\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Powers of Two as Sums of\\ \vskip .1in Three Lucas Numbers } \vskip 1cm \large Bilizimb\'ey\'e~Edjeou\\ Universit\'e Gaston Berger de Saint-Louis\\ BP: 234, Saint-Louis\\ Senegal\\ \href{mailto:olivieredjeou@gmail.com}{\tt olivieredjeou@gmail.com} \\ \ \\ Amadou~Tall\\ Universit\'e Cheikh Anta Diop de Dakar\\ BP $5005$, Dakar Fann\\ Senegal\\ \href{mailto:amadou7.tall@ucad.edu.sn}{\tt amadou7.tall@ucad.edu.sn}\\ \ \\ Mohamed Ben Fraj Ben~Maaouia\\ Universit\'e Gaston Berger de Saint-Louis\\ BP $234$, Saint-Louis\\ Senegal\\ \href{mailto:maaouiaalg@hotmail.com}{\tt maaouiaalg@hotmail.com}\\ \end{center} \vskip .2 in \begin{abstract} In this paper, we find all positive integer solutions of the Diophantine equation $L_k +L_l + L_t = 2^d$ in non-negative integers $k,l,t$, and $d$, where $(L_n)_{n\geq0}$ is the Lucas sequence. The tools used to solve our main theorem are linear forms in logarithms, properties of continued fractions, and a version of the Baker-Davenport reduction method in Diophantine approximation. \end{abstract} \section{Introduction} The Lucas sequence $(L_k)_{k\geq0}$ is a linear recurrence given by $L_0 = 2, L_1 = 1$ and $$L_{k+2} = L_{k+1} + L_k, \quad \hbox{for $k\geq 0$}.$$ It satisfies the same recurrence as the Fibonacci sequence $(F_k)_{k\geq0}$ given by $F_0 = 0, F_1 = 1$ and $$F_{k+2} = F_{k+1} + F_k, \quad \hbox{for $k\geq 2$},$$ whose numbers are found everywhere in nature. The Fibonacci numbers are famous for possessing many wonderful and amazing properties. In 2014, Bravo and Luca \cite{2} studied the Diophantine equation $$L_k + L_l = 2^t$$ in positive integers $k,l$ and $t$. Similar equations involving Fibonacci and Padovan sequences are solved in \cite{5,7}. E. Bravo and J. Bravo \cite{4} found also all powers of $2$ which are sums of three Fibonacci numbers. Specifically, they proved the following theorems. \begin{theorem} \label{thm1} The only solutions $(k,l,t)$ of the Diophantine equation $L_k + L_l = 2^t$ in positive integers $k,l, t $ and with $k \geq l$ are $${(0,0,2);(1,1,1);(3,3,3);(2,1,2);(4,1,3);(7,2,5)}.$$ \end{theorem} \begin{theorem} \label{thm2} All solutions $(k,l,t,d)$ of the Diophantine equation $$F_k + F_l + F_t = 2^d$$ in non-negative integers $k,l, t $, with $k \geq l \geq t $ and $d$ are $${(3,1,1,2);(3,2,2,2);(3,2,1,2);(4,4,3,3);(5,3,1,3);(5,3,2,3);(6,5,4,4)};$$ $${(7,3,1,4); (7,3,2,4); (8,6,4,5); (10,6,1,6); (10,6,2,6); (11,9,5,7); (13,8,3,8); (16,9,4,10)}.$$ \end{theorem} In this paper, we prove an extension of Theorem \ref{thm1} when the two Lucas numbers are replaced by three Lucas numbers and determine all the solutions of the Diophantine equation $$ L_k + L_l+ L_t = 2^d$$ in non-negative integers $k,l,t$ and $d$. We prove the following result. \begin{theorem} \label{thm3} All solutions $(k,l,t,d)$ of the Diophantine equation \begin{equation} \label{eq2} L_k + L_l + L_t = 2^d \end{equation} in non-negative integers $k\geq l\geq t$ and $d$, are $$(1, 1, 0, 2),(2, 2, 0, 3),(3, 0, 0, 3),(3, 2, 1, 3),(4, 4, 0, 4),(5, 2, 0, 4),(5, 3, 1, 4),(6, 4, 4, 5),$$ $$(6, 5, 2, 5),(7, 1, 0, 5),(10, 2, 0, 7),(10, 3, 1, 7),(17, 13, 3, 12).