Prime powers in sums of terms of binary recurrence sequences

Abstract

Let $(u_{n})_{n \ge 0}$ be a non-degenerate binary recurrence sequence with positive discriminant and p be a fixed prime number. In this paper, we are interested in finding a finiteness result for the solutions of the Diophantine equation $u_{n_{1}} + u_{n_{2}} + \cdots + u_{n_{t}} = p^{z}$ with $n_1> n_2> \cdots > n_t\ge 0$. Moreover, we explicitly find all the powers of three which are sums of three balancing numbers using lower bounds for linear forms in logarithms. Further, we use a variant of the Baker–Davenport reduction method in Diophantine approximation due to Dujella and Pethő.

1 Introduction

A large number of interesting Diophantine equations arise when one studies the intersection of two sequences of positive integers. For instance, one can ask when the terms of a fixed binary recurrence sequence are perfect powers, factorials or combinatorial numbers etc. As an example, one can consider the solvability of the Diophantine equation

$$\begin{aligned} u_{n} = x^{z} \end{aligned}$$

(1.1)

in integers n, x, z with $z\ge 2$, where $(u_{n})_{n \ge 0}$ is a linear recurrence sequence. Pethő [15, Corollary] and Shorey-Stewart [21, Theorem 2] independently proved that, Eq. (1.1) has only finitely many solutions under certain natural assumptions. The problem of finding all perfect powers in the Fibonacci sequence has a very rich history [9, 16] and this problem was open for a quite long time. In 2006, Bugeaud, Mignotte, and Siksek [8, Theorem 1] by using both the classical and the modular approach, proved that

$$\begin{aligned}(n,x,z) \in \{(0, 0, z), (1, 1, z), (2, 1, z), (6, 2, 3), (12, 12, 2)\} \end{aligned}$$

are the only solutions of Eq. (1.1) where $u_{n}$ is the Fibonacci sequence. In the same paper [8, Theorem 2], they also studied Eq. (1.1) for the Lucas number sequence. In 2008, Pethő [18, Theorem 2] (see also [10]) solved Eq. (1.1) when $u_{n}$ is the Pell sequence and proved that 0, 1 and 169 are the only perfect powers.

Now one can extend Eq. (1.1) by taking two terms of a binary recurrence sequence and ask the same question, i.e., solvability of the Diophantine equation

$$\begin{aligned} u_{n} + u_m = x^{z} \end{aligned}$$

(1.2)

in integers n, m, x, z with $z\ge 2$. In this direction, several authors studied the problem of finding (n, m, z) such that

$$\begin{aligned} u_{n} + u_{m} = 2^{z}, \end{aligned}$$

(1.3)

where $(u_{n})_{n \ge 0}$ is a fixed recurrence sequence. In particular, Bravo and Luca considered the case when $u_{n}$ is the Fibonacci sequence [7, Theorem 2] and the Lucas number sequence [6, Theorem 2]. Further, variants of Eq. (1.3) were studied independently by Bravo et al. [5, Theorem 1] and Marques [13, Theorem 1], where $u_{n}$ is replaced by the generalized Fibonacci sequence. Bravo et al. [4, Theorem 1] investigated the case when the power of two can be expressed as sum of three Fibonacci numbers. Recently, Pink and Ziegler [19, Theorem 1] have generalized the results of Bravo and Luca [6, 7], and considered a more general Diophantine equation

$$\begin{aligned} u_{n} + u_{m} = wp_{1}^{z_{1}} \cdots p_{s}^{z_{s}} \end{aligned}$$

(1.4)

in non-negative integer unknowns $n, m, z_{1}, \ldots , z_{s}$, where $(u_{n})_{n \ge 0}$ is a binary non-degenerate recurrence sequence, $p_{1}, \ldots , p_{s}$ are distinct primes and w is a non-zero integer with $p_{i} \not \mid w$ for $1\le i \le s$. They proved that under certain assumptions, Eq. (1.4) has finitely many solutions using lower bounds for linear forms of p-adic logarithms. In [2, Theorem 2.1], Bertók et al. solved completely equations of the form $u_n = 2^a + 3^b + 5^c$, where $u_n$ is one of the Fibonacci, Lucas, Pell and associated Pell sequences, respectively.

The purpose of this paper is twofold. On one hand, we give a general finiteness result for the solutions of the equation

$$\begin{aligned} u_{n_{1}} + u_{n_{2}} + \cdots + u_{n_{t}} = p^{z} \end{aligned}$$

(1.5)

in non-negative integer unknowns $n_{1},\ldots , n_{t}, z$, where $(u_{n})_{n \ge 0}$ is a binary non-degenerate recurrence sequence with $n_{1}> n_{2}> \cdots > n_{t} \ge 0$ and p is a given prime. On the other hand, we completely solve Eq. (1.5) where $u_{n}$ is a balancing number sequence and $(t, p) = (3, 3)$. To prove our main results, we use lower bounds for linear forms in logarithms of algebraic numbers and a version of the Baker–Davenport reduction method.

2 Notation and main results

The sequence $(u_{n})_{n \ge 0} = u_{n}(P, Q, u_{0}, u_{1})$ is called a binary linear recurrence sequence if the relation

$$\begin{aligned} u_{n} = Pu_{n-1} + Qu_{n-2} \;\;(n\ge 2) \end{aligned}$$

(2.1)

holds, where $P, Q\in \mathbb {Z}$ with $PQ\ne 0$ and $u_{0}, u_{1}$ are fixed rational integers with $|u_{0}| + |u_{1}| > 0$. The polynomial $f(x) = x^2 -Px -Q$ is called the companion polynomial of the sequence $u_n$. Let $\Delta = P^2 + 4Q$ be the discriminant of f. We call $\Delta$ the discriminant of the sequence $u_n$. The roots of the companion polynomial are denoted by $\alpha$ and $\beta$. Then for $n\ge 0$

$$\begin{aligned} u_{n} = \frac{a\alpha ^{n}-b\beta ^{n}}{\alpha -\beta }\quad (\alpha \ne \beta ), \end{aligned}$$

(2.2)

where $a = u_{1} - u_{0}\beta ,\, b = u_{1} - u_{0}\alpha$. The sequence $(u_{n})_{n\ge 0}$ is called non-degenerate, if $ab\alpha \beta \ne 0$ and $\alpha /\beta$ is not a root of unity.

Throughout the paper, we assume that $u_{n}$ is non-degenerate, $\sqrt{\Delta } = (\alpha -\beta ) >0$ (that is $\alpha$ and $\beta$ are real). We label the roots in such a way that $|\alpha | > |\beta |$. Up to changing the signs to $(\alpha , \beta )$ simultaneously (that is, replacing $(\alpha , \beta )$ by $(-\alpha , -\beta )$, which has as effect replacing (P, Q) by $(-P, Q)$ and the sequence $(u_n)_{n\ge 0}$ by the sequence$((-1)^nu_n)_{n\ge 0}$, we may assume that $\alpha$ is positive. From now on, we will assume these conditions.

With these notations, we have the following theorem.

