$\displaystyle=\left(\frac{-t\overline{y}}{1-a^{2}-s}\right)\left(\frac{ty}{1-a^{2}-s}\right)+\overline{y}\left(\frac{-t^{2}y}{s(1-a^{2}-s)}\right)$
(119)
$\displaystyle=-t^{2}\left(\frac{1}{1-a^{2}-s}+\frac{1}{s}\right)=-\frac{t^{2}(1-a^{2})}{s(1-a^{2}-s)}.$
(120)

The latter implies $a=1$, and from (114), we deduce a contradiction:
$0<|y|^{2}=-s<0$.

In conclusion, the only solutions for this type of curvature tensors are
obtained for

$\displaystyle a=1,$ $\displaystyle b=0,$ $\displaystyle\sigma=0,$ or
$\displaystyle a=1,$ $\displaystyle b=1,$
$\displaystyle\sigma=\frac{\sqrt{3}}{2}e^{i\alpha}(\theta^{2})^{3},$
$\displaystyle\textrm{for }\alpha\in\mathcal{C}^{\infty}{\left({M}\right)}.$

In Table 4, these results correspond to: VI for $a=1$ and
$\sigma=\frac{\sqrt{3}}{2}e^{i\alpha}(\theta^{2})^{3}$ for
$\alpha\in\mathcal{C}^{\infty}{\left({M}\right)}$; VII for $a=1$ and
$\sigma=0$; VIII and IX for $a=\frac{1}{\sqrt{\delta}}$, $\delta>0$ and
$\sigma=0$.

Table 5 summarises (up to isomorphisms) the cases satisfying the curvature
condition, showing the non vanishing differentials of the coframe and the
Levi-Civita connection.

Case | Structure constants | Levi-Civita connection | PSK
---|---|---|---
III |  | $du^{2}=-\sqrt{2}u^{1,2}$
---
$du^{4}=-2u^{3,4}$
$\begin{pmatrix}&\sqrt{2}u^{2}&\vline&&\\\ -\sqrt{2}u^{2}&&\vline&&\\\ \hline\cr&&\vline&&2u^{4}\\\ &&\vline&-2u^{4}&\end{pmatrix}$ | ✓
VI |  | $du^{1}=2u^{1,2}$
---
$du^{3}=u^{2,3}$
$du^{4}=-2u^{1,3}-u^{2,4}$
$\begin{pmatrix}0&-2u^{1}&\vline&u^{4}&u^{3}\\\ 2u^{1}&0&\vline&-u^{3}&u^{4}\\\ \hline\cr-u^{4}&u^{3}&\vline&0&-u^{1}\\\ -u^{3}&-u^{4}&\vline&u^{1}&0\end{pmatrix}$ |
VII |  | $du^{1}=u^{1,3}$
---
$du^{2}=u^{2,3}$
$du^{4}=-2u^{1,2}-2u^{3,4}$
$\begin{pmatrix}0&u^{4}&\vline&-u^{1}&u^{2}\\\ -u^{4}&0&\vline&-u^{2}&-u^{1}\\\ \hline\cr u^{1}&u^{3}&\vline&0&2u^{4}\\\ -u^{2}&-u^{4}&\vline&-2u^{4}&0\end{pmatrix}$ | ✓
VIII |  | $du^{1}=u^{1,3}+\frac{2}{\delta}u^{2,3}$
---
$du^{2}=-\frac{2}{\delta}u^{1,3}+u^{2,3}$
$du^{4}=-2u^{1,2}-2u^{3,4}$
$\delta>0$
$\begin{pmatrix}0&\frac{2}{\delta}u^{3}+u^{4}&\vline&-u^{1}&u^{2}\\\ -\frac{2}{\delta}u^{3}-u^{4}&0&\vline&-u^{2}&-u^{1}\\\ \hline\cr u^{1}&u^{2}&\vline&0&2u^{4}\\\ -u^{2}&u^{1}&\vline&-2u^{4}&0\end{pmatrix}$ | ✓
IX |  | $du^{1}=u^{1,4}-\frac{2}{\delta}u^{2,4}$
---
$du^{2}=\frac{2}{\delta}u^{1,4}+u^{2,4}$
$du^{3}=2u^{1,2}+2u^{3,4}$
$\delta>0$
$\begin{pmatrix}0&-\frac{2}{\delta}u^{4}-u^{3}&\vline&-u^{2}&-u^{1}\\\ \frac{2}{\delta}u^{4}+u^{3}&0&\vline&u^{1}&-u^{2}\\\ \hline\cr u^{2}&-u^{1}&\vline&0&-2u^{3}\\\ u^{1}&u^{2}&\vline&2u^{3}&0\end{pmatrix}$ | ✓
Table 5: Cases satisfying the curvature condition.

Now we must check whether condition D2enumi holds for the cases left. Notice
that for cases III, VII, VIII, IX, the Kähler form is exact with invariant
potentials; respectively $-\frac{1}{\sqrt{2}}u^{2}-\frac{1}{2}u^{4}$,
$-\frac{1}{2}u^{4}$, $-\frac{1}{2}u^{4}$, $\frac{1}{2}u^{3}$. We can
immediately say that cases VII, VIII, IX are all projective special Kähler
because $\sigma=0$, so the differential condition is trivially satisfied.

Concerning case III, we can compute $d^{LC}\sigma$ by understanding how the
Levi-Civita connection behaves on the unitary complex coframe $\theta$.

$\displaystyle\nabla^{LC}\theta^{1}$
$\displaystyle=\nabla^{LC}u^{1}+i\nabla^{LC}u^{2}=-(\omega^{LC})^{1}_{k}\otimes
u^{k}-i(\omega^{LC})^{2}_{k}\otimes u^{k}$
$\displaystyle=-\sqrt{2}u^{2}\otimes u^{2}+i\sqrt{2}u^{2}\otimes
u^{1}=\sqrt{2}iu^{2}\otimes\theta^{1};$ $\displaystyle\nabla^{LC}\theta^{2}$
$\displaystyle=\nabla^{LC}u^{3}+i\nabla^{LC}u^{4}=-(\omega^{LC})^{3}_{k}\otimes
u^{k}-i(\omega^{LC})^{4}_{k}\otimes u^{k}$ $\displaystyle=-2u^{4}\otimes
u^{4}+i2u^{4}\otimes u^{3}=2iu^{4}\otimes\theta^{2}.$

Now we can compute

$\displaystyle\nabla^{LC}\sigma$
$\displaystyle=\nabla^{LC}\left(\frac{3}{2}e^{i\alpha}(\theta^{1})^{2}\theta^{2}\right)$
$\displaystyle=\frac{3}{2}id\alpha\otimes
e^{i\alpha}(\theta^{1})^{2}\theta^{2}+3\sqrt{2}iu^{2}e^{i\alpha}(\theta^{1})^{2}\theta^{2}+\frac{3}{2}2iu^{4}\otimes
e^{i\alpha}(\theta^{1})^{2}\theta^{2}$
$\displaystyle=-4i\left(-\frac{1}{4}d\alpha-\frac{1}{\sqrt{2}}u^{2}-\frac{1}{2}u^{4}\right)\otimes\sigma.$

If we define
$\lambda:=-\frac{1}{4}d\alpha-\frac{1}{\sqrt{2}}u^{2}-\frac{1}{2}u^{4}$, we
have that $d\lambda=\omega$ and $d^{LC}\sigma=-4i\lambda\wedge\sigma$. Thanks
to Corollary 7.9, we have proven that also case III has a projective special
Kähler structure for every choice of
$\alpha\in\mathcal{C}^{\infty}{\left({M}\right)}$.

Suppose that VI is projective special Kähler, than by Theorem 7.6, locally we
must have the differential condition 3D2. Consider the unitary global complex
coframe $\theta$.

