1 Introduction

Let \((F_n)_{n\ge 0}\) be the Fibonacci sequence given by \(F_{n+2}=F_{n+1}+F_n\), for \(n\ge 0\), where \(F_0=0\) and \(F_1=1\). These numbers are well known for possessing wonderful and amazing properties (consult [7] together with their very extensive annotated bibliography for additional references and history).

Diophantine equations related to sums of powers of two terms of a given linear recurrence sequence were studied by several authors. For instance, motivated by the naive identity

$$\begin{aligned} F_n^2+F_{n+1}^2=F_{2n+1}, \end{aligned}$$
(1.1)

which tells us that the sum of the square of two consecutive Fibonacci numbers is still a Fibonacci number, Marques and Togbé [12] questioned what about such sums with higher powers, and showed that, if \(x \ge 1\) is an integer such that \(F_{n}^x + F_{n+1}^x\) is a Fibonacci number for all sufficiently large n, then \(x \in \{1, 2 \}.\) In 2011, Luca and Oyono [10] solved this problem completely by showing that the Diophantine equation

$$\begin{aligned} F_m^s+F_{m+1}^s=F_n \end{aligned}$$
(1.2)

has no solutions (m, n, s) with \(m\ge 2\) and \(s \ge 3\).

Subsequently, their result have been extended by considering k-generalized Fibonacci numbers [4, 15]. Recently, Bednařík et al. [1] has proved that the Diophantine equation \(( F_{n}^{(k)})^2 + ( F_{n+1}^{(k)})^2 = F_{m}^{(l)}\) has no solution in positive integers n, m, k, l with \(2 \le k < l\) and \(n > 1\). Hirata-Kohno and Luca [6] was revisited (1.2) and found that the solutions of the Diophantine equation \(F_{n}^x + F_{n+1}^x = F_{m}^y\), in positive integers (m, n, x, y), are \((3,1,x,1),(n+2,n,1,1),(2n+1,n,1,1),(3,4,1,3),(4,2,3,2)\). Subsequently, Luca and Oyono [11] reversed the role of two exponents of the previous equation and studied whether \(F_{n}^x + F_{n+1}^y = F_{m}^x\) or \(F_{n}^y + F_{n+1}^x = F_{m}^x\). They proved that the only positive integer solution (m, n, x, y) of one of the mentioned equations with \(n \ge 3\) and \(x \ne y\), is (5, 3, 2, 4), for which \(F_4^3 + F_2^4 = F_2^5\). Miyazaki [14] showed that the only positive integer solutions (x, y, z, n) of the equation \(F_{n}^x + F_{n+1}^{y} = F_{2n+1}^z\) are for \((x, y, z) = (2, 2, 1)\) (and for all positive integers n).

Among the several pretty algebraic identities involving Fibonacci numbers, we are interested in the following one:

$$\begin{aligned} F_{2n} = F_{n+1}^{2} - F_{n-1}^{2},\quad \text{ for } \text{ all }\ n\ge 0. \end{aligned}$$
(1.3)

In particular, the above identity tell us that, for two Fibonacci numbers whose positions in the sequence differ by two, the difference of their squares is still a Fibonacci number. So, in the same spirit of the previous works done on Eq. (1.1), one could ask: Does this property still holds for \(F_{n+1}^{3} - F_{n-1}^{3}\)? And for \(F_{n+1}^{4} - F_{n-1}^{4}\)? And so on?

The aim of this paper is to answer such questions, i.e. to know when \(F_{n+1}^{x} - F_{n-1}^{x}\) is a Fibonacci number. More precisely, our main result is the following.

Theorem 1.1

The only non-negative integer solutions (m, n, x) of the Diophantine equation

$$\begin{aligned} F_{n+1}^{x} - F_{n-1}^{x} = F_{m}, \end{aligned}$$
(1.4)

are (2n, n, 2), (1, 1, x), (2, 1, x), (0, n, 0).

