On the Diophantine equation $\sum_{j=1}^kjF_j^p=F_n^q$

TL;DR

This paper investigates a Fibonacci-based Diophantine equation involving sums of Fibonacci powers, providing complete solutions for certain exponents and conjecturing only trivial solutions exist beyond specific known cases.

Contribution

The paper offers a complete solution for the equation when exponents are 1 or 2, and conjectures that only trivial solutions exist for other cases based on computational evidence.

Findings

Complete solutions for exponents 1 and 2.

Conjecture that only trivial solutions exist beyond specific known cases.

Computational search supports the conjecture for p,q,k ≤ 100.

Abstract

Let $F_n$ denote the $n^{th}$ term of the Fibonacci sequence. In this paper, we investigate the Diophantine equation $F_1^p+2F_2^p+\cdots+kF_{k}^p=F_{n}^q$ in the positive integers $k$ and $n$, where $p$ and $q$ are given positive integers. A complete solution is given if the exponents are included in the set $\{1,2\}$. Based on the specific cases we could solve, and a computer search with $p,q,k\le100$ we conjecture that beside the trivial solutions only $F_8=F_1+2F_2+3F_3+4F_4$, $F_4^2=F_1+2F_2+3F_3$, and $F_4^3=F_1^3+2F_2^3+3F_3^3$ satisfy the title equation.

Equations

$$F_{n+2}=F_{n+1}+F_n\text{and}{L}_{n+2}=L_{n+1}+L_n\text{for all}n\ge 0,$$

$$F_n=\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }\text{and}{L}_n={\alpha }^n+{\beta }^n$$

$$F_1$$

$$F_1+2F_2$$

$$F_1+2F_2+3F_3$$

$$F_1+2F_2+3F_3+4F_4$$

$$a_1{F}_{n+1}^{s_1}+a_2{F}_{n+2}^{s_2}+...+a_l{F}_{n+l}^{s_l}$$

$$F_1^p+F_2^p+\cdots +F_{k-1}^p=F_{k+1}^q+\cdots +F_{k+r}^q,$$

$$F_1+F_2+\cdots +F_k=F_n^q,\mathrm{and}{F}_1^2+F_2^2+\cdots +F_k^2=F_n^q$$

$$F_{k+2}-1=F_n^q,F_k{F}_{k+1}=F_n^q.$$

$$x^2+2(x+1{)}^2+\cdots +n(x+n-1{)}^2=y^2$$

$$F_1^p+2F_2^p+\cdots +k{F}_k^p=F_n^q$$

$$F_1^p=1=F_1^q=F_2^q,\mathrm{and}{F}_1^p+2F_2^p=3=F_4$$

$$F_4^2$$

$$F_8$$

$$F_4^3$$

$$F_1+2F_2+\cdots +k{F}_k=F_n,$$

$$F_1^2+2F_2^2+\cdots +k{F}_k^2=F_n^2$$

$$F_1+2F_2+\cdots +k{F}_k=F_n^2,$$

$$F_1^2+2F_2^2+\cdots +k{F}_k^2=F_n$$

$$F_{k}-1=\begin{cases}F_{(k+2)/2}L_{(k-2)/2},\quad{\rm if}&\mbox{$k\equiv 0\pmod{4}$}\\ F_{(k-1)/2}L_{(k+1)/2},\quad{\rm if}&\mbox{$k\equiv 1\pmod{4}$}\\ F_{(k-2)/2}L_{(k+2)/2},\quad{\rm if}&\mbox{$k\equiv 2\pmod{4}$}\\ F_{(k+1)/2}L_{(k-1)/2},\quad{\rm if}&\mbox{$k\equiv 3\pmod{4}$}\end{cases};$$

$$F_{k}+1=\begin{cases}F_{(k-2)/2}L_{(k+2)/2},\quad{\rm if}&\mbox{$k\equiv 0\pmod{4}$}\\ F_{(k+1)/2}L_{(k-1)/2},\quad{\rm if}&\mbox{$k\equiv 1\pmod{4}$}\\ F_{(k+2)/2}L_{(k-2)/2},\quad{\rm if}&\mbox{$k\equiv 2\pmod{4}$}\\ F_{(k-1)/2}L_{(k+1)/2},\quad{\rm if}&\mbox{$k\equiv 3\pmod{4}$}\end{cases}.$$