$\displaystyle\nabla^{LC}\theta^{2}$
$\displaystyle=\nabla^{LC}u^{3}+i\nabla^{LC}u^{4}$ $\displaystyle=u^{4}\otimes
u^{1}-u^{3}\otimes u^{2}+u^{1}\otimes u^{4}+i(u^{3}\otimes u^{1}+u^{4}\otimes
u^{2}-u^{1}\otimes u^{3})$
$\displaystyle=u^{4}\otimes\theta^{1}+iu^{3}\otimes\theta^{1}-iu^{1}\otimes\theta^{2}=i\overline{\theta^{2}}\otimes\theta^{1}-iu^{1}\otimes\theta^{2}.$

Thus,

$\displaystyle\begin{split}\nabla^{LC}\sigma&=\nabla^{LC}\left(\frac{\sqrt{3}}{2}e^{i\alpha}(\theta^{2})^{3}\right)\\\
&=id\alpha\otimes\frac{\sqrt{3}}{2}e^{i\alpha}(\theta^{2})^{3}+3\frac{\sqrt{3}}{2}e^{i\alpha}(\nabla^{LC}\theta^{2})(\theta^{2})^{2}\\\
&=id\alpha\otimes\sigma+3\frac{\sqrt{3}}{2}e^{i\alpha}(i\overline{\theta^{2}}\otimes\theta^{1}-iu^{1}\otimes\theta^{2})(\theta^{2})^{2}\\\
&=i(d\alpha-3u^{1})\otimes\sigma+3i\overline{\theta^{2}}\otimes\frac{\sqrt{3}}{2}e^{i\alpha}\theta^{1}(\theta^{2})^{2};\\\
d^{LC}\sigma&=i(d\alpha-3u^{1})\wedge\sigma+3i\overline{\theta^{2}}\wedge\frac{\sqrt{3}}{2}e^{i\alpha}\theta^{1}(\theta^{2})^{2}.\end{split}$
(121)

Notice that this is never of the form required by condition 3D2 for any
available choice of $\sigma$, since evaluating the last component at
$\theta_{1}$, we obtain
$i\frac{\sqrt{3}}{2}\overline{\theta^{2}}\wedge\theta^{2}\otimes\theta^{2}$,
whereas the same operation on a form of type $i\tau\wedge\sigma$ would
evaluate to zero. We deduce that VI does not admit a projective special Kähler
structure.

We are now left with cases III, VII, VIII, IX. At the level of Lie groups,
case III corresponds to the connected simply connected Lie group
$\mathcal{H}_{\sqrt{2}}\times\mathcal{H}_{2}$ with
$\sigma=\frac{3}{2}(\theta^{1})^{2}\theta^{2}$ up to isomorphism. The other
deviances are in fact obtained by taking $e^{i\alpha}\sigma$ and thus we are
in the situation noted in Remark 8.4. The Lie groups corresponding to the
cases VII, VIII and IX, are in particular homogeneous, and they all have zero
deviance, so by Proposition 9.5 we deduce that they are all isomorphic to
$\mathcal{H}_{\mathbb{C}}^{2}$ as projective special Kähler manifolds. ∎

######  Remark 10.3.

It is striking that in case III, which is obtained via the r-map from the
polynomial $x^{2}y$, the deviance is a global tensor which is a multiple of
this polynomial with respect to a Kähler holomorphic coframe.

It turns out that all $4$-dimensional projective special Kähler Lie groups are
simply connected, so this theorem already presents all possible cases.

######  Proposition 10.4.

Let $(\pi\colon\widetilde{M}\to M,\nabla)$ be a projective special Kähler
manifold, then the universal cover $p\colon U\to M$ admits a projective
special Kähler structure. In particular, if $\gamma\colon
S\to\sharp_{2}S_{3,0}M$ is the intrinsic deviance for $M$, then $p^{*}S\to U$
is an $S^{1}$-bundle and if we call $p^{\prime}$ the canonical map $p^{*}S\to
S$, then $U$ has deviance $p^{*}\circ\gamma\circ p^{\prime}\colon
p^{*}S\to\sharp_{2}S_{3,0}U$ on $U$.

If $M$ is a projective special Kähler Lie group, then so is $U$.

###### Proof.

Since $p\colon U\to M$ is a cover, we can lift the whole Kähler structure of
$M$ to $U$ by pullback $(U,p^{*}g,p^{*}I,p^{*}\omega)$ (the pullback of $I$
makes sense, since $p$ is a local diffeomorphism). We will now use Theorem
7.6. The $S^{1}$-bundle $S$ lifts to an $S^{1}$-bundle $\pi_{p^{*}S}\colon
p^{*}S\to U$, where the right action can be defined locally, since $p$ is a
local diffeomorphism. The principal connection $\varphi$ on $S$ lifts to
$\varphi^{\prime}=p^{\prime*}\varphi$ and its curvature is, as expected,
$d\varphi^{\prime}=p^{\prime*}d\varphi=-2p^{\prime}\pi_{S}^{*}\omega=-2\pi_{p^{*}S}^{*}p^{*}\omega$.
Let $\gamma^{\prime}=p^{*}\circ\gamma\circ p^{\prime}\colon
p^{*}S\to\sharp_{2}S_{3,0}U$, then
$\gamma^{\prime}(ua)=a^{2}\gamma^{\prime}(u)$ holds, as the action is defined
on the fibres, which are preserved by the pullback. The remaining properties
also follow from the fact $p$ is a local diffeomorphism.

Finally, if $M$ is a Lie group with left invariant Kähler structure, then $U$
is a Lie group and its Kähler structure is also left invariant. ∎

Given a universal cover $p\colon U\to M$ of a projective special Kähler Lie
group, $\ker(p)$ is a discrete subgroup and when $M$ is connected, $\ker(p)$
is in the centre $Z(U)$ of $U$.

From this observation we obtain the following corollary

######  Corollary 10.5.

A connected $4$-dimensional projective special Kähler Lie group is isomorphic
to one of the following:

  * •

$\mathcal{H}_{\sqrt{2}}\times\mathcal{H}_{2}$ with deviance
$\sharp_{2}(\frac{3}{2}(\theta^{1})^{2}\theta^{2})$ in the standard complex
unitary coframe $\theta$;

  * •

complex hyperbolic $2$-space with zero deviance.

###### Proof.

The proof follows from Theorem 10.2 with Proposition 10.4, as a connected
group $M$ with universal cover $p\colon U\to M$ is isomorphic to $U/\ker(p)$
and, if $M$ is a projective special Kähler Lie group, so is $U$ by Proposition
10.4. Since $U$ is also simply connected, Theorem 10.2 provides all the
possibilities up to isomorphisms preserving the Lie structure. The statement
follows from the fact that these possibilities for $U$ have trivial centre. ∎

##

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# On solutions of the Diophantine equation
$F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^{a}$

Pagdame TIEBEKABE $\&$ Ismaïla DIOUF

###### Abstract

Let $(F_{n})_{n\geq 0}$ be the Fibonacci sequence given by $F_{0}=0,F_{1}=1$
and $F_{n+2}=F_{n+1}+F_{n}$ for $n\geq 0$. In this paper, we solve all powers
of two which are sums of four Fibonacci numbers with a few exceptions that we
characterize.

Keywords: Linear forms in logarithm; Diophantine equations; Fibonacci
sequence; Lucas sequence; perfect powers.
2020 Mathematics Subject Classification: 11B39, 11J86, 11D61.

##  1 Introduction

The equation $F_{n}+F_{m}=y^{a}$, contrary to $F_{n}-F_{m}=y^{a}$ has been
less studied. For instance Z. Siar and R. Keskin [1] have found all the
solutions for $y=2$, B. Demirtürk et al [2] and P. Tiebekabe et al [3] have
independently determined all the solutions for y=3 and finally F. Erduvan and
R. Keskin [4] for the case $y=5$ and they presumed that there are no solutions
for $y>7$ prime. Concerning the works on the the sum of Fibonacci numbers, the
credit goes to Bravo and Luca [5] who were the first to tackle this equation
by resolving the equation $F_{n}+F_{m}=2^{a}$. The result has been generalized
by Pink and Ziegler in [6]. In 2015, J. J Bravo and E. Bravo have determined
in [8] the powers of $2$ as the sums of three Fibonacci numbers and in the
comments, they said that they hope it is possible to come to the end of the
equation

$F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^{a}$ (1)

in positive integers $n_{1},n_{2},n_{3},n_{4}\quad\text{and}\quad a$ with
$n_{1}\geqslant n_{2}\geqslant n_{3}\geqslant n_{4}$, using the same method.
But since the number of cases to be considered is difficult, they leave it for
other researchers. It is in that idea that we have decided to tackle this
difficult case.

