ≤ h(d\sqrt{5}) + h(9) + h(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)
≤ h(9\sqrt{5}) + h(9) + log 2 + h(α^{m_{3}−m_{4}}(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1))
≤ h(9) + h(\sqrt{5}) + h(9) + 2 log 2 + h(α^{m_{3}−m_{4}}) + h(α^{m_{2}−m_{3}}(α^{m_{1}−m_{2}} + 1))
≤ h(\sqrt{5}) + 2h(9) + 3 log 2 + h(α^{m_{3}−m_{4}}) + h(α^{m_{2}−m_{3}}) + h(α^{m_{1}−m_{2}})

≤ h(\sqrt{5}) + 2h(9) + 3 log 2 + (m_{3} − m_{4})h(α) + (m_{2} − m_{3})h(α) + (m_{1} − m_{2})h(α)

= \frac{1}{2} log 5 + 2h(9) + 3 log 2 + \frac{1}{2}(m_{1} − m_{4}) log α.

Hence, 2h(α_{3}) ≤ 15 + (m_{1} − m_{4}) log α. Therefore, we get

max{2h(α_{3}), | log α_{3}|, 0.16} ≤ 15 + (m_{1} − m_{4}) log α =: A_{3}.

By applying Lemma 3 to Γ_{1} given by (16), and using (15) we have that

exp(−(15 + (m_{1} − m_{4}) log α)C_{1}(1 + log m_{1})) < α^{4−m_{1}}.

Thus,

m_{1} log α < 4 log α + (15 + (m_{1} − m_{4}) log α)C_{1}(1 + log m_{1}).  (17)

Step 2: We have that

\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) − \frac{d × 10^{n}}{9} = −\frac{d}{9} + \frac{1}{\sqrt{5}}(β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}} − α^{m_{4}}). (18)

Consequently, we get

|\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + \frac{1}{\sqrt{5}}(|β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}} + α^{m_{4}}),

and so

|\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) − \frac{d × 10^{n}}{9} | < \frac{α^{m_{4}+5}}{\sqrt{5}} .  (19)

We multiply both sides of inequality (19) by \frac{\sqrt{5}α^{−m_{1}}}{1+α^{m_{2}−m_{1}}+α^{m_{3}−m_{1}}} to get

|1 − α^{−m_{3}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)})| < \frac{α^{m_{4}−m_{1}+5}}{1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}} ,

which gives us

|1 − α^{−m_{3}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)})| < α^{m_{4}−m_{1}+5}.  (20)

Put

Γ_{2} := 1 − α^{−m_{3}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) .  (21)

Suppose that Γ_{2} = 0. Then, we get

α^{m_{1}} + α^{m_{2}} + α^{m_{3}} = \frac{10^{n} × d\sqrt{5}}{9}.

Taking the conjugate of this in ℚ(\sqrt{5}), we get

β^{m_{1}} + β^{m_{2}} + β^{m_{3}} = −\frac{10^{n} × d\sqrt{5}}{9}.

Consequently, we obtain

\frac{10^{125} × \sqrt{5}}{9} ≤ \frac{10^{n} × d\sqrt{5}}{9} = |β^{m_{1}} + β^{m_{2}} + β^{m_{3}}| ≤ |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} < 3,

which means that \frac{10^{125}×\sqrt{5}}{9} < 3. This is false. We conclude that Γ_{2} ≠ 0.

To apply Lemma 3 to Γ_{2} given by (21), we set

α_{1} = α, α_{2} = 10, α_{3} = \frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}, b_{1} = −m_{3}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. We have B = max\{m_{3}, n, 1\} ≤ m_{1}. We proceed to compute A_{3} by first observing that

α_{3} = \frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)} < \sqrt{5}

and

α_{3}^{−1} = \frac{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}{d\sqrt{5}} ≤ \frac{27}{\sqrt{5}}α^{m_{1}−m_{3}}.

Hence, | log α_{3}| < 3 + (m_{1} − m_{3}) log α. Additionally, we get

h(α_{3}) ≤ h(d\sqrt{5}) + h(9) + log 2 + h(α^{m_{2}−m_{3}}(α^{m_{1}−m_{2}} + 1))

≤ \frac{1}{2} log 5 + 2h(9) + 2 log 2 + h(α^{m_{2}−m_{3}}) + h(α^{m_{1}−m_{2}})

≤ \frac{1}{2} log 5 + 2h(9) + 2 log 2 + (m_{2} − m_{3})h(α) + (m_{1} − m_{2})h(α)

= \frac{1}{2} log 5 + 2h(9) + 2 log 2 + \frac{1}{2}(m_{1} − m_{3}) log α.

Hence, 2h(α_{3}) ≤ 14 + (m_{1} − m_{3}) log α. As a result, we find that

max{2h(α_{3}), | log α_{3}|, 0.16} ≤ 14 + (m_{1} − m_{3}) log α =: A_{3}.

By applying Lemma 3 to Γ_{2} given by (21) and using (20), we deduce that

exp(−(14 + (m_{1} − m_{3}) log α)C_{1}(1 + log m_{1})) < α^{m_{4}−m_{1}+5}.

Thus, we get

(m_{1} − m_{4}) log α < 5 log α + (14 + (m_{1} − m_{3}) log α)C_{1}(1 + log m_{1}).  (22)

Step 3: We begin with (12) written in the form

\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}}) − \frac{d × 10^{n}}{9} = −\frac{d}{9} + \frac{1}{\sqrt{5}}(β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}} − α^{m_{3}} − α^{m_{4}}). (23)

Equation (23) leads us to

|\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}}) − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + \frac{1}{\sqrt{5}}(|β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}} + α^{m_{3}} + α^{m_{4}})

< 1 + \frac{4}{\sqrt{5}} + \frac{2α^{m_{3}}}{\sqrt{5}}

≤ \frac{1}{\sqrt{5}} (\sqrt{5} + 6) α^{m_{3}},

from which we obtain

|\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}}) − \frac{d × 10^{n}}{9} | < \frac{α^{m_{3}+5}}{\sqrt{5}} .  (24)

Multiplying both sides of inequality (24) by \frac{\sqrt{5}α^{−m_{1}}}{1+α^{m_{2}−m_{1}}} gives us

|1 − α^{−m_{2}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)})| < \frac{α^{m_{3}−m_{1}+5}}{1 + α^{m_{2}−m_{1}}} ,

which yields

|1 − α^{−m_{2}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)})| < α^{m_{3}−m_{1}+5}.  (25)