Our method follows roughly the following steps: First, we use Matveev’s result [13] on linear forms in logarithms to obtain an upper bound for x in terms of m and n. When n is small, say \(n \le 112\), we use Dujella and Pethö’s result [5] to decrease the range of possible values and then let the computer check the non-existence of solutions in this case. To the case where \(n \ge 113\), we use again linear forms in logarithms to obtain an absolute upper bound for x. In the final step, we use continued fractions to lower the bounds and then let the computer cover the range of possible values, showing that there are no solutions also in this case, which completes the proof.

2 Auxiliary Results

In this section, the main results used to “attack” our problem are presented. One basic fact about Fibonacci numbers is their closed-form, given by Binet’s formula, which states that

$$\begin{aligned} F_{n} = \frac{\alpha ^{n} - \beta ^n}{\sqrt{5}}, \quad \text{ for } \text{ all } n \ge 0, \end{aligned}$$
(2.1)

where \(\alpha := (1 + \sqrt{5})/2\), is the Golden Ratio, and \(\beta = - \alpha ^{-1}\), are the roots of \(x^2 - x - 1.\) This implies on the following well-known inequalities:

$$\begin{aligned} \alpha ^{n-2} \le F_n \le \alpha ^{n-1}, \end{aligned}$$
(2.2)

which holds for all \(n \ge 1\). Also, it is easy to show that

$$\begin{aligned} \frac{F_{n-1}}{F_{n+1}} \le \frac{2}{5},\quad \text {holds for any}~~n \ge 3. \end{aligned}$$
(2.3)

2.1 Lower Bounds for Linear Forms in Logarithms of Algebraic Numbers

     To prove Theorem 1.1, we use a few times a Baker-type lower bound for a non-zero linear forms in logarithms of algebraic numbers. We state a result of Matveev [13] about the general lower bound for linear forms in logarithms, but first, recall some basic notations from algebraic number theory.

Let \(\eta \) be an algebraic number of degree d with minimal primitive polynomial

$$\begin{aligned} f(X):= a_0 X^d+a_1 X^{d-1}+ \cdots +a_d = a_0 \prod _{i=1}^{d}(X- \eta ^{(i)}) \in \mathbb {Z}[X], \end{aligned}$$

where the \(a_i\)s are relatively prime integers, \(a_0 >0\), and the \(\eta ^{(i)}\)s are conjugates of \(\eta \). Then

$$\begin{aligned} h(\eta )=\frac{1}{d}\left( \log a_0+\sum _{i=1}^{d}\log \left( \max \{|\eta ^{(i)}|,1\}\right) \right) \end{aligned}$$
(2.4)

is called the logarithmic height of \(\eta \). Some properties of the logarithmic height, which will be used in the next section, are the following:

$$\begin{aligned} h(\eta \pm \gamma )&~\le ~ h(\eta ) + h(\gamma )+\log 2.\\ h(\eta \gamma ^{\pm 1})&~ \le ~ h(\eta )+h(\gamma ).\\ h(\eta ^{s})&~=~|s|h(\eta )~~~(s \in \mathbb {Z}). \end{aligned}$$

With the established notations, Matveev (see [13] or [2, Theorem 9.4]) proved the ensuing result.

Theorem 2.1

Assume that \(\gamma _1, \ldots , \gamma _t\) are positive real algebraic numbers in a real algebraic number field \(\mathbb {K}\) of degree D, \(b_1, \ldots , b_t\) are rational integers, and

$$\begin{aligned} \Lambda :=\gamma _1^{b_1}\cdots \gamma _t^{b_t}-1, \end{aligned}$$

is not zero. Then

$$\begin{aligned} |\Lambda | \ge \exp \left( -1.4\cdot 30^{t+3}\cdot t^{4.5}\cdot D^2(1+\log D)(1+\log B)A_1\cdots A_t\right) , \end{aligned}$$

where

$$\begin{aligned} B\ge \max \{|b_1|,\ldots ,|b_t|\}, \end{aligned}$$

and

$$\begin{aligned} A_i\ge \max \{Dh(\gamma _i),|\log \gamma _i|, 0.16\}, ~ \text {for all} ~ i=1,\ldots ,t. \end{aligned}$$

2.2 Reduction Algorithm

Another result which will play an important role in our proof is due to Dujella and Pethö [5, Lemma 5 (a)]. It will be used to reduce the upper bounds of the variables on Eq. (1.4).