Theorem 1

Suppose $(u_{n})_{n \ge 0}$ is a non-degenerate binary recurrence sequence of integers satisfying recurrence (2.1). Let us assume that $\alpha > |\beta |$ and $\Delta = r^2s$ where $r, s \in \mathbb {Z}_{ >0}, s\ne 1, s$ is square-free. Then there exists an effectively computable constant C depending on $(u_{n})_{n\ge 0}, p, t$ such that all solutions $(n_{1},\ldots , n_{t},z)$ to Eq. (1.5) satisfy

$$\begin{aligned} \max \{n_1,\ldots ,n_t,z\} < C. \end{aligned}$$

Remark 2.1

Suppose $u_n = 2^n - 1$ and $t = p = 2$. In this case $\alpha = 2,\; \beta = a = b =1$ and $\Delta = s = 1$. Then (1.5) becomes

$$\begin{aligned} \frac{2^{n_1} - 1}{2 -1} + \frac{2^{n_2} - 1}{2 -1} = 2^z, \end{aligned}$$

which has infinitely many solutions given by $n_2 = 1$ and $n_1 = z$. This shows that the assumption $s \ne 1$ is necessary in Theorem 1.

Balancing numbersn are solutions of the Diophantine equation

$$\begin{aligned} \sum _{i=1}^{n-1}i=\sum _{j=n+1}^{m}j, \end{aligned}$$

for some natural number m [1]. Let us denote the nth balancing number by $B_{n}$. Also, the balancing numbers satisfy the recurrence relation $B_{n+1}=6B_{n}-B_{n-1}$ with initial conditions $B_0 = 0, B_1 = 1$ for $n\ge 1$. Therefore, the sequence in Eq. (2.1) for $(P, Q, u_{0}, u_{1}) = (6, -1, 0, 1)$ is a balancing sequence. For more details regarding balancing numbers one can refer [1, 11, 20]. We prove the following theorem as an example for the explicit computations of constants.

Theorem 2

There is no solution of the Diophantine equation

$$\begin{aligned} B_{n_{1}} + B_{n_{2}} + B_{n_{3}} = 3^{z} \end{aligned}$$

(2.3)

in integers $n_{1}, n_{2}, n_{3}, z$ with $n_{1}> n_{2} > n_{3} \ge 0$.

3 Auxiliary results

In the following lemma we find an upper bound for $|u_{n}|$ which is useful in the proof of Theorem 1 and also we find a relation between $n_1$ and z when Eq. (1.5) holds.

Lemma 3.1

There exist constants $d_0$ and $d_1$ depending only on $(u_n)_{n\ge 0}$ such that the following holds.

• (1) $|u_{n}|< d_{0}\alpha ^{n}$.

• (2) If Eq. (1.5) holds with $n_1> n_2> \cdots > n_t$, then $z\le d_1 n_1$.

Proof

From Eq. (2.2), we have

$$\begin{aligned} |u_{n}|&\le \frac{\alpha ^{n}}{\sqrt{\Delta }}\left( |a| + |b| \frac{|\beta |^{n}}{\alpha ^{n}}\right) . \end{aligned}$$

Since $\alpha > |\beta |$, the above inequality becomes

$$\begin{aligned} |u_{n}|< \alpha ^{n} \frac{|a| + |b|}{\sqrt{\Delta }} = d_{0}\alpha ^{n}, \end{aligned}$$

(3.1)

where $d_{0} := (|a| + |b|)/\sqrt{\Delta }$. This proves (1).

From Eqs. (1.5) and (3.1)

$$\begin{aligned} p^z < d_0(\alpha ^{n_1} + \cdots + \alpha ^{n_t}). \end{aligned}$$

As p is given, we can always choose a suitable integer $d_1$ such that $d_{0}\alpha ^{n_i} < p^{n_id_1} \, (1\le i\le t)$. Thus, the above inequality becomes

$$\begin{aligned} p^z&< \left( p^{n_1d_1} + \cdots + p^{n_{t}d_{1}}\right) \\&<p^{n_1d_1}\left( \frac{p}{p-1}\right) \le p^{d_{1}n_{1}+1}. \end{aligned}$$

Therefore, we get $z \le d_{1}n_{1}$. $\square$

The following lemma is due to Pethő and de Weger [17, Lemma 2.2].

Lemma 3.2

[17] Let $u, v \ge 0, h\ge 1$ and $x\in \mathbb {R}$ be the largest solution of $x = u + v (\log x)^{h}$. Then

$$\begin{aligned} x < \max \{2^{h}(u^{1/h} + v^{1/h} \log (h^{h}v))^{h}, 2^{h}(u^{1/h} + 2e^{2})^{h}\}. \end{aligned}$$

Let $\eta$ be an algebraic number of degree d with minimal polynomial

$$\begin{aligned} a_{0}x^d + a_1x^{d-1} + \cdots + a_d = a_0 \prod _{i=1}^{d}\left( X - \eta ^{(i)}\right) , \end{aligned}$$

where $a_0$ is the leading coefficient of the minimal polynomial of $\eta$ over $\mathbb {Z}$ and the $\eta ^{(i)}$’s are conjugates of $\eta$ in $\mathbb {C}$. Define the absolute logarithmic height of an algebraic number $\eta$ by

$$\begin{aligned} h(\eta ) = \frac{1}{d} \left( \log |a_0| + \sum _{i=1}^d \log \max ( 1, |\eta ^{(i)}| ) \right) . \end{aligned}$$

In particular, if $\eta = p/q$ is a rational number with $\gcd (p, q) = 1$ and $q >0$, then $h(\eta ) = \log \max \{|p|, q\}$. Some important properties of logarithmic height which we use in our further investigation are as follows (see e.g. [22]).

• (1) $h(\eta \pm \gamma ) \le h(\eta ) + h(\gamma ) + \log 2$,

• (2) $h(\eta \gamma ^{\pm 1}) \le h(\eta ) + h(\gamma )$,

• (3) $h(\eta ^{t}) \le |t|h(\eta ),\;$ for $t \in \mathbb {Z}$.

Generally it is a very hard problem to find lower bounds for the height of elements in a number field of given degree. For the quadratic fields, we have the following result due to Pink and Ziegler [19, Lemma 3].

Lemma 3.3

[19] For an algebraic number $\alpha$ of degree two we have $h(\alpha ) \ge 0.24$ or $\alpha$ is a root of unity.

To prove our theorem, we use lower bounds for linear forms in logarithms to bound the index $n_1$ appearing in Eq. (1.5). Generically, we need the following general lower bound for linear forms in logarithms due to Matveev [14, Theorem 2.2] (see also [8, Theorem 9.4]).

Lemma 3.4

[8] Let $\gamma _1,\ldots ,\gamma _t$ be positive real algebraic numbers and let $b_{1},\ldots , b_{t}$ be non-zero rational integers. Let $\mathbb {Q}(\gamma _1,\ldots ,\gamma _t)$ be the quadratic number field over $\mathbb {Q}$ and let $A_{j}$ be real numbers satisfying

$$\begin{aligned} A_j \ge \max \left\{ 2h(\gamma _j) , |\log \gamma _j|, 0.16 \right\} , \quad j= 1, \ldots ,t. \end{aligned}$$

(3.2)

Assume that $B\ge \max \{|b_1|, \ldots , |b_{t}|\}$ and $\Lambda :=\gamma _{1}^{b_1}\cdots \gamma _{t}^{b_t} - 1$. If $\Lambda \ne 0$, then

$$\begin{aligned} |\Lambda | \ge \exp \left( -1.4\times 30^{t+3}\times t^{4.5}\times 2^{2}(1 + \log 2)(1 + \log B)A_{1}\cdots A_{t}\right) . \end{aligned}$$

The upper bound on $n_1$ of the Diophantine equation (2.3) is too large for practical purpose, thus the next step is to reduce it. We present a lemma from [3, Lemma 4] and it is a variant of the famous Dujella and Pethő [12, Lemma 5a] reduction lemma. Now for a real number x, let $||x|| := \min \{|x - n| : n \in \mathbb {Z}\}$ denote the distance from x to the nearest integer.

