Powers of Two as Sums of Three Lucas Numbers

Bilizimbéyé Edjeou Université Gaston Berger de Saint-Louis BP: 234, Saint-Louis Senegal

                olivieredjeou@gmail.com

Amadou Tall Université Cheikh Anta Diop de Dakar BP 5005, Dakar Fann Senegal

               amadou7.tall@ucad.edu.sn

Mohamed Ben Fraj Ben Maaouia Université Gaston Berger de Saint-Louis BP 234, Saint-Louis Senegal

                maaouiaalg@hotmail.com

Abstract

In this paper, we find all positive integer solutions of the Diophantine equation L_{k} + L_{l}+L_{t} = 2^{d} in non-negative integers k, l, t, and d, where (L_{n})_{n≥0} is the Lucas sequence. The tools used to solve our main theorem are linear forms in logarithms, properties of continued fractions, and a version of the Baker-Davenport reduction method in Diophantine approximation.

1 Introduction

The Lucas sequence (L_{k})_{k≥0} is a linear recurrence given by L_{0} = 2, L_{1} = 1 and

L_{k+2} = L_{k+1} + L_{k}, for k ≥ 0.

It satisfies the same recurrence as the Fibonacci sequence (F_{k})_{k≥0} given by F_{0} = 0, F_{1} = 1 and

F_{k+2} = F_{k+1} + F_{k}, for k ≥ 2,

whose numbers are found everywhere in nature. The Fibonacci numbers are famous for possessing many wonderful and amazing properties.

In 2014, Bravo and Luca [3] studied the Diophantine equation

L_{k} + L_{l} = 2^{t}

in positive integers k, l and t. Similar equations involving Fibonacci and Padovan sequences are solved in [5, 7]. E. Bravo and J. Bravo [2] found also all powers of 2 which are sums of three Fibonacci numbers. Specifically, they proved the following theorems.

Theorem 1. The only solutions (k, l, t) of the Diophantine equation L_{k} + L_{l} = 2^{t} in positive integers k, l, t and with k ≥ l are

(0, 0, 2); (1, 1, 1); (3, 3, 3); (2, 1, 2); (4, 1, 3); (7, 2, 5).

Theorem 2. All solutions (k, l, t, d) of the Diophantine equation

F_{k} + F_{l} + F_{t} = 2^{d}

in non-negative integers k, l, t, with k ≥ l ≥ t and d are

(3, 1, 1, 2); (3, 2, 2, 2); (3, 2, 1, 2); (4, 4, 3, 3); (5, 3, 1, 3); (5, 3, 2, 3); (6, 5, 4, 4);

(7, 3, 1, 4); (7, 3, 2, 4); (8, 6, 4, 5); (10, 6, 1, 6); (10, 6, 2, 6); (11, 9, 5, 7); (13, 8, 3, 8); (16, 9, 4, 10).

In this paper, we prove an extension of Theorem 1 when the two Lucas numbers are replaced by three Lucas numbers and determine all the solutions of the Diophantine equation

L_{k} + L_{l} + L_{t} = 2^{d}

in non-negative integers k, l, t and d. We prove the following result.

Theorem 3. All solutions (k, l, t, d) of the Diophantine equation

L_{k} + L_{l} + L_{t} = 2^{d}  (1)

in non-negative integers k ≥ l ≥ t and d, are

(1, 1, 0, 2), (2, 2, 0, 3), (3, 0, 0, 3), (3, 2, 1, 3), (4, 4, 0, 4), (5, 2, 0, 4), (5, 3, 1, 4), (6, 4, 4, 5),

(6, 5, 2, 5), (7, 1, 0, 5), (10, 2, 0, 7), (10, 3, 1, 7), (17, 13, 3, 12).

Our method of proof is similiar to the method described in [3, 2].

2 Preliminaries

Before proceeding further, we recall the Binet formula for the Lucas numbers (L_{k})_{k≥0}, namely

L_{k} = α^{k} + β^{k}, for k ≥ 0,

where

α = \frac{1 + \sqrt{5}}{2} and β = \frac{1 − \sqrt{5}}{2}

are the roots of the characteristic equation x^{2} − x − 1 = 0. In particular, the inequality

α^{k−1} ≤ L_{k} ≤ 2α^{k}  (2)

holds for all k ≥ 0.