Another similar problems as the one discussed in this paper were investigated
for the Fibonacci and Lucas sequences. For example, repdigits which are sums
of at most three Fibonacci numbers were found in [10]; repdigits as sums of
four Fibonacci or Lucas numbers were found in [13]; Fibonacci numbers which
are sums of two repdigits were obtained in [12], while factorials which are
sums of at most three Fibonacci numbers were found in [11].

After a deep analysis of works on powers of $2,3$ and $5$ as sums/differences
of Lucas or Fibonacci numbers, we think that it would be interesting to
generalize all perfect powers which are sums or differences of linear re-
currences. We should admit that recent work suggests that this will be
diffcult and may not happen any time soon. When solving $F_{n}\pm F_{m}=y^{a}$
in [7], the authors were compelled to use the $abc-$conjecture to show that
the equation admits finitely-many integer solutions $(n,m,y,a)$ with
$\min\\{y,a\\}\geqslant 2$.

Let’s recall that this conjecture is still an open problem and so is not yet
resolved. Recall that the Zeckendorf representation [9] of a positive integer
N is the representation

$N=F_{m_{1}}+F_{m_{2}}+\ldots+F_{m_{t}};\quad\text{with}\quad
m_{i}-m_{i+1}\geqslant 2\quad\text{for}\quad i=1,\ldots,t-1.$

Equation (1) is a particular case of Zeckendorf representation with $N=2^{a}$
and $t=4$.

This paper is subdivided as follows: In Section $2$, we introduce auxyliary
results used in Section $3$ to prove the main theorem of this paper stated
below.

######  Theorem 1.1.

All non-trivial solutions of the Diophantine equation (1) in positive integers
$n_{1},n_{2},n_{3},n_{4}$ and $a$ with $n_{1}\geqslant n_{2}\geqslant
n_{3}\geqslant n_{4}$ are:

$F_{5}+3F_{2}=2^{3}$ | $F_{7}+3F_{2}=2^{4}$ | $2F_{4}+2F_{2}=2^{3}$ | $F_{13}+F_{8}+2F_{2}=2^{8}$
---|---|---|---
$F_{4}+2F_{3}+F_{2}=2^{3}$ | $F_{6}+F_{5}+F_{3}+F_{2}=2^{4}$ | $F_{8}+F_{6}+F_{3}+F_{2}=2^{5}$ | $F_{16}+F_{9}+F_{3}+F_{2}=2^{10}$
$F_{10}=F_{5}+F_{4}+F_{2}=2^{6}$ | $3F_{5}+F_{2}=2^{4}$ | $F_{8}+2F_{5}+F_{2}=2^{5}$ | $2F_{7}+F_{5}+F_{2}=2^{5}$
$F_{9}+F_{8}+F_{6}+F_{2}=2^{6}$ | $3F_{8}+F_{2}=2^{6}$ | $F_{10}+F_{5}+2F_{3}=2^{6}$ | $F_{6}+2F_{4}+F_{3}=2^{4}$
$F_{11}+F_{9}+F_{4}+F_{3}=2^{7}$ | $F_{13}+F_{7}+F_{6}+F_{3}=2^{8}$ | $F_{12}+F_{11}+F_{8}+F_{3}=2^{8}$ | $F_{12}+2F_{10}+F_{3}=2^{8}$
$F_{10}+3F_{4}=2^{6}$ | $2F_{5}+2F_{4}=2^{4}$ | $F_{8}+F_{5}+2F_{4}=2^{5}$ | $2F_{7}+2F_{4}=2^{5}$
$F_{7}+2F_{6}+F_{4}=2^{5}$ | $F_{16}+F_{8}+F_{7}+F_{4}=2^{10}$ | $F_{15}+F_{14}+F_{9}+F_{4}=2^{10}$ | $F_{13}+F_{7}+2F_{5}=2^{8}$
$F_{11}+F_{8}+F_{7}+F_{5}=2^{7}$ | $2F_{10}+F_{7}+F_{5}=2^{7}$ | $F_{10}+2F_{9}+F_{5}=2^{7}$ | $F_{16}+F_{8}+2F_{6}=2^{10}$
$F_{11}+3F_{7}=2^{7}$ | $4F_{6}=2^{5}$ |  |

##  2 Auxiliary results

In this section, we give some important known definitions, proprieties,
theorem and lemmas.

######  Definition 2.1.

For all algebraic numbers $\gamma$, we define its measure by the following
identity :

${\rm M}(\gamma)=|a_{d}|\prod\limits_{i=1}^{d}\max\\{1,|\gamma_{i}|\\},$

where $\gamma_{i}$ are the roots of
$f(x)=a_{d}\prod\limits_{i=1}^{d}(x-\gamma_{i})$ is the minimal polynomial of
$\gamma$.

Let us define now another height, deduced from the last one, called the
absolute logarithmic height.

######  Definition 2.2.

( Absolute logarithmic height)

For a non-zero algebraic number of degree $d$ on $\mathbb{Q}$ where the
minimal polynomial on $\mathbb{Z}$ is
$f(x)=a_{d}\prod\limits_{i=1}^{d}(x-\gamma_{i})$, we denote by

$h(\gamma)=\dfrac{1}{d}\left(\log|a_{d}|+\sum\limits_{i=1}^{d}\log\max\\{1,|\gamma_{i}|\\}\right)=\dfrac{1}{d}\log{\rm
M}(\gamma).$

the usual logarithmic absolute height of $\gamma$.

The following properties of the logarithmic height are well-known:

  * •

$h(\gamma\pm\eta)\leq h(\gamma)+h(\eta)+\log 2$;

  * •

$h(\gamma\eta^{\pm 1})\leq h(\gamma)+h(\eta)$;

  * •

$h(\gamma^{k})=|k|h(\gamma)\quad k\in\mathbb{Z}$.

The $n$th Fibonacci number can be represented as

$F_{n}=\dfrac{\alpha^{n}-\beta^{n}}{\sqrt{5}}\quad\text{for all}\quad
n\geqslant 0.$

where $(\alpha,\beta):=((1+\sqrt{5})/2,(1-\sqrt{5})/2)$. The following
inequalities

$\alpha^{n-2}\leqslant F_{n}\leqslant\alpha^{n-1}$

are well-known to hold for all $n\geqslant 1$ and can be proved by induction
on $n$. The following theorem is deduced from Corollary $2.3$ of Matveev [15].

######  Theorem 2.1.

(Matveev [15])

Let $n\geq 1$ an integer. Let $\mathbb{L}$ be a field of algebraic number of
degree $D$. Let $\eta_{1}$, …, $\eta_{l}$ non-zero elements of $\mathbb{L}$
and let $b_{1}$, $b_{2}$, …, $b_{l}$ integers,

$B:=\max\\{|b_{1}|,...,|b_{l}|\\},$

and

$\Lambda:=\eta_{1}^{b_{1}}\cdots\eta_{l}^{b_{l}}-1=\left(\prod\limits_{i=1}^{l}\eta_{i}^{b_{i}}\right)-1.$

Let $A_{1}$, …, $A_{l}$ reals numbers such that

$A_{j}\geq\max\\{Dh(\eta_{j}),|\log(\eta_{j})|,0.16\\},1\leq j\leq l.$

Assume that $\Lambda\neq 0$, So we have

$\log|\Lambda|>-3\times 30^{l+4}\times(l+1)^{5.5}\times d^{2}\times
A_{1}...A_{l}(1+\log D)(1+\log nB).$

Further, if $\mathbb{L}$ is real, then

$\log|\Lambda|>-1.4\times 30^{l+3}\times(l)^{4.5}\times d^{2}\times
A_{1}...A_{l}(1+\log D)(1+\log B).$

The two following Lemmas are due Dujella and Pethő, and to Legendre
respectively.

For a real number $X$, we write $\left\lVert X\right\rVert:=\min\\{\mid
X-n\mid:n\in\mathbb{Z}\\}$ for the distance of $X$ to the nearest integer.