Put

Γ_{3} := 1 − α^{−m_{2}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}) .  (26)

Suppose that Γ_{3} = 0. Then

α^{m_{1}} + α^{m_{2}} = \frac{10^{n} × d\sqrt{5}}{9},

giving us

β^{m_{1}} + β^{m_{2}} = −\frac{10^{n} × d\sqrt{5}}{9}

by conjugating in ℚ (\sqrt{5}). It follows that

\frac{10^{125} × \sqrt{5}}{9} ≤ \frac{10^{n} × d\sqrt{5}}{9} = |β^{m_{1}} + β^{m_{2}}| ≤ |β|^{m_{1}} + |β|^{m_{2}} < 2,

which is false. Hence, Γ_{3} ≠ 0. Using the notations in Lemma 3, we put

α_{1} = α, α_{2} = 10, α_{3} = \frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}, b_{1} = −m_{2}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. We have B = max\{m_{2}, n, 1\} ≤ m_{1}. Now, we deduce that

α_{3} = \frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)} ≤ \sqrt{5} and α_{3}^{−1} = \frac{9(α^{m_{1}−m_{2}} + 1)}{d\sqrt{5}} ≤ \frac{18}{\sqrt{5}}α^{m_{1}−m_{2}}.

So | log α_{3}| < 3 + (m_{1} − m_{2}) log α. Furthermore,

h(α_{3}) ≤ h(d\sqrt{5}) + h(9) + log 2 + h(α^{m_{1}−m_{2}})

≤ h(\sqrt{5}) + 2h(9) + log 2 + (m_{1} − m_{2})h(α)

= \frac{1}{2} log 5 + 2h(9) + log 2 + \frac{1}{2}(m_{1} − m_{2}) log α.

Thus, 2h(α_{3}) ≤ 12 + (m_{1} − m_{2}) log α and so

max{2h(α_{3}), | log α_{3}|, 0.16} < 12 + (m_{1} − m_{2}) log α =: A_{3}.

Applying Lemma 3 to Γ_{3} given by (26), and using (25) we produce

exp(−(12 + (m_{1} − m_{2}) log α)C_{1}(1 + log m_{1})) < α^{m_{3}−m_{1}+5},

from which we obtain

(m_{1} − m_{3}) log α < 5 log α + (12 + (m_{1} − m_{2}) log α)C_{1}(1 + log m_{1}).  (27)

Step 4: Using equation (12) in the form

\frac{α^{m_{1}}}{\sqrt{5}} − \frac{d × 10^{n}}{9} = −\frac{d}{9} + \frac{1}{\sqrt{5}} (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}} − α^{m_{2}} − α^{m_{3}} − α^{m_{4}}) , (28)

we get

|\frac{α^{m_{1}}}{\sqrt{5}} − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + \frac{1}{\sqrt{5}} (|β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}} + α^{m_{2}} + α^{m_{3}} + α^{m_{4}})

< 1 + \frac{4}{\sqrt{5}} + \frac{3}{\sqrt{5}}α^{m_{2}}

≤ \frac{1}{\sqrt{5}}(\sqrt{5} + 7)α^{m_{2}},

which means that

|\frac{α^{m_{1}}}{\sqrt{5}} − \frac{d × 10^{n}}{9} | < \frac{α^{m_{2}+5}}{\sqrt{5}} .  (29)

Multiplying both sides of (29) by \sqrt{5}α^{−m_{1}} yields

|1 − α^{−m_{1}}10^{n} (\frac{d\sqrt{5}}{9})| < α^{m_{2}−m_{1}+5}.  (30)

Put

Γ_{4} := 1 − α^{−m_{1}}10^{n} (\frac{d\sqrt{5}}{9}) .  (31)

Suppose that Γ_{4} = 0. Then

α^{m_{1}} = \frac{d\sqrt{5} × 10^{n}}{9} ,

which implies that

β^{m_{1}} = −\frac{d\sqrt{5} × 10^{n}}{9} .

Consequently,

\frac{\sqrt{5} × 10^{125}}{9} ≤ \frac{d\sqrt{5} × 10^{n}}{9} = |β|^{m_{1}} < 1,

which is impossible. Hence, Γ_{4} ≠ 0. In order to apply Lemma 3 to Γ_{4} given by (31), we take

α_{1} = α, α_{2} = 10, α_{3} = \frac{\sqrt{5}d}{9}, b_{1} = −m_{1}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. To compute A_{3}, we observe that

α_{3} = \frac{d\sqrt{5}}{9} ≤ \sqrt{5} and α_{3}^{−1} = \frac{9}{d\sqrt{5}} ≤ \frac{9}{\sqrt{5}},

so | log α_{3}| < 1.4. In addition, we have

h(α_{3}) ≤ h(d\sqrt{5}) + h(9) ≤ \frac{1}{2} log 5 + 2h(9).

As a result, we have that 2h(α_{3}) < 10.4. We see that

max{2h(α_{3}), | log α_{3}|, 0.16} < 10.4 =: A_{3}.

By applying Lemma 3 to Γ_{4} given by (31) and using (30), we obtain

exp(−10.4C_{1}(1 + log m_{1})) < |1 − α^{−m_{1}}10^{n} (\frac{d\sqrt{5}}{9})| < α^{m_{2}−m_{1}+5}.

This means that

(m_{1} − m_{2}) log α < 5 log α + 10.4C_{1}(1 + log m_{1}) < 10.5C_{1}(1 + log m_{1}).  (32)

Putting together (32) and (27) yields

(m_{1} − m_{3}) log α < 5 log α + (12 + 10.5C_{1}(1 + log m_{1}))C_{1}(1 + log m_{1})
< 5 log α + 10.6C_{1}^{2}(1 + log m_{1})^{2}
< 10.7C_{1}^{2}(1 + log m_{1})^{2}.

That is

(m_{1} − m_{3}) log α < 10.7C_{1}^{2}(1 + log m_{1})^{2}.  (33)

Combining (33) and (22), we obtain

(m_{1} − m_{4}) log α < 5 log α + (14 + 10.7C_{1}^{2}(1 + log m_{1})^{2}) C_{1}(1 + log m_{1})
< 5 log α + 10.8C_{1}^{3}(1 + log m_{1})^{3}
< 10.9C_{1}^{3}(1 + log m_{1})^{3}.