To prove Theorem 3, using a result on linear forms in two logarithms., we require some notation. Let δ be an algebraic number of degree d with minimal polynomial

a_{0}x^{d} + a_{1}x^{d−1} + ⋯ + a_{d} = a_{0} ∏_{i=1}^{d}(X − δ^{(i)})

where the a_{i} are relatively prime integers with a_{0} > 0 and the δ^{(i)} denotes the conjugates of δ. Then

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

is called the logarithmic height of δ. In particular, if δ = p/q is a rational number with gcd(p, q) = 1 and q > 0, then

h(δ) = log max{|p|, q}.

The following properties of the logarithmic height, will be used in the next section. Let δ, ν be algebraic numbers and r ∈ ℤ. Then

• h(δ ± ν) ≤ h(δ) + h(ν) + log 2,

• h(δν^{±1}) ≤ h(δ) + h(ν),

• h(δ^{r}) = |r|h(δ).

Using the above notation, we restate Laurent, Mignotte, and Nesterenko’s result [6, Cor. 1].

Theorem 4. Let δ_{1}, δ_{2} be two non-zero algebraic numbers, and let log δ_{1} and log δ_{2} be any determinations of their logarithms. Set

D = [ℚ(δ_{1}, δ_{2}) : ℚ]/[ℝ(δ_{1}, δ_{2}) : ℝ]

and

Γ := b_{2} log δ_{2} − b_{1} log δ_{1},

where b_{1} and b_{2} are positive integers. Further, let A_{1}, A_{2} > 1 be real numbers such that

log A_{i} ≥ max\{h(δ_{i}), \frac{|h(δ_{i})|}{D}, \frac{1}{D}\}, i = 1, 2.

Then, assuming that δ_{1} and δ_{2} are multiplicatively independent, we have

log |Γ| > −30.9 · D^{4}(max \{log b′, \frac{21}{D}, \frac{1}{2}\})^{2} log A_{1} log A_{2},

where

b′ = \frac{b_{1}}{D log A_{2}} + \frac{b_{2}}{D log A_{1}} .

We also need the following general lower bound for linear forms in logarithms due to Matveev [8].

Theorem 5. Assume that δ_{1}, …, δ_{t} are positive real algebraic numbers in a real algebraic number field 𝕂 of degree D. Let b_{1}, ⋯, b_{n} be rational integers, and

Λ := δ_{1}^{b_{1}} ⋯ δ_{t}^{b_{t}} − 1

be not zero. Then

|Λ| > exp (−1.4 · 30^{t+3} · t^{4.5} · D^{2}(1 + log D)(1 + log B)A_{1} ⋯ A_{t}),

where

B ≥ max {|b_{1}|, …, |b_{t}|},

and

A_{i} ≥ max {Dh(δ_{i}), | log δ_{i}|, 0.16}, for all i = 1, ⋯, t.

Finally, we present a version of the reduction method based on the Baker-Davenport Lemma [1], from Dujella and Pethő [4]. This will be one of the key tools used to reduce the upper bounds on the variables of the equation (1).

Lemma 6. Let N be a positive integer, let p/q be a convergent of the irrational number γ such that q > 6N, and let A, B, μ be real numbers with A > 0 and B > 1. Define

ξ := ∥μq∥ − N∥γq∥,

where ∥ · ∥ denotes the distance to the nearest integer. If ξ > 0, then there is no solution to the inequality

0 < uγ − v + μ < AB^{−w},

in positive integers u, v, and w, with u ≤ N and w ≥ \frac{log (Aq/ξ)}{log B}.

3 The Proof of Theorem 3

First of all, observe that if k = l = t, then equation (1) becomes 3L_{k} = 2^{d}. Since 3 ∤ 2, then equation (1) has no solution. Subsequently, we assume that either k > l or l > t.