$$F_{k}^{2}-1=\begin{cases}F_{k-1}F_{k+1},\quad{\rm if}&\mbox{$k\equiv 1\pmod{2}$}\\ F_{k-2}F_{k+2},\quad{\rm if}&\mbox{$k\equiv 0\pmod{2}$}\end{cases};$$

$$F_{k}^{2}+1=\begin{cases}F_{k-1}F_{k+1},\quad{\rm if}&\mbox{$k\equiv 0\pmod{2}$}\\ F_{k-2}F_{k+2},\quad{\rm if}&\mbox{$k\equiv 1\pmod{2}$}\end{cases}.$$

$$2F_{j-1}^{\phi (F_j)-1}\equiv F_{j-3}(\mathrm{mod}{F}_j).$$

$$2\equiv F_{j-3}{F}_{j-1}(\mathrm{mod}{F}_j).$$

$$F_{j-1}{F}_{j-3}=(F_{j+1}-F_j)F_{j-3}\equiv F_{j+1}{F}_{j-3}=F_j{F}_{j-2}+(-1{)}^{j-2}{F}_3\equiv 2(\mathrm{mod}{F}_j)$$

$$F_{j-1}^{\phi (F_j)-1}\equiv F_{j-2}(\mathrm{mod}{F}_j).$$

$$1\equiv F_{j-2}{F}_{j-1}(\mathrm{mod}{F}_j).$$

$$F_{j-2}{F}_{j-1}=(F_j-F_{j-1})F_{j-1}\equiv -F_{j-1}^2=-(F_{j-2}{F}_j+(-1{)}^j{F}_{-1})\equiv 1(\mathrm{mod}{F}_j).$$

$${\delta }_i={\log }_{\alpha }\left(\frac{1+(-1{)}^{i-1}{\left(|\beta |/\alpha \right)}^{k_0}}{\sqrt{5}}\right),$$

$${\alpha }^{k+{\delta }_0}\le F_k\le {\alpha }^{k+{\delta }_1}$$

$${\alpha }^{k-2}\le F_k\le {\alpha }^{k-1},$$

Full text

*Key Words: Fibonacci sequence, Diophantine equation.

AMS Classification: 11B39, 11D45.*

1 Introduction

As usual, let $(F_{n})_{n\geq 0}$ and $(L_{n})_{n\geq 0}$ denote the sequences of Fibonacci and Lucas numbers, respectively, given by the initial values $F_{0}=0,~{}F_{1}=1,~{}L_{0}=2,~{}L_{1}=1$, and by the recurrence relations

[TABLE]

respectively. Putting $\alpha=(1+\sqrt{5})/2$ and $\beta=(1-\sqrt{5})/2=-1/\alpha$ for the two roots of the common characteristic equation $x^{2}-x-1=0$ of the two sequences, the formulae

[TABLE]

hold for all $n\geq 0$. These numbers are well-known for possessing amazing and wonderful properties (consult, for instance, [13] and [5] together with their very rich annotated bibliography for history and additional references). Observing

[TABLE]

the question arises naturally: is there any rule for $F_{1}+2F_{2}+3F_{3}+\cdots+kF_{k}$? We study this question more generally, according to the title equation.

Diophantine equations among the terms of Fibonacci numbers have a very extensive literature. Here we quote a few results that partially motivated us.

By the defining equality (1) of the Fibonacci numbers and the identity $F_{n}^{2}+F_{n+1}^{2}=F_{2n+1}$ (Lemma 1.8), we see that $F_{n}^{s}+F_{n+1}^{s}$ ($n\geq 0$) is a Fibonacci number for $s\in\{1,2\}$. For larger $s$ Marques and Togbé [8] proved in 2010 that if $F_{n}^{s}+F_{n+1}^{s}$ is a Fibonacci number for all sufficiently large $n$ then $s=1$ or $2$. Next year Luca and Oyono [7] completed the solution of the question by showing that apart from $F_{1}^{s}+F_{2}^{s}=F_{3}$ there is no solution $s\geq 3$ to the equation $F_{n}^{s}+F_{n+1}^{s}=F_{m}$.

