1 Introduction

The Fibonacci sequence \((F_{n})\) is given by the recurrence relation

$$\begin{aligned} F_{n}=F_{n-1}+F_{n-2},\quad n\ge 2 \end{aligned}$$

with initial conditions \(F_{0}=0,F_{1}=1\). The Lucas sequence \((L_{n})\) is given by the recurrence relation

$$\begin{aligned} L_{n}=L_{n-1}+L_{n-2},\quad n\ge 2 \end{aligned}$$

with initial conditions \(L_{0}=2,\) \(L_{1}=1.\) \(F_{n}\) and \(L_{n}\) are called the n-th terms of Fibonacci and Lucas sequences, respectively. The Binet formulas for these sequences are given by

$$\begin{aligned} F_{n}=\dfrac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta }\quad \text { and }\quad L_{n}=\alpha ^{n}+\beta ^{n}, \end{aligned}$$

where \(\alpha =\dfrac{1+\sqrt{5}}{2}\) and \(\beta =\dfrac{1-\sqrt{5}}{2}.\) It is easy to see that \(L_{n}=F_{n-1}+F_{n+1}\) and \(5F_{n}=L_{n-1}+L_{n+1}.\) Many properties of these recurrence sequences are given in [6, 12].

In [2], Bravo and Luca determined all non-negative solutions (n, m, a) of the diophantine equation \(F_{n}+F_{m}=2^{a}\) with \(n\ge m.\) Then in [10], Pink and Zeigler considered a more general form \( u_{n}+u_{m}=wp_{1}^{z_{1}}\cdots p_{s}^{z_{s}}\) in non-negative integers \( n,m,z_{1},\ldots ,z_{s}\), where \(\left( u_{n}\right) _{n\ge 0}\) is a binary nondegenerate recurrence sequence, \(\ p_{1},\ldots ,p_{s}\) are the distinct primes and w is a non-zero integer with \(p_{i}\not \mid w\) for all \(1\le i\le s\). After noticing Pink and Zeigler’s more general diophantine equation for sums of terms of recurrence sequence \(( u_{n}) _{n\ge 0}\), Şiar and Keskin [11] proved that all non-negative integer solutions of the diophantine equation \(F_{n}-F_{m}=2^{a}\) are given by

$$\begin{aligned}&(n,m,a)\in \{ ( 1,0,0) ,( 2,0,0),(3,0,1),(6,0,3),(3,1,0),(4,1,1),\\&\quad (5,1,2),(3,2,0)\} \end{aligned}$$

and

$$\begin{aligned}&(n,m,a)\in \{ ( 4,3,0), (4,2,1),(5,2,2),(9,3,5),(5,4,1),(7,5,3),\\ {}&\quad (8,5,4),(8,7,3)\} . \end{aligned}$$

From this point of [11], we consider the solutions of diophantine equation of the form \(F_{n}-F_{m}=3^{a}\) in non-negative integers. Moreover, Erduvan et al. [5] showed that the solutions of the equation \(F_{n}-F_{m}=5^{a}\) is given by

$$\begin{aligned} F_{1}-F_{0}=F_{2}-F_{0}=F_{3}-F_{2}=F_{3}-F_{1}=5^{0} \end{aligned}$$

and

$$\begin{aligned} F_{5}-F_{0}=F_{4}-F_{3}=F_{6}-F_{4}=F_{7}-F_{6}=5, \end{aligned}$$

in nonnegative integers m, n and a.

Now we can give the logarithmic height definition from [8].

DEFINITION 1

Let \(\alpha \) be an algebraic number of degree d and

$$\begin{aligned} f(x)=\underset{i=0}{\overset{d}{\sum }}a_{i}x^{d-i}\in \mathbb {Z[}x\mathbb {]} \end{aligned}$$

be the minimal polynomial of \(\alpha \) with \(a_{0}>0\) and \(\gcd (a_{0},\ldots ,a_{d})=1.\) The logarithmic height of \(\alpha \) is given by

$$\begin{aligned} h(\alpha )=\frac{1}{d}\left( \log \left| a_{0}\right| +\underset{i=1}{\overset{d}{\sum }}\log \max \{ \vert \alpha ^{(i)}\vert ,1\}\right) , \end{aligned}$$

where \(\alpha ^{(i)}\)’s are the conjugates of \(\alpha \).