If k ≤ 250, then a brute force search using Sagemath in the range 0 ≤ t ≤ l ≤ k ≤ 250 produces the solutions

(1, 1, 0, 2), (2, 2, 0, 3), (3, 0, 0, 3), (3, 2, 1, 3), (4, 4, 0, 4), (5, 2, 0, 4), (5, 3, 1, 4), (6, 4, 4, 5), (6, 5, 2, 5),

(7, 1, 0, 5), (10, 2, 0, 7), (10, 3, 1, 7), (17, 13, 3, 12).

Thus, for the remainder of the paper, we assume that k > 250. Let us now establish a relation between k and d.

Combining (1) with the right inequality of (2), one gets that

2^{d} ≤ 2α^{k} + 2α^{l} + 2α^{t} < 6α^{k} < 6 · 2^{k} < 2^{k+3},

which leads to d ≤ k + 2.

3.1 Bounding k − l and k − t in terms of k

We rewrite (1) as

α^{k} − 2^{d} = −β^{k} − L_{l} − L_{t}.

Now taking absolute values, we obtain

|α^{k} − 2^{d}| ≤ |β|^{k} + L_{l} + L_{t} < \frac{1}{2} + 2α^{l} + 2α^{t}.

Dividing both sides of the above expression by α^{k} and taking into account that k ≥ l ≥ t, we get

|1 − 2^{d}α^{−k}| < \frac{1}{2}α^{−k} + 2α^{−k+l} + 2α^{−k+t} < 5α^{−k+l}.

Thus

|1 − 2^{d}α^{−k}| < \frac{5}{α^{k−l}} .  (3)

We apply Theorem 4 to

Γ := d log α − k log 2.

Therefore the estimate (3) can be rewritten as

|1 − e^{Γ}| < \frac{5}{α^{k−l}} .  (4)

The algebraic number field containing 2, α is ℚ(\sqrt{5}), so we can take D := 2. By using (1) and the Binet formula for the Lucas sequence, we have

α^{k} = L_{k} − β^{k} < L_{k} + 1 ≤ L_{k} + L_{l} + L_{t} = 2^{d}.  (5)

Consequently, 1 < 2^{d}α^{−k} and so Γ > 0. Using the fact that log(1 + x) ≤ x for all x ∈ ℝ^{+}, together with (4), gives

0 < Γ < \frac{5}{α^{k−l}} ,  (6)

Hence,

log Γ < log 5 − (k − l) log α.  (7)

Note further that h(α) = log α/2 and h(2) = log 2. Thus, we can choose

log A_{1} := log α and log A_{2} := log 2.

Finally, recall that d ≤ k + 2, and so

b′ = \frac{k}{2 log 2} + \frac{d}{2 log α} < 4k.

Since α and 2 are multiplicatively independent, we have, by Theorem 4, that

log Γ ≥ −30.9 · 2^{4} · (max{log (4k), 21/2, 1/2})^{2} · log α · log 2.

Thus

log Γ > −174 · (max{log (4k), 21/2, 1/2})^{2}.  (8)

Combining (7) and (8), we obtain

(k − l) log α < 180 · (max{log(4k), 21/2})^{2}.  (9)

Let us now establish a second linear form in logarithms. To this end, we rewrite equation (1) as follows

α^{k}(1 + α^{(l−k)}) − 2^{d} = −β^{k} − β^{l} − L_{t}.

Taking absolute values in the above relation and using the fact that β = (1 − \sqrt{5})/2 we get

|α^{k}(1 + α^{(l−k)}) − 2^{d}| = | − β^{k} − β^{l} − L_{t}| < 2 + 2α^{t}

for all k > 250 and l ≥ t ≥ 0. Dividing both sides of the above inequality by the first term of the left-hand side, we obtain

|1 − 2^{d}α^{−k}(1 + α^{(l−k)})^{−1}| < \frac{2}{α^{k}(1 + α^{(l−k)})} + \frac{2}{α^{k−t}(1 + α^{(l−k)})} < \frac{4}{α^{k−t}} .  (10)

We are now ready to apply Matveev’s result given in Theorem 5. To do this, we take the parameters n := 3 and

δ_{1} := 2, δ_{2} := α, δ_{3} := (1 + α^{(l−k)}).