$$ \end{theorem} Our method of proof is similiar to the method described in \cite{2, 4}. \section{Preliminaries} Before proceeding further, we recall the Binet formula for the Lucas numbers $(L_k)_{k\geq0}$, namely $$L_k = \alpha^k + \beta^k, \quad\hbox{for}\quad k \geq 0,$$ where $$\alpha = \frac{1 + \sqrt{5}}{2}\quad\hbox{and} \quad \beta = \frac{1 - \sqrt{5}}{2}$$ are the roots of the characteristic equation $x^2 - x - 1 = 0$. In particular, the inequality \begin{equation} \label{eq3} \alpha^{k - 1} \leq L_k \leq 2\alpha^k \end{equation} holds for all $k \geq 0$. To prove Theorem \ref{thm3}, using a result on linear forms in two logarithms., we require some notation. Let $\delta$ be an algebraic number of degree $d$ with minimal polynomial $$a_0x^d + a_1 x^{d-1} + \cdots + a_d = a_0\prod_{i = 1}^d(X - \delta^{(i)} )$$ where the $a_i$ are relatively prime integers with $a_0 > 0$ and the $\delta^{(i)}$ denotes the conjugates of $\delta$. Then $$h(\delta) = \frac{1}{d} (\log a_0 + \sum_{i = 1}^d\log(\max\lbrace\vert\delta^{(i)}\vert,1\rbrace))$$ is called the logarithmic height of $\delta$. In particular, if $\delta = p/q$ is a rational number with $\gcd(p,q) = 1$ and $q > 0$, then $$h(\delta) = \log \max \lbrace{\vert p\vert,q}\rbrace.$$ The following properties of the logarithmic height, will be used in the next section. Let $\delta$, $\nu$ be algebraic numbers and $r\in \mathbb{Z}$. Then \begin{itemize} \item $h(\delta \pm \nu) \leq h(\delta) + h(\nu) + \log2$, \item $h(\delta\nu^{\pm 1}) \leq h(\delta) + h(\nu)$, \item $h(\delta^r ) = \vert r \vert h(\delta)$. \end{itemize} Using the above notation, we restate Laurent, Mignotte, and Nesterenko's result \cite[Cor.\ 1]{6}. \begin{theorem} \label{thm4} Let $\delta_1 ,\delta_2$ be two non-zero algebraic numbers, and let $\log\delta_1$ and $\log\delta_2$ be any determinations of their logarithms. Set $$D = [\mathbb{Q}(\delta_1 ,\delta_2 ) :\mathbb{Q} ]/[\mathbb{R}(\delta_1 ,\delta_2 ) : \mathbb{R}]$$ and $$\Gamma := b_2\log\delta_2 - b_1\log\delta_1,$$ where $b_1$ and $b_2$ are positive integers. Further, let $A_1, A_2 > 1$ be real numbers such that $$\log{A_i} \geq \max\lbrace h(\delta_i) , \frac{\vert h(\delta_i)\vert} {D}, \frac{1}{D}\rbrace,\quad i = 1,2.$$ Then, assuming that $\delta_1$ and $\delta_2$ are multiplicatively independent, we have $$\log{ \vert \Gamma \vert} > -30.9\cdot D^4(\max{\lbrace \log{b'}, \frac{21}{D},\frac{1}{2} \rbrace)^2} \log{A_1} \log{A_2}, $$ where $$b' = \frac{b_1}{D\log{A_2}} + \frac{b_2}{D\log{A_1}}.$$ \end{theorem} We also need the following general lower bound for linear forms in logarithms due to Matveev \cite{8}. \begin{theorem} \label{thm5} Assume that $\delta_1 ,\ldots, \delta_t$ are positive real algebraic numbers in a real algebraic number field $\mathbb{K}$ of degree $D$. Let $b_1 ,\cdots,b_n$ be rational integers, and $$\Lambda := \delta_1^{b_1} \cdots \delta_t^{b_t} - 1$$ be not zero. Then $$\vert \Lambda \vert > \exp{(-1.4\cdot 30^{t+3}\cdot t^{4.5}\cdot D^2(1 + \log D)(1 + \log B)}A_1 \cdots A_t),$$ where $$B \geq \max{\lbrace \vert b_1\vert,\ldots,\vert b_t\vert\rbrace},$$ and $$A_i \geq \max{\lbrace Dh(\delta_i ),\vert\log\delta_i \vert,0.16\rbrace}, \quad \hbox{for all} \quad i = 1,\cdots,t.