Lemma 2.2

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

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

in positive integers u and v with

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

The following result is known as Legendre’s criterion of a rational r/s to be a convergent of \(\alpha \).

Lemma 2.3

[16, Lemma 5C] If \(\alpha \) is irrational and r/s is a rational number with \(s>0\) such that

$$\begin{aligned} \Bigl |\alpha - \frac{r}{s} \Bigr |< \frac{1}{2s^2}, \end{aligned}$$

then r/s is a convergent to \(\alpha \).

3 The Proof of Theorem 1.1

Since the case \(nx=0\) is trivial, we assume \(n, x \ge 1\). Observe that, for \(x=1\), the Diophantine equation (1.4) becomes \(F_n = F_{n+1} - F_{n-1} =F_m \), which gives \(m=n\), and for \(x=2\), \(F_{2n} = F_{n+1}^2 - F_{n-1}^2=F_m\), gives \(m=2n\). When \(n=1\), we have that \(F_{2}^x - F_{0}^x = 1= F_1 = F_2\) holds for all \(x \ge 1\). For \(n=2,\) we get \(F_{m} + 1= 2^x\), which has only the solutions \((m , x) = (0, 0)\) and (4, 2), as it was showed in [3, Theorem 2].

So, assume that \(n \ge 3\). Since \(x \ge 3,\) we get that \(F_{m} \ge F_{4}^3 - F_{2}^3=26\), so \(m > 8\). Using (2.1) for \(F_m\), we rewrite Eq. (1.4) as

$$\begin{aligned} \frac{\alpha ^m}{\sqrt{5}} - F_{n+1}^x = -F_{n-1}^x + \frac{\beta ^m}{\sqrt{5}} \in [-F_{n}^{x}, -F_{n}^{x}+1]. \end{aligned}$$
(3.1)

Dividing by \(F_{n+1}^x\) on both sides of (3.1) and taking absolute values, we get

$$\begin{aligned} \left| \alpha ^{m} 5^{-1/2} F_{n+1}^{-x} -1\right|< 2 \left( \frac{F_{n-1}}{F_{n+1}} \right) ^x < \frac{2}{2.5^x}. \end{aligned}$$
(3.2)

To apply Theorem 2.1, we take

$$\begin{aligned} \gamma _1:= & {} \alpha ,~~ \gamma _2 := \sqrt{5},~~ \gamma _3 := F_{n+1},\\ b_1:= & {} m,~~ b_2 := -1, ~~ b_3 := -x, \end{aligned}$$

and also \(\Lambda _1 := \alpha ^{m} 5^{-1/2} F_{n+1}^{-x} -1\). Note that \(\mathbb {Q}(\sqrt{5})\) is the algebraic number field containing \(\gamma _1, \gamma _2\) and \(\gamma _{3}\), so \(D := 2\). If \(\alpha ^m = \sqrt{5} F_{n+1}^x\), then \(\alpha ^{2m} \in \mathbb {Z}\) which is false for all positive integers m, therefore \(\Lambda _1 \ne 0\).