That is

(m_{1} − m_{4}) log α < 10.9C_{1}^{3}(1 + log m_{1})^{3}.  (34)

We now combine (34) and (17) to obtain

m_{1} log α < 4 log α + (15 + 10.9C_{1}^{3}(1 + log m_{1})^{3})C_{1}(1 + log m_{1})
< 4 log α + 11.0C_{1}^{4}(1 + log m_{1})^{4}
< 11.1C_{1}^{4}(1 + log m_{1})^{4}
< 11.1 (2.3 × 10^{12})^{4} (1 + log m_{1})^{4}.

That is

m_{1} log α < 11.1 (2.3 × 10^{12})^{4} (1 + log m_{1})^{4}.  (35)

Inequality (35) gives rise to the inequality m_{1} < 2.3 × 10^{59}. Now, we lower the bound. Let

Λ_{1} = −m_{1} log α + n log 10 + log (\frac{d\sqrt{5}}{9}) .  (36)

Equation (28) leads us to

\frac{α^{m_{1}}}{\sqrt{5}} − \frac{d × 10^{n}}{9} = \frac{α^{m_{1}}}{\sqrt{5}} (1 − α^{−m_{1}}10^{n} (\frac{d\sqrt{5}}{9}))

= \frac{α^{m_{1}}}{\sqrt{5}} (1 − e^{Λ_{1}})

= −\frac{d}{9} + \frac{β^{m_{1}}}{\sqrt{5}} − F_{m_{2}} − F_{m_{3}} − F_{m_{4}}

≤ −\frac{1}{9} + \frac{|β|^{600}}{\sqrt{5}}

< 0,

as m_{1} ≥ 600. Thus, Λ_{1} > 0 and so from (30) we obtain

0 < Λ_{1} < e^{Λ_{1}} − 1 = |1 − α^{−m_{1}}10^{n} (\frac{d\sqrt{5}}{9})| < α^{m_{2}−m_{1}+5}.

This means that

log (\frac{d\sqrt{5}}{9}) + m_{1}(− log α) + n log 10 < α^{5}α^{−(m_{1}−m_{2})}
< α^{5.1} exp(−0.48(m_{1} − m_{2})),

which leads to

|Λ_{1}| < α^{5.1} exp(−0.48(m_{1} − m_{2})),  (37)

with X = max{m_{1}, n} = m_{1} ≤ 2.3 × 10^{59}. We also have that

\frac{Λ_{1}}{log 10} = \frac{log(d\sqrt{5}/9)}{log 10} − m_{1} \frac{log α}{log 10} + n.

Thus, we take

c = α^{5.1}, δ = 0.48, X_{0} = 2.3 × 10^{59}, ψ = \frac{log(d\sqrt{5}/9)}{log 10}, Y = m_{1} − m_{2},

ϑ = \frac{log α}{log 10}, ϑ_{1} = − log α, ϑ_{2} = log 10, β = log(d\sqrt{5}/9).

The smallest value of q such that q > X_{0} is q = q_{125}. We find that q = q_{128} satisfies the hypothesis of Lemma 6 for d = 1, …, 9. Applying Lemma 6, we get m_{1} − m_{2} ≤ 310, and hence m_{2} ≥ 290.

Taking 1 ≤ d ≤ 9 and 0 ≤ m_{1} − m_{2} ≤ 310, we let

Λ_{2} = −m_{2} log α + n log 10 + log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}) .  (38)

We see from equation (23) that

\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}}) (1 − e^{Λ_{2}}) = −\frac{d}{9} + \frac{β^{m_{1}}}{\sqrt{5}} + \frac{β^{m_{2}}}{\sqrt{5}} − F_{3} − F_{4}

≤ −\frac{1}{9} + \frac{|β|^{600}}{\sqrt{5}} + \frac{|β|^{290}}{\sqrt{5}}

< 0,

making use of m_{1} ≥ 600 and m_{2} ≥ 290. Hence, Λ_{2} > 0, and so from (25) we see that

0 < Λ_{2} < e^{Λ_{2}} − 1 = |1 − α^{−m_{2}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)})| < α^{m_{3}−m_{1}+5}.

Thus, we have

log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}) + m_{2}(− log α) + n log 10 < α^{m_{3}−m_{1}+5}
< α^{5.1} exp(−0.48(m_{1} − m_{3})),

which gives us

|Λ_{2}| < α^{5.1} exp(−0.48(m_{1} − m_{3})),  (39)

where X = max{m_{2}, n} ≤ m_{1} ≤ 2.3 × 10^{59}. We also have that

\frac{Λ_{2}}{log 10} = \frac{1}{log 10} log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}) − m_{1} \frac{log α}{log 10} + n.

Thus, we consider

c = α^{5.1}, δ = 0.48, X_{0} = 2.3 × 10^{59}, ψ = \frac{1}{log 10} log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}) ,

Y = m_{1} − m_{3}, ϑ = \frac{log α}{log 10}, ϑ_{1} = − log α, ϑ_{2} = log 10, β = log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{2}} + 1)}) .

We find that q = q_{132} satisfies the hypothesis of Lemma 6 for d = 1, …, 9 and 0 ≤ m_{1}−m_{2} ≤ 310. Applying Lemma 6, we get m_{1} − m_{3} ≤ 328. Hence, m_{3} ≥ 272.

Taking 1 ≤ d ≤ 9, 0 ≤ m_{2} − m_{3} ≤ m_{1} − m_{3} ≤ 328, we let

Λ_{3} = −m_{3} log α + n log 10 + log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) .  (40)

Equation (18) ensures that

\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) (1 − e^{Λ_{3}}) = −\frac{d}{9} + \frac{1}{\sqrt{5}} (β^{m_{1}} + β^{m_{2}} + β^{m_{3}}) − F_{4}

≤ −\frac{1}{9} + \frac{1}{\sqrt{5}} (|β|^{600} + |β|^{290} + |β|^{272})

< 0,

where we use m_{1} ≥ 600, m_{2} ≥ 290, and m_{3} ≥ 272. Hence, Λ_{3} > 0, and so from (20) we see that

0 < Λ_{3} < e^{Λ_{3}} − 1 = |1 − α^{−m_{3}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)})| < α^{m_{4}−m_{1}+5}.