We take b_{1} := d, b_{2} := −k and b_{3} := −1. As before, 𝕂 := ℚ(\sqrt{5}) contains δ_{1}, δ_{2}, δ_{3} and has D := [𝕂 : ℚ] = 2. To see why the left-hand side of (10) is not zero, note that otherwise, we would get the relation

α^{k} + α^{l} = 2^{d}.  (11)

Conjugating the above relation in ℚ(\sqrt{5}), we get

β^{k} + β^{l} = 2^{d}.  (12)

Further, combining (11) and (12), we obtain

α^{k} < α^{k} + α^{l} = |β^{k} + β^{l}| < 2.

This is impossible because k > 250. Thus,

1 − 2^{d}α^{−k}(1 + α^{(l−k)})^{−1}

is not zero.

In this application of Theorem 5, we take A_{1} := 2 log 2 and A_{2} := log α. Since t ≤ k + 2, it follows that we can take B := k + 2. Let us now estimate h(δ_{3}). We begin by observing that

δ_{3} = (1 + α^{(l−k)}) < 2 and δ_{3}^{−1} < 1.

So that

0 < log δ_{3} < 1.

Next, notice that

h(δ_{3}) ≤ (k − l) log α + log 2.

Hence, we can take

A_{3} := 2 + (k − l) log α > max{2h(δ_{3}), | log δ_{3}|, 0.16}.

Now Theorem 5 implies that a lower bound on the left-hand side of (10) is

log |Λ| > −1.4 · 30^{6} · 3^{4.5} · 2^{2}(1 + log 2)(1 + log(k + 2)) · 2 log 2 · 2 log α · (2 + (k − l) log α).

So, inequality (10) yields

k − t < 2.8 · 10^{12} log(k + 2) · (2 + (k − l) log α),  (13)

where we used the inequality 1 + log(k + 2) < 2 log(k + 2), which holds because k > 250.

Now using (9) in the right-most term of inequality (13) and performing the respective calculations, we obtain

k − t < 5.1 · 10^{14} log(k + 2)(max{log(4k), 21/2})^{2}.  (14)

3.2 Bounding k

Finally, we consider a third linear form in logarithms. We now rewrite equation (1) as follows

α^{k}(1 + α^{(l−k)} + α^{(t−k)}) − 2^{d} = −β^{k} − β^{l} − β^{t}.

Taking absolute values in the above relation and using the fact that β = (1 − \sqrt{5})/2, we get

|α^{k}(1 + α^{(l−k)} + α^{(t−k)}) − 2^{d}| = | − β^{k} − β^{l} − β^{t}| < 3

for all k > 250 and l ≥ t ≥ 0. Dividing both sides of the above inequality by the first term of the left-hand side, we obtain

|1 − 2^{d}α^{−k}(1 + α^{(l−k)} + α^{(t−k)})^{−1}| < \frac{3}{α^{k}(1 + α^{(l−k)} + α^{(t−k)})} < \frac{3}{α^{k}} .  (15)

We apply Theorem 5 to

Λ = 1 − 2^{d}α^{−k}(1 + α^{(l−k)} + α^{(t−k)})^{−1},

with the parameters n := 3, δ_{1} := 2, δ_{2} := α, δ_{3} := (1 + α^{(l−k)} + α^{(t−k)}), b_{1} := d, b_{2} := −k and b_{3} := −1, 𝕂 := ℚ(\sqrt{5}) contains δ_{1}, δ_{2}, δ_{3}, D := [𝕂 : ℚ] = 2. To see why the left-hand side of (15) is not zero, note that otherwise, we would get the relation

α^{k} + α^{l} + α^{t} = 2^{d}.  (16)

Conjugating the above relation in ℚ(\sqrt{5}), we get

β^{k} + β^{l} + β^{t} = 2^{d}.  (17)

Furthermore, combining (16) and (17), we obtain

α^{k} < α^{k} + α^{l} + α^{t} = |β^{k} + β^{l} + β^{t}| < 3.

This is impossible because k > 250. Thus,

1 − 2^{d}α^{−k}(1 + α^{(l−k)} + α^{(t−k)})^{−1}

is not zero. We now apply Theorem 5 with A_{1} := 2 log 2, A_{2} := log α. Since d ≤ k + 2; it follows that we can take B := k + 2. Let us now estimate h(δ_{3}). We begin by observing that

γ_{3} = (1 + α^{(l−k)} + α^{(t−k)}) < 3

and

0 < log γ_{3} < log 3.