Since \(h(\gamma _{1}) = \log \alpha /2\), we can choose \(A_{1} := 0.5 > \log \alpha \). Furthermore, since \(h(\gamma _{2}) = \log 5/2\), and \(h(\gamma _{3}) = \log F_{n+1}\), it follows that we can take \(A_{2} := 1.61 > \log 5\), and \(A_{3} := 2n \log \alpha > 2 \log F_{n+1}\). Lastly, using (2.2), the following chains of inequalities holds

$$\begin{aligned} \alpha ^{(n-2)x+2}< & {} \alpha ^{(n-2)x}\cdot \underbrace{(\alpha ^x-1)}_{> \alpha ^3-1> \alpha ^2}< F_{n+1}^x-F_{n-1}^x = F_m < \alpha ^{m-1},\\ \alpha ^{nx}> & {} F_{n+1}^x> F_{n+1}^x - F_{n-1}^x = F_m > \alpha ^{m-2}. \end{aligned}$$

Thus, since \(n \ge 3\) and \(x>2\),

$$\begin{aligned} x< (n-2)x+3< m< nx+2 < (n+1)x, \end{aligned}$$
(3.3)

so we can take \(B := m\).

From Matveev’s theorem, we have a lower bound for \(|\Lambda _1|\), which combined with (3.2) gives

$$\begin{aligned} \exp \left( -1.4 \cdot 30^6 \cdot 3^{4.5} \cdot 2^2 \cdot (1+\log 2) \cdot (1+\log m) \cdot 0.5 \cdot 1.61 \cdot 2n \log \alpha \right) > \frac{2}{2.5^x} ~ . \end{aligned}$$

Hence, using (3.2), (3.3) and that \(1 + \log m < 1.5 \log m\), is true for all \(m \ge 8\), we get

$$\begin{aligned} x < \frac{\log 2}{\log 2.5} + 1.23 \cdot 10^{12} \cdot n\log m , \end{aligned}$$

therefore,

$$\begin{aligned} x < 1.24 \cdot 10^{12} n \log ((n+1)x). \end{aligned}$$
(3.4)

3.1 The Case \(n \in [3,112]\)

In this case,

$$\begin{aligned} x < 1.24 \cdot 10^{12} n \log ((n+1)x) \le 1.4 \cdot 10^{14} \log (113x), \end{aligned}$$

providing \(x < 5.8\cdot 10^{15}\).

We set

$$\begin{aligned} \Gamma _{1} := - x\log F_{n+1} + m\log \alpha - \log \sqrt{5}. \end{aligned}$$

Thus, \(\Lambda _1 = e^{\Gamma _1}-1\). Recall that, from (3.1), we have \(\Lambda _1 < 0\), which implies \(\Gamma _1 < 0\). Now, since \(|\Lambda _1| < 2/(2.5)^x \le 0.128\), for \(x \ge 3\), it follows that \(e^{|\Gamma _1|} < 1.15\). Hence, we get

$$\begin{aligned} 0< |\Gamma _1|< e^{|\Gamma _1|}-1 \le e^{|\Gamma _1|}|e^{\Gamma _1}-1| < \frac{2.3}{2.5^x}. \end{aligned}$$

Dividing the last inequality by \(\log \alpha \), and using from (3.3) that \(m < (n+1)x\), we have the following

$$\begin{aligned} 0< x\left( \frac{\log F_{n+1}}{\log \alpha }\right) - m + \left( \frac{\log \sqrt{5}}{\log \alpha }\right) < 4.78 \cdot (2.5^{\frac{1}{113}})^{-m}. \end{aligned}$$
(3.5)

To use the Reduction Method, take

$$\begin{aligned} \gamma _n := \frac{\log F_{n+1}}{\log \alpha },~~\mu := \frac{\log \sqrt{5}}{\log \alpha },~~A:= 4.78,~~B:= 2.5^{\frac{1}{113}}. \end{aligned}$$

We claim that \(\gamma _n\) is irrational, for every \(n \in [3,112]\). Indeed, if \(\gamma _n = p/q\) for some positive integers p and q, we would have \(\alpha ^q=F_{n+1}^p \in \mathbb {Z}\), which is an absurd. Let \(q_{(146,n)}\) be the denominator of the 146th convergent of the continued fraction of \(\gamma _{n}\).