Hence, we have

log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) + m_{3}(− log α) + n log 10 < α^{m_{4}−m_{1}+5}
< α^{5.1} exp(−0.48(m_{1} − m_{4})),

leading to

|Λ_{3}| < α^{5.1} exp(−0.48(m_{1} − m_{4})),  (41)

where X = max{m_{3}, n} ≤ m_{1} ≤ 2.3 × 10^{59}. Furthermore, we obtain

\frac{Λ_{3}}{log 10} = \frac{1}{log 10} log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) − m_{3} \frac{log α}{log 10} + n.

Thus, we take

c = α^{5.1}, δ = 0.48, X_{0} = 2.3 × 10^{59}, ψ = \frac{1}{log 10} log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) ,

Y = m_{1} − m_{4}, ϑ = \frac{log α}{log 10}, ϑ_{1} = − log α, ϑ_{2} = log 10,

β = log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) .

We find that q = q_{135} satisfies the hypothesis of Lemma 6 for 1 ≤ d ≤ 9, 0 ≤ m_{2} − m_{3} ≤ m_{1} − m_{3} ≤ 328. Applying Lemma 6, we get m_{1} − m_{4} ≤ 335, and hence m_{4} ≥ 265.

Taking 1 ≤ d ≤ 9, 0 ≤ m_{3} − m_{4} ≤ m_{2} − m_{4} ≤ m_{1} − m_{4} ≤ 335, we let

Λ_{4} = −m_{4} log α + n log 10 + log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) .  (42)

Using equation (13), we have that

\frac{α^{m_{1}}}{\sqrt{5}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) (1 − e^{Λ_{4}}) = −\frac{d}{9} + \frac{1}{\sqrt{5}} (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}})

≤ −\frac{1}{9} + \frac{1}{\sqrt{5}} (|β|^{600} + |β|^{290} + |β|^{272} + |β|)

< 0

making use of m_{1} ≥ 600, m_{2} ≥ 290, m_{3} ≥ 272 and m_{4} ≥ 265. Hence, Λ_{4} > 0, and so from (15) we see that

0 < Λ_{4} < e^{Λ_{4}} − 1 = |1 − α^{−m_{4}}10^{n} (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)})| < α^{4−m_{1}}.

Hence, we have

log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) + m_{4}(− log α) + n log 10 < α^{4−m_{1}},

which implies that

|Λ_{4}| < α^{4.1} exp(−0.48m_{1}),  (43)

where X = max{m_{4}, n} ≤ m_{1} < 2.3 × 10^{59}. In addition,

\frac{Λ_{4}}{log 10} = \frac{1}{log 10} log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) − m_{4} \frac{log α}{log 10} + n.

Thus,

c = α^{4.1}, δ = 0.48, X_{0} = 2.3 × 10^{59}, Y = m_{1}

ψ = \frac{1}{log 10} log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) , ϑ = \frac{log α}{log 10},

ϑ_{1} = − log α, ϑ_{2} = log 10, β = log (\frac{d\sqrt{5}}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) .

We find that q = q_{141} satisfies the hypothesis of Lemma 6 for 1 ≤ d ≤ 9, 0 ≤ m_{3} − m_{4} ≤ m_{2} − m_{4} ≤ m_{1} − m_{4} ≤ 335. Applying Lemma 6, we get m_{1} ≤ 392, which contradicts the assumption that m_{1} ≥ 600. This proves the result.

4 Proof of Theorem 2

Although this is similar to the proof of the previous theorem, we give the details for the convenience of the reader. We use the fact that

L_{m} = α^{m} + β^{m} holds for all m ≥ 0, where (α, β) = (\frac{1 + \sqrt{5}}{2}, \frac{1 − \sqrt{5}}{2}) .

In equation (2), we suppose that m_{1} ≥ m_{2} ≥ m_{3} ≥ m_{4}. A search with Maple in the range 0 ≤ m_{1} ≤ 599 yielded only the solutions shown in the statement of Theorem 2.

Let us suppose that solutions of equation (2) exist for m_{1} ≥ 600. We observe that

L_{600} ≤ L_{m_{1}} ≤ L_{m_{1}} + L_{m_{2}} + L_{m_{3}} + L_{m_{4}} = d (\frac{10^{n} − 1}{9}) ≤ 10^{n} − 1.

This leads us to

125 ≤ \frac{log(1 + L_{600})}{log 10} ≤ n,

and so n ≥ 125. We further observe that

10^{n−1} ≤ d (\frac{10^{n} − 1}{9}) = L_{m_{1}} + L_{m_{2}} + L_{m_{3}} + L_{m_{4}} ≤ 4 (α^{m_{1}} + |β|^{m_{1}}) < α^{m_{1}+4.33}.

Hence, we obtain

4.78(n − 1) < (n − 1)\frac{log 10}{log α} < m_{1} + 4.33,

which gives us

n < 4.78n − 9.11 < m_{1},

for n ≥ 125. Therefore, 125 ≤ n < m_{1}.

We can put equation (2) in the form

α^{m_{1}} + β^{m_{1}} + α^{m_{2}} + β^{m_{2}} + α^{m_{3}} + β^{m_{3}} + α^{m_{4}} + β^{m_{4}} − \frac{d × 10^{n}}{9} = −\frac{d}{9}.  (44)

Equation(44) is treated in four different ways in the steps that follow.

Step 1: We express (44) in the form

α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) − \frac{d × 10^{n}}{9} = −\frac{d}{9} − (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}}) . (45)

It follows that

|α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + (|β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}}) ,

leading to

|α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) − \frac{d × 10^{n}}{9} | < α^{3.35}.  (46)

Multiplication of both sides of (46) by \frac{α^{−m_{1}}}{1+α^{m_{2}−m_{1}}+α^{m_{3}−m_{1}}+α^{m_{4}−m_{1}}} gives us

|1 − α^{−m_{4}}10^{n} (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)})| < \frac{α^{3.35−m_{1}}}{1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}} ,

from which we get

|1 − α^{−m_{4}}10^{n} (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)})| < α^{3.35−m_{1}}.  (47)

Put

Γ_{1} := 1 − α^{−m_{4}}10^{n} (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) .  (48)

Suppose that Γ_{1} = 0. Then, we have that

α^{m_{1}} + α^{m_{2}} + α^{m_{3}} + α^{m_{4}} = \frac{10^{n} × d}{9}.