Some known properties of the logarithmic heights are as follows:

$$\begin{aligned} h(\alpha \pm \beta )\le & {} h(\alpha )+h(\beta )+\log 2, \end{aligned}$$
(1.1)
$$\begin{aligned} h(\alpha \beta ^{\pm 1})\le & {} h(\alpha )+h(\beta ), \end{aligned}$$
(1.2)
$$\begin{aligned} h(\alpha ^{k})= & {} \left| k\right| h(\alpha ). \end{aligned}$$
(1.3)

Now we can give the following lemmas from [7], that will be useful in the proof of Theorem 1.

Lemma 1

If \(n\equiv m\pmod {2}\), then

$$\begin{aligned} F_{n}-F_{m}=\left\{ \begin{array}{c} F_{(n-m)/2}L_{(n+m)/2}, \,\,n\equiv m\pmod {4} \\ F_{(n+m)/2}L_{(n-m)/2},\,\, n\equiv m+2\pmod {4}. \end{array} \right. \end{aligned}$$

Lemma 2

Let \(L_{n}=3^{s}\cdot y^{b}\) for some integers \(n\ge 1,y\ge 1,b\ge 2\) and \(s\ge 0.\) The solutions of this equation are given by \(n\in \left\{ 1,2,3\right\} .\)

The following lemma was an interesting problem and was completely proved by Bugeaud et al. [3]. This lemma will be used in the proof of Theorem 1.

Lemma 3

The only perfect powers in the Fibonacci sequence are \(F_{0}=0,\ F_{1}=F_{2}=1,~F_{6}=8\) and \(F_{12}=144.\)

In [1], Baker gave an effective lower bound for a non-zero expression of the form \(c_{1}\log \alpha _{1}+\cdots +c_{n}\log \alpha _{n}\), where \(\alpha _{i}\) are the algebraic numbers and \(c_{i}\) are the integers for all \( 1\le i\le n.\) We will use the reduction method of Baker–Davenport in the proof of Theorem 1.

2 Preliminaries

Before proving the main theorem, we shall state a useful inequality associated with Fibonacci sequence.

Lemma 4

Let \(n\ge 1.\) Then \(\alpha ^{n-2}\le F_{n}\le \alpha ^{n-1}.\)

Since the proof of the following lemma is given firstly in [9] and then in [3], we will omit its proof.

Lemma 5

[3, 9]. Let \(\mathbb {L}\) be a number field of degree D and \(\alpha _{1},\alpha _{2},\ldots ,\alpha _{n}\) be non-zero elements of \(\mathbb {L}\), and let \(b_{1},b_{2},\ldots ,b_{n}\) be rational integers such that

$$\begin{aligned} \Lambda =a_{1}^{b_{1}}\cdots a_{n}^{b_{n}}-1 \end{aligned}$$

and

$$\begin{aligned} B=\max \left\{ \left| b_{1}\right| ,\ldots ,\left| b_{n}\right| \right\} \text {.} \end{aligned}$$

Let h denote the absolute logarithmic height and \(A_{1},\ldots ,A_{n}\) be real numbers with

$$\begin{aligned} A_{j}\ge \max \{ Dh(\alpha _{j}),\vert \log \alpha _{j}\vert ,0.16\} \quad \text { for all }1\le j\le n. \end{aligned}$$

If \(\Lambda \ne 0,\) then

$$\begin{aligned} \log \vert \Lambda \vert >-3\cdot 30^{n+4}(n+1)^{5.5}D^{2}(1+\log D)(1+\log nB)A_{1}\cdots A_{n}. \end{aligned}$$

Furthermore, if \(\mathbb {L}\) is real, then

$$\begin{aligned} \log \vert \Lambda \vert >-1.4\cdot 30^{n+3}n^{4.5}D^{2}(1+\log D)(1+\log B)A_{1}\cdots A_{n}. \end{aligned}$$

We give the following lemma from [4].