Lemma 3.5

[3] Suppose that M is a positive integer, and A, B are positive reals with $B > 1$. Let p / q be the convergent of the continued fraction expansion of the irrational number $\gamma$ such that $q > 6M$, and let $\epsilon := ||\mu q|| - M||\gamma q||$, where $\mu$ is a real number. If $\epsilon > 0$, then there is no solution of the inequality

$$\begin{aligned} 0< |u \gamma - n + \mu | < AB^{-m} \end{aligned}$$

in positive integers u, m and n with

$$\begin{aligned} u \le M\quad \text{ and }\quad m\ge \frac{\log (Aq/\epsilon )}{\log B}. \end{aligned}$$

To prove Theorem 1, we apply linear forms in logarithms several times. Every time, we find an upper bound for $(n_1 - n_i)$ in terms of $n_1$ for $2\le i \le t$. In order to apply the linear forms in logarithms to bound $(n_1 - n_i)$ for a fixed i, we require upper bounds of $(n_1 - n_j)$ for all $1\le j < i$. Finally, using these upper bounds for $(n_1 - n_i),\;2\le i \le t$, we get an upper bound for $n_1$. In order to apply Lemma 3.4 we must ensure that $\Lambda _i$ does not vanish. In this regard, we have the following lemma.

Lemma 3.6

Suppose $\Lambda _i := p^{z}\alpha ^{-n_1}|a|^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})^{-1} - 1$ for all $1\le i \le t$. If $\Lambda _i = 0$, then $n_1 < \ell _i$, where

$$\begin{aligned} \ell _i:= {\left\{ \begin{array}{ll} \frac{\log \left( i(|b|+|a|)/|a|\right) }{\log \alpha } &{} \quad \text {if } |\beta | \le 1,\\ \frac{\log (i(|b|+|a|)/|a|)}{\log (\alpha /|\beta |)} &{} \quad \text {if } |\beta | > 1. \end{array}\right. } \end{aligned}$$

Proof

Now $\Lambda _i = 0$ imply

$$\begin{aligned} p^z r\sqrt{s} = |a| (\alpha ^{n_1} + \alpha ^{n_2 } + \cdots + \alpha ^{n_i }). \end{aligned}$$

(3.3)

Since s is a square-free integer other than one, by taking the conjugate of (3.3), we get

$$\begin{aligned} - p^z r\sqrt{s} = |b| (\beta ^{n_1} + \beta ^{n_2 } + \cdots + \beta ^{n_i }). \end{aligned}$$

(3.4)

Thus, from (3.3) and (3.4)

$$\begin{aligned} |a| (\alpha ^{n_1} + \alpha ^{n_2 } + \cdots + \alpha ^{n_i }) = |b (\beta ^{n_1} + \beta ^{n_2 } + \cdots + \beta ^{n_i })|. \end{aligned}$$

Since $\alpha > 0$

$$\begin{aligned} |a|\alpha ^{n_1}&< |a|(\alpha ^{n_1} + \alpha ^{n_2} + \cdots + \alpha ^{n_i}) = |b (\beta ^{n_1} + \beta ^{n_2 } + \cdots + \beta ^{n_i })|. \end{aligned}$$

If $|\beta | \le 1$, then $|a|\alpha ^{n_1}< |b|i< i (|b|+|a|)$ which implies that $n_1 < \frac{\log (i(|b|+|a|)/|a|)}{\log \alpha }$. Now for $|\beta | > 1$,

$$\begin{aligned} |a|\alpha ^{n_1} \le i |b||\beta |^{n_1} < i|\beta |^{n_1}(|b|+|a|) \end{aligned}$$

which implies that $n_1 < \frac{\log (i(|b|+|a|)/|a|)}{\log (\alpha /|\beta |)}.$$\square$

The following lemma gives a relationship between height of an algebraic number and its logarithm. We can suitably choose a constant $d_2$ such that

$$\begin{aligned} d_2 > \max \{\max (|a|, 1/|a|), \max (|b|, 1/|b|)\}. \end{aligned}$$

We also denote $K = \mathbb {Q}(\alpha , \beta ) = \mathbb {Q}(\sqrt{\Delta }) = \mathbb {Q}(\sqrt{s})$ for the quadratic number field corresponding to binary recurrence sequence $(u_n)_{\ge 0}$.

Lemma 3.7

Let $\gamma _3(i) := |a|^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})^{-1}$ are the algebraic numbers in the quadratic number field ${\mathbb {Q}}(\sqrt{s})$ for $1 \le i \le t$. Then there exists a constant $d_2$ depending on $\alpha$ and $\beta$ such that

$$\begin{aligned}&\max \left\{ 2h(\gamma _3(i)), |\log \gamma _3(i)|, 0.16 \right\} \\&\quad \le 2\left( \log d_2 + \log (r \sqrt{s}) + \left( |n_2 - n_1| + \cdots + |n_i - n_1| \right) h(\alpha )\right) \\&\qquad + (i + 1)\log 4 . \end{aligned}$$

Proof

First observe that $r\sqrt{s} \ge 1$, as $r\sqrt{s}=\sqrt{\Delta } = \alpha - \beta = \sqrt{P^2 + 4 Q} > 0$ and P, Q are rational integers. Since $\gamma _3(i) = |a|^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})^{-1}$ and $\alpha > 0$, we have

$$\begin{aligned} \gamma _3(i)&= \frac{r\sqrt{s}}{|a|(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})} \\&\le \max (|a|, 1/|a|)r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})^{-1} \end{aligned}$$

and this implies

$$\begin{aligned} \begin{aligned} \log \gamma _3(i)&\le \log \max (|a|, 1/|a|) + \log (r\sqrt{s}) - \log (1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})\\&\le \log \max (|a|, 1/|a|) + \log (r\sqrt{s}) + \log i \\&< \log d_2 + \log (r\sqrt{s}) + \log i. \end{aligned} \end{aligned}$$

(3.5)

Since $\log i < i\log 2$ for $i\ge 1$, we have

$$\begin{aligned} |\log \gamma _3(i)| \le \log d_2 + \log (r\sqrt{s}) + i\log 2. \end{aligned}$$

(3.6)

Furthermore, a and b are conjugate algebraic numbers which are roots of the quadratic polynomial

$$\begin{aligned} x^2 - (2u_1 - u_0 P)x + u_1^2 - u_0u_1P - u_0^2Q . \end{aligned}$$

Thus,

$$\begin{aligned} h(a) = \frac{1}{2}\left( \log \max \{|a|, 1\} + \log \max \{|b|, 1\}\right) < \log d_2. \end{aligned}$$

(3.7)

Now we estimate height of $\gamma _3(i)$. We have

$$\begin{aligned} h(\gamma _3(i))&= h\left( |a|^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_i - n_1})^{-1}\right) \nonumber \\&\le h(|a|)+ h(r\sqrt{s}) + \left( |n_2 - n_1| + \cdots + |n_i - n_1| \right) h(\alpha ) + i\log 2. \end{aligned}$$

(3.8)

Since s is square-free we have

$$\begin{aligned} h(r\sqrt{s}) = \log (r\sqrt{s}). \end{aligned}$$

(3.9)

By Lemma 3.3, we have $h(\alpha ) \ge 0.24$. Finally the lemma follows by comparing (3.6), (3.7), (3.8) and (3.9).$\square$

Suppose $\ell := \max \{\ell _1, \cdots , \ell _{t}\}$, where $\ell _i \;(1\le i\le t)$ are defined in Lemma 3.6. If $n_1 \le \ell$, then the conclusion of Theorem 1 follows trivially (see Lemma 3.1(2)). Henceforth, we assume $n_1 > \ell$.