Let $l,s_{1},...,s_{l},a_{1},...,a_{l}$ be integers with $l\geq 1$ and $s_{j}\geq 1$. Suppose that there exists $1\leq t\leq l$ such that $a_{t}\neq 0$ and $s_{t}>s_{j}$, for all $j\neq t$. Chaves, Marques and Togbé [4], showed that if either $s_{t}$ is even or $a_{t}$ is not a positive power of $5$, then the sum

[TABLE]

does not belong to the Fibonacci sequence for all sufficiently large $n$.

A balancing problem having similar flavor has been considered by Behera et al. [3]. They studied the equation

[TABLE]

and solved it for the cases $(p,q)=(2,1),(3,1),(3,2)$, and for $2\leq p\leq q$ by showing the non-existence of any solution. Further the authors conjectured that only the quadruple $(k,r,p,q)=(4,3,8,2)$ of positive integers satisfies (2). The conjecture was completely justified by Alvarado et al. [1]. Note that if $(p,q)=(1,1)$ we obtain the problem of sequence balancing numbers handled by Panda [9].

Recalling the formulae $F_{1}+F_{2}+\cdots+F_{k}=F_{k+2}-1$ and $F_{1}^{2}+F_{2}^{2}+\cdots+F_{k}^{2}=F_{k}F_{k+1}$, it is obvious that the problems

[TABLE]

are rather simple. Indeed, the equations above lead to the lightsome ones

[TABLE]

However the equation $F_{1}^{p}+F_{2}^{p}+\cdots+F_{k}^{p}=F_{n}^{q}$ might be taken an interest if $p\geq 3$.

The last motivation of our examination was the Diophantine equation

[TABLE]

to determine the values of $n$ for which it has finitely or infinitely many positive integer solutions $(x,y)$ (see Wulczyn [14], and for details, see also [2]). For variations of the equation (3), we refer the reader to [12].

In this paper, we investigate the Diophantine equation

[TABLE]

in the positive integers $k$ and $n$, where $p$ and $q$ are fixed positive integers. We consider

[TABLE]

as trivial solutions to (4). We have the following conjecture based upon the specific cases we could solve, and a computer search with $p,q,k\leq 100$.

Conjecture 1**.**

The non-trivial solutions to (4) are only

[TABLE]

This work handles the particular cases $p,q\in\{1,2\}$ (hence the first two solutions above will be obtained), the precise results proved are described as follow.

Theorem 1**.**

If

[TABLE]

then $(k,n)=(1,1),~{}(1,2),~{}(2,4),~{}(4,8)$, among them only the last one is non-trivial solution.

Theorem 2**.**

The Diophantine equation

[TABLE]

possesses only the trivial solutions $(k,n)=(1,1),~{}(1,2)$.

Theorem 3**.**

If

[TABLE]

then $(k,n)=(1,1),~{}(1,2),~{}(3,4)$, among them only the last one is non-trivial solution.

Theorem 4**.**

The Diophantine equation

[TABLE]

possesses only the trivial solutions $(k,n)=(1,1),~{}(1,2),~{}(2,4)$.

2 Lemmata

In this section, we present the lemmata that are needed in the proofs of the theorems. The first lemma is a collection of a few well-known results, we state them without proof, and in the proof of the theorems sometimes we do not refer to them.

Lemma 1**.**

Let $k$ and $n$ be arbitrary integers.

$\sum_{j=1}^{k}jF_{j}=kF_{k+2}-F_{k+3}+2$. 2. 2.

$\sum_{j=1}^{k}jF_{j}^{2}=F_{k}(kF_{k+1}-F_{k})+\tau$, where $\tau=0$ if $k$ is even, and $\tau=1$ otherwise. 3. 3.

For $k\geq 0$ we have $F_{-k}=(-1)^{k+1}F_{k}$, further $L_{-k}=(-1)^{k}L_{k}$ (extension of the sequences for negative subscripts). 4. 4.

$\gcd(F_{k},F_{n})=F_{\gcd(k,n)}$. 5. 5.

$\gcd(F_{k},L_{n})=1$* or $2$ or $L_{\gcd(k,n)}$.* 6. 6.

$F_{k}\mid F_{n}$* if and only if $k\mid n$.* 7. 7.

$F_{k+1}F_{n}-F_{k}F_{n+1}=(-1)^{n+1}F_{k-n}$* (d’ Ocagne’s identity).* 8. 8.