Conjugating in ℚ (\sqrt{5}) yields

β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}} = \frac{10^{n} × d}{9}.

Thus,

\frac{10^{125}}{9} ≤ \frac{10^{n} × d}{9} = |β^{m_{1}} + β^{m_{2}} + β^{m_{3}}β^{m_{4}}| ≤ |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}} < 4.

This implies that \frac{10^{125}}{9} < 4, which is false. Hence, it follows that Γ_{1} ≠ 0.

In the notation of Lemma 3, we set

α_{1} = α, α_{2} = 10, α_{3} = \frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)},

b_{1} = −m_{4}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. We get B = max\{m_{4}, n, 1\} ≤ m_{1}. The minimal polynomial of α over ℤ is x^{2} − x − 1, and so d(α) = 2 and a_{0}(α) = 1. It is known that

h(α) = \frac{1}{2} log α.

We have

max{2h(α_{1}), | log α_{1}|, 0.16} = log α < 0.49 =: A_{1},

max{2h(α_{2}), | log α_{2}|, 0.16} = 2 log 10 < 4.61 =: A_{2}.

Set

C_{1} = 2.4 × 10^{12} > 1.4 × 30^{6} × 3^{4.5} × D^{2} × (1 + log D) × A_{1} × A_{2}.

Next, we compute A_{3}. We find that,

α_{3} = \frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)} < 1,

and

α_{3}^{−1} = \frac{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}{d} ≤ 36α^{m_{1}−m_{4}},

hence, | log α_{3}| < 4 + (m_{1} − m_{4}) log α. Also, we have that

h(α_{3}) ≤ h(d) + h(9) + log 2 + h(α^{m_{3}−m_{4}}(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1))

≤ 2h(9) + 2 log 2 + h(α^{m_{3}−m_{4}}) + h(α^{m_{2}−m_{3}}(α^{m_{1}−m_{2}} + 1)) ≤ 2h(9) + 3 log 2 + h(α^{m_{3}−m_{4}}) + h(α^{m_{2}−m_{3}}) + h(α^{m_{1}−m_{2}}) ≤ 2h(9) + 3 log 2 + (m_{3} − m_{4})h(α) + (m_{2} − m_{3})h(α) + (m_{1} − m_{2})h(α)

= 2h(9) + 3 log 2 + \frac{1}{2}(m_{1} − m_{4}) log α.

Hence, 2h(α_{3}) ≤ 13 + (m_{1} − m_{4}) log α. Therefore, we get

max{2h(α_{3}), | log α_{3}|, 0.16} ≤ 13 + (m_{1} − m_{4}) log α =: A_{3}.

By applying Lemma 3 to Γ_{1} given by (48), and using (47) we have that

exp(−(13 + (m_{1} − m_{4}) log α)C_{1}(1 + log m_{1})) < α^{3.35−m_{1}}.

Thus,

m_{1} log α < 3.35 log α + (13 + (m_{1} − m_{4}) log α)C_{1}(1 + log m_{1}).  (49)

Step 2: Writing equation (44) as

α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) − \frac{d × 10^{n}}{9} = −\frac{d}{9} − α^{m_{4}} − (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}}), (50)

we get

|α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + α^{m_{4}} + |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}},

and so

|α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) − \frac{d × 10^{n}}{9} | < α^{m_{4}+3.73}.  (51)

By multiplying both sides of inequality (51) by \frac{α^{−m_{1}}}{1+α^{m_{2}−m_{1}}+α^{m_{3}−m_{1}}} we obtain

|1 − α^{−m_{3}}10^{n} (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)})| < \frac{α^{m_{4}−m_{1}+3.73}}{1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}} ,

which leads to

|1 − α^{−m_{3}}10^{n} (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)})| < α^{m_{4}−m_{1}+3.73}.  (52)

Put

Γ_{2} := 1 − α^{−m_{3}}10^{n} (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) .  (53)

Suppose that Γ_{2} = 0. Then, we get

α^{m_{1}} + α^{m_{2}} + α^{m_{3}} = \frac{10^{n} × d}{9}.

Taking the conjugate of this in ℚ(\sqrt{5}), we get

β^{m_{1}} + β^{m_{2}} + β^{m_{3}} = \frac{10^{n} × d}{9},

which implies that

\frac{10^{125}}{9} ≤ \frac{10^{n} × d}{9} = |β^{m_{1}} + β^{m_{2}} + β^{m_{3}}| ≤ |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} < 3.

Thus, \frac{10^{125}}{9} < 3, which is false. We conclude that Γ_{2} ≠ 0.

To apply Lemma 3 to Γ_{2} given by (53), we set

α_{1} = α, α_{2} = 10, α_{3} = \frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}, b_{1} = −m_{3}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. Also, we obtain B = max\{m_{3}, n, 1\} ≤ m_{1}. We proceed to compute A_{3} by first observing that

α_{3} = \frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)} < 1,

and

α_{3}^{−1} = \frac{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}{d} ≤ 27α^{m_{1}−m_{3}}.

Hence, | log α_{3}| < 4 + (m_{1} − m_{3}) log α. Additionally, we get

h(α_{3}) ≤ h(d) + h(9) + log 2 + h(α^{m_{2}−m_{3}}(α^{m_{1}−m_{2}} + 1))

≤ 2h(9) + 2 log 2 + h(α^{m_{2}−m_{3}}) + h(α^{m_{1}−m_{2}})

≤ 2h(9) + 2 log 2 + (m_{2} − m_{3})h(α) + (m_{1} − m_{2})h(α)

= 2h(9) + 2 log 2 + \frac{1}{2}(m_{1} − m_{3}) log α.

Hence, 2h(α_{3}) ≤ 12 + (m_{1} − m_{3}) log α. As a result, we find that

max{2h(α_{3}), | log α_{3}|, 0.16} ≤ 12 + (m_{1} − m_{3}) log α =: A_{3}.

By applying Lemma 3 to Γ_{2} given by (53) and using (52), we deduce that

exp(−(12 + (m_{1} − m_{3}) log α)C_{1}(1 + log m_{1})) < α^{m_{4}−m_{1}+3.73}.