Taking \(M:= 5.8 \cdot 10^{15}\), we use Mathematica to get

$$\begin{aligned} \min _{3 \le n \le 112} q_{(146,n)}> 10^{22} > 6M\quad \text{ and }\quad \max _{3 \le n \le 112} q_{(146,n)} < 1.33 \cdot 10^{84}. \ \end{aligned}$$

Also, for \(\epsilon _n:=||\mu \cdot q_{(146,n)}||-5.8\cdot 10^{15}||\gamma _n \cdot q_{(146,n)}||\), we obtain that

$$\begin{aligned} \min _{3 \le n \le 112} \epsilon _{n} > 0.004 , \end{aligned}$$

which means that \(\epsilon _{n}\) is always positive (this is not true if we consider the denominator of the 145th convergent). Notice that the conditions to apply Lemma 2.2 are fulfilled, and hence there is no solution to inequality (3.5) (and consequently no solution do the Diophantine equation (1.2)) for x and m satisfying

$$\begin{aligned} \frac{\log (A\cdot q_{(146,n)}/\epsilon _{n})}{\log (B)} \le m \quad \text{ and }\quad x \le M , \end{aligned}$$

for all \(n \in [3,112]\). Then, the solutions, in this case, must be when

$$\begin{aligned} m< & {} \frac{\log (A\cdot q_{(146,n)}/\epsilon _{n})}{\log (B)} \\< & {} \frac{\log (4.78\cdot 1.33 \cdot 10^{84}/0.004)}{2.5^{\frac{1}{113}}} \\< & {} 24761. \end{aligned}$$

Therefore, \(m \le 24760\), and so \(x< (m-3)/(n-2) < 24757/(n-2)\). Now, we prepare a simple routine in Mathematica in the range \(n \in [3,112]\), \(m \in [8,24761]\) and \(x \in [3,24757/(n-2)]\), which returns no solutions for (1.2). This completes the case \(n \in [3,112]\).

3.2 An Upper Bound for x in Terms of n

From now on, assume \(n \ge 113\). Our short term goal is to find an explicit lower bound for x in terms of n. If \(x \le n+1\), we are done. Otherwise, if \(x > n+2\), then from (3.4), we get

$$\begin{aligned} x < 2.48 \cdot 10^{12} n \log x, \end{aligned}$$

which can be rewritten as

$$\begin{aligned} \frac{x}{\log x} < 2.48 \cdot 10^{12}n. \end{aligned}$$
(3.6)

From the useful fact that, for all \(x > e\), \(x/ \log x< A \Rightarrow x < 2A \log A\), holds whenever \(A \ge 3\), and since \(\log (2.48 \cdot 10^{12} n) < 7.05 \log n,\) holds for all \(n \ge 113\), it follows

$$\begin{aligned} x&< 2\cdot (2.48 \cdot 10^{12} n)\log (2.48 \cdot 10^{12}n)\\&< 4.96 \cdot 10^{12}n \cdot (7.04\log n)\\&< 3.5 \cdot 10^{13} n \log n. \end{aligned}$$

So finally, we have

$$\begin{aligned} x < 3.5 \cdot 10^{13} n \log n \end{aligned}$$
(3.7)

for \(n \ge 113\).

3.3 An Absolute Upper Bound on x

Set \(y := x/\alpha ^{2n}\). From (3.7) and \(n \ge 113\), we have

$$\begin{aligned} y< \frac{3.5 \cdot 10^{13}n \log n}{\alpha ^{2n}} < \frac{1}{\alpha ^{n}}. \end{aligned}$$
(3.8)