######  Lemma 2.1.

(Dujella and Pethő, [14])

Let $M$ a positive integer, let $p/q$ the convergent of the continued fraction
expansion of $\kappa$ such that $q>6M$ and let $A$, $B$, $\mu$ real numbers
such that $A>0$ and $B>1$. Let $\varepsilon:=\left\lVert\mu
q\right\rVert-M\left\lVert\kappa q\right\rVert$.
If $\varepsilon>0$ then there is no solution of the inequality

$0<m\kappa-n+\mu<AB^{-m}$

in integers $m$ and $n$ with

$\dfrac{\log(Aq/\varepsilon)}{\log B}\leqslant m\leqslant M.$

######  Lemma 2.2.

(Legendre)

Let $\tau$ real number such that $x$, $y$ are integers such that

$\left|\tau-\dfrac{x}{y}\right|<\dfrac{1}{2y^{2}},$

then $\dfrac{x}{y}=\dfrac{p_{k}}{q_{k}}$ is the convergence of $\tau$.

Further,

$\left|\tau-\dfrac{x}{y}\right|>\dfrac{1}{(q_{k+1}+2)y^{2}}.$

##  3 Main result

###### Proof.

Assume that

$F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^{a}$

holds.
Let us first find relation between $n_{1}$ and $a$.

Combining equation(1) with the well-known inequality
$F_{n}\leqslant\alpha^{n-1}$ for all $n\geqslant 1$, one gets that

$\displaystyle F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^{a}\leqslant$
$\displaystyle\alpha^{n_{1}-1}+\alpha^{n_{2}-1}+\alpha^{n_{3}-1}+\alpha^{n_{4}-1}$
$\displaystyle<$ $\displaystyle
2^{n_{1}-1}+2^{n_{2}-1}+2^{n_{3}-1}+2^{n_{4}-1}\quad\because\alpha<2$
$\displaystyle<$ $\displaystyle
2^{n_{1}-1}\left(1+2^{n_{2}-n_{1}}+2^{n_{3}-n_{1}}+2^{n_{4}-n_{1}}\right)$
$\displaystyle\leqslant$ $\displaystyle
2^{n_{1}-1}\left(1+1+2^{-1}+2^{-2}\right)=2^{n_{1}-1}\left(2+2^{-1}+2^{-2}\right)$
$\displaystyle<$ $\displaystyle 2^{n_{1}+1}.$

Hence

$2^{a}<2^{n_{1}+1}\Longrightarrow a<n_{1}+1\Longrightarrow a\leqslant n_{1}.$

This inequality will help us to calculate some parameters.

Rewriting equation (1), we get

$\dfrac{\alpha^{n_{1}}}{\sqrt{5}}-2^{a}=\dfrac{\beta^{n_{1}}}{\sqrt{5}}-(F_{n_{2}}+F_{n_{3}}+F_{n_{4}}).$

Taking absolute values on the above equation, we obtain

$\left|\dfrac{\alpha^{n_{1}}}{\sqrt{5}}-2^{a}\right|\leqslant\left|\dfrac{\beta^{n_{1}}}{\sqrt{5}}\right|+(F_{n_{2}}+F_{n_{3}}+F_{n_{4}})<\dfrac{|\beta|^{n_{1}}}{\sqrt{5}}+(\alpha^{n_{2}}+\alpha^{n_{3}}+\alpha^{n_{4}}),$

and

$\left|\dfrac{\alpha^{n_{1}}}{\sqrt{5}}-2^{a}\right|<\dfrac{1}{2}+\left(\alpha^{n_{2}}+\alpha^{n_{3}}+\alpha^{n_{4}}\right)\quad\text{where
we used}\quad F_{n}\leqslant\alpha^{n-1}.$

Dividing both side of the above equation by $\alpha^{n_{1}}/\sqrt{5}$, we get

$\left|1-2^{a}\cdot\alpha^{-n_{1}}\cdot\sqrt{5}\right|<\dfrac{\sqrt{5}}{2\alpha^{n_{1}}}+\left(\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}+\alpha^{n_{4}-n_{1}}\right)\sqrt{5}<\dfrac{\sqrt{5}}{2\alpha^{n_{1}}}+\dfrac{\sqrt{5}}{\alpha^{n_{1}-n_{2}}}+\dfrac{\sqrt{5}}{\alpha^{n_{1}-n_{3}}}+\dfrac{\sqrt{5}}{\alpha^{n_{1}-n_{4}}}.$

Taking account the assumption $n_{4}\leqslant n_{3}\leqslant n_{3}\leqslant
n_{2}\leqslant n_{1}$, we get

$|\Lambda_{1}|=\left|1-2^{a}\cdot\alpha^{-n_{1}}\cdot\sqrt{5}\right|<\dfrac{9}{\alpha^{n_{1}-n_{2}}}$
(2)

Let apply Matveev’s theorem, with the following parameters $t:=3$ and

$\gamma_{1}:=2,\quad\gamma_{2}:=\alpha,\quad\gamma_{3}:=\sqrt{5},\quad
b_{1}:=a,\quad b_{2}:=-n,\quad\text{and}\quad b_{3}:=1.$

Since $\gamma_{1},\gamma_{2},\gamma_{3}\in\mathbb{K}:=\mathbb{Q}(\sqrt{5}),$
we can take $D:=2$. Before to apply Matveev’s theorem, we have to check the
last condition: the left-hand side of (2) is not zero. Indeed, if it were
zero, we would then get that $2^{a}\sqrt{5}=\alpha^{n}$. Squaring the previous
relation, we get $\alpha^{2n}=5\cdot 2^{2a}=5\cdot 4^{a}$. This implies that
$\alpha^{2n}\in\mathbb{Z}$. Which is impossible. Then $\Lambda_{1}\neq 0$. The
logarithmic height of $\gamma_{1},\gamma_{2}$ and $\gamma_{3}$ are:

$h(\gamma_{1})=\log 2=0.6931\ldots$, so we can choose $A_{1}:=1.4$.

$h(\gamma_{2})=\dfrac{1}{2}\log\alpha=0.2406\ldots$, so we can choose
$A_{2}:=0.5$.

$h(\gamma_{3})=\log\sqrt{5}=0.8047\ldots$, it follows that we can choose
$A_{3}:=1.7$.

Since $a<n_{1}+1$, $B:=\max\\{|b_{1}|,|b_{2}|,|b_{3}|\\}=n_{1}.$ Matveev’s
result informs us that

$\left|1-2^{a}\cdot\alpha^{n_{1}}\cdot\sqrt{5}\right|>\exp\left(-c_{1}\cdot(1+\log
n)\cdot 1.4\cdot 0.5\cdot 1.7\right),$ (3)

where $c_{1}:=1.4\cdot 30^{6}\cdot 3^{4.5}\cdot 2^{2}\cdot(1+\log 2)<9.7\times
10^{11}.$

Taking $\log$ in inequality (2), we get

$\log|\Lambda_{1}|<\log 9-(n_{1}-n_{2})\log\alpha.$

Taking $\log$ in inequality (3), we get

$\log|\Lambda_{1}|>2.31\times 10^{12}\log n_{1}.$

Comparing the previous two inequality, we get

$(n_{1}-n_{2})\log\alpha-\log 9<2.31\times 10^{12}\log n_{1},$

where we used $1+\log n_{1}<2\log n_{1}$ which holds for all $n_{1}\geqslant
3$. Then we have

$(n_{1}-n_{2})\log\alpha<2.32\times 10^{12}\log n_{1}.$ (4)

Let us now consider a second linear form in logarithms. Rewriting equation (1)
as follows

$\dfrac{\alpha^{n_{1}}}{\sqrt{5}}+\dfrac{\alpha^{n_{2}}}{\sqrt{5}}-2^{a}=\dfrac{\beta^{n_{1}}}{\sqrt{5}}+\dfrac{\beta^{n_{2}}}{\sqrt{5}}-\left(F_{n_{3}}+F_{n_{4}}\right).$

Taking absolute values on the above equation and the fact that
$\beta=(1-\sqrt{5})/2$, we get

$\displaystyle\left|\dfrac{\alpha^{n_{1}}}{\sqrt{5}}\left(1+\alpha^{n_{2}-n_{1}}\right)-2^{a}\right|\leqslant$
$\displaystyle\dfrac{|\beta|^{n_{1}}+|\beta|^{n_{2}}}{\sqrt{5}}+F_{n_{3}}+F_{n_{4}}$
$\displaystyle<$
$\displaystyle\dfrac{1}{3}+\alpha^{n_{3}}+\alpha^{n_{4}}\quad\text{for
all}\quad n_{1}\geqslant 5\quad\text{and}\quad n_{2}\geqslant 5.$