$$ \end{theorem} Finally, we present a version of the reduction method based on the Baker-Davenport Lemma \cite{1}, from Dujella and Peth\H{o} \cite{3}. This will be one of the key tools used to reduce the upper bounds on the variables of the equation \eqref{eq2}. \begin{lemma} \label{lem1} Let $N$ be a positive integer, let $p/q$ be a convergent of the irrational number $\gamma$ such that $q > 6N$, and let ${A,B,\mu}$ be real numbers with ${A > 0}$ and ${B > 1}$. Define $$\xi := \Vert \mu q \Vert- N\Vert\gamma q\Vert,$$ where $\Vert \cdot \Vert$ denotes the distance to the nearest integer. If $\xi > 0$, then there is no solution to the inequality $$0 < u\gamma - v + \mu < AB^{-w},$$ in positive integers $u$, $v$, and $w$, with $u \leq N$ and $w \geq \frac{\log{(Aq/\xi)}}{\log{B}}$. \end{lemma} \section{The Proof of Theorem~\ref{thm3}} First of all, observe that if $k = l = t$, then equation \eqref{eq2} becomes $3L_k = 2^d$. Since $3\nmid 2$, then equation \eqref{eq2} has no solution. Subsequently, we assume that either $k > l$ or $l > t$. If $k \leq 250$, then a brute force search using {\tt Sagemath} in the range $0 \leq t \leq l \leq k \leq 250$ produces the solutions $$(1, 1, 0, 2),(2, 2, 0, 3),(3, 0, 0, 3),(3, 2, 1, 3),(4, 4, 0, 4),(5, 2, 0, 4),(5, 3, 1, 4),(6, 4, 4, 5),(6, 5, 2, 5),$$ $$(7, 1, 0, 5),(10, 2, 0, 7),(10, 3, 1, 7),(17, 13, 3, 12).$$ Thus, for the remainder of the paper, we assume that $k > 250$. Let us now establish a relation between $k$ and $d$. Combining \eqref{eq2} with the right inequality of \eqref{eq3}, one gets that $$2^d \leq 2\alpha^{k} + 2\alpha^{l} + 2\alpha^{t} < 6\alpha^{k} < 6\cdot 2^{k} < 2^{k+3},$$ which leads to $d \leq k+2$. \subsection{Bounding $k-l$ and $k-t$ in terms of $k$} We rewrite \eqref{eq2} as $$\alpha^{k} - 2^d = -\beta^{k} - L_l - L_t.$$ Now taking absolute values, we obtain $$\vert\alpha^k - 2^d \vert \leq \vert\beta\vert^k + L_l + L_t < \frac{1}{2} + 2\alpha^l + 2\alpha^t.$$ Dividing both sides of the above expression by $\alpha^k$ and taking into account that $k \geq l \geq t$, we get $$\vert1 - 2^d\alpha^{-k} \vert < \frac{1}{2}\alpha^{-k} + 2\alpha^{-k+l} + 2\alpha^{-k+t} < 5\alpha^{-k+l}.$$ Thus \begin{equation} \label{eq6} \vert1 - 2^d\alpha^{-k} \vert < \frac{5}{\alpha^{k-l}}. \end{equation} We apply Theorem \ref{thm4} to $$\Gamma := d \log\alpha - k\log2.$$ Therefore the estimate \eqref{eq6} can be rewritten as \begin{equation} \label{eq7} \vert1 - e^\Gamma \vert < \frac{5}{\alpha^{k-l}}. \end{equation} The algebraic number field containing $2 ,\alpha$ is $\mathbb{Q}(\sqrt{5})$, so we can take $D := 2$. By using \eqref{eq2} and the Binet formula for the Lucas sequence, we have \begin{equation} \label{eq8} \alpha^k = L_k - \beta^k < L_k + 1 \leq L_k + L_l + L_t = 2^d. \end{equation} Consequently, $1 < 2^d\alpha^{-k}$ and so $\Gamma > 0$. Using the fact that $\log(1 + x) \leq x$ for all $x \in \mathbb{R}^+$, together with \eqref{eq7}, gives \begin{equation} \label{eq9} 0 < \Gamma < \frac{5}{\alpha^{k-l}}, \end{equation} Hence, \begin{equation} \label{eq10} \log\Gamma < \log5 - (k-l)\log\alpha. \end{equation} Note further that $h(\alpha) = \log\alpha/2$ and $h(2) = \log2$. Thus, we can choose $$\log A_1 := \log\alpha \quad \hbox{and} \quad \log A_2 := \log2.