In particular, \(y< \alpha ^{-31} < 10^{-23}\). Now, we need to make a few algebraic manipulations apply Theorem 2.1 a second time. Rewrite Eq. (1.4) as follows:

$$\begin{aligned} \frac{\alpha ^{m}}{\sqrt{5}} - \frac{\beta ^{m}}{\sqrt{5}} = \frac{\alpha ^{(n+1)x}}{5^{x/2}} - \frac{\alpha ^{(n-1)x}}{5^{x/2}} + \left( F_{n+1}^x - \frac{\alpha ^{(n+1)x}}{5^{x/2}} \right) - \left( F_{n-1}^x - \frac{\alpha ^{(n-1)x}}{5^{x/2}} \right) . \end{aligned}$$
(3.9)

Since \(y < 10^{-23}\) is very small, we have

$$\begin{aligned} \text {max} \left\{ \left| F_{n+1}^{x} - \frac{\alpha ^{(n+1)x}}{5^{x/2}} \right| , \left| F_{n-1}^{x} - \frac{\alpha ^{(n-1)x}}{5^{x/2}} \right| \right\} < \frac{2y\alpha ^{(n+1)x}}{5^{x/2}}~~~~~\left( \text {Eq. (16)}, [10]\right) . \end{aligned}$$

Taking absolute values, after a slight modification of (3.9), we obtain

$$\begin{aligned} \left| \frac{\alpha ^m}{\sqrt{5}} - \frac{\alpha ^{(n-1)x}}{5^{x/2}} (\alpha ^{2x} - 1)\right|&= \left| \frac{\beta ^m}{\sqrt{5}} + \left( F_{n+1}^x - \frac{\alpha ^{(n+1)x}}{5^{x/2}} \right) - \left( F_{n-1}^x - \frac{\alpha ^{(n-1)x}}{5^{x/2}} \right) \right| \\&< \frac{1}{\alpha ^m} + 2y \left( \frac{\alpha ^{(n-1)x}(\alpha ^{2x} - 1)}{5^{x/2}} \right) . \end{aligned}$$

Now, dividing both sides of the above inequality by \(\alpha ^{(n+1)x}/5^{x/2}\), we get

$$\begin{aligned} \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - (1 - \alpha ^{-2x}) \right|< & {} \frac{5^{x/2}}{\alpha ^{m+(n+1)x}} + 2y(1 - \alpha ^{-2x}) \nonumber \\< & {} \frac{1}{2\alpha ^{n}} + \frac{17y}{9} \nonumber \\< & {} \frac{43}{18 \alpha ^{n}}, \end{aligned}$$
(3.10)

where we used that \(5^{x/2}/\alpha ^{(n+1)x} \le \left( \sqrt{5}/\alpha ^{113} \right) ^x < 1/2\), and \(\alpha ^{2x} \ge \alpha ^6 > 18\). Thus,

$$\begin{aligned} \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1 \right|&< \frac{43}{18 \alpha ^n} + \frac{1}{\alpha ^{2x}} \\&\le \frac{61}{18 \alpha ^l}, \end{aligned}$$

where \(l:= \text {min}\{n, 2x\}.\) Hence, we deduce that

$$\begin{aligned} \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1 \right| < \frac{61}{18 \alpha ^l}. \end{aligned}$$
(3.11)

Observe that, by the same argument used previously, \(\alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1\) is non-zero. Since \(x \ge 3\) and \(n \ge 113\), we get

$$\begin{aligned} \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1 \right| \le \frac{1}{\alpha ^{3}} + \frac{1}{\alpha ^{113}} < \frac{1}{2}, \end{aligned}$$
(3.12)

so that \(\alpha ^{m-(n+1)x} 5^{(x-1)/2} \in [1/2, 2]\). In particular,

$$\begin{aligned} 1.6x - 4< (n+1)x-m < 1.7x. \end{aligned}$$
(3.13)

Now, we are about to use of Matveev’s result one more time, to obtain a lower bound for the left-hand side of (3.11). For this, we take

$$\begin{aligned} t:= & {} 2, \gamma _{1} := \alpha , \gamma _{2} := \sqrt{5}, b_{1}:= m-(n+1)x, b_{2} := x-1,\\ D:= & {} 2, A_{1}:= \log \alpha , A_{2}:= \log 5,~ \text {and}~ B:=1.7x > \text {max}\{|b_1|, |b_2|\}. \end{aligned}$$