Thus, we get

(m_{1} − m_{4}) log α < 3.73 log α + (12 + (m_{1} − m_{3}) log α)C_{1}(1 + log m_{1}).  (54)

Step 3: Writing (44) as

α^{m_{1}} (1 + α^{m_{2}−m_{1}}) − \frac{d × 10^{n}}{9} = −\frac{d}{9} − (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}}) − (α^{m_{3}} + α^{m_{4}}), (55)

gives us

|α^{m_{1}} (1 + α^{m_{2}−m_{1}}) − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}} + α^{m_{3}} + α^{m_{4}} ≤ 7α^{m_{3}},

which leads to

|α^{m_{1}} (1 + α^{m_{2}−m_{1}}) − \frac{d × 10^{n}}{9} | < α^{m_{3}+4.05}.  (56)

Multiplying both sides of (56) by \frac{α^{−m_{1}}}{1+α^{m_{2}−m_{1}}} gives us

|1 − α^{−m_{2}}10^{n} (\frac{d}{9(α^{m_{1}−m_{2}} + 1)})| < \frac{α^{m_{3}−m_{1}+4.05}}{1 + α^{m_{2}−m_{1}}} ,

which yields

|1 − α^{−m_{2}}10^{n} (\frac{d}{9(α^{m_{1}−m_{2}} + 1)})| < α^{m_{3}−m_{1}+4.05}.  (57)

Put

Γ_{3} := 1 − α^{−m_{2}}10^{n} (\frac{d}{9(α^{m_{1}−m_{2}} + 1)}) .  (58)

Suppose that Γ_{3} = 0. Then

α^{m_{1}} + α^{m_{2}} = \frac{10^{n} × d}{9},

giving us

β^{m_{1}} + β^{m_{2}} = \frac{10^{n} × d}{9}

by conjugating in ℚ (\sqrt{5}). We see that

\frac{10^{125}}{9} ≤ \frac{10^{n} × d}{9} = |β^{m_{1}} + β^{m_{2}}| ≤ |β|^{m_{1}} + |β|^{m_{2}} < 2,

which is false. Hence, Γ_{3} ≠ 0. Using the notations in Lemma 3, we put

α_{1} = α, α_{2} = 10, α_{3} = \frac{d}{9(α^{m_{1}−m_{2}} + 1)}, b_{1} = −m_{2}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. We get B = max\{m_{2}, n, 1\} ≤ m_{1}. It is easily seen that

α_{3} = \frac{d}{9(α^{m_{1}−m_{2}} + 1)} ≤ 1 and α_{3}^{−1} = \frac{9(α^{m_{1}−m_{2}} + 1)}{d} ≤ 18α^{m_{1}−m_{2}}.

So | log α_{3}| < 3 + (m_{1} − m_{2}) log α. Additionally, we have

h(α_{3}) ≤ h(d) + h(9) + log 2 + h(α^{m_{1}−m_{2}})

≤ 2h(9) + log 2 + (m_{1} − m_{2})h(α)

= 2h(9) + log 2 + \frac{1}{2}(m_{1} − m_{2}) log α.

Thus, 2h(α_{3}) ≤ 11 + (m_{1} − m_{2}) log α and so

max{2h(α_{3}), | log α_{3}|, 0.16} < 11 + (m_{1} − m_{2}) log α =: A_{3}.

Applying Lemma 3 to Γ_{3} given by (58), and using (57) we produce

exp(−(11 + (m_{1} − m_{2}) log α)C_{1}(1 + log m_{1})) < α^{m_{3}−m_{1}+4.05},

from which we obtain

(m_{1} − m_{3}) log α < 4.05 log α + (11 + (m_{1} − m_{2}) log α)C_{1}(1 + log m_{1}).  (59)

Step 4: Writing equation (44) as

α^{m_{1}} − \frac{d × 10^{n}}{9} = −\frac{d}{9} − (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}}) − (α^{m_{2}} + α^{m_{3}} + α^{m_{4}}),  (60)

we get

|α^{m_{1}} − \frac{d × 10^{n}}{9} | ≤ \frac{d}{9} + |β|^{m_{1}} + |β|^{m_{2}} + |β|^{m_{3}} + |β|^{m_{4}} + α^{m_{2}} + α^{m_{3}} + α^{m_{4}} ≤ 8α^{m_{2}},

which means that

|α^{m_{1}} − \frac{d × 10^{n}}{9} | < α^{m_{2}+4.33}.  (61)

Multiplying both sides of (61) by α^{−m_{1}} yields

|1 − α^{−m_{1}}10^{n} (\frac{d}{9})| < α^{m_{2}−m_{1}+4.33}.  (62)

Put

Γ_{4} := 1 − α^{−m_{1}}10^{n} (\frac{d}{9}) .  (63)

Suppose that Γ_{4} = 0. Then

α^{m_{1}} = \frac{d × 10^{n}}{9} ,

and by conjugation

β^{m_{1}} = \frac{d × 10^{n}}{9} .

Consequently,

\frac{10^{125}}{9} ≤ \frac{d × 10^{n}}{9} = |β|^{m_{1}} < 1,

which is impossible. Hence, Γ_{4} ≠ 0. In order to apply Lemma 3 to Γ_{4} given by (63), we take

α_{1} = α, α_{2} = 10, α_{3} = \frac{d}{9}, b_{1} = −m_{1}, b_{2} = n, b_{3} = 1,

where α_{1}, α_{2}, α_{3} ∈ ℚ (\sqrt{5}) and b_{1}, b_{2}, b_{3} ∈ ℤ. To compute A_{3}, we observe that

α_{3} = \frac{d}{9} ≤ 1 and α_{3}^{−1} = \frac{9}{d} ≤ 9,

so | log α_{3}| < 2.2. In addition, we have

h(α_{3}) ≤ h(d) + h(9) ≤ 2h(9).

This gives us 2h(α_{3}) < 8.79. And so we have

max{2h(α_{3}), | log α_{3}|, 0.16} < 8.79 =: A_{3}.

By applying Lemma 3 to Γ_{4} given by (63) and using (62), we obtain

exp(−8.79C_{1}(1 + log m_{1})) < |1 − α^{−m_{1}}10^{n} (\frac{d}{9})| < α^{m_{2}−m_{1}+4.33}.