Next, notice that

h(δ_{3}) ≤ (k − l) log α + (k − t) log α + 2 log 2 ≤ 2(k − t) log α + 2 log 2.

Hence, we can take

A_{3} := 4 + 2(k − t) log α > max{2h(δ_{3}), | log δ_{3}|, 0.16}.

Now, from Theorem 5 we have

log |Λ| > −1.4 · 30^{6} · 3^{4.5} · 2^{2}(1 + log 2)(1 + log(k + 2)) · 2 log 2 · 2 log α · (4 + 2(k − t) log α).

So, inequality (15) gives

k < 10^{13} log(k + 2) · (2 + (k − t) log α),  (18)

where we used the inequality 1 + log(k + 2) < 2 log(k + 2), which holds because k > 250.

Now using (13) in the rightmost term of the above inequality (18) and performing the respective calculations, we obtain

k < 2.6 · 10^{27}(log(k + 2))^{2}(max{log(4k), 21/2})^{2}.  (19)

If max{log(4k), 21/2} = 21/2, it then follows from (19) that

k < 287 · 10^{27}(log(k + 2))^{2},

giving

k < 15 · 10^{32}.

If on the other hand we have that max{log(4k), 21/2} = log(4k), then inequality (19) gives that

k < 2.6 · 10^{27}(log(k + 2))^{2}(log(4k))^{2},

and so

k < 12 · 10^{34}.

In any case, we have that

k < 12 · 10^{34}

always holds. We summarize what we have so far in the following lemma.

Lemma 7. If (k, l, t, d) is a solution in positive integers of equation (1) with k ≥ l ≥ t and k > 250, then inequalities

d ≤ k + 2 and k < 12 · 10^{34}

hold.

4 The final computations

In this section, we will reduce the upper bound on k. Firstly, we determine a suitable upper bound on k − l, k − t, and later we use Lemma 6 to conclude that k must be smaller than 250.

Turning back to inequality (6), we obtain

0 < d log 2 − k log α < \frac{5}{α^{k−l}} .

Dividing across by log α, we get

0 < dγ − k < \frac{11}{α^{k−l}} ,  (20)

where

γ := \frac{log 2}{log α}.

Let [a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, …] = [1, 2, 3, 1, 2, 3, 2, 4 …] be the continued fraction expansion of γ, and let denote p_{n}/q_{n} its nth convergent. Recall also that d < 12 · 10^{34} by Lemma 7. A quick inspection using Sagemath reveals that

37527245802242661673724926130723830 = q_{73} < 12 · 10^{34} <
q_{74} = 175184858909722330004986691804684639.

Furthermore, a_{N} := max{a_{i}; i = 0, 1, …, 44} = a_{17} = 134. So, from the known properties of continued fractions, we obtain that

|dγ − k| > \frac{1}{(a_{N} + 2)d}.  (21)

Comparing estimates (20) and (21), we get right away that

α^{k−l} < 11 · 136 · d < 18 · 10^{37},

leading to k − l < 184.

Let us now go back to (10) and determine an improved upper bound on k − t. Put

ω_{1} := d log 2 − k log α − log(1 + α^{−(k−l)}).  (22)

Therefore, (10) implies that

|1 − e^{ω_{1}}| < \frac{4}{α^{k−t}} .  (23)

Note that ω_{1} ≠ 0, by using (1) and the Binet formula for the Lucas sequence, we have

α^{k} + α^{l} = L_{t} − β^{k} − β^{l} < L_{k} + L_{l} + L_{t} = 2^{d}.

Therefore,

1 < 2^{d}α^{−k}(1 + α^{(l−k)})^{−1}

and so ω_{1} > 0. Thus

0 < ω_{1} ≤ e^{ω_{1}} − 1 < \frac{4}{α^{k−t}} .  (24)

Replacing ω_{1} in the above inequality by its formula (22) and dividing both sides of the resulting inequality by log α, we get

0 < d (\frac{log 2}{log α}) − k − \frac{log(1 + α^{−(k−l)})}{log α} < \frac{9}{α^{k−t}} .  (25)

We now put

γ := \frac{log 2}{log α}, μ := −\frac{log(1 + α^{−(k−l)})}{log α}, A := 9 and B := α.