Dividing both sides of the above inequality by
$\dfrac{\alpha^{n_{1}}}{\sqrt{5}}\left(1+\alpha^{n_{2}-n_{1}}\right)$, we
obtain

$|\Lambda_{2}|=\left|1-2^{a}\cdot\alpha^{n_{1}}\cdot\sqrt{5}\left(1+\alpha^{n_{2}-n_{1}}\right)^{-1}\right|<\dfrac{6}{\alpha^{n_{2}-n_{1}}}.$
(5)

Let apply Matveev’s theorem second time with the following data

$t:=3,\quad\gamma_{1}:=2,\quad\gamma_{2}:=\alpha,\quad\gamma_{3}:=\sqrt{5}\left(1+\alpha^{n_{2}-n_{1}}\right)^{-1},\quad
b_{1}:=a,\quad b_{2}:=-n_{1},\quad\text{and}\quad b_{3}:=1.$

Since $\gamma_{1},\gamma_{2},\gamma_{3}\in\mathbb{K}:=\mathbb{Q}(\sqrt{5}),$
we can take $D:=2$. The left hand side of (5) is not zero, otherwise, we would
get the relation

$2^{a}\sqrt{5}=\alpha^{n_{1}}+\alpha^{n_{2}}.$ (6)

Conjugating (6) in the field $\mathbb{Q}(\sqrt{5})$, we get

$-2^{a}\sqrt{5}=\beta^{n_{1}}+\beta^{n_{2}}.$ (7)

Combining (6) and (7), we get

$\alpha^{n_{1}}<\alpha^{n_{1}}+\alpha^{n_{2}}=|\beta^{n_{1}}+\beta^{n_{2}}|\leqslant|\beta|^{n_{1}}+|\beta|^{n_{2}}<1$

which is impossible for $n_{1}>350$. Hence $\Lambda_{2}\neq 0$. We know that,
$h(\gamma_{1})=\log 2$ and $h(\gamma_{2})=\dfrac{1}{2}\log\alpha$. Let us now
estimate $h(\gamma_{3})$ by first observing that

$\gamma_{3}=\dfrac{\sqrt{5}}{1+\alpha^{n_{2}-n_{1}}}<\sqrt{5}\quad\text{and}\quad\gamma_{3}^{-1}=\dfrac{1+\alpha^{n_{2}-n_{1}}}{\sqrt{5}}<\dfrac{2}{\sqrt{5}},$

so that $|\log\gamma_{3}|<1.$ Using proprieties of logarithmic height stated
in Section $2$, we have

$h(\gamma_{3})\leqslant\log\sqrt{5}+|n_{2}-n_{1}|\left(\dfrac{\log\alpha}{2}\right)+\log
2=\log(2\sqrt{5})+(n_{1}-n_{2})\left(\dfrac{\log\alpha}{2}\right).$

Hence, we can take
$A_{3}:=3+(n_{1}-n_{2})\log\alpha>\max\\{2h(\gamma_{3}),|\log\gamma_{3}|,0.16\\}$.

Matveev’s theorem implies that

$\exp\left(-c_{2}(1+\log n_{1})\cdot 1.4\cdot
0.5\cdot(3+(n_{1}-n_{2})\log\alpha)\right)$

where $c_{2}:=1.4\cdot 30^{6}\cdot 3^{4.5}\cdot 2^{2}\cdot(1+\log 2)<9.7\times
10^{11}.$

Since $(1+\log n_{1})<2\log n_{1}$ hold for $n_{1}\geqslant 3,$ from (5), we
have

$(n_{1}-n_{3})\log\alpha-\log 6<1.4\times 10^{12}\log
n_{1}(3+(n_{1}-n_{2})\log\alpha).$ (8)

Putting relation (4) in the right-hand side of (8), we get

$(n_{1}-n_{3})\log\alpha<3.29\times 10^{24}\log^{2}n_{1}.$ (9)

Let us consider a third linear form in logarithms. To this end, we again
rewrite (1) as follows

$\dfrac{\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}}}{\sqrt{5}}-2^{a}=\dfrac{\beta^{n_{1}}+\beta^{n_{2}}+\beta^{n_{3}}}{\sqrt{5}}-F_{n_{4}}.$

Taking absolute values on both sides, we obtain

$\displaystyle\left|\dfrac{\alpha^{n_{1}}}{\sqrt{5}}\left(1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}\right)-2^{a}\right|\leqslant$
$\displaystyle\dfrac{|\beta|^{n_{1}}+|\beta|^{n_{2}}+|\beta|^{n_{3}}}{\sqrt{5}}+F_{n_{4}}$
$\displaystyle<$ $\displaystyle\dfrac{3}{4}+\alpha^{n_{4}}\quad\text{for
all}\quad n_{1}>350,\quad\text{and}\quad n_{2},n_{3},n_{4}\geqslant 1.$

Thus we have

$|\Lambda_{3}|=\left|1-2^{a}\cdot\alpha^{-n_{1}}\cdot\sqrt{5}\left(1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}\right)^{-1}\right|<\dfrac{3}{\alpha^{n_{1}-n_{4}}}.$
(10)

In a third application of Matveev’s theorem, we can take parameters

$t:=3,\quad\gamma_{1}:=2,\quad\gamma_{2}:=\alpha,\quad\gamma_{3}:=\sqrt{5}\left(1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}\right)^{-1},\quad
b_{1}:=a,\quad b_{2}:=-n,\quad\text{and},\quad b_{3}:=1.$

Since $\gamma_{1},\gamma_{2},\gamma_{3}\in\mathbb{K}:=\mathbb{Q}(\sqrt{5}),$
we can take $D:=2$. The left hand side of (10) is not zero. The proof is done
by contradiction. Suppose the contrary. Then

$2^{a}\sqrt{5}=\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}}.$

Taking the conjugate in the field $\mathbb{Q}(\sqrt{5})$, we get

$-2^{a}\sqrt{5}=\beta^{n_{1}}+\beta^{n_{2}}+\beta^{n_{3}},$

which leads to

$\alpha^{n_{1}}<\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}}=|\beta^{n_{1}}+\beta^{n_{2}}+\beta^{n_{3}}|\leqslant|\beta|^{n_{1}}+|\beta|^{n_{2}}+|\beta|^{n_{3}}<1$

and leads to a contradiction since $n_{1}>350.$ Hence $\Lambda_{3}\neq 0$.

As before we did, we can take $A_{1}:=1.4,A_{2}:=0.5$ and $B:=n_{1}.$ We can
also see that

$\gamma_{3}=\dfrac{\sqrt{5}}{1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}}<\sqrt{5}\quad\text{and}\quad\gamma_{3}^{-1}=\dfrac{1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}}{\sqrt{5}}<\dfrac{3}{\sqrt{5}},$

so $|\log\gamma_{3}|<1$. Applying proprieties on logarithmic height, we
estimate $h(\gamma_{3})$. Hence

$\displaystyle h(\gamma_{3})\leqslant$
$\displaystyle\log\sqrt{5}+|n_{2}-n_{1}|\left(\dfrac{\log\alpha}{2}\right)+|n_{3}-n_{1}|\left(\dfrac{\log\alpha}{2}\right)+\log
3$ $\displaystyle=$
$\displaystyle\log(3\sqrt{5})+(n_{1}-n_{2})\left(\dfrac{\log\alpha}{2}\right)+(n_{1}-n_{3})\left(\dfrac{\log\alpha}{2}\right);$

so we can take

$A_{3}:=4+(n_{1}-n_{2})\log\alpha+(n_{1}-n_{3})\log\alpha>\max\\{2h(\gamma_{3}),|\log\gamma_{3}|,0.16\\}.$

A lower bound on the left-hand side of (10) is

$\exp(-c_{3}\cdot(1+\log n_{1})\cdot 1.4\cdot
0.5\cdot(4+(n_{1}-n_{2})\log\alpha+(n_{1}-n_{3})\log\alpha))$

where $c_{3}=1.4\cdot 30^{6}\cdot 3^{4.5}\cdot 2^{2}\cdot(1+\log 2)<9.7\times
10^{11}.$