Lemma 6

Let M be a positive integer and p / q be a convergent of the continued fraction of the irrational number \(\gamma \) such that \( q>6M\). Let \(A,B,\mu \) be some real numbers with \(A>0\) and \(B>1.\) Let \( \varepsilon :=\Vert \mu q\Vert -M\Vert \gamma q\Vert \), where \(\left\| \cdot \right\| \) denotes the distance from the nearest integer. If \(\varepsilon >0,\) then there exist no solutions of the inequality

$$\begin{aligned} 0<\left| u\gamma -v+\mu \right| <AB^{-w}, \end{aligned}$$

in positive integers u, v and w with

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

3 Main theorem

Theorem 1

Let n, m and a be non-negative integers with \(n>m.\) Then all solutions of the equation

$$\begin{aligned} F_{n}-F_{m}=3^{a} \end{aligned}$$
(3.1)

are given by

$$\begin{aligned}&( n,m,a) \in \{ ( 1,0,0) ,(2,0,0),(4,0,1),(3,1,0),(3,2,0),(4,3,0),(5,3,1),\\ {}&\quad (6,5,1),(11,6,4)\} . \end{aligned}$$

Proof

If \(m=0,\) then from Lemma 3, we get \((n,m,a) =\) (1, 0, 0) , (2, 0, 0), (4, 0, 1). Let \(1\le m<n\le 100.\) Then the solutions of (3.1) are \(( n,m,a) \in \{(3,1,0), (3,2,0),(4,3,0)\),\((5,3,1),(6,5,1),(11,6,4)\}.\) Thus, from now on, we will assume that \(n>100.\) If \(n-m=1\), then we get \( 3^{a}=F_{n}-F_{m}=F_{m-1}.\) This implies that \(m-1=1,2,4\), by Lemma 3. So \(m=2,3,5.\) If \(n-m=2,\) then \(3^{a}=F_{n}-F_{m}=F_{m+1}.\) This implies that \(m+1=2,4\), by Lemma 3. So \(m=1,3.\) But these solutions are given in the theorem for \(1\le m<n\le 100.\) Therefore, we may suppose \(n-m\ge 3.\) Since \(3^{a}=F_{n}-F_{m}<F_{n}\le \alpha ^{n-1}<3^{n-1}\) by Lemma 4, we get \(a<n.\)

Now recalling \(F_{n}-F_{m}=3^{a},\) we get

$$\begin{aligned} \frac{\alpha ^{n}}{\sqrt{5}}-3^{a}=\frac{\beta ^{n}}{\sqrt{5}}+F_{m}. \end{aligned}$$
(3.2)

Taking absolute value of (3.2), we have

$$\begin{aligned} \left| \frac{\alpha ^{n}}{\sqrt{5}}-3^{a}\right| \le \frac{ \vert \beta \vert ^{n}}{\sqrt{5}}+F_{m}<\frac{\vert \beta \vert ^{n}}{\sqrt{5}}+\alpha ^{m}<\frac{1}{2}+\alpha ^{m}, \end{aligned}$$
(3.3)

where we take into account \(\vert \beta \vert \,\widetilde{=}\, 0.6<1\) and \(\dfrac{\vert \beta \vert ^{n}}{\sqrt{5}}<\dfrac{1}{2} . \) Dividing both sides of (3.3) by \(\dfrac{\alpha ^{n}}{\sqrt{5}}\), we obtain

$$\begin{aligned} \vert 1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\vert <\sqrt{5}\cdot \alpha ^{m-n}\cdot \left( \frac{1}{2}\alpha ^{-m}+1\right) , \end{aligned}$$

which implies that

$$\begin{aligned} \vert 1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\vert<\sqrt{5}\cdot \alpha ^{m-n}\cdot \left( \frac{1}{2}+1\right) =\frac{3}{2}\sqrt{5}\cdot \alpha ^{m-n}< \frac{4}{\alpha ^{n-m}}. \end{aligned}$$
(3.4)