$$ Finally, recall that $d \leq k+2 $, and so $$b' = \frac{k}{2\log2} + \frac{d}{2\log\alpha} < 4k.$$ Since $\alpha$ and $2$ are multiplicatively independent, we have, by Theorem \ref{thm4}, that $$\log{\Gamma} \geq {-30.9\cdot 2^4}\cdot (\max \lbrace\log{(4k)},21/2,1/2\rbrace)^2\cdot \log\alpha\cdot \log2.$$ Thus \begin{equation} \label{eq11} \log{\Gamma} > {-174}\cdot (\max\lbrace\log{(4k)},21/2,1/2\rbrace)^2. \end{equation} Combining \eqref{eq10} and \eqref{eq11}, we obtain \begin{equation} \label{eq12} (k-l)\log\alpha < 180\cdot (\max\lbrace\log(4k),21/2\rbrace)^2. \end{equation} Let us now establish a second linear form in logarithms. To this end, we rewrite equation \eqref{eq2} as follows $$\alpha^k(1 + \alpha^{(l-k)} ) - 2^d = -\beta^k- \beta^l - L_t.$$ Taking absolute values in the above relation and using the fact that $\beta = (1 - \sqrt{5})/2$ we get $$\vert \alpha^k(1 + \alpha^{(l-k)}) - 2^d\vert = \vert-\beta^k- \beta^l - L_t\vert < 2+ 2\alpha^t$$ for all $k > 250$ and $l \geq t \geq 0$. Dividing both sides of the above inequality by the first term of the left-hand side, we obtain \begin{equation} \label{eq15} \vert 1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)})^{-1}\vert < \frac{2}{\alpha^{k}(1 + \alpha^{(l-k)})} + \frac{2}{\alpha^{k-t}(1 + \alpha^{(l-k)})} < \frac{4}{\alpha^{k-t}}. \end{equation} We are now ready to apply Matveev's result given in Theorem \ref{thm5}. To do this, we take the parameters $n := 3$ and $$\delta_1 := 2,\quad \delta_2 := \alpha, \quad \delta_3 := (1 + \alpha^{(l-k)}).$$ We take $b_1 := d$, $b_2 := -k$ and $b_3 := -1$. As before, $\mathbb{K} := \mathbb{Q}(\sqrt{5})$ contains $ \delta_1 ,\delta_2 ,\delta_3$ and has $D := [\mathbb{K} : \mathbb{Q}] = 2$. To see why the left-hand side of \eqref{eq15} is not zero, note that otherwise, we would get the relation \begin{equation} \label{eq16} \alpha^k + \alpha^l = 2^d. \end{equation} Conjugating the above relation in $\mathbb{Q}({\sqrt{5}})$, we get \begin{equation} \label{eq17} \beta^k + \beta^l = 2^d. \end{equation} Further, combining \eqref{eq16} and \eqref{eq17}, we obtain $$\alpha^k<\alpha^k + \alpha^l = \vert \beta^k + \beta^l\vert<2 .$$ This is impossible because $k>250$. Thus, $$1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)})^{-1}$$ is not zero. In this application of Theorem \ref{thm5}, we take $A_1 := 2\log2$ and $A_2 := \log\alpha$. Since $t \leq k+2$, it follows that we can take $B := k+2$. Let us now estimate $h(\delta_3 )$. We begin by observing that $$\delta_3 = (1 + \alpha^{(l-k)}) < 2\quad \hbox{and} \quad \delta_3^{-1} < 1.$$ So that $$0 <\log\delta_3 < 1.$$ Next, notice that $$ h(\delta_3 ) \leq (k-l)\log\alpha + \log2.$$ Hence, we can take $$A_3 := 2 +(k-l)\log\alpha > \max\lbrace{2h(\delta_3 ),\vert\log\delta_3\vert,0.16}\rbrace.$$ Now Theorem \ref{thm5} implies that a lower bound on the left-hand side of \eqref{eq15} is $$\log\vert\Lambda\vert > -1.4\cdot 30^6\cdot 3^{4.5}\cdot 2^2(1 + \log2)(1 + \log(k+2))\cdot 2\log2\cdot 2\log\alpha\cdot (2 +(k-l)\log\alpha).