From inequality (10), we have

$(n_{1}-n_{4})\log\alpha<1.4\times 10^{12}\log
n_{1}\cdot(4+(n_{1}-n_{2})\log\alpha+(n_{1}-n_{3})\log\alpha).$ (11)

Combining equation (4) and (9) in the right-most terms of equation (11) and
performing the respective calculations, we get

$(n_{1}-n_{4})\log\alpha<9.3\times\log^{3}n_{1}.$ (12)

Let us now consider a forth and last linear form in logarithms. Rerwriting (1)
once again by separating large terms and small terms, we get

$\dfrac{\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}}+\alpha^{n_{4}}}{\sqrt{5}}-2^{a}=\dfrac{\beta^{n_{1}}+\beta^{n_{2}}+\beta^{n_{3}}+\beta^{n_{4}}}{\sqrt{5}}.$

Taking absolute values on both sides, we get

$\displaystyle\left|\dfrac{\alpha^{n_{1}}}{\sqrt{5}}\left(1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}+\alpha^{n_{4}-n_{1}}\right)-2^{a}\right|\leqslant$
$\displaystyle\dfrac{|\beta|^{n_{1}}+|\beta|^{n_{2}}+|\beta|^{n_{3}}+|\beta|^{n_{4}}}{\sqrt{5}}$
$\displaystyle<$ $\displaystyle\dfrac{4}{5}\quad\text{for all}\quad
n_{1}>350,\quad\text{and}\quad n_{2},n_{3},n_{4}\geqslant 1.$

Dividing both sides of the above relation by the fist term of the RHS of the
previous equation, we get

$|\Lambda_{4}|=\left|1-2^{a}\cdot\alpha^{-n_{1}}\cdot\sqrt{5}\left(1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}+\alpha^{n_{4}-n_{1}}\right)^{-1}\right|<\dfrac{2}{\alpha^{n_{1}}}.$
(13)

In the last application of Matveev’s theorem, we have the following parameters

$\gamma_{1}:=2,\quad\gamma_{2}:=\alpha,\quad\gamma_{3}:=\sqrt{5}\left(1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}+\alpha^{n_{4}-n_{1}}\right)^{-1},$

and we can also take $b_{1}:=a,\quad b_{2}:=-n$ and $b_{3}:=1.$ Since
$\gamma_{1},\gamma_{2},\gamma_{3}\in\mathbb{K}:=\mathbb{Q}(\sqrt{5}),$ we can
take $D:=2$. The left hand side of (13) is not zero. The proof is done by
contradiction. Suppose the contrary. Then

$2^{a}\sqrt{5}=\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}}+\alpha^{n_{4}}.$

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

$-2^{a}\sqrt{5}=\beta^{n_{1}}+\beta^{n_{2}}+\beta^{n_{3}}+\beta^{n_{4}}.$

Combining the above two equations, we get

$\alpha^{n_{1}}<\alpha^{n_{1}}+\alpha^{n_{2}}+\alpha^{n_{3}}+\alpha^{n_{4}}=|\beta^{n_{1}}+\beta^{n_{2}}+\beta^{n_{3}}+\beta^{n_{4}}|\leqslant|\beta|^{n_{1}}+|\beta|^{n_{2}}+|\beta|^{n_{3}}+|\beta|^{n_{4}}<1,$

and leads to contradiction since $n_{1}>350.$

As before, here, we can take $A_{1}:=1.4,A_{2}:=0.5$ and $B:=n_{1}$. Let us
estimate $h(\gamma_{3})$. We can see that,

$\gamma_{3}=\dfrac{\sqrt{5}}{1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}+\alpha^{n_{4}-n_{1}}}<\sqrt{5}\quad\text{and}\quad\gamma_{3}^{-1}=\dfrac{1+\alpha^{n_{2}-n_{1}}+\alpha^{n_{3}-n_{1}}+\alpha^{n_{4}-n_{1}}}{\sqrt{5}}<\dfrac{4}{\sqrt{5}}.$

Hence $|\log\gamma_{3}|<1.$ Then

$\displaystyle h(\gamma_{3})\leqslant$
$\displaystyle\log(4\sqrt{5})+|n_{2}-n_{1}|\left(\dfrac{\log\alpha}{2}\right)+|n_{3}-n_{1}|\left(\dfrac{\log\alpha}{2}\right)+|n_{4}-n_{1}|\left(\dfrac{\log\alpha}{2}\right)$
$\displaystyle=$
$\displaystyle\log(4\sqrt{5})+(n_{1}-n_{2})\left(\dfrac{\log\alpha}{2}\right)+(n_{1}-n_{3})\left(\dfrac{\log\alpha}{2}\right)+(n_{1}-n_{4})\left(\dfrac{\log\alpha}{2}\right);$

so we can take

$A_{3}:=5+(n_{1}-n_{2})\log\alpha+(n_{1}-n_{3})\log\alpha+(n_{1}-n_{4})\log\alpha.$

Then a lower bound on the left-hand side of (13) is

$\exp(-c_{4}\cdot(1+\log n_{1})\cdot 1.4\cdot
0.5\cdot(5+(n_{1}-n_{2})\log\alpha+(n_{1}-n_{3})\log\alpha+(n_{1}-n_{4})\log\alpha)),$

where $c_{4}=1.4\cdot 30^{6}\cdot 3^{4.5}\cdot 2^{2}\cdot(1+\log 2)<9.7\times
10^{11}.$

So, inequality (13) yields

$n_{1}\log\alpha<1.4\times 10^{12}\log
n_{1}\cdot(5+(n_{1}-n_{2})\log\alpha+(n_{1}-n_{3})\log\alpha+(n_{1}-n_{4})\log\alpha).$
(14)

Using now (4), (9) and (12) in the right-most terms of the above inequality
(14) and performing the respective calculation, we find that

$n_{1}\log\alpha<40.32\times 10^{48}\log^{4}n_{1}.$

With the help of Mathematica, we get from the previous inequality

$n<2.8\times 10^{58}.$

We record what we have proved.

######  Lemma 3.1.

If $(n_{1},n_{2},n_{3},n_{4},a)$ is a positive solution of (1) with
$n_{1}\geqslant n_{2}\geqslant n_{3}\geqslant n_{4}$, then

$a\leqslant n_{1}<2.8\times 10^{58}.$

##  4 Reduction the bound on $n$

The goal of this section is reduce the upper bound on $n$ to a size that can
be handled. To do this, we shall use Lemma 2.1 four times. Let us consider

$z_{1}:=a\log 2-n_{1}\log\alpha+\log\sqrt{5}.$ (15)

From equation (15), (2) can be written as

$\left|1-e^{z_{1}}\right|<\dfrac{9}{\alpha^{n_{1}-n_{2}}}.$ (16)

Associating (1) and Binet’s formula for the Fibonacci sequence, we have

$\dfrac{\alpha^{n_{1}}}{\sqrt{5}}=F_{n_{1}}+\dfrac{\beta^{n_{1}}}{\sqrt{5}}<F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^{a},$

hence

$\dfrac{\alpha^{n_{1}}}{\sqrt{5}}<2^{a},$

which leads to $z_{1}>0.$ This result together with (16), give

$0<z_{1}<e^{z_{1}}-1<\dfrac{9}{\alpha^{n_{1}-n_{2}}}.$

Replacing (15) in the inequality and dividing both sides of the resulting
inequality by $\log\alpha$, we get

$0<a\left(\dfrac{\log
2}{\log\alpha}\right)-n+\left(\dfrac{\log\sqrt{5}}{\log\alpha}\right)<\dfrac{9}{\log\alpha}\cdot\alpha^{n_{1}-n_{2}}<19\cdot\alpha^{n_{1}-n_{2}}.$
(17)

We put

$\tau:=\dfrac{\log
2}{\log\alpha},\quad\mu:=\dfrac{\log\sqrt{5}}{\log\alpha},\quad
A:=19,\quad\text{and}\quad B:=\alpha.$

$\tau$ is an irrational number. We also put $M:=2.8\times 10^{58}$, which is
an upper bound on $a$ by Lemma 2.1 applied to inequality, that

$n_{1}-n_{2}<\dfrac{\log(Aq/\varepsilon)}{\log B},$

where $q>6M$ is a denominator of a convergent of the continued fraction of
$\tau$ such that $\varepsilon:=\left\lVert\mu q\right\rVert-M\left\lVert\tau
q\right\rVert>0.$ A computation with SageMath revealed that if
$(n_{1},n_{2},n_{3},n_{4},a)$ is a possible solution of the equation 1, then

$n_{1}-n_{2}\in[0,314].$

Let us now consider a second function, derived from (5) in order to find an
improved upper bound on $n_{1}-n_{2}$.