This means that

(m_{1} − m_{2}) log α < 4.33 log α + 8.79C_{1}(1 + log m_{1}) < 8.80C_{1}(1 + log m_{1}).  (64)

Putting together (64) and (59) yields

(m_{1} − m_{3}) log α < 4.05 log α + (11 + 8.80C_{1}(1 + log m_{1}))C_{1}(1 + log m_{1})

= 4.05 log α + 11C_{1}(1 + log m_{1}) + 8.80C_{1}^{2}(1 + log m_{1})^{2}

< 8.81C_{1}^{2}(1 + log m_{1})^{2},

since 4.05 log α + 11C_{1}(1 + log m_{1}) < 0.01C_{1}^{2}(1 + log m_{1})^{2}. Hence,

(m_{1} − m_{3}) log α < 8.81C_{1}^{2}(1 + log m_{1})^{2}.  (65)

Combining (65) and (54), we obtain

(m_{1} − m_{4}) log α < 3.73 log α + (12 + 8.81C_{1}^{2}(1 + log m_{1})^{2}) C_{1}(1 + log m_{1})

= 3.73 log α + 12C_{1}(1 + log m_{1}) + 8.81C_{1}^{3}(1 + log m_{1})^{3}

< 8.82C_{1}^{3}(1 + log m_{1})^{3}.

since 3.73 log α + 12C_{1}(1 + log m_{1}) < 0.01C_{1}^{3}(1 + log m_{1})^{3}. Thus,

(m_{1} − m_{4}) log α < 8.82C_{1}^{3}(1 + log m_{1})^{3}.  (66)

We now combine (66) and (49) to obtain

m_{1} log α < 3.35 log α + (13 + 8.82C_{1}^{3}(1 + log m_{1})^{3})C_{1}(1 + log m_{1})
= 3.35 log α + 13C_{1}(1 + log m_{1}) + 8.82C_{1}^{4}(1 + log m_{1})^{4}
< 8.83C_{1}^{4}(1 + log m_{1})^{4}
< 8.83 (2.4 × 10^{12})^{4} (1 + log m_{1})^{4}.

That is

m_{1} log α < 8.83 (2.4 × 10^{12})^{4} (1 + log m_{1})^{4}.  (67)

Inequality (67) gives rise to the inequality m_{1} < 2.2 × 10^{59}. Now, we need to lower the bound.

Let

Λ_{1} = −m_{1} log α + n log 10 + log (\frac{d}{9}) .  (68)

Making use of equation (60), we have that

α^{m_{1}} − \frac{d × 10^{n}}{9} = α^{m_{1}} (1 − α^{−m_{1}}10^{n} (\frac{d}{9})) = α^{m_{1}} (1 − e^{Λ_{1}})

= −\frac{d}{9} − β^{m_{1}} − L_{m_{2}} − L_{m_{3}} − L_{m_{4}}

≤ −\frac{1}{9} + |β|^{600}

< 0,

as m_{1} ≥ 600. Thus, Λ_{1} > 0 and so from (62) we obtain

0 < Λ_{1} < e^{Λ_{1}} − 1 = |1 − α^{−m_{1}}10^{n} (\frac{d}{9})| < α^{m_{2}−m_{1}+4.33}.

log (\frac{d}{9}) + m_{1}(− log α) + n log 10 < α^{4.33}α^{−(m_{1}−m_{2})} < α^{4.34} exp(−0.48(m_{1} − m_{2})),

which leads to

|Λ_{1}| < α^{4.34} exp(−0.48(m_{1} − m_{2})),  (69)

where X = max{m_{1}, n} = m_{1} ≤ 2.2 × 10^{59}. We also have that

\frac{Λ_{1}}{log 10} = \frac{log(d/9)}{log 10} − m_{1} \frac{log α}{log 10} + n.

Hence, we set

c = α^{4.34}, δ = 0.48, X_{0} = 2.2 × 10^{59}, ψ = \frac{log(d/9)}{log 10}, Y = m_{1} − m_{2}

ϑ = \frac{log α}{log 10}, ϑ_{1} = − log α, ϑ_{2} = log 10, β = log(d/9).

When d = 9, β = 0. Substituting X_{0} = 2.2 × 10^{59} into inequality (8) yields 0 ≤ k ≤ 284. In the notation of Lemma 5 we find that A = a_{138} = 770, from the continued fraction expansion of \frac{log α}{log 10}. Applying Lemma 5, we get m_{1} − m_{2} ≤ 301. We now consider the case β ≠ 0. The smallest value of q such that q > X_{0} is q = q_{125}. We find that q = q_{127} satisfies the hypothesis of Lemma 6 for d = 1, …, 8. Applying Lemma 6, we get m_{1} − m_{2} ≤ 309. We see that m_{1} − m_{2} ≤ 309 for d = 1, …, 9 and hence m_{2} ≥ 291.

Taking 1 ≤ d ≤ 9 and 0 ≤ m_{1} − m_{2} ≤ 309, we let

Λ_{2} = −m_{2} log α + n log 10 + log (\frac{d}{9(α^{m_{1}−m_{2}} + 1)}) .  (70)

We use equation (55) to arrive at

α^{m_{1}}(1 + α^{m_{2}−m_{1}}) (1 − e^{Λ_{2}}) = −\frac{d}{9} − β^{m_{1}} − β^{m_{2}} − L_{3} − L_{4}

≤ −\frac{1}{9} + |β|^{600} + |β|^{291}

< 0,

making use of m_{1} ≥ 600 and m_{2} ≥ 291. Hence, Λ_{2} > 0, and so from (57) we see that

0 < Λ_{2} < e^{Λ_{2}} − 1 = |1 − α^{−m_{2}}10^{n} (\frac{d}{9(α^{m_{1}−m_{2}} + 1)})| < α^{m_{3}−m_{1}+4.05}.

Thus, we have

log (\frac{d}{9(α^{m_{1}−m_{2}} + 1)}) + m_{2}(− log α) + n log 10 < α^{m_{3}−m_{1}+4.05}
< α^{4.06} exp(−0.48(m_{1} − m_{3})),

which gives us

|Λ_{2}| < α^{4.06} exp(−0.48(m_{1} − m_{3})),  (71)

where X = max{m_{2}, n} ≤ m_{1} ≤ 2.2 × 10^{59}. In addition, we have

\frac{Λ_{2}}{log 10} = \frac{1}{log 10} log (\frac{d}{9(α^{m_{1}−m_{2}} + 1)}) − m_{1} \frac{log α}{log 10} + n.