Hence,

$$\begin{aligned} \log \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1 \right|> & {} -1.4 \cdot 30^5 \cdot 2^{4.5} \cdot 2^2 \nonumber \\&(1+\log 2) (1+\log (1.7x)) (\log \alpha ) (\log 5). \end{aligned}$$
(3.14)

From (3.11) and (3.14), we deduce

$$\begin{aligned} l < 3.5 \times 10^9 \log x. \end{aligned}$$

Treating separately the case \(l= n\) and \(l=2x\), and performing the respective calculations, we arrive that \(x < 5 \times 10^{36}\) and \(x < 4.29 \times 10^{10}\) respectively. In any case we have that

$$\begin{aligned} x < 5 \times 10^{36}. \end{aligned}$$

3.4 Reducing the Bound on x

Next, we take \(\Gamma _{2} := (x-1) \log \sqrt{5} - ((n+1)x-m) \log \alpha \). Observe that (3.12) becomes

$$\begin{aligned} \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1 \right| = |e^{\Gamma _{2}} - 1| < \frac{1}{2}. \end{aligned}$$

Thus, we have that \(e^{|\Gamma _{2}|} < 2.\) Hence,

$$\begin{aligned} \left| \Gamma _{2} \right| \le e^{|\Gamma _{2}|} \left| e^{\Gamma _{2}} -1 \right|< 2 \cdot \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} - 1 \right| < 2 \left( \frac{43}{18 \alpha ^n} + \frac{1}{\alpha ^{2x}} \right) . \end{aligned}$$

This leads to

$$\begin{aligned} \left| \frac{\log \sqrt{5}}{\log \alpha } - \frac{(n+1)x - m}{x-1} \right| < \frac{1}{(x-1) \log \alpha } \left( \frac{43}{9 \alpha ^n} + \frac{2}{\alpha ^{2x}} \right) . \end{aligned}$$
(3.15)

Assume that \(x > 100.\) Then \(\alpha ^{2x}> \alpha ^{200}> 10^{41} > 10^{4}x.\) Hence, we deduce that

$$\begin{aligned} \frac{1}{(x-1) \log \alpha } \left( \frac{43}{9 \alpha ^{n}} + \frac{2}{\alpha ^{2x}}\right)&< \frac{8}{x(x-1)10^4 \log \alpha } \nonumber \\&< \frac{1}{600 (x-1)^{2}}. \end{aligned}$$
(3.16)

Estimates (3.15) and (3.16) lead to the following inequality:

$$\begin{aligned} \left| \frac{\log \sqrt{5}}{\log \alpha } - \frac{(n+1)x - m}{x-1} \right| < \frac{1}{600 (x-1)^{2}}. \end{aligned}$$
(3.17)

By virtue of Lemma 2.3, inequality (3.17) implies that the rational number

$$\begin{aligned} \frac{(n+1)x-m}{(x-1)} \end{aligned}$$

is a convergent to \(\gamma := \log (\sqrt{5})/\log \alpha \). Let \([a_0, a_1, a_2, a_3, \dots ] = [1, 1, 2, 19, \dots ]\) be the continued fraction of \(\gamma \) and \(p_t/q_t\) its tth convergent. Assume that \(((n+1)x-m)/(x-1) = p_k/q_k\) for some k. Then \(x-1 = dq_{k}\) for some positive integer d, where \(d = \gcd [(n+1)x-m, x-1]\). On the other hand, using the Mathematica, we get that

$$\begin{aligned} q_{54} = 14014190203160504083256905054> 1.4 \times 10^{28}> 3 \times 10^{27}-1 > x-1; \end{aligned}$$

therefore, \(0 \le k \le 53\). Again, using Mathematica, we observe that \(a_{t} \le 29\) for \(t \in \{0, \dots , 53\}\). From the properties of continued fractions, we have

$$\begin{aligned} \left| \gamma - \frac{(n+1)x - m}{x-1} \right| = \left| \gamma - \frac{p_k}{q_k} \right| > \frac{1}{(a_{k}+2)q_{k}^2} \ge \frac{1}{31(x-1)^2}, \end{aligned}$$

which contradicts (3.17). Hence, \(x \le 100\).