Put

$z_{2}:=a\log 2-n_{1}\log\alpha+\log\Upsilon_{1}(n_{1}-n_{2})$

where $\Upsilon$ is the function given by the formula
$\Upsilon(t):=\sqrt{5}\left(1+\alpha^{-t}\right)^{-1}.$ From (5), we have

$\left|1-e^{z_{2}}\right|<\dfrac{6}{\alpha^{n_{1}-n_{3}}}.$ (18)

Using (1) and the Binet’s formula for the Fibonacci sequence, we have

$\dfrac{\alpha^{n_{1}}}{\sqrt{5}}+\dfrac{\alpha^{n_{2}}}{\sqrt{5}}=F_{n_{1}}+F_{n_{2}}+\dfrac{\beta^{n_{1}}}{\sqrt{5}}+\dfrac{\beta^{n_{2}}}{\sqrt{5}}<F_{n_{1}}+F_{n_{2}}+1\leqslant
F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^{a}.$

Therefore
$1<2^{a}\sqrt{5}\alpha^{-n_{1}}\left(1+\alpha^{n_{2}-n_{1}}\right)^{-1}$ and
so $z_{2}>0$. This with (18) give

$0<z_{2}\leqslant e^{z_{2}}-1<\dfrac{6}{\alpha^{n_{1}-n_{3}}}.$

Putting expression of $z_{2}$ in the above inequality and arguing as in (17),
we obtain

$0<a\left(\dfrac{\log
2}{\log\alpha}\right)-n_{1}+\dfrac{\log\Upsilon_{1}(n_{1}-n_{2})}{\log\alpha}<13\cdot\alpha^{-(n_{1}-n_{3})}.$
(19)

As before, we take again $M:=2.8\times 10^{58}$ which is the upper bound on
$a$, and, as explained before, we apply Lemma 2.1 to inequamity (19) for all
choices $n_{1}-n_{2}\in[0,314]$ except when $n_{1}-n_{2}=2,6$. With the help
of SageMath, we find that if $(n_{1},n_{2},n_{3},n_{4},a)$ is a possible
solution of the equation (1) with $n_{1}-n_{2}\neq 2$ and $n_{1}-n_{2}\neq 6$,
then $n_{1}-n_{3}\in[0,314]$.

Study of the cases $n_{1}-n_{2}\in\\{2,6\\}$. For these cases, when we apply
Lemma 2.1 to the expression (19), the corresponding parameter $\mu$ appearing
in Lemma 2.1 is

$\dfrac{\log\Upsilon_{1}(t)}{\log\alpha}=\left\\{\begin{array}[]{lllll}1&\text{if}&t=2;\\\
3-\dfrac{\log 2}{\log\alpha}&\text{if}&t=6.\end{array}\right.$

In both case, the parameters $\tau$ and $\mu$ are linearly dependent, which
yields that the corresponding value of $\varepsilon$ from Lemma 2.1 is always
negative and therefore the reduction method is not useful for reducing the
bound on $n$ in these instances. For this, we need to treat these case
differently.

However, we can see that if $t=2$ and $6$, then resulting inequality from (19)
has the shape $0<|x\tau-y|<13\cdot\alpha^{-(n_{1}-n_{3})}$ with $\tau$ being
irrational number and $x,y\in\mathbb{Z}$. Then, using the known proprieties of
the convergents of the continued fractions to obtain a nontrivial lower bound
for $|x\tau-y|$. Let see how to do.

For $n_{1}-n_{2}=2$, from (19), we get that

$0<a\tau-(n_{1}-1)<13\cdot\alpha^{-(n_{1}-n_{3})},\quad\text{where}\quad\tau=\dfrac{\log
2}{\log\alpha}.$ (20)

Let $[a_{1},a_{2},a_{3},a_{4},\ldots]=[1,2,3,1,\ldots]$ be the continued
fraction of $\tau$, and let denote $p_{k}/q_{k}$ it $k$th convergent. By Lemma
2.2, we know that $a<2.8\times 10^{58}$. An inspection in SageMath reveals
that

$1207471144047491451512110092657730332808809199105354185685=q_{113}<2.8\times
10^{58}<$
$q_{114}=28351096929195187169517686575841899309129196859170938821667.$

Furthermore, $a_{M}:=\max\\{a_{i}:i=0,1,\ldots,114\\}=134.$ So, from the
proprieties of continued fractions, we obtain that

$|a\tau-(n_{1}-1)|>\dfrac{1}{(a_{M}+2)a}.$ (21)

Comparing (20) and (21), we get

$\alpha^{n_{1}-n_{3}}<13\cdot(134+2)a.$

Taking $\log$ on both sides of above equation and the divide the obtained
result by $\log\alpha$, we get

$n_{1}-n_{3}<296.$

In order to avoid repetition, we freely omits the details for the case
$n_{1}-n_{2}=6.$ Here, we get $n_{1}-n_{3}<314$.

This completes the analysis of the two special cases $n_{1}-n_{2}=2$ and
$n_{1}-n_{2}=6.$ Consequently $n_{1}-n_{3}\leqslant 314$ always holds.

Now let us use (10) in order to find improved upper bound on $n_{1}-n_{4}.$
Put

$z_{3}:=a\log 2-n_{1}\log\alpha+\log\Upsilon_{2}(n_{1}-n_{2},n_{1}-n_{3}),$

where $\Upsilon_{2}$ is the function given by the formula
$\Upsilon_{2}(t,s):=\sqrt{5}\left(1+\alpha^{-t}+\alpha^{-s}\right)^{-1}.$ From
(10), we have

$|1-e^{z_{3}}|<\dfrac{3}{\alpha^{n_{1}-n_{4}}}.$ (22)

Note that, $z_{3}\neq 0$; thus, two cases arises: $z_{3}>0$ and $z_{3}<0$.

If $z_{3}>0,$ then

$0<z_{3}\leqslant e^{z_{3}}-1<\dfrac{3}{\alpha^{n_{1}-n_{4}}}.$

Suppose now $z_{3}<0.$ It is easy to check that $3/\alpha^{n_{1}-n_{4}}<1/2$
for all $n_{1}>350$ and $n_{4}\geqslant 2$. From (22), we have that

$|1-e^{z_{3}}|<1/2\quad\text{and therefore}\quad e^{|z_{3}|}<2.$

Since $z_{3}<0$, we have:

$0<|z_{3}|\leqslant
e^{|z_{3}|}-1=e^{|z_{3}|}\left|e^{|z_{3}|}-1\right|<\dfrac{6}{\alpha^{n_{1}-n_{4}}}$

which give

$0<|z_{3}|<\dfrac{6}{\alpha^{n_{1}-n_{4}}}$

holds for $z_{3}<0$, $z_{3}>0$ and for all for all $n_{1}>350$, and
$n_{4}\geqslant 2$. Replacing the expression of $z_{3}$ in the above
inequality and arguing again as before, we conclude that

$0<\left|a\left(\dfrac{\log
2}{\log\alpha}\right)-n_{1}+\dfrac{\log\Upsilon_{2}(n_{1}-n_{2},n_{1}-n_{3})}{\log\alpha}\right|<13\cdot\alpha^{-(n_{1}-n_{4})}.$
(23)

Here, we also take, $M:=2.8\times 10^{58}$ and we apply Lemma 2.1 in
inequality (23) for all choices $n_{1}-n_{2}\in\\{0,314\\}$ and
$n_{1}-n_{3}\in\\{0,314\\}$ except when

$(n_{1}-n_{2},n_{1}-n_{3})\in\\{(0,3),(1,1),(1,5),(3,0),(3,4),(4,3),(5,1),(7,8),(8,7)\\}.$

Indeed, with the help of SageMath we find that if
$(n_{1},n_{2},n_{3},n_{4},a)$ is a possible solution of the equation (1)
excluding theses cases presented before. Then $n_{1}-n_{4}\leqslant 314.$

SPECIAL CASES. We deal with the cases when

$(n_{1}-n_{2},n_{1}-n_{3})\in\\{(1,1),(3,0),(4,3),(5,1),(8,7)\\}.$

It is easy to check that

$\dfrac{\log\Upsilon_{2}(t,s)}{\log\alpha}=\left\\{\begin{array}[]{lllll}0,&\text{if}&(t,s)=(1,1);\\\
0,&\text{if}&(t,s)=(3,0);\\\ 1,&\text{if}&(t,s)=(4,3);\\\ 2-\dfrac{\log
2}{\log\alpha},&\text{if}&(t,s)=(5,1);\\\ 3-\dfrac{\log
2}{\log\alpha},&\text{if}&(t,s)=(8,7).\end{array}\right.$

As we explained before, when we apply Lemma 2.1 to the expression (23), the
parameters $\tau$ and $\mu$ are linearly dependent, so the corresponding value
of $\varepsilon$ from Lemma 2.1 is always negative in all cases. For this
reason, we shall treat these cases differently.