$F_{k+n}=F_{k}F_{n+1}+F_{k-1}F_{n}$. 9. 9.

$F_{2k}=F_{k}L_{k}$. 10. 10.

$F_{k+n}^{2}-F_{k-n}^{2}=F_{2k}L_{2n}$.

Lemma 2**.**

[TABLE]

[TABLE]

Proof.

See, for instance, [6] and [10]. ∎

Lemma 3**.**

[TABLE]

[TABLE]

Proof.

See Lemma 3 in [11]. ∎

Lemma 4**.**

If $j\geq 4$ is even, then

[TABLE]

Proof.

Since $\gcd(F_{j-1},F_{j})=1$, and $F_{j-1}^{\varphi(F_{j})}\equiv 1\pmod{F_{j}}$, it is sufficient to show that

[TABLE]

But

[TABLE]

follows from the definition of the Fibonacci numbers, d’ Ocagne’s identity (Lemma 1.7), and the parity of $j$. ∎

Lemma 5**.**

If $j\geq 3$ is odd, then

[TABLE]

Proof.

Similarly to the proof of the previous lemma, the statement is equivalent to

[TABLE]

And it is easy to see that

[TABLE]

∎

Lemma 6**.**

Let $k_{0}$ be a positive integer, and for $i\in\{0,1\}$ put

[TABLE]

where $\log_{\alpha}$ is the logarithm in base $\alpha=(1+\sqrt{5})/2$. Then for all integers $k\geq k_{0}$, the two inequalities

[TABLE]

hold.

Proof.

This is a part of Lemma 5 in [6]. ∎

In order to make the application of Lemma 6 more convenient, we shall suppose that $k_{0}\geq 1$. Then we have

Corollary 5**.**

If $k\geq 1$, then

[TABLE]

and equality holds if and only if $k=2$, and $k=1$, respectively.

Now, we are ready to justify the theorems.

3 Proofs

Proof of Theorem 1.

Verifying the cases $k=1,\dots,5$ by hand we found the solutions listed in Theorem 1. Put $\kappa=k+2$, and suppose that $\kappa\geq 8$. Consequently, $F_{\kappa-3}\geq 5$ and $F_{\kappa}\geq 21$. If equation (5) holds, then $n>\kappa$, and then by Lemma 1.1 we conclude

[TABLE]

In the sequel, we study the sequence $(F_{u})_{u=0}^{\infty}$ modulo $F_{\kappa}$ if $\kappa$ is fixed. Note that we indicate a suitable value congruent to $F_{u}$ modulo $F_{\kappa}$, not always the smallest non-negative remainders. The period can be deduced from the range

[TABLE]

of length $2\kappa$ if $\kappa$ is even, since then, by Lemma 1.7 we have $F_{\kappa-1}^{2}\equiv 1\pmod{F_{\kappa}}$ and then

[TABLE]

In case of odd $\kappa$ we have $F_{\kappa-1}^{2}\equiv-1\pmod{F_{\kappa}}$, therefore the length of the period is $4\kappa$ coming from

[TABLE]

Based on the length of the period we distinguish two cases.

Case I: $\kappa$ is even. Either $F_{n}\equiv F_{j}$ or $F_{n}\equiv F_{j}F_{\kappa-1}$ modulo $\kappa$ holds for some $j=0,1,\dots,\kappa-1$. Hence

[TABLE]

We will show that none of them is congruent to 0 modulo $F_{\kappa}$. In the first branch

[TABLE]

further if $j\neq\kappa-1$, then

[TABLE]

Thus $F_{j}+F_{\kappa-1}-2\not\equiv 0\pmod{F_{\kappa}}$, hence (9) does not hold. Assume now, that $j=\kappa-1$. Then, together with the definition of the Fibonacci sequence we have

[TABLE]

But $3\leq F_{\kappa-3}-2<F_{\kappa}$ contradicts to (9).

Choosing the second branch of (10), suppose that $F_{\kappa-1}(F_{j}+1)-2$ is congruent to 0 modulo $F_{\kappa}$. Then

[TABLE]

Subsequently, by Lemma 4, it leads to

[TABLE]

Since $j=0,1,\dots,\kappa-1$, ($\kappa\geq 8$) it follows that $F_{j}=F_{\kappa-3}-1$, a contradiction.