$$ So, inequality \eqref{eq15} yields \begin{equation} \label{eq20} k-t < 2.8\cdot 10^{12}\log(k+2)\cdot (2 + (k-l)\log\alpha), \end{equation} where we used the inequality $1 + \log(k+2) < 2\log(k+2)$, which holds because $k > 250$. Now using \eqref{eq12} in the right-most term of inequality \eqref{eq20} and performing the respective calculations, we obtain \begin{equation} \label{eq21} k-t < 5.1\cdot 10^{14} \log(k+2)(\max\lbrace{\log(4k),21/2}\rbrace)^2. \end{equation} \subsection{ Bounding $k$} Finally, we consider a third linear form in logarithms. We now rewrite equation \eqref{eq2} as follows $$\alpha^k(1 + \alpha^{(l-k)} + \alpha^{(t-k)} ) - 2^d = -\beta^k- \beta^l -\beta^t.$$ Taking absolute values in the above relation and using the fact that $\beta = (1 - \sqrt{5})/2,$ we get $$\vert \alpha^k(1 + \alpha^{(l-k)} + \alpha^{(t-k)}) - 2^d \vert = \vert-\beta^k- \beta^l -\beta^t\vert < 3$$ for all $k > 250$ and $l \geq t \geq 0$. Dividing both sides of the above inequality by the first term of the left-hand side, we obtain \begin{equation} \label{eq24} \vert 1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})^{-1}\vert < \frac{3}{\alpha^{k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})} < \frac{3}{\alpha^{k}}. \end{equation} We apply Theorem \ref{thm5} to $$\Lambda = 1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})^{-1},$$ with the parameters $n := 3$, $\delta_1 := 2$, $\delta_2 := \alpha$, $\delta_3 := (1 + \alpha^{(l-k)} + \alpha^{(t-k)})$, $b_1 := d$, $b_2 := -k$ and $b_3 := -1$, $\mathbb{K} := \mathbb{Q}(\sqrt{5})$ contains $ \delta_1 ,\delta_2 , \delta_3$, $D := [\mathbb{K} : \mathbb{Q}] = 2$. To see why the left-hand side of \eqref{eq24} is not zero, note that otherwise, we would get the relation \begin{equation} \label{eq25} \alpha^k + \alpha^l + \alpha^t = 2^d. \end{equation} Conjugating the above relation in $\mathbb{Q}({\sqrt{5}})$, we get \begin{equation} \label{eq26} \beta^k + \beta^l + \beta^t = 2^d. \end{equation} Furthermore, combining \eqref{eq25} and \eqref{eq26}, we obtain $$\alpha^k<\alpha^k + \alpha^l + \alpha^t = \vert \beta^k + \beta^l+\beta^t\vert<3.$$ This is impossible because $k>250$. Thus, $$1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})^{-1}$$ is not zero. We now apply Theorem \ref{thm5} with $A_1 := 2\log2$, $A_2 := \log\alpha$. Since $d \leq k+2$; it follows that we can take $B := k+2$. Let us now estimate $h(\delta_3 )$. We begin by observing that $$\gamma_3 = (1 + \alpha^{(l-k)} + \alpha^{(t-k)}) < 3$$ and $$0 <\log\gamma_3 < \log3.$$ Next, notice that $$h(\delta_3 ) \leq (k-l)\log\alpha + (k-t)\log\alpha + 2\log2 \leq 2(k-t)\log\alpha + 2\log2.$$ Hence, we can take $$A_3 := 4 +2(k-t)\log\alpha > \max\lbrace{2h(\delta_3 ),\vert\log \delta_3\vert,0.16}\rbrace.$$ Now, from Theorem \ref{thm5} we have $$\log\vert\Lambda\vert > -1.4\cdot 30^6\cdot 3^{4.5}\cdot 2^2(1 + \log2)(1 + \log(k+2))\cdot 2\log2\cdot 2\log\alpha\cdot (4 +2(k-t)\log\alpha).$$ So, inequality \eqref{eq24} gives \begin{equation} \label{eq29} k < 10^{13}\log(k+2)\cdot (2 + (k-t)\log\alpha), \end{equation} where we used the inequality $1 + \log(k+2) < 2\log(k+2)$, which holds because $k > 250$. Now using \eqref{eq20} in the rightmost term of the above inequality \eqref{eq29} and performing the respective calculations, we obtain \begin{equation} \label{eq30} k < 2.6\cdot 10^{27} (\log(k+2))^2(\max\lbrace{\log(4k),21/2}\rbrace)^2. \end{equation} If $\max\lbrace{\log(4k),21/2}\rbrace = 21/2$, it then follows from \eqref{eq30} that $$k < 287\cdot 10^{27}(\log(k+2))^2,$$ giving $$k < 15\cdot 10^{32}.