Take the parameters \(t:=3\), \(\gamma _{1}=3,\gamma _{2}=\alpha ,\gamma _{3}= \sqrt{5},b_{1}=a,b_{2}=-n\) and \(b_{3}=1\) in Lemma 5. We also notice that \(D=2\) and \(\gamma _{1},\gamma _{2},\gamma _{3}\) are positive real numbers and belong to \( \mathbb {Q} (\sqrt{5}).\)

Now it is necessary to show that \(\Lambda =1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\) is non-zero. Assume that \(\Lambda =1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}=0\). Then we get \(\alpha ^{2n}=5\cdot 3^{a},\) which is impossible since \(\alpha ^{2n}\notin \mathbb {Z} .\)

Thus we have \(h(\gamma _{1})=\log 3,h(\gamma _{2})=h(\alpha )=\dfrac{\log \alpha }{2}\) and \(h(\gamma _{3})=h(\sqrt{5})=\log \sqrt{5}.\) Therefore, we can choose

$$\begin{aligned} A_{1}:=2.2\ge \max \{ 2\log 3,\vert \log \gamma _{1}\vert ,0.16\} =2.1972\text {, }A_{2}:=0.5\text { , }A_{3}:=1.7. \end{aligned}$$

Now by considering \(a<n,\) we can take

$$\begin{aligned} B:=\max \{ \vert a\vert ,\vert -n\vert ,1\} =n. \end{aligned}$$

According to Lemma 5, we get

$$\begin{aligned}&\vert 1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\vert >\exp \{ -1.4\times 30^{6}\times 3^{4.5}\times 4\times (1+\log 2)(1+\log n)\nonumber \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \times 2.2\times 0.5\times 1.7\} . \end{aligned}$$
(3.5)

If we take logarithms in equality (3.4) and combine the calculation of the right-hand side of (3.5), we get

$$\begin{aligned} (n-m)\log \alpha <18.139\times 10^{11}(1+\log n)+\log 4. \end{aligned}$$
(3.6)

Using the fact that \(1+\log n<2\log n\) for all \(n\ge 3,\) we have

$$\begin{aligned} (n-m)\log \alpha <3.63\times 10^{12}\log n. \end{aligned}$$
(3.7)

Moreover, we can obtain a second linear form by using equation (3.1) as follows:

$$\begin{aligned} \left| \frac{\alpha ^{n}}{\sqrt{5}}(1-\alpha ^{m-n})-3^{a}\right| =\left| \frac{\beta ^{n}}{\sqrt{5}}-\frac{\beta ^{m}}{\sqrt{5}} \right| \le \frac{\left| \beta \right| ^{n}+\left| \beta \right| ^{m}}{\sqrt{5}}<0.445. \end{aligned}$$
(3.8)

Now dividing both sides of (3.8) by \(\dfrac{\alpha ^{n}}{\sqrt{5}} (1-\alpha ^{m-n})\), we get

$$\begin{aligned} \vert 1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\cdot (1-\alpha ^{m-n})^{-1}\vert \le \dfrac{0.445\times \sqrt{5}}{\alpha ^{n}(1-\alpha ^{m-n})}. \end{aligned}$$

Since \(\alpha ^{m-n}=\dfrac{1}{\alpha ^{n-m}}<\dfrac{1}{\alpha }<0.67\), it is obvious that \(\dfrac{1}{1-\alpha ^{m-n}}<2.71.\) Thus it follows that

$$\begin{aligned} \vert 1-3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\cdot (1-\alpha ^{m-n})^{-1}\vert <\dfrac{2.71}{\alpha ^{n}}. \end{aligned}$$
(3.9)

Let \(\Lambda =3^{a}\cdot \alpha ^{-n}\cdot \sqrt{5}\cdot (1-\alpha ^{m-n})^{-1}-1.\) Now, in order to apply Lemma 5 again, we take parameters \(t:=3\) and

$$\begin{aligned} \gamma _{1}=3,\text { }\gamma _{2}=\alpha ,\text { }\gamma _{3}=\sqrt{5} \cdot (1-\alpha ^{m-n})^{-1},\text { }b_{1}=a\text {, }b_{2}=-n,\text { }b_{3}=1. \end{aligned}$$