Case II: $\kappa$ is odd. Now $\kappa\geq 9$, and either $F_{n}\equiv\pm F_{j}\;\pmod{F_{\kappa}}$ or $F_{n}\equiv\pm F_{j}F_{\kappa-1}\pmod{F_{\kappa}}$ holds for some $j=0,1,\dots,\kappa-1$. Hence

[TABLE]

First, obviously, if $j\neq\kappa-1$, then

[TABLE]

so dividing $\pm F_{j}+F_{\kappa-1}-2$ by $F_{\kappa}$, the result is not an integer. If $j=\kappa-1$, then the treatment of the “$+$” case coincides the treatment when $\kappa$ was even. The “$-$” case leads to $F_{n}+F_{\kappa-1}-2\equiv-2\pmod{F_{\kappa}}$, a contradiction.

Assume now that $F_{n}+F_{\kappa-1}-2\equiv\pm F_{j}F_{\kappa-1}+F_{\kappa-1}-2\pmod{F_{\kappa}}$. Thus $F_{\kappa-1}(1\pm F_{j})\equiv 2\pmod{F_{\kappa}}$. Multiplying both sides by $F_{\kappa-1}^{\varphi(F_{\kappa})-1}$, by Lemma 5 it gives

[TABLE]

First let $F_{j}=2F_{\kappa-2}-1$, which leads immediately a contradiction via $0<2F_{\kappa-2}-1=F_{\kappa-1}+F_{\kappa-4}-1<F_{\kappa}$. If $F_{\kappa}-F_{j}+1=2F_{\kappa-2}$, then $F_{j}=F_{\kappa-3}+1$ follows, a contradiction again. The proof of Theorem 1 is complete.

Proof of Theorem 2.

For the range $k=1,2,\dots,20$ we checked (6) by hand. From now we assume $k\geq 21$. Based on Lemma 1.2, we must distinguish two cases.

Case I: $k$ is even. Consider the equation

[TABLE]

Trivially, $n>k$. Put $\nu=\gcd(k,n)$.

If $\nu=k$, then $F_{k}\mid F_{n}$ by Lemma 1.6. Consequently,

[TABLE]

is integer. But $F_{k}$ and $F_{k+1}$ are coprime, hence $F_{k}\mid k$, and it results $k\leq 5$, a contradiction.

Examine the possibility $\nu=k/2$. Put $\kappa=k/2$. Now $F_{\kappa}L_{\kappa}(kF_{k+1}-F_{\kappa}L_{\kappa})=F_{n}^{2}$ leads to

[TABLE]

This is an equality of integers, which together with $\gcd(F_{\kappa},F_{k+1})$ and $\gcd(F_{\kappa},L_{\kappa})=1,2$ (see Lemma 1.5) shows that $2k/F_{\kappa}$ is integer. Thus $k\leq 14$, a contradiction.

Finally, we have $3\leq\nu\leq k/3$. Since $\gcd(F_{k}/F_{\nu},F_{n}/F_{\nu})=1$, then from the equation

[TABLE]

we conclude

[TABLE]

Subsequently, $F_{k}\mid F_{\nu}F_{n}$ and $F_{n}\mid F_{\nu}(kF_{k+1}-F_{k})$. Thus $F_{k}\mid F_{\nu}^{2}(kF_{k+1}-F_{k})$, and then $F_{k}\mid kF_{\nu}^{2}$ holds since $\gcd(F_{k},F_{k+1})=1$. Applying Corollary 5, we obtain

[TABLE]

which implies $k<19$.

Case II: $k$ is odd. In this part, we follow the idea of the previous case. Recall that $k\geq 21$. Now

[TABLE]

where $\varepsilon=1$ or $2$ depending on the parity of $n$ (see Lemma 3). Clearly, $\gcd(n-\varepsilon,n+\varepsilon)=2$ or 4. Thus $\gcd(F_{n-\varepsilon},F_{n+\varepsilon})=1$ or 3, respectively.