$$ If on the other hand we have that $\max\lbrace{\log(4k),21/2}\rbrace = \log(4k)$, then inequality \eqref{eq30} gives that $$k < 2.6\cdot 10^{27}(\log(k+2))^2(\log(4k))^2,$$ and so $$k < 12\cdot 10^{34}.$$ In any case, we have that $$k < 12\cdot 10^{34}$$ always holds. We summarize what we have so far in the following lemma. \begin{lemma} \label{lem2} If $(k,l,t,d)$ is a solution in positive integers of equation \eqref{eq2} with $k \geq l \geq t$ and $k > 250$, then inequalities $$d \leq k+2 \quad\hbox{and}\quad k < 12\cdot 10^{34}$$ hold. \end{lemma} \section{The final computations} In this section, we will reduce the upper bound on $k$. Firstly, we determine a suitable upper bound on $k-l, k-t$, and later we use Lemma \ref{lem1} to conclude that $k$ must be smaller than $250$. Turning back to inequality \eqref{eq9}, we obtain $$0 < d\log2 - k\log \alpha < \frac{5}{\alpha^{k-l}}.$$ Dividing across by $\log\alpha$, we get \begin{equation} \label{eq31} 0 < d\gamma - k < \frac{11}{\alpha^{k-l}}, \end{equation} where $$\gamma := \frac{\log2}{\log\alpha}.$$ Let $[a_0 ,a_1 ,a_2 ,a_3 ,a_4,a_5,a_6,a_7,\ldots] = [1,2,3,1,2,3,2,4\ldots]$ be the continued fraction expansion of $\gamma$, and let denote $p_n /q_n$ its $n$th convergent. Recall also that $d < 12\cdot10^{34}$ by Lemma \ref{lem2}. A quick inspection using {\tt Sagemath} reveals that \begin{align*} & 37527245802242661673724926130723830 = q_{73} < 12\cdot10^{34} < \\ & q_{74} = 175184858909722330004986691804684639. \end{align*} Furthermore, $a_N := \max\lbrace{a_i; i = 0,1,\ldots ,44}\rbrace = a_{17} = 134$. So, from the known properties of continued fractions, we obtain that \begin{equation} \label{eq32} \vert d\gamma - k\vert > \frac{1}{(a_N + 2)d}. \end{equation} Comparing estimates \eqref{eq31} and \eqref{eq32}, we get right away that $$\alpha^{k-l} < 11\cdot 136\cdot d < 18\cdot 10^{37},$$ leading to $k-l < 184$. Let us now go back to \eqref{eq15} and determine an improved upper bound on $k-t$. Put \begin{equation} \label{eq34} \omega_1 := d\log2 - k\log\alpha - \log(1 + \alpha^{-(k-l)}). \end{equation} Therefore, \eqref{eq15} implies that \begin{equation} \label{eq35} \vert1 - e^{\omega_1}\vert < \frac{4}{\alpha^{k-t}}. \end{equation} Note that $\omega_1 \neq 0$, by using \eqref{eq2} and the Binet formula for the Lucas sequence, we have $$\alpha^k + \alpha^l = L_t -\beta^k -\beta^l < L_k+L_l+L_t=2^d.$$ Therefore, $$1 < 2^d\alpha^{-k}(1 + \alpha^{(l-k)})^{-1}$$ and so $\omega_1>0$. Thus \begin{equation} \label{eq36} 0 < \omega_1 \leq e^{\omega_1} - 1 < \frac{4}{\alpha^{k-t}}. \end{equation} Replacing $\omega_1$ in the above inequality by its formula \eqref{eq34} and dividing both sides of the resulting inequality by $\log\alpha$, we get \begin{equation} \label{eq37} 0 < d\left(\frac{\log2}{\log\alpha}\right) - k - \frac{\log(1 + \alpha^{-(k-l)})}{\log\alpha} < \frac{9}{\alpha^{k-t}}. \end{equation} We now put $$\gamma := \frac{\log2}{\log\alpha}, \quad \mu := - \frac{\log(1 + \alpha^{-(k-l)})}{\log\alpha}, \quad A := 9 \quad\hbox{and}\quad B := \alpha.