As before, we have \(D:=[ \mathbb {Q} (\sqrt{5}): \mathbb {Q} ]=2.\) It is obvious that \(\Lambda \ne 0.\) Because, if \(\Lambda =0,\) we get \(\alpha ^{n}-\alpha ^{m}=3^{a}\sqrt{5}.\) Taking the conjugate of this equation, we have \(\beta ^{n}-\beta ^{m}=-3^{a}\sqrt{5}.\) Addition of these two conjugate equations gives \(L_{n}-L_{m}=0,\) which contradicts the fact that \(n>m.\)

Hence the left-hand side of (3.9) is non-zero. As before, \(A_{1}=2.2,\) \(A_{2}=0.5\) and \(B=n\) in Lemma 5.

Now we calculate \(A_{3}.\) For all \(n-m>3,\) we have \(\vert \log \gamma _{3}\vert <1.\) Taking the identities (1.1), (1.2) and (1.3) into account, we get

$$\begin{aligned} h(\gamma _{3})= & {} h(\sqrt{5}\cdot (1-\alpha ^{m-n})^{-1}) \\\le & {} h(\sqrt{5})+h(1)+h(\alpha ^{m-n})+\log 2 \\\le & {} \log \sqrt{5}+\left| m-n\right| h(\alpha )+\log 2 \\= & {} \log \sqrt{5}+(n-m)\dfrac{\log \alpha }{2}+\log 2 \\= & {} (n-m)\dfrac{\log \alpha }{2}+\log 2\sqrt{5}. \end{aligned}$$

Thus, it follows that

$$\begin{aligned} A_{3}\ge \max \{ 2h(\gamma _{3}),\vert \log \gamma _{3}\vert ,0.16\} \ge \log 20+(n-m)\log \alpha . \end{aligned}$$

Hence, we can take \(A_{3}=\log 20+(n-m)\log \alpha .\)

By Lemma 5, we obtain a lower bound for the left-hand side of (3.9) as

$$\begin{aligned} \dfrac{3.03}{\alpha ^{n}}> & {} \vert \Lambda \vert >\exp \{ -1.4\times 30^{6}\times 3^{4.5}\times 4\times (1+\log 2)(1+\log n)\\&\times 2.2\times 0.5\times [ \log 20+(n-m)\log \alpha ] \} . \end{aligned}$$

Taking logarithms on both sides of this inequality and considering the fact that \(1+\log n<2\log n\) for all \(n>1,\) we get

$$\begin{aligned} n\log \alpha <2.134\times 10^{12}\log n\times \left[ \log 20+(n-m)\log \alpha \right] . \end{aligned}$$
(3.10)

By a fast calculation with Mathematica, we obtain \(n<7.09616\times 10^{28}.\)

If (n, m, a) is a positive integer solution of equation (3.1) with \(n>m\), then we have \(a<n<7.09616\times 10^{28}.\) We have obtained an upper bound for n and now we will reduce this bound to a size that can be easily dealt with. For doing this, we will use Lemma 4 again.

Let \(z_{1}:=a\log 3-n\log \alpha +\log \sqrt{5}.\) By considering equation (3.4), we get

$$\begin{aligned} \left| 1-\mathrm{e}^{z_{1}}\right| <\dfrac{4}{\alpha ^{n-m}}. \end{aligned}$$

Thus, by using equation (3.1) and the Binet formula, we obtain

$$\begin{aligned} \frac{\alpha ^{n}}{\sqrt{5}}=F_{n}+\frac{\beta ^{n}}{\sqrt{5}}>F_{n}-1>F_{n}-F_{m}=3^{a}. \end{aligned}$$

Hence \(z_{1}=\log (3^{a}\sqrt{5}/\alpha ^{n})<0.\) It is obvious that \(\dfrac{ 4}{\alpha ^{n-m}}<0.945\) for all \(n-m\ge 3.\) Therefore we get \( \mathrm{e}^{\vert z_{1}\vert }<18.2\) and therefore, it follows that

$$\begin{aligned} 0<\left| z_{1}\right|<\mathrm{e}^{\vert z_{1}\vert }-1\le \mathrm{e}^{\vert z_{1}\vert }\vert 1-\mathrm{e}^{z_{1}}\vert <\frac{73}{ \alpha ^{n-m}}. \end{aligned}$$