Now we are ready to prove Theorem 1. The proof is motivated by the ideas of Bravo and Luca [6, 7].

4 Proof of Theorem 1

4.1 Bounding $(n_1 - n_i)$ in terms of $n_1$ for $2\le i \le t$

Here, we claim that for $2\le i \le t$, we have

$$\begin{aligned} n_1 - n_i \le C_i (\log n_1)^{i - 1}, \end{aligned}$$

(4.1)

where $C_i$’s are effectively computable constants depending on $P, Q, u_0, u_1, t, p$. We use induction on i to find an upper bound of $n_1 - n_i$ for $2\le i \le t$. First, we calculate the upper bound of $n_1 - n_2$. We can rewrite Eq. (1.5) as

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{\sqrt{\Delta }} - p^{z}\right|&= \left| \frac{b\beta ^{n_1}}{\sqrt{\Delta }} - (u_{n_2} + \cdots + u_{n_t})\right| \le \left| \frac{b\beta ^{n_1}}{\sqrt{\Delta }}\right| + |u_{n_2} + \cdots + u_{n_t}|. \end{aligned}$$

From Lemma 3.1, $|u_{n}|\le d_{0}\alpha ^{n}$. Also we have assumed that $n_1> n_2> \cdots > n_{t}$. Thus,

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{r\sqrt{s}} - p^{z}\right| \le \left| \frac{b\beta ^{n_1}}{r\sqrt{s}}\right| + (t-1)d_{0}\alpha ^{n_{2}}. \end{aligned}$$

(4.2)

If $|\beta | < 1$, then the above inequality becomes

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{r\sqrt{s}} - p^{z}\right| < c_{1}\alpha ^{n_{2}}, \end{aligned}$$

(4.3)

where $c_1$ is a suitable constant. Now dividing both sides of the inequality (4.3) by $|a|\alpha ^{n_1}/r\sqrt{s}$,

$$\begin{aligned} \begin{aligned} \left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s} \right|&\le \frac{c_{1}r\sqrt{s}}{|a|\alpha ^{n_1-n_2}} < \frac{c_2}{\alpha ^{n_1-n_2}}, \end{aligned} \end{aligned}$$

(4.4)

where $c_2$ is a suitable constant. For the case $|\beta | > 1$, we divide both sides of the inequality (4.2) by $|a|\alpha ^{n_1}/r\sqrt{s}$,

$$\begin{aligned} \left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s} \right|&\le \frac{|b\beta ^{n_1}|}{|a| \alpha ^{n_1}}+ \frac{(t-1)d_{0}r\sqrt{s}}{|a|}\alpha ^{n_{2} -n_1}\\&< \frac{|b|}{|a|}\left( \frac{\alpha }{|\beta |}\right) ^{n_2 - n_1} + \frac{(t-1)d_{0}r\sqrt{s}}{|a|}\alpha ^{n_{2} -n_1}\\&< \frac{|b|}{|a|}\left( \frac{|\beta |}{\alpha }\right) ^{n_1 - n_2} + \frac{(t-1)d_{0}r\sqrt{s}}{|a|}\left( \frac{|\beta |}{\alpha }\right) ^{n_{1} -n_2}\\&< \left( \frac{|b|}{|a|} + \frac{(t-1)d_{0}r\sqrt{s}}{|a|}\right) \left( \frac{|\beta |}{\alpha }\right) ^{n_{1} -n_2}. \end{aligned}$$

Thus, for any $\beta$ it follows that

$$\begin{aligned} \left| 1 - p^{z}\alpha ^{-n_1} |a|^{-1}r\sqrt{s} \right| \le \left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s} \right| < \frac{c_3}{\min \left( \frac{\alpha }{|\beta |}, \alpha \right) ^{n_{1} -n_2}}, \end{aligned}$$

(4.5)

where $c_3 := \max \left\{ c_2, \left( \frac{|b|}{|a|} + \frac{(t-1)d_{0}r\sqrt{s}}{|a|}\right) \right\}$ and the left inequality in (4.5) is obtained using triangle inequality.

In order to apply Lemma 3.4, we take

$$\begin{aligned} \gamma _1 := p,\; \gamma _2 := \alpha ,\; \gamma _3 := |a|^{-1}r\sqrt{s},\; b_1 := z,\; b_2 := -n_1,\; b_3= 1. \end{aligned}$$

Thus our first linear form is $\Lambda _1 := \gamma _{1}^{b_{1}}\gamma _{2}^{b_{2}}\gamma _{3}^{b_{3}} - 1$ . Further, to apply Lemma 3.4 we must ensure that $\Lambda _1 \ne 0$. Suppose on contrary $\Lambda _1 = 0$. Then from Lemma 3.6, we have $n_1< \ell _1$ and hence $n_1\le \ell$ (see the discussion towards the end of the Sect. 3). Thus, this is a contradiction as we have assumed $n_1 > \ell$. Hence, $\Lambda _1 \ne 0$. Here we are taking the quadratic number field ${\mathbb {Q}}(\sqrt{s})$ over ${\mathbb {Q}}$ and $t=3$. Finally, recall $z\le d_1n_1$ and so we deduce

$$\begin{aligned} \max \{|b_1|, |b_2|, |b_3|\} = \max \{z, n_1, 1\} \le d_1n_1. \end{aligned}$$

Hence we can take $B := d_1n_1$. Also $h(\gamma _1)= \log p, h(\gamma _2)\le \log \alpha$. Thus, we can take $A_1 := 2\log p, A_2 = 2\log \alpha$. Further, using Lemma 3.7, we have

$$\begin{aligned} 2\left( \log d_2+ \log (r\sqrt{s}) + \log 4 \right) \ge \max \{2h(\gamma _3), \log \gamma _3, 0.16\}. \end{aligned}$$

Hence, $A_3 = 2\left( \log d_2+ \log (r\sqrt{s}) + \log 4 \right)$. Using Lemma 3.4, we have

$$\begin{aligned} |\Lambda _1| >&\exp \left( -1.4\times 30^{5}\times 2^{4.5}\right. \\&\left. \times 4\times (1+\log 2)(2\log p) (2\log \alpha ) A_3(1+ \log d_1n_1)\right) . \end{aligned}$$