Here, we have to solve the equations

$F_{n_{2}+1}+2F_{n_{2}}+F_{n_{4}}=2^{a},\quad
2F_{n_{2}+3}+F_{n_{2}}+F_{n_{4}}=2^{a},\quad
F_{n_{2}+4}+F_{n_{2}}+F_{n_{2}+1}+F_{n_{4}}=2^{a},$
$F_{n_{2}+5}+F_{n_{2}}+F_{n_{2}+4}+F_{n_{4}}=2^{a},\quad\text{and}\quad
F_{n_{2}+8}+F_{n_{2}}+F_{n_{2}+1}+F_{n_{4}}=2^{a}$ (24)

in positive integers $n_{2},n_{4}$ and $a$. To do so, we recall the following
well-known relation between the Fibonacci and the Lucas numbers:

$L_{k}=F_{k-1}+F_{k+1}\quad\text{for all}\quad k\geqslant 1.$ (25)

From (25) and (24), we have the following identities

$F_{n_{2}+1}+2F_{n_{2}}+F_{n_{4}}=F_{n_{2}+2}+F_{n_{2}}+F_{n_{4}}=F_{k+2}+F_{k}+F_{m},$
$2F_{n_{2}+3}+F_{n_{2}}+F_{n_{4}}=F_{n_{2}+2}+F_{n_{2}+4}+F_{n_{4}}=F_{k+2}+F_{k+4}+F_{m},$
$F_{n_{2}+4}+F_{n_{2}}+F_{n_{2}+1}+F_{n_{4}}=F_{n_{2}+2}+F_{n_{2}+4}+F_{n_{4}}=F_{k+2}+F_{k+4}+F_{m},$
(26)
$F_{n_{2}+5}+F_{n_{2}}+F_{n_{2}+4}+F_{n_{4}}=2F_{n_{2}+2}+2F_{n_{2}+4}+F_{n_{4}}=2F_{k+2}+2F_{k+4}+F_{m},$
$\text{and}\quad
F_{n_{2}+8}+F_{n_{2}}+F_{n_{2}+1}+F_{n_{4}}=2F_{n_{2}+6}+2F_{n_{2}+4}+F_{n_{4}}=2F_{k+6}+2F_{k+4}+F_{m},$

hold for all $k,m\geqslant 0$.

Equation (24) are transformed into the equations

$L_{k+1}+F_{m}=2^{a},\quad L_{k+3}+F_{m}=2^{a},\quad
2L_{k+3}+F_{m}=2^{a},\quad 2L_{k+5}+F_{m}=2^{a},$ (27)

to be resolved in positive integers $k,m$ and $a$.

A quick search in SageMath and analytical resolution leads to :

$(k,m,a)\in\\{(4,5,4),(4,8,5)\\}\quad\text{for}\quad L_{k+1}+F_{m}=2^{a},$

$(k,m,a)\in\\{(2,5,4),(2,8,5),(4,4,5)\\}\quad\text{for}\quad
L_{k+3}+F_{m}=2^{a},$

$(k,m,a)=(5,9,7)\quad\text{for}\quad 2L_{k+3}+F_{m}=2^{a},$

$(k,m,a)=(3,9,7)\quad\text{for}\quad 2L_{k+5}+F_{m}=2^{a}.$

A completes resolution and analysis gives solutions that are already listed in
Theorem 1.1. This completes the analysis of the special cases.

Finally let us use (13) in order to find improved upper bound on $n_{1}.$ Put

$z_{4}:=a\log
2-n_{1}\log\alpha+\log\Upsilon_{3}(n_{1}-n_{2},n_{1}-n_{3},n_{1}-n_{4}),$

where $\Upsilon_{3}$ is the function given by the formula

$\Upsilon_{3}(t,u,v):=\sqrt{5}\left(1+\alpha^{-t}+\alpha^{-u}+\alpha^{-v}\right)^{-1}$

with $t=n_{1}-n_{2},u=n_{1}-n_{3}$ and $v=n_{1}-n_{4}.$ From (13), we get

$|1-e^{z_{3}}|<\dfrac{2}{\alpha^{n_{1}}}.$ (28)

Since $z_{3}\neq 0$, as before, two cases arise: $z_{4}<0$ and $z_{4}>0$.

If $z_{4}>0$, then

$0<z_{4}\leqslant e^{z_{4}}-1<\dfrac{2}{\alpha^{n_{1}}}.$

Suppose now that $z_{4}<0$. We have $2/\alpha^{n_{1}}<1/2$ for all
$n_{1}>350$. Then, from (28), we have

$|1-e^{z_{4}}|<\dfrac{1}{2}$

and therefore $e^{|z_{3}|}<2$.

Since $z_{3}<0$, we have :

$0<|z_{3}|\leqslant
e^{|z_{3}|}-1=e^{|z_{3}|}\left|e^{|z_{3}|}-1\right|<\dfrac{4}{\alpha^{n_{1}}}$

which gives

$0<|z_{3}|<\dfrac{4}{\alpha^{n_{1}}}$

for the both cases ($z_{3}<0$ and $z_{3}>0$ ) and holds for all $n_{1}>350.$

Replacing the expression of $z_{3}$ in the above inequality and arguing again
as before, we conclude that

$0<\left|a\left(\dfrac{\log
2}{\log\alpha}\right)-n_{1}+\dfrac{\log\Upsilon_{3}(n_{1}-n_{2},n_{1}-n_{3},n_{1}-n_{4})}{\log\alpha}\right|<9\cdot\alpha^{-n_{1}}.$
(29)

Here, we also take, $M:=2.8\times 10^{58}$ and we apply Lemma 2.1 last time in
inequality (29) for all choices $n_{1}-n_{2}\in\\{0,314\\}$,
$n_{1}-n_{3}\in\\{0,314\\}$ and $n_{1}-n_{4}\in\\{0,314\\}$ with
$(n_{1},n_{2},n_{3},n_{4},a)$ a possible solution of equation (1), and by
omitting study of special cases (because it give a solution presented in
Theorem 1.1 ), we get:

$n_{1}<320.$

This is false because our assumption that $n_{1}>350$.

This ends the proof of our main theorem.

∎

##  5 Comments

In this paper, we found all instances in which a power of two can be expressed
as a sum of four Fibonacci numbers. Given the results obtained, we can make
the following conjecture.

######  Conjecture 5.1.

Consider the Diophantine equation

$F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=p^{a},p\geqslant 2,a\geqslant 2$ (30)

where $n_{1},n_{2},n_{3},n_{4},a$ are positive integers with $n_{1}\geqslant
n_{2}\geqslant n_{3}\geqslant n_{4}$ and $p$ is prime, then $p=2,3,5,7.$

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Pagdame TIEBEKABE
Université Cheikh Anta Diop (UCAD), Laboratoire d’Algèbre, de Cryptologie, de
Géométrie Algébrique et Applications (LACGAA)
Dakar, Sénégal

Ismaïla DIOUF
Université Cheikh Anta Diop (UCAD),  Laboratoire d’Algèbre, de Cryptologie, de
Géométrie Algébrique et Applications (LACGAA)
Dakar, Sénégal