Thus it can be seen that

$$\begin{aligned} 0<\vert a\log 3-n\log \alpha +\log \sqrt{5}\vert <\frac{73}{ \alpha ^{n-m}}. \end{aligned}$$
(3.11)

Dividing both sides of the inequality (3.11) by \(\log \alpha \), we get

$$\begin{aligned} 0<\left| a\frac{\log 3}{\log \alpha }-n+\frac{\log \sqrt{5}}{\log \alpha }\right| <\frac{73}{\log \alpha }\cdot \alpha ^{-(n-m)}. \end{aligned}$$
(3.12)

Now considering Lemma 6, we have the irrational \(\gamma =\dfrac{\log 3}{ \log \alpha }\) with

$$\begin{aligned} \mu =\frac{\log \sqrt{5}}{\log \alpha },\ A=\frac{73}{\log \alpha },\ B=\alpha ,\ w=n-m. \end{aligned}$$

Also, we know that \(a<n<7.09616\times 10^{28}.\) So it follows that \( M:=7.09616\times 10^{28}\), according to Lemma 6 and \(q>6M\) is the denominator of a convergent of the continued fraction of \(\gamma \) such that \(\varepsilon =\left\| \mu q\right\| -M\left\| \gamma q\right\| >0.\) Considering the denominator of the 61-st convergence of \(\dfrac{\log 3}{\log \alpha },\) we have \(q=10.52\times 10^{29}.\) By some calculations with Mathematica, we obtain \(\varepsilon =0.154453.\)

According to Lemma 6, we know that there is no solution of the inequality (3.12) for the values \(n-m\) with \(n-m\ge \dfrac{\log (Aq/\varepsilon )}{\log B}.\) Therefore, it follows that inequality (3.12) has no solutions for \(n-m\ge 157.972.\) This means that a bound for \( n-m \) is \(n-m\le 157.\) Considering this fact in inequality (3.10), we get \(n<1.29184\times 10^{16}.\)

Let us work on (3.9) for finding an upper bound on n. Now take

$$\begin{aligned} z_{2}:=a\log 3-n\log \alpha +\log (\sqrt{5}(1-\alpha ^{m-n})^{-1}). \end{aligned}$$

Thus (3.9) implies that

$$\begin{aligned} \vert 1-\mathrm{e}^{z_{2}}\vert <\dfrac{2.71}{\alpha ^{n}}. \end{aligned}$$

It is obvious that \(\dfrac{2.71}{\alpha ^{n}}<\dfrac{1}{2}.\) If \(z_{2}>0\), then \(0<z_{2}<\mathrm{e}^{z_{2}}-1<\dfrac{2.71}{\alpha ^{n}}.\) If \(z_{2}<0,\) then \( \vert 1-\mathrm{e}^{z_{2}}\vert =1-\mathrm{e}^{z_{2}}<\dfrac{2.71}{\alpha ^{n}}< \dfrac{1}{2}.\) Thus, we get \(1-\dfrac{1}{2}<\mathrm{e}^{z_{2}},\) so that \( \mathrm{e}^{\vert z_{2}\vert }<2.\) Therefore, we have

$$\begin{aligned} 0<\vert z_{2}\vert<\mathrm{e}^{\vert z_{2}\vert }-1\le \mathrm{e}^{\vert z_{2}\vert }\cdot \vert 1-\mathrm{e}^{z_{2}}\vert <2\times \dfrac{2.71}{\alpha ^{n}}. \end{aligned}$$

Thus it follows that

$$\begin{aligned} 0<\left| a\dfrac{\log 3}{\log \alpha }-n+\dfrac{\log \sqrt{5}\cdot (1-\alpha ^{m-n})^{-1}}{\log \alpha }\right| <\dfrac{5.42}{\log \alpha } \cdot \alpha ^{-n}. \end{aligned}$$
(3.13)