So the above inequality can be rewritten as,

$$\begin{aligned} \log |\Lambda _1| > - C_{1} \log n_1, \end{aligned}$$

(4.6)

where $C_1$ is a suitable constant. Taking logarithms in both sides of inequality (4.5) and comparing the resulting inequality with inequality (4.6),

$$\begin{aligned} (n_1-n_2)\log \min \left( \frac{\alpha }{|\beta |}, \alpha \right) < C_2' \log n_1, \end{aligned}$$

(4.7)

where $C_2'$ is a suitable constant. This implies that

$$\begin{aligned} (n_1-n_2)< C_2\log n_1. \end{aligned}$$

Therefore, for $i = 2$, the statement is true. We assume that inequality (4.1) is true for $i = k$ with $2\le k \le t-1$. We want to show that inequality (4.1) is true for $i = k+1$. To formulate the next linear form we rewrite Eq. (1.5) as follows

$$\begin{aligned} \frac{a\alpha ^{n_1}}{\sqrt{\Delta }} + \cdots + \frac{a\alpha ^{n_{k}}}{\sqrt{\Delta }} - p^{z} = \frac{b\beta ^{n_1}}{\sqrt{\Delta }} + \cdots + \frac{b\beta ^{n_{k}}}{\sqrt{\Delta }} - u_{n_{k+1}}. \end{aligned}$$

Taking absolute value and using inequality (3.1), we have

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{r\sqrt{s}}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1}) - p^{z}\right| \le \left| \frac{b(\beta ^{n_1} +\cdots +\beta ^{n_{k}})}{r\sqrt{s}}\right| + d_{0}\alpha ^{n_{k+1}}.\nonumber \\ \end{aligned}$$

(4.8)

If $|\beta | < 1$, then (4.8) becomes

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{r\sqrt{s}}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1}) - p^{z}\right| < c_4 \alpha ^{n_{k+1}}, \end{aligned}$$

(4.9)

where $c_4$ is a suitable constant. Dividing both sides of the inequality (4.9) by $|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{k}})/r\sqrt{s}$,

$$\begin{aligned} \begin{aligned} \left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1})^{-1} \right|&< \frac{c_4\alpha ^{n_{k+1}}r\sqrt{s}}{|a|(\alpha ^{n_1}+ \cdots + \alpha ^{n_k})}\\&< \frac{c_4\alpha ^{n_{k+1}}r\sqrt{s}}{|a|\alpha ^{n_1}} < \frac{c_{5}}{\alpha ^{n_1-n_{k+1}}}, \end{aligned} \end{aligned}$$

(4.10)

where $c_5 = c_4 r\sqrt{s}/|a|$. For the case $|\beta | > 1$, we divide both sides of the inequality (4.8) by $|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{k}})/r\sqrt{s}$,

$$\begin{aligned}&\left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1})^{-1} \right| \\&\quad \le \left| \frac{b(\beta ^{n_1} +\cdots +\beta ^{n_{k}})}{|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{k}})}\right| + \frac{d_{0}r\sqrt{s}\alpha ^{n_{k+1}}}{|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{k}})}\le \frac{|kb\beta ^{n_1}|}{|a|\alpha ^{n_1}} + \frac{d_{0}r\sqrt{s}\alpha ^{n_{k+1}}}{|a|\alpha ^{n_{1}} }\\&\quad< \frac{k|b|}{|a|}\left( \frac{\alpha }{|\beta |}\right) ^{n_{k+1} - n_1} + \frac{d_{0}r\sqrt{s}}{|a|}\alpha ^{n_{k+1} -n_1}\\&\quad < \left( \frac{k|b|}{|a|} + \frac{d_{0}r\sqrt{s}}{|a|}\right) \left( \frac{|\beta |}{\alpha }\right) ^{n_{1} -n_{k+1}}. \end{aligned}$$

Thus, for any $\beta$ it follows that

$$\begin{aligned} \begin{aligned}&\left| 1 - p^{z}\alpha ^{-n_1} |a|^{-1}r\sqrt{s} (1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1})^{-1} \right| \\&\quad \le \left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1})^{-1} \right| \\&\quad < \frac{c_6}{\min \left( \frac{\alpha }{|\beta |}, \alpha \right) ^{n_{1} -n_{k+1}}}, \end{aligned} \end{aligned}$$

(4.11)

where $c_6 := \max \left\{ c_5, \left( \frac{k|b|}{|a|} + \frac{d_{0}r\sqrt{s}}{|a|}\right) \right\}$. To apply Lemma 3.4 we take,

$$\begin{aligned} \gamma _1 := p,\;&\gamma _2 := \alpha ,\; \gamma _3 := |a|^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{k} - n_1})^{-1},\\&b_1 := z,\;\; b_2 := -n_1,\;\; b_3= 1. \end{aligned}$$

Our next linear form is $\Lambda _2 := \gamma _{1}^{b_{1}}\gamma _{2}^{b_{2}}\gamma _{3}^{b_{3}} - 1$. Suppose $\Lambda _2 = 0$. Then from Lemma 3.6, $n_1<\ell _k$. That is $n_1\le \ell$ and this is a contradiction as we have assumed $n_1 > \ell$. Thus, $\Lambda _2 \ne 0$. Like in the previous case, here we take $B= d_1n_1$. We already estimate $A_1, A_2$ in previous case. Using Lemma 3.7 we can take $A_3^k = c_{7} + 2(|n_2 - n_1| + \cdots +|n_{k} - n_1|) \log \alpha$ where $c_7 = 2\left( \log d_2 + \log (r\sqrt{s}) \right) + (k+1)\log 4$.

From Lemma 3.4, we have

$$\begin{aligned} |\Lambda _2|> & {} \exp \left( - 1.4\times 30^{5}\times 2^{4.5}\times 4\right. \nonumber \\&\left. \times (1+\log 2)(1+ \log d_1n_1)(2\log p) (2\log \alpha ) A_3^k\right) . \end{aligned}$$

(4.12)

Now from the inequalities (4.11) and (4.12), we get

$$\begin{aligned}&\exp \left( -c_{8}(\log n_{1})(c_{7} + 2(|n_2 - n_1| + \cdots + |n_k - n_1|)\log \alpha ) \right) \nonumber \\&\quad < \frac{c_{6}}{\min \left( \frac{\alpha }{|\beta |}, \alpha \right) ^{n_{1}-n_{k+1}}}, \end{aligned}$$

(4.13)

where $c_8$ is a suitable constant. Now taking logarithm on both sides of (4.13)

$$\begin{aligned}&(n_1- n_{k+1}) \log \min \left( \frac{\alpha }{|\beta |}, \alpha \right) < \log c_6 \nonumber \\&\quad + c_{8}(\log n_{1})(c_{7} + 2(|n_2 - n_1| + \cdots + |n_k - n_1|)\log \alpha ). \end{aligned}$$

(4.14)

By induction hypothesis, we have $(n_1 - n_i) < C_i (\log n_1)^{i-1}$ for $2\le i \le k$. Thus from (4.14), we obtain

$$\begin{aligned} (n_1 - n_{k+1}) < C_{k+1} (\log n_1)^{k}. \end{aligned}$$

(4.15)