So, we take

c = α^{4.06}, δ = 0.48, X_{0} = 2.2 × 10^{59}, ψ = \frac{1}{log 10} log (\frac{d}{9(α^{m_{1}−m_{2}} + 1)}) ,

Y = m_{1} − m_{3}, ϑ = \frac{log α}{log 10}, ϑ_{1} = − log α, ϑ_{2} = log 10, β = log (\frac{d}{9(α^{m_{1}−m_{2}} + 1)}) .

We find that q = q_{132} satisfies the hypothesis of Lemma 6 for d = 1, …, 9 and 0 ≤ m_{1}−m_{2} ≤ 309. Applying Lemma 6, we get m_{1} − m_{3} ≤ 327. Hence, m_{3} ≥ 273.

Taking 1 ≤ d ≤ 9, 0 ≤ m_{2} − m_{3} ≤ m_{1} − m_{3} ≤ 327, we let

Λ_{3} = −m_{3} log α + n log 10 + log (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) .  (72)

Equation (50) allows us to write

α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}}) (1 − e^{Λ_{3}}) = −\frac{d}{9} − (β^{m_{1}} + β^{m_{2}} + β^{m_{3}}) − L_{4}

≤ −\frac{1}{9} + |β|^{600} + |β|^{291} + |β|^{273}

< 0,

where we use m_{1} ≥ 600, m_{2} ≥ 291, and m_{3} ≥ 273. Hence, Λ_{3} > 0, and so from (52) we see that

0 < Λ_{3} < e^{Λ_{3}} − 1 = |1 − α^{−m_{3}}10^{n} (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)})| < α^{m_{4}−m_{1}+3.73}.

Hence, we have

log (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) + m_{3}(− log α) + n log 10 < α^{m_{4}−m_{1}+3.73}
< α^{3.74} exp(−0.48(m_{1} − m_{4})),

leading to

|Λ_{3}| < α^{3.74} exp(−0.48(m_{1} − m_{4})),  (73)

where X = max{m_{3}, n} ≤ m_{1} ≤ 2.2 × 10^{59}. Furthermore, we obtain

\frac{Λ_{3}}{log 10} = \frac{1}{log 10} log (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) − m_{3} \frac{log α}{log 10} + n.

Thus, we take

c = α^{3.74}, δ = 0.48, X_{0} = 2.2 × 10^{59}, Y = m_{1} − m_{4},

ψ = \frac{1}{log 10} log (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) , ϑ = \frac{log α}{log 10}, ϑ_{1} = − log α,

ϑ_{2} = log 10, β = log (\frac{d}{9(α^{m_{1}−m_{3}} + α^{m_{2}−m_{3}} + 1)}) .

We find that q = q_{138} satisfies the hypothesis of Lemma 6 for 1 ≤ d ≤ 9, 0 ≤ m_{2} − m_{3} ≤ m_{1} − m_{3} ≤ 327. Applying Lemma 6, we get m_{1} − m_{4} ≤ 371, and hence m_{4} ≥ 229.

Taking 1 ≤ d ≤ 9, 0 ≤ m_{3} − m_{4} ≤ m_{2} − m_{4} ≤ m_{1} − m_{4} ≤ 371, we let

Λ_{4} = −m_{4} log α + n log 10 + log (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) .  (74)

Using equation (45), we get

α^{m_{1}} (1 + α^{m_{2}−m_{1}} + α^{m_{3}−m_{1}} + α^{m_{4}−m_{1}}) (1 − e^{Λ_{4}}) = −\frac{d}{9} − (β^{m_{1}} + β^{m_{2}} + β^{m_{3}} + β^{m_{4}})

≤ −\frac{1}{9} + (|β|^{600} + |β|^{291} + |β|^{273} + |β|^{229})

< 0

making use of m_{1} ≥ 600, m_{2} ≥ 291, m_{3} ≥ 273 and m_{4} ≥ 229. Hence, Λ_{4} > 0, and so from (47) we see that

0 < Λ_{4} < e^{Λ_{4}} − 1 = |1 − α^{−m_{4}}10^{n} (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)})| < α^{3.35−m_{1}}.

Hence, we have

log (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) + m_{4}(− log α) + n log 10 < α^{3.35−m_{1}},

which implies that

|Λ_{4}| < α^{3.36} exp(−0.48m_{1}),  (75)

where X = max{m_{4}, n} ≤ m_{1} < 2.2 × 10^{59}. In addition,

\frac{Λ_{4}}{log 10} = \frac{1}{log 10} log (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) − m_{4} \frac{log α}{log 10} + n.

Thus,

c = α^{3.36}, δ = 0.48, X_{0} = 2.2 × 10^{59}, Y = m_{1},

ψ = \frac{1}{log 10} log (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) , ϑ = \frac{log α}{log 10},

ϑ_{1} = − log α, ϑ_{2} = log 10, β = log (\frac{d}{9(α^{m_{1}−m_{4}} + α^{m_{2}−m_{4}} + α^{m_{3}−m_{4}} + 1)}) .

We find that q = q_{145} satisfies the hypothesis of Lemma 6 for 1 ≤ d ≤ 9, 0 ≤ m_{3} − m_{4} ≤ m_{2} − m_{4} ≤ m_{1} − m_{4} ≤ 371. Applying Lemma 6, we get m_{1} ≤ 403, which contradicts the assumption that m_{1} ≥ 600. This proves the result.

5 Acknowledgments

We thank the referee for comments which improved the quality of this paper. F. L. was supported in parts by Grants CPRR160325161141 of NRF and the Focus Area Number Theory Grant from CoEMaSS at Wits (South Africa) and CGA 17-02804S (Czech Republic). This paper was completed during a visit of F. L. at Purdue University Northwest in February, 2018. This author thanks this institution for the hospitality and the support.

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2010 Mathematics Subject Classification: Primary 11A25; Secondary 11B39, 11J86. Keywords: Fibonacci number, Lucas number, linear form.

(Concerned with sequences \underline{A000032} and \underline{A000045}.)

Received May 26 2018; revised versions received May 27 2018; August 23 2018; August 29 2018. Published in Journal of Integer Sequences, September 8 2018.

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