Put $\nu_{1}=\gcd(k,n-\varepsilon)$ and $\nu_{2}=\gcd(k,n+\varepsilon)$. Obviously, $\gcd(\nu_{1},\nu_{2})$ divides $\gcd(n-\varepsilon,n+\varepsilon)$. Hence $\nu=\gcd(\nu_{1},\nu_{2})=1$ or 2 or 4, and then $F_{\nu}=\gcd(F_{\nu_{1}},F_{\nu_{2}})=1$ or 3. Thus $F_{\nu_{1}}F_{\nu_{2}}\mid F_{\nu}F_{k}$. The terms of both the left and right sides of

[TABLE]

are integers, and

[TABLE]

Combining them, $F_{k}\mid F_{\nu_{1}}F_{\nu_{2}}(kF_{k+1}-F_{k})$ follows, and then $F_{k}\mid kF_{\nu_{1}}F_{\nu_{2}}$. The remaining part of the proof consists of three cases.

Suppose first that $\nu_{1}=k$, i.e. $F_{k}\mid F_{n-\varepsilon}$. Observe, that $n-\varepsilon$ and $n+\varepsilon$ are even, and $k\mid(n-\varepsilon)/2$. Thus $k+1\leq(n-\varepsilon)/2+1$, which does not exceed $(n+\varepsilon)/2$. Then

[TABLE]

Simplifying by $F_{k+1}$ we conclude

[TABLE]

and we arrived at a contradiction since $21\leq k<n$. Note that the same machinery works when $\nu_{2}=k$, i.e. $F_{k}\mid F_{n+\varepsilon}$.

If none of the two conditions above holds, we can assume $\nu_{1}\leq k/3$ and $\nu_{2}\leq k/3$. Indeed, $k$ is odd, so the largest non-trivial divisor of $k$ is at most $k/3$. The application of Corollary 5 gives

[TABLE]

and then $k<19$.

The proof of the theorem is complete.

Proof of Theorem 3. The proof partially follows the proof of Theorem 1. The small cases of (7) can be verified by hand. Suppose $\kappa=k+2\geq 9$. Similarly to (9), we have

[TABLE]

Now we study the sequence $(F_{u}^{2})_{u=0}^{\infty}$ modulo $F_{\kappa}$, and we again indicate the most suitable values by modulo $F_{\kappa}$, not always the smallest non-negative remainders. Lemma 1.10, together with Lemma 1.3 implies

[TABLE]

where $j=0,1,\dots,\kappa-1$. Hence the period having length $2\kappa$ can be given by

[TABLE]

Let us distinguish two cases according to the parity of $\kappa$.

Case I: $\kappa$ is even. Put $\kappa=2\ell$. Again by Lemma 1.10, together with $\kappa-i=2\ell-i=\ell+(\ell-i)$ and $i=\ell-(\ell-i)$ admits $F_{\kappa-i}^{2}\equiv F_{i}^{2}\pmod{F_{\kappa}}$. It reduces the possibilities to $j=0,1,\dots,\ell$.

If $j\leq\ell-1=(\kappa-2)/2$, then

[TABLE]

hold since $F_{(\kappa-2)/2}^{2}\leq F_{(\kappa-2)/2}L_{(\kappa-2)/2}=F_{\kappa-2}$.

Suppose now that $j=\ell=\kappa/2$. Repeating the previous idea we find $F_{\kappa/2}^{2}+F_{\kappa-1}-2\leq F_{\kappa}+F_{\kappa-1}-2<2F_{\kappa}$. Consequently, $F_{\kappa/2}^{2}+F_{\kappa-1}-2=F_{\kappa}$ might be fulfilled. Thus $F_{\kappa/2}^{2}-2=F_{\kappa-2}$. Recalling $\kappa=2\ell$ we equivalently obtain

[TABLE]

Both sides have decomposition described in Lemma 3 and Lemma 2, respectively, providing

[TABLE]

if $\ell$ is odd or even, respectively. Firstly, $F_{\ell-2}\mid F_{\ell-1}F_{\ell+1}$, and then $F_{\ell-2}\mid F_{\ell+1}$ holds only for small $\ell$ values. Secondly, $F_{\ell}\mid F_{\ell-2}F_{\ell+2}$ contradicts to $\ell\geq 5$.

Case II: $\kappa$ is odd. Now we have $F_{\kappa-i}^{2}\equiv-F_{i}^{2}\pmod{F_{\kappa}}$. Indeed, Lemma 1.8 admits

[TABLE]

and then Lemma 3 justifies the statement. It makes possible to split the proof into a few parts.