4.2 Bounding $n_1$

To bound $n_1$, we write the Eq. (1.5) as

$$\begin{aligned} \frac{a\alpha ^{n_1}}{\sqrt{\Delta }} + \cdots + \frac{a\alpha ^{n_t}}{\sqrt{\Delta }} - p^{z} = \frac{b\beta ^{n_1}}{\sqrt{\Delta }} + \cdots + \frac{b\beta ^{n_t}}{\sqrt{\Delta }}, \end{aligned}$$

which implies that

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{\sqrt{\Delta }}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_t - n_1}) - p^{z}\right| = \left| \frac{b(\beta ^{n_1} + \cdots +\beta ^{n_t})}{r\sqrt{s}}\right| . \end{aligned}$$

(4.16)

If $|\beta | < 1$, then (4.16) becomes

$$\begin{aligned} \left| \frac{a\alpha ^{n_1}}{r\sqrt{s}}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{t} - n_1}) - p^{z}\right| < \frac{t|b|}{r\sqrt{s}}. \end{aligned}$$

(4.17)

Dividing both sides of the inequality (4.17) by $|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{t}})/r\sqrt{s}$,

$$\begin{aligned}&\left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{t} - n_1})^{-1} \right| \nonumber \\&\quad < \frac{t|b|}{|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{t}})}\le \frac{c_{9}}{\alpha ^{n_1}}, \end{aligned}$$

(4.18)

where $c_{9} = t|b|/ |a|$. For the case $|\beta | > 1$, we divide both sides of the inequality (4.17) by $|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{t}})/r\sqrt{s}$,

$$\begin{aligned} \left| 1 - p^{z}\alpha ^{-n_1} a^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{t} - n_1})^{-1} \right|&\le \left| \frac{b(\beta ^{n_1} +\cdots +\beta ^{n_{t}})}{|a|(\alpha ^{n_1} +\cdots +\alpha ^{n_{t}})}\right| \\&< \left( \frac{t|b|}{|a|}\right) \left( \frac{|\beta |}{\alpha }\right) ^{n_{1} }. \end{aligned}$$

Thus, for any $\beta$ we have

$$\begin{aligned} \begin{aligned}&\left| 1 - p^{z}\alpha ^{-n_1} |a|^{-1} r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{t} - n_1})^{-1} \right| \\&\quad \le \left| 1 - p^{z}\alpha ^{-n_1} a^{-1} r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_{t} - n_1})^{-1} \right| < \frac{c_{10}}{\min \left( \frac{\alpha }{|\beta |}, \alpha \right) ^{n_{1}}}, \end{aligned} \end{aligned}$$

(4.19)

where $c_{10}:= \max \left\{ c_9, \left( \frac{t|b|}{|a|} \right) \right\}$. To apply Lemma 3.4, we take

$$\begin{aligned}&\gamma _1 := p, \gamma _2 := \alpha , \gamma _3 := |a|^{-1}r\sqrt{s}(1 + \alpha ^{n_2 - n_1} + \cdots + \alpha ^{n_t - n_1})^{-1}\\&b_1 := z,\;\; b_2 := -n_1,\;\; b_3= 1. \end{aligned}$$

Thus, the final linear form is $\Lambda _t := \gamma _{1}^{b_{1}}\gamma _{2}^{b_{2}}\gamma _{3}^{b_{3}} - 1$. Suppose $\Lambda _t = 0$. Then from Lemma 3.6, $n_1< \ell _t$. That is $n_1\le \ell$ and this is a contradiction as we have assumed $n_1 > \ell$. Thus, $\Lambda _t \ne 0$. Here we take $B= d_1n_1, A_1 = 2\log p, A_2 = 2\log \alpha$. From the conclusions of Lemma 3.7, we can take $A_3^t= c_{11} + 2[(n_1 - n_2) + \cdots + (n_1 - n_t)]\log \alpha$ where $c_{11} = 2\left( \log d_2+ \log (r\sqrt{s})\right) + (t+1)\log 4$. From Lemma 3.4, we have

$$\begin{aligned} |\Lambda _t|> & {} \exp \left( - 1.4\times 30^{5}\times 2^{4.5}\right. \nonumber \\&\left. \times 4\times (1+\log 2)(1+ \log d_1n_1)(2\log p) (2\log \alpha ) A_3^t\right) . \end{aligned}$$

(4.20)

Now from the inequalities (4.19) and (4.20), we get

$$\begin{aligned}&\exp \left( -c_{12}(\log n_{1})(c_{10} + 2(|n_2 - n_1| + \cdots + |n_t - n_1|)\log \alpha ) \right) \nonumber \\&\quad < \frac{c_{10}}{\min \left( \frac{\alpha }{|\beta |}, \alpha \right) ^{n_{1}}}, \end{aligned}$$

(4.21)

where $c_{12}$ is a suitable constant. Now taking logarithm on both sides of (4.21)

$$\begin{aligned}&n_1\log \min \left( \frac{\alpha }{|\beta |}, \alpha \right) < \log c_{10} + c_{{11}}(\log n_{1})(c_{10} \nonumber \\&\quad + 2(|n_2 - n_1| + \cdots + |n_t - n_1|)\log \alpha ) \end{aligned}$$

(4.22)

Since $(n_1 - n_i) < C_i (\log n_1)^{i-1}$ for $2\le i \le t$, we deduce from (4.22)

$$\begin{aligned} n_1 < C (\log n_1)^{t}, \end{aligned}$$

(4.23)

where C is a suitable constant. Theorem 1 follows by applying Lemma 3.2 and Lemma 3.1(2) to the inequality (4.23).

5 Proof of Theorem 2

We give the computational details of the resolution of Diophantine Eq. (2.3). Let $(B_{n})_{n \ge 0}$ be the balancing sequence given by $B_{0} = 0, B_1 = 1$ and $B_{n+1} = 6B_{n} - B_{n-1}$ for all $n\ge 1$. Now one can easily see from (2.2), $a =1, b=1, \alpha =3+2\sqrt{2},\; \beta =3-2\sqrt{2}$ and the nth term of balancing sequence is

$$\begin{aligned} B_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta } \quad \text{ for } \text{ all }\;\; n=0,1,\ldots . \end{aligned}$$

(5.1)

5.1 The case $n_3 =0$

In this case $n_2 > 0$, thus we have

$$\begin{aligned} B_{n_1} + B_{n_2} = 3^z, \end{aligned}$$

(5.2)

which is a reduced form of Eq. (2.3). Hence without loss of generality, we assume $n_3 > 0$.