Clearly γ is an irrational number. We also put N := 12·10^{34} , which is an upper bound on d by Lemma 7. We therefore apply Lemma 6 to inequality (25) for all choices k −l ∈ {1, …, 184} except when k − l = 1, 3 and get that

k − t < \frac{log(Aq/ξ)}{log B},

where q > 6N is a denominator of a convergent of the continued fraction of γ such that ξ = ∥μq∥ − N∥γq∥ > 0. Indeed, using Sagemath, we have that

q = q_{75} = 1439006117080021301713618460568200942.

We find that if (k, l, t, d) is a possible solution of the equation (1) with ω_{1} > 0 and k − l ∈ {1, …, 184} except when k − l = 1, 3, then

k − t < 178.

Let us now treat the cases where k − l = 1 and 3. The discussion of these cases will be different from the previous ones, because when applying Lemma 6 to the expression (25), the corresponding parameter μ appearing in Lemma 6 is

\frac{log(1 + α^{−(k−l)})}{log α} = \begin{cases}−1, & if k − l = 1; \\ 1 − \frac{log 2}{log α}, & if k − l = 3. \end{cases}

In both cases, the parameters γ and μ are linearly dependent, which yields that the corresponding value of ξ from Lemma 6 is always negative and therefore the reduction method is not useful for reducing the bound on k −t in these instances. One can see that if k −l = 1, 3, then the resulting inequality from (25) has the shape

0 < |aγ − b| < \frac{9}{α^{k−t}} ,

with γ being an irrational number and a, b ∈ ℤ. So, one can appeal to the known properties of the convergents of the continued fractions to obtain a nontrivial lower bound for

|aγ − b|.

When k − l = 1, from (25), we get that

0 < dγ − (k + 1) < \frac{9}{α^{k−t}} .  (26)

Let [a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, ...] = [1, 2, 3, 1, 2, 3, 2, 4 …] be the continued fraction expansion of γ, and let denote p_{n}/q_{n} its nth convergent. Recall also that d < 12 · 10^{34} by Lemma 4. Furthermore, a_{N} := max{a_{i} : i = 0, 1, …, 44} = a_{17} = 134. So, from the known properties of continued fractions, we obtain that

|dγ − (k + 1)| > \frac{1}{(a_{N} + 2)d}.  (27)

Comparing estimates (26) and (27), we get right away that

α^{k−t} < 9 · 136 · d < 15 · 10^{37},

leading to k − t < 184.

By the same argument as the one we did before, we get that k − t < 184 in the case when k − l = 3. This completes the analysis of the cases when k − l = 1, 3. Consequently, k − t < 184 always holds.

Finally, we shall use (15) to reduce the upper bound on k. Put

ω_{2} = d log 2 − k log α − log φ(u, v)  (28)

where φ is the function given by the formula φ(u, v) = 1+α^{−u}+α^{−v}, where u = k−l, v = k−t. Note that ω_{2} ≠ 0. Thus, we distinguish the following cases. If ω_{2} > 0 then, from (15), we obtain

0 < ω_{2} ≤ e^{ω_{2}} − 1 < \frac{3}{α^{k}} .

Replacing ω_{2} in the above inequality by its formula (28) and dividing both sides of the resulting inequality by log α, we get

0 < d (\frac{log 2}{log α}) − k − \frac{log(1 + α^{−u} + α^{−v})}{log α} < \frac{7}{α^{k}} .  (29)

We now put

γ := \frac{log 2}{log α}, μ := −\frac{log(1 + α^{−u} + α^{−v})}{log α}, A := 7 and B := α.