3.5 The Final Step

From (3.10), we have

$$\begin{aligned} \left| \alpha ^{m-(n+1)x} 5^{(x-1)/2} (1 - \alpha ^{-2x})^{-1} - 1 \right|< \frac{43}{18 \alpha ^{n} (1 - \alpha ^{-2x})} < \frac{3}{\alpha ^{n}}. \end{aligned}$$

Put \( s := (n+1)x -m\). We computed all the numbers \(\left| {\alpha }^{-s} 5^{{(x-1)}{/}{2}} ({1}{-}{\alpha }^{-2x})^{-1} {-1} \right| \) for all \(x \in [3, 100]\) and all \(s \in [\lfloor 1.6x -4 \rfloor , \lfloor 1.7x \rfloor ]\). None of them ended up being zero, since if it were we would get the Diophantine equation

$$\begin{aligned} \alpha ^{2x} - 1 = 5^{(x-1)/2}\alpha ^{2x - s}. \end{aligned}$$
(3.18)

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

$$\begin{aligned} \beta ^{2x} - 1 = 5^{(x-1)/2}\beta ^{2x - s}. \end{aligned}$$
(3.19)

Multiplying (3.18) and (3.19), we get

$$\begin{aligned} - \alpha ^{2x} - \beta ^{2x} + (-1)^s + 1&= (\alpha ^{2x} -1) (\beta ^{2x-1} - 1) \nonumber \\&= (\alpha \beta )^{2x-s} 5^{(x-1)} \nonumber \\&= (-1)^s 5^{(x-1)}. \end{aligned}$$
(3.20)

If s is even, then (3.20) implies that \(L_{2x} = 2 - 5^{x-1},\) where \((L_{n})_{n \ge 0}\) be the Lucas sequence given by \(L_{n+2} = L_{n+1} + L_{n},\) for \(n \ge 0\), with \((L_0, L_1) = (2, 1).\) However, \(5^{x-1} > 2\) for any \(x \ge 2.\) Thus, s is odd, then (3.20) implies that \(L_{2x} = 5^{(x-1)}.\) Using the identity \(L_{k}^2 - 5F_{k}^2 = 4(-1)^k\), it is easy to check that \(5 \not \mid L_k\) for any positive integer k. This theorem is, therefore, proved.

4 Comments

It is evident from the title Diophantine equation that one can revisit the equation under some more general forms. A number of directions for future research are discussed below.

  1. (i)

    The only positive integer solution (m, n, x, y) of \(F_{n+1}^{x} - F_{n-1}^y = F_{m}^x\) with \(m \ge 1, n \ge 3, x \ge 2, y \ge 1\) and \(x \ne y\) is (4, 4, 2, 4) for which \(F_{5}^2 - F_{3}^4 = F_{4}^2\).

  2. (ii)

    The only positive integer solutions (m, n, x, y) of \(F_{n+1}^{x} - F_{n-1}^x = F_{m}^y\) with \(n >1\) are (n, n, 1, 1),  (n, 2n, 2, 1),  (3, 3, 2, 3), (6, 3, 1, 3).

  3. (iii)

    The only positive integer solutions (n, x, y, z) of \(F_{n+1}^{x} - F_{n-1}^y = F_{2n}^z\) are \((x, y, z) = (2, y, 1)\) for \(n= 2, 3\) and \((x, y, z) = (2, 2, 1)\) for \(n \ge 4.\)