5.2 Bounding $(n_1 - n_2)$ and $(n_1 - n_3)$ in terms of $n_1$

If $1\le n_1 \le 100$, then a brute force search with Mathematica in the range $1\le n_{3}< n_2 < n_1 \le 100$ gives no solution of Eq. (2.3). Hence from this point onward, we assume that $n_1 > 100$. Using Lemma 3.1, for balancing sequence we have $|B_n| \le \alpha ^n$. Now from (2.3) we have

$$\begin{aligned} 3^z \le \alpha ^{n_1} + \alpha ^{n_2}+\alpha ^{n_3}= \alpha ^{n_1}(1+ \alpha ^{n_2-n_1}+ \alpha ^{n_3-n_1})\le 3\alpha ^{n_1} \le 3^{2n_1} \end{aligned}$$

and this implies

$$\begin{aligned} z \le 2n_1. \end{aligned}$$

(5.3)

We record this estimate for future referencing. We rewrite (2.3) as

$$\begin{aligned} \frac{\alpha ^{n_1}}{4\sqrt{2}} -3^z = \frac{\beta ^{n_1}}{4\sqrt{2}} - B_{n_2} - B_{n_3}. \end{aligned}$$

Now take the absolute values on the both hand side of the above relation obtaining

$$\begin{aligned} \left| \frac{\alpha ^{n_1}}{4\sqrt{2}} -3^z\right| \le \frac{|\beta |^{n_1}}{4\sqrt{2}} + |B_{n_2}| +| B_{n_3}| \le \frac{|\beta |^{n_1}}{4\sqrt{2}} +2\alpha ^{n_2}<2. 5\alpha ^{n_2}. \end{aligned}$$

(5.4)

Dividing both sides of the above inequality by $\frac{\alpha ^{n_1}}{4\sqrt{2}}$, we obtain the first linear form

$$\begin{aligned} \left| 1 - 3^{z}\alpha ^{-n_1} 4\sqrt{2} \right| < \frac{16}{\alpha ^{n_1-n_2}}. \end{aligned}$$

(5.5)

In order to apply Lemma 3.4, we take

$$\begin{aligned} t= 3, \;\; \gamma _1 = 3,\;\;\gamma _2= \alpha ,\;\;\gamma _3 = 4\sqrt{2}, b_1= z, \; b_2 = -n_1, \; b_3 = 1. \end{aligned}$$

Thus, $B = 2n_1, h(\gamma _1)= \log 3= 1.0986\cdots , h(\gamma _2) = (\log \alpha )/2 = 0.8813\cdots , h(\gamma _3) \le 0.8664\cdots$. We can choose $A_1 = 2.4, A_2 = 1.9, A_3 = 1.8$. By Lemma 3.4, we obtain the lower bound for the linear form $\Lambda _1 := \gamma _{1}^{b_{1}}\gamma _{2}^{b_{2}}\gamma _{3}^{b_{3}} - 1$ is

$$\begin{aligned} \log |\Lambda _1| > -C_1 (1 + \log 2n_1), \end{aligned}$$

(5.6)

where $C_1 = 1.4 \times 30^{5}\times 2^{4.5}\times 4\times (1+ \log 2) \times 2.4 \times 1.9\times 1.8 = 7.9\times 10^{12}$. Further, since $(1+\log 2n_1) < 2\log n_1$ for $n_1 > 100$, we get from (5.5) and (5.6) that

$$\begin{aligned} (n_1-n_2)\log \alpha< \log 16 + 15.8\times 10^{12} \log n_1 < 15.9\times 10^{12}\log n_1. \end{aligned}$$

(5.7)

We now construct a second linear form in logarithms by rewriting Eq. (2.3) as follows:

$$\begin{aligned} \frac{\alpha ^{n_1}}{4\sqrt{2}} + \frac{\alpha ^{n_2}}{4\sqrt{2}} - 3^{z} = \frac{\beta ^{n_1}}{4\sqrt{2}} + \frac{\beta ^{n_2}}{4\sqrt{2}} - B_{n_3}. \end{aligned}$$

Taking absolute values in the above relation and using the fact that $|\beta | <1$

$$\begin{aligned} \left| \frac{\alpha ^{n_1}}{4\sqrt{2}}(1 + \alpha ^{n_2 - n_1}) - 3^{z}\right| \le \frac{|\beta |^{n_1}}{4\sqrt{2}} + \frac{|\beta |^{n_2}}{4\sqrt{2}} +| B_{n_3}| < \frac{1}{2\sqrt{2}}+ \alpha ^{n_3}. \end{aligned}$$

Dividing both sides of the above inequality by $\frac{\alpha ^{n_1}}{4\sqrt{2}}(1 + \alpha ^{n_2 - n_1})$, we obtain

$$\begin{aligned} | 1- 3^z 4\sqrt{2}\alpha ^{-n_1}(1+\alpha ^{n_2-n_1})^{-1}|< \frac{7}{\alpha ^{n_1-n_3}}. \end{aligned}$$

(5.8)

Here our second linear form is $\Lambda _2 := \gamma _{1}^{b_{1}}\gamma _{2}^{b_{2}}\gamma _{3}^{b_{3}} - 1$ where

$$\begin{aligned} t= 3, \;\; \gamma _1 = 3,\;\;\gamma _2 = \alpha ,\gamma _3=4\sqrt{2} (1 + \alpha ^{n_2- n_1})^{-1},\;\; b_1= z, \; b_2 = -n_1, \; b_3 = 1. \end{aligned}$$

In this case $A_1$ and $A_2$ are same as previous case but $A_3 = \log 8\sqrt{2} + (n_1 - n_2)\log \alpha$. Thus by applying Lemma 3.4 we get

$$\begin{aligned} \log |\Lambda _2| > - 4.4 \times 10^{12}\times (1 + \log 2n_1)\left( \log 8\sqrt{2} + (n_1 - n_2)\log \alpha \right) . \end{aligned}$$

(5.9)

From inequalities (5.8) and (5.9),

$$\begin{aligned} (n_1-n_3)\log \alpha < 1.4\times 10^{26} (\log n_1)^2. \end{aligned}$$

(5.10)

5.3 Bounding $n_1$

To bound $n_1$, we consider the analogue equation for Eq. (4.18) which is

$$\begin{aligned} | 1- 3^z 4\sqrt{2}\alpha ^{-n_1}(1+\alpha ^{n_2-n_1} + \alpha ^{n_3-n_1})^{-1}|< \frac{c_{9}}{\alpha ^{n_1}}, \end{aligned}$$

(5.11)

with $c_{9} = t|b|/|a| = 3$. Thus, final linear form is $\Lambda _3 := \gamma _{1}^{b_{1}}\gamma _{2}^{b_{2}}\gamma _{3}^{b_{3}} - 1$, where

$$\begin{aligned} t= 3, \;\; \gamma _1 = 3,\;\;\gamma _2 = \alpha ,\gamma _3=4\sqrt{2} (1+\alpha ^{n_2-n_1} + \alpha ^{n_3-n_1})^{-1}. \end{aligned}$$

Here by taking $A_3 = \log 4\sqrt{2} + (n_1 - n_2)\log \alpha + (n_1 - n_3)\log \alpha + \log 4$ and applying Lemma 3.4, we get

$$\begin{aligned} \log |\Lambda _3| > - 4.4 \times 10^{12}\times (1 + \log 2n_1)(6 + (n_1 - n_2)\log \alpha + (n_1 - n_3)\log \alpha ).\nonumber \\ \end{aligned}$$

(5.12)

From inequalities (5.11) and (5.12),

$$\begin{aligned} n_1 \log \alpha < 22\times 10^{38} (\log n_1)^3. \end{aligned}$$

(5.13)

Further, from Lemma 3.2, we have

$$\begin{aligned} n_1 < 1.5\times 10^{45}. \end{aligned}$$

(5.14)

We summarize the above discussion in the following proposition.