Clearly γ is an irrational number. We also put N := 12 · 10^{34} , which is an upper bound on d by Lemma 7. We therefore apply Lemma 6 to inequality (29) for all choices of u ∈ {1, …, 184}, v ∈ {1, …, 184} except when

(u, v) ∈ ϖ_{1} = {(4, 2), (13, 1), (17, 4), (17, 5), (18, 7), (19, 4), (19, 5), (22, 2), (23, 1), (23, 2), (24, 1),

(24, 5), (25, 1), (28, 2), (30, 4), (31, 1), (32, 7), (33, 1), (33, 4), (33, 5), (36, 1), (36, 2), (37, 2), (38, 2), (39, 1), (43, 4), (44, 1), (45, 4), (48, 1), (48, 5), (50, 1), (51, 9), (52, 1), (52, 2), (53, 4), (54, 2), (55, 2), (56, 2), (57, 2), (59, 2), (61, 1), (62, 4), (64, 1), (66, 1), (67, 1), (69, 2), (70, 5), (75, 1), (75, 2), (83, 5), (87, 1), (87, 7), (89, 2), (90, 2), (95, 1), (95, 7), (97, 1), (98, 1), (99, 5), (100, 1), (102, 2), (107, 2), (110, 4), (111, 2), (112, 1),

(112, 2), (113, 1), (113, 5), (113, 6), (114, 1), (118, 1), (122, 1), (122, 4), (122, 6),
(128, 2), (129, 1), (129, 2), (130, 1), (130, 4), (130, 5), (132, 2), (133, 2), (136, 4),
(138, 2), (139, 1), (139, 2), (141, 1), (141, 2), (142, 4), (147, 1), (148, 1), (158, 7)}

and get that

k < \frac{log(Aq/ξ)}{log B},

where q > 6N is a denominator of a convergent of the continued fraction of γ such that ξ = ∥μq∥ − N∥γq∥ > 0. Indeed, using Sagemath, we have that

q = q_{82} = 12054118444825786260212254516320106583249.

We find that if (k, l, t, d) is a possible solution of the equation (1) with ω_{2} > 0 and (u, v) ∉ ϖ_{1}, then k < 196. This is false because our assumption is that k > 250.

Let us now work with the cases when (u, v) ∈ ϖ_{1}. We cannot study these cases as before because when applying Lemma 6 to the expression (29), the corresponding quantity ∥μq∥ appearing in Lemma 6 is zero. In these cases, the parameters γ and μ are linearly dependent, which yields that the corresponding value of ξ from Lemma 6 is always negative and therefore the reduction method is not useful for reducing the bound on k in these instances. However, one can see that if (u, v) = (2, 4), then the resulting inequality from (29) has the shape

0 < |xγ − y| < \frac{7}{α^{k}} ,

with γ being an irrational number and x, y ∈ ℤ. So, one can use to the known properties of the convergents of the continued fractions to obtain a nontrivial lower bound for

|xγ − y|.

This clearly gives us an upper bound for k. For example, when (u, v) = (2, 4),

−\frac{log(1 + α^{−u} + α^{−v})}{log α} = 2 − 2\frac{log 2}{log α}

and from (29), we get that

0 < (d − 2)γ − (k − 2) < \frac{7}{α^{k}} .  (30)

Let [a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, …] = [1, 2, 3, 1, 2, 3, 2, 4 …] be the continued fraction expansion of γ, and let denote p_{n}/q_{n} its nth convergent. Recall also that d < 12 · 10^{34} by Lemma 7.

Furthermore, a_{N} := max{a_{i}; i = 0, 1, …, 44} = a_{17} = 134. So, from the known properties of continued fractions, we obtain that

|(d − 2)γ − (k − 2)| > \frac{1}{(a_{N} + 2)d}.  (31)

Comparing estimates (30) and (31), we get that

α^{k} < 7 · 136 · d < 12 · 10^{37},  (32)

leading to k < 184. Using the above argument, we obtain k < 184 in the case when (u, v) ∈ ϖ_{1} except (2, 4). We omit the details in order to avoid unnecessary repetitions. This completes the analysis of the cases when (u, v) ∈ ϖ_{1}. Consequently, k < 196 always holds.

Suppose now that ω_{2} < 0. First, note that \frac{7}{α^{k}} < \frac{1}{2} since k > 250. Then, from (15), we have that

|1 − e^{ω_{2}}| < \frac{1}{2},

thus

\frac{1}{2} < e^{ω_{2}} < \frac{3}{2}.

Therefore

e^{|ω_{2}|} < 2.