Now considering Lemma 6, we obtain

$$\begin{aligned} \gamma =\dfrac{\log 3}{\log \alpha },\ \mu =\dfrac{\log \sqrt{5}\cdot (1-\alpha ^{m-n})^{-1}}{\log \alpha },\ A=\dfrac{5.42}{\log \alpha },\ B=\alpha ,\ w=n. \end{aligned}$$

It is obvious that \(\gamma \) is irrational. Also, \(3\le n-m\le 157.\) Firstly, calculate the denominator q of continued fraction of \( \gamma .\) Since \(M=1.29184\times 10^{16}\), we must choose the 39-th denominator \(q=48,9\times 10^{16}\) such that \(q>6M=7.75107\times 10^{16}.\) Hence by applying Lemma 6 to (3.13) with \(3\le n-m\le 157\) except for \(n-m=4\) or 8,  we obtain

$$\begin{aligned} \varepsilon =\Vert \mu q_{40}\Vert -M\Vert \gamma q_{40}\Vert \ge 0.492868 \end{aligned}$$

by a fast computation with Mathematica. Furthermore, according to Lemma , we know that there is no solution of the inequality (3.13) for values n with \(n\ge \dfrac{\log (Aq/\varepsilon )}{\log B}=91.1453\). Thus, an upper bound for n must be \(n\le 91\). This contradicts our assumption that \(n>100.\)

Finally, we consider the cases \(n-m=4\) or 8. According to Lemma 1, when \(n\equiv m\pmod {4}\), we have \(F_{n}-F_{m}=F_{(n-m)/2}L_{(n+m)/2}\). Thus, it follows that \(F_{n}-F_{m}=F_{2}L_{m+2}=L_{m+2}\) for \(n-m=4\). This gives \(L_{m+2}=3^{a}\). According to Lemma 2, the possible values for \(m+2\) are 1, 2, 3. Since \(4\not \mid 3^{a}\) and m is a non-negative integer, we have that \(m+2\ne 3\) and \(m+2\ne 1.\) If \(m+2=2\), then we get \(L_{2}=3^{a}\). Thus, it follows that \((n,m,a)=(4,0,1)\) is a solution to (3.1). Moreover, the case \(n-m=8\) gives \(F_{n}-F_{m}=F_{4}L_{m+4}\) by Lemma 1 . Hence we get \(L_{m+4}=3^{a-1}\), which implies that \(m+4=1,2,3\) by Lemma 2. This is impossible since m is a non-negative integer. \(\square \)

4 Conclusion

In [7], it is shown that if \(n\equiv m\pmod {2}\), then all the solutions of the equation

$$\begin{aligned} F_{n}-F_{m}=y^{p},p\ge 2,y\ge 1 \end{aligned}$$
(4.1)

satisfy max\(\left\{ n,m\right\} \le 36\). Then the authors conjectured that all the solutions of equation (4.1) are

$$\begin{aligned}&F_{1}-F_{0}=1,\ F_{2}-F_{0}=1,\ F_{3}-F_{1}=1,\ F_{3}-F_{2}=1,\ F_{4}-F_{3}=1,\\&F_{5}-F_{1}=2^{2},\ F_{5}-F_{2}=2^{2},\ F_{6}-F_{4}=5,\ F_{7}-F_{5}=2^{3},\\&F_{7}-F_{6}=5,\ F_{8}-F_{5}=2^{4}, F_{8}-F_{7}=2^{3},\ F_{9}-F_{3}=2^{5},\\&F_{11}-F_{6}=9^{2},\ F_{13}-F_{6}=15^{2},\ F_{13}-F_{11}=12^{2},\ F_{14}-F_{9}=7^{3},\\&F_{14}-F_{13}=12^{2},\ F_{15}-F_{9}=24^{2}. \end{aligned}$$

Consequently, it is true that the above conjecture is valid for \(y=2,3\) by our result and the results in [5, 11]. It is reasonable to conjecture that if \(F_{n}-F_{m}=p^{a}\) for some prime p and positive integer a,  then \(p=2,3,5,7.\)