Proposition 3

Let us assume that $n_1> n_2 > n_3$ and $n_1 > 100$. If $(n_1, n_2, n_3, z)$ is a positive integral solution of Eq. (2.3), then

$$\begin{aligned} z \le 2n_1 < 3\times 10^{45}. \end{aligned}$$

(5.15)

5.4 Reducing the size of $n_1$

From Proposition 3, we can see that the bound we have obtained for $n_{1}$ is very large. Now our job is to reduce this upper bound to a certain minimal range. We return to inequality (5.5). Put

$$\begin{aligned} \Lambda _1 := z\log 3 -n_1\log \alpha + \log 4\sqrt{2}. \end{aligned}$$

(5.16)

Then (5.5) implies

$$\begin{aligned} |1- e^{\Lambda _1}|< \frac{16}{\alpha ^{n_1-n_2}}. \end{aligned}$$

Note that $\Lambda _1 > 0$, otherwise $3^z \le \alpha ^{n_1}/4\sqrt{2}$. But we always have

$$\begin{aligned} \frac{\alpha ^{n_1}}{4\sqrt{2}}< B_{n_1}+1\le B_{n_1}+B_{n_2}+B_{n_3}= 3^z. \end{aligned}$$

Using the fact that $1+x< e^x$ holds for all positive real numbers x, we get

$$\begin{aligned} 0< \Lambda _1 \le e ^{\Lambda _1} - 1 < \frac{16}{\alpha ^{n_1-n_2}}. \end{aligned}$$

Dividing Eq. (5.16) by $\log \alpha$, we have

$$\begin{aligned} 0< z\left( \frac{\log 3}{\log \alpha }\right) -n_1 +\left( \frac{\log 4\sqrt{2}}{\log \alpha }\right) < \frac{10}{\alpha ^{n_1-n_2}}. \end{aligned}$$

(5.17)

We are now ready to use Lemma 3.5 with parameters

$$\begin{aligned} \gamma := \frac{\log 3}{\log \alpha },\;\; \mu := \frac{\log 4\sqrt{2}}{\log \alpha },\;\; A:= 10,\;\; B:= \alpha . \end{aligned}$$

Let $[a_0, a_1, a_2,\ldots ] = [0, 1, 1, 1, 1, 1, 8, 4, 17,\ldots ]$ be a continued fraction expansion of $\gamma$ and let $p_k/q_k$ be its kth convergent. We take $M:= 3\times 10^{45}$, then using Mathematica, we can see that

$$\begin{aligned} 6M < q_{99}. \end{aligned}$$

To apply Lemma 3.5, consider $\epsilon := \Vert \mu q_{99}\Vert - M\Vert \gamma q_{99}\Vert$ which is positive. If (2.3) has a solution $(n_1, n_2, n_3, z)$, then $(n_1 - n_2) \in [0,70]$. Next we look into inequality (5.8) to estimate the upper bound for $(n_1 - n_3)$. Now putting

$$\begin{aligned} \Lambda _2:= z\log 3 - n_1 \log \alpha + \log \phi (n_1-n_2), \end{aligned}$$

(5.18)

where we take $\phi (x)= 4\sqrt{2}(1+\alpha ^{-x})^{-1}$, it implies

$$\begin{aligned} | 1- e^{\Lambda _2} | < \frac{7}{\alpha ^{n_1-n_3}}. \end{aligned}$$

Using the Binet formula of the balancing sequence, one can show that $\Lambda _2 >0$ since

$$\begin{aligned} \frac{\alpha ^{n_1}}{4\sqrt{2}}+\frac{\alpha ^{n_2}}{4\sqrt{2}}< B_{n_1}+B_{n_2}+1\le B_{n_1}+B_{n_2}+B_{n_3}= 3^z. \end{aligned}$$

Altogether we get

$$\begin{aligned} 0< \Lambda _2 <\frac{7}{\alpha ^{n_1 - n_3}}. \end{aligned}$$

(5.19)

Replacing $\Lambda _2$ in inequality (5.19) by its formula in Eq. (5.18) and arguing as in inequality (5.17), we get

$$\begin{aligned} 0< z\left( \frac{\log 3}{\log \alpha }\right) -n_1 +\left( \frac{\log \phi (n_1-n_2)}{\log \alpha }\right) < \frac{4}{\alpha ^{n_1-n_3}}. \end{aligned}$$

(5.20)

Again we use Lemma 3.5 here with the following parameters

$$\begin{aligned} \gamma := \frac{\log 3}{\log \alpha },\;\; \mu := \frac{\log \phi (n_1-n_2)}{\log \alpha },\;\; A:= 4,\;\; B:= \alpha . \end{aligned}$$

(5.21)

Proceeding like before with $M := 3\times 10^{45}$ and applying Lemma 3.5 to the inequality (5.20) for all possible choices of $n_1-n_2 \in [0,70]$ we find that if Eq. (2.3) has a solution $(n_1, n_2, n_3, z)$, then $(n_1 - n_3) \in [0,72]$. Finally, in order to obtain a better upper bound on $n_1$, we can put

$$\begin{aligned} \Lambda _3:= z\log 3 - n_1 \log \alpha + \log \psi (n_1-n_2, n_1-n_3), \end{aligned}$$

(5.22)

with $\psi (x_1,x_2):= 4\sqrt{2}(1+ \alpha ^{-x_1}+\alpha ^{-x_2})^{-1}$, which implies

$$\begin{aligned} |1 - e^{\Lambda _3}|< \frac{3}{\alpha ^{n_1}}. \end{aligned}$$

One can observe $\Lambda _3 \ne 0$. So, for rest of the cases we consider $\Lambda _3 > 0$ and $\Lambda _3< 0$ separately. If $\Lambda _3 > 0$, then

$$\begin{aligned} 0< \Lambda _3 < \frac{3}{\alpha ^{n_1}}. \end{aligned}$$

(5.23)

Suppose $\Lambda _3< 0$. Since $\frac{3}{\alpha ^{n_1}}< \frac{1}{2}$ for $n_1> 2$, we get that $|e^{\Lambda _3}-1|< 1/2$, therefore $e^{|\Lambda _3|}< 2$. Thus,

$$\begin{aligned} 0< |\Lambda _3| \le e^{|\Lambda _3|}-1=e^{|\Lambda _3|}|e^{\Lambda _3}-1|<\frac{6}{\alpha ^{n_1}}. \end{aligned}$$

Thus for both cases we have

$$\begin{aligned} 0< |\Lambda _3| \le e^{|\Lambda _3|} - 1< \frac{6}{\alpha ^{n_1}}. \end{aligned}$$

(5.24)

Putting the value of $\Lambda _3$ from Eq. (5.22) in inequality (5.24) and arguing as previously we obtain

$$\begin{aligned} 0< \left| z\frac{\log 3}{\log \alpha } - n_1 + \frac{\log \psi (n_1-n_2, n_1-n_3)}{\log \alpha }\right| < \frac{4}{\alpha ^{n_1}}. \end{aligned}$$

(5.25)

Now, we repeat the same procedure as earlier with $M:= 3\times 10^{45}$ for inequality (5.25). For all possible choices of $n_1-n_2 \in [0,70]$ and $n_1-n_3 \in [0,72]$, we apply Lemma 3.5 to inequality (5.25). If Eq. (2.3) has a solution $(n_1, n_2, n_3, z)$, then $n_1 \in [0,75]$. This leads to a contradiction to our assumption that $n_1 > 100$, which completes the proof of Theorem 2.