Since ω_{2} < 0, we have

0 < |ω_{2}| ≤ e^{|ω_{2}|} − 1 = e^{|ω_{2}|}|e^{−|ω_{2}|} − 1| = e^{|ω_{2}|}|e^{ω_{2}} − 1| < \frac{6}{α^{k}} .

Then we obtain

0 < −d log 2 + k log α + log(1 + α^{−u} + α^{−v}) < \frac{6}{α^{k}} .

By the same arguments used for proving (15), we obtain

0 < k (\frac{log α}{log 2}) − d + \frac{log(1 + α^{−u} + α^{−v})}{log 2} < \frac{9}{α^{k}} .  (33)

We now put

γ := \frac{log α}{log 2}, μ := \frac{log(1 + α^{−u} + α^{−v})}{log 2}, A := 9 and B := α.

(u, v) ∈ ϖ_{2} = {(2, 1), (3, 3), (4, 2), (5, 2), (7, 7), (8, 2), (9, 5), (9, 6), (10, 2), (11, 4), (12, 1), (13, 1),

(14, 2), (17, 4), (19, 5), (19, 6), (20, 9), (22, 1), (23, 1), (24, 1), (24, 5), (25, 1), (25, 2), (28, 2), (30, 1), (31, 4), (32, 7), (33, 1), (36, 3), (37, 2), (38, 1), (40, 2), (44, 1), (47, 2), (48, 1), (48, 6), (50, 1), (50, 2), (51, 9), (53, 1), (54, 2), (56, 2), (57, 2), (58, 5), (58, 6),

(58, 8), (62, 1), (62, 4), (64, 1), (66, 1), (69, 2), (70, 6), (75, 1), (75, 2), (76, 1),

(76, 4), (77, 2), (77, 3), (78, 2), (79, 2), (83, 6), (87, 1), (87, 7), (89, 1), (89, 2), (91, 1), (91, 2), (92, 1), (94, 1), (95, 1), (95, 7), (96, 2), (97, 1), (98, 1), (99, 5), (99, 6), (102, 2),

(102, 3), (112, 2), (113, 1), (113, 5), (113, 6), (120, 1), (122, 1), (122, 5), (122, 6),
(123, 2), (129, 2), (129, 3), (130, 4), (130, 5), (134, 7), (135, 2), (138, 1), (139, 1),
(141, 1), (141, 2), (142, 4), (143, 1), (147, 1), (148, 1), (149, 2)},

and get that

k < \frac{log(Aq/ξ)}{log B},

where q > 6N is a denominator of a convergent of the continued fraction of γ such that ξ = ∥μq∥ − N∥γq∥ > 0. Indeed, using Sagemath, we have hat

q = q_{82} = 1234165504911193651820557190855668171489.

We find that if (k, l, t, d) is a possible solution of the equation (1) with ω_{2} < 0 and (u, v) ∉ ϖ_{2}, then k < 192. This is false because our assumption is that k > 250. With the same arguments as in the case ω_{2} > 0, when (u, v) ∈ ϖ_{2}, we obtain that k < 192. Thus, Theorem 3 is proven.

5 Acknowledgments

This work was partially supported by a grant from the Simons Foundation. Part of this work was done during a very enjoyable visit of B. Edjeou at AIMS Sénégal. This author thanks AIMS Sénégal for the hospitality and support. We also thank Bernadette Faye and Eva Goedhart for useful discussions.

References

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[5] A. C. G. Lomelí and S. H. Hernández, Powers of two as sums of two Padovan numbers, INTEGERS 18 (2018), A84.

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[7] F. Luca and S. Siksek, Factorials expressible as sums of two and three Fibonacci numbers, Proc. Edinb. Math. Soc. 53 (2010), 747–763.

[8] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers. II, Izv. Ran. Ser. Mat. 64 (2000), 125–180; Izv. Math. 64 (2000), 1217–1269.

2010 Mathematics Subject Classification: Primary 11B39, Secondary 11J86. Keywords: Lucas number, Diophantine equation, linear form in logarithms.

(Concerned with sequence \underline{A000032}.)

Received August 21 2019; revised versions received August 9 2020; August 26 2020. Published in Journal of Integer Sequences, October 13 2020.

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