If $j=(\kappa-1)/2$, then $F_{(\kappa-1)/2}^{2}+F_{\kappa-1}-2<2F_{\kappa-1}-2<2F_{\kappa}$. Thus $F_{(\kappa-1)/2}^{2}+F_{\kappa-1}-2=F_{\kappa}$ is the only one chance to fulfill (11). Then apply Lemma 6 with $k_{0}=4\leq(\kappa-1)/2$ for $F_{\kappa-2}<F_{(\kappa-1)/2}^{2}$ to reach a contradiction.

If $j\leq(\kappa-3)/2$, then

[TABLE]

holds since $F_{(\kappa-3)/2}^{2}\leq F_{(\kappa-3)/2}L_{(\kappa-3)/2}=F_{\kappa-3}$.

Finally, if $(\kappa+1)/2\leq j\leq\kappa-1$, then

[TABLE]

where $1\leq J=\kappa-j\leq(\kappa-1)/2$. Thus we need to check the equation $F_{J}^{2}+2=F_{\kappa-1}$, since $-2\leq-F_{J}^{2}+F_{\kappa-1}-2<F_{\kappa-1}<F_{\kappa}$. When $J\leq(\kappa-3)/2$ holds then $F_{J}^{2}+2<F_{\kappa-3}+2<F_{\kappa-1}$. Lastly, $J=(\kappa-1)/2$ leads to $F_{(\kappa-1)/2}^{2}+2=F_{\kappa-1}$, and then to $2=F_{J}\left(L_{J}-F_{J}\right)$. Obviously, it gives $J\leq 3$. Thus $\kappa\leq 7$, a contradiction.

Proof of Theorem 4. The statement for (8) is obviously true if $k\leq 12$. So we may assume $k\geq 13$. Let $\tau\in\{0,1\}$. The formula of the summation, together with Corollary 5 implies

[TABLE]

and then

[TABLE]

Similarly,

[TABLE]

subsequently

[TABLE]

that is $n-k\geq k+4$. Putting together the two estimates it gives

[TABLE]

Case I: $k$ is even. Clearly, $k+2<n-k-1\leq 3k/2-2$ holds. By Lemma 1.8, we conclude

[TABLE]

and equivalently

[TABLE]

Thus $\gcd(F_{k},F_{k+1})=1$ admits $F_{k+1}\mid F_{k}+F_{n-k-1}$. The periodicity of $(F_{u})_{u=0}^{\infty}$ modulo $F_{k+1}$ guarantees, together with the bounds on $n-k-1$ that

[TABLE]

holds for some $j=1,2,\dots,k/2-3$. Consequently, $F_{k+1}\mid F_{j}+1$, a contradiction.

Case II: $k$ is odd. Again $k+2<n-k-1\leq 3k/2-2$ holds. Lemma 2 and Lemma 1.8 imply

[TABLE]

where $\varepsilon\in\{\pm 1,\pm 2\}$ according to the modular property of $n$. It leads to

[TABLE]

Thus $F_{k}\mid F_{(n-\varepsilon)/2-k}L_{(n+\varepsilon)/2}$.

By (12) it is obvious that

[TABLE]

which exclude $\gcd(k,(n+\varepsilon)/2)=k$. Thus $\gcd(k,(n+\varepsilon)/2)\leq k/3$ since $k$ is odd, subsequently $\gcd(F_{k},L_{(n+\varepsilon)/2})\leq L_{k/3}$. On the other hand

[TABLE]

which implies $\gcd(F_{k},F_{(n-\varepsilon)/2-k})\leq F_{k/4+1}$.

Thus, $F_{k}\mid F_{(n-\varepsilon)/2-k}L_{(n+\varepsilon)/2}$, together with the previous arguments entails $F_{k}\leq F_{k/4+1}L_{k/3}$, but it leads to a contradiction since $L_{k/3}=F_{k/3-1}+F_{k/3+1}<2F_{k/3+1}$, and the application of Corollary 5 on $F_{k}\leq 2F_{k/4+1}F_{k/3+1}$ returns with $k<9$.

Acknowledgments. This paper was partially written when the third author visited the Department of Mathematics of Uludağ University in Bursa. He would like to thank the Turkish colleagues of the department for the kind hospitality. The first author was supported by the Research Fund of Uludağ University under project numbers: 2015/23, 2016/9.