$$ Clearly $\gamma$ is an irrational number. We also put $N := 12\cdot 10^{34}$ , which is an upper bound on $d$ by Lemma \ref{lem2}. We therefore apply Lemma \ref{lem1} to inequality \eqref{eq37} for all choices $k-l \in \lbrace1,\ldots,184\rbrace$ except when $k-l = 1,3$ and get that $$k-t < \frac{\log(Aq/\xi)}{\log B},$$ where $q > 6N$ is a denominator of a convergent of the continued fraction of $\gamma$ such that $\xi = \Vert \mu q\Vert - N\Vert \gamma q\Vert > 0$. Indeed, using {\tt Sagemath}, we have that $$q = q_{75} = 1439006117080021301713618460568200942.$$ We find that if $(k,l,t,d)$ is a possible solution of the equation \eqref{eq2} with $\omega_1 > 0$ and $k-l \in \lbrace1,\ldots,184\rbrace$ except when $k-l = 1,3$, then $$k-t < 178.$$ Let us now treat the cases where $k-l = 1$ and $3$. The discussion of these cases will be different from the previous ones, because when applying Lemma \ref{lem1} to the expression \eqref{eq37}, the corresponding parameter $\mu$ appearing in Lemma \ref{lem1} is \begin{displaymath} \frac{\log(1 + \alpha^{-(k-l)})}{\log\alpha} = \begin{cases} -1, & \text{if $k-l=1$;} \\ 1-\frac{\log2}{\log\alpha}, & \text{if $k-l=3$.} \end{cases} \end{displaymath} In both cases, the parameters $\gamma$ and $\mu$ are linearly dependent, which yields that the corresponding value of $\xi$ from Lemma \ref{lem1} is always negative and therefore the reduction method is not useful for reducing the bound on $k-t$ in these instances. One can see that if $k-l = 1, 3,$ then the resulting inequality from \eqref{eq37} has the shape $$0 < \vert a \gamma - b \vert < \frac{9}{\alpha^{k-t}},$$ with $\gamma$ being an irrational number and $a,b \in \mathbb{Z}$. So, one can appeal to the known properties of the convergents of the continued fractions to obtain a nontrivial lower bound for $$\vert a \gamma - b \vert.$$ When $k-l= 1$, from \eqref{eq37}, we get that \begin{equation} \label{eq38} 0 < d\gamma - (k+1) < \frac{9}{\alpha^{k-t}}. \end{equation} Let $[a_0 ,a_1 ,a_2 ,a_3 ,a_4,a_5,a_6,a_7,...] = [1,2,3,1,2,3,2,4\ldots]$ be the continued fraction expansion of $\gamma$, and let denote $p_n /q_n$ its $n$th convergent. Recall also that $d < 12\cdot10^{34}$ by Lemma \ref{eq7}. Furthermore, $a_N := \max\lbrace{a_i: i = 0,1,\ldots,44}\rbrace = a_{17} = 134$. So, from the known properties of continued fractions, we obtain that \begin{equation} \label{eq39} \vert d\gamma - (k+1)\vert > \frac{1}{(a_N + 2)d}. \end{equation} Comparing estimates \eqref{eq38} and \eqref{eq39}, we get right away that $$\alpha^{k-t} < 9\cdot 136\cdot d < 15\cdot 10^{37},$$ leading to $k-t < 184$. By the same argument as the one we did before, we get that $k-t<184$ in the case when $k-l = 3$. This completes the analysis of the cases when $k-l = 1,3$. Consequently, $k-t< 184$ always holds. Finally, we shall use \eqref{eq24} to reduce the upper bound on $k$. Put \begin{equation} \label{eq41} \omega_2 = d\log2 -k\log\alpha - \log{\varphi(u,v)} \end{equation} where $\varphi$ is the function given by the formula $\varphi(u,v) = 1+\alpha^{-u}+\alpha^{-v}$, where $u=k-l, v=k-t$. Note that $\omega_2 \neq 0$. Thus, we distinguish the following cases. If $\omega_2 > 0$ then, from \eqref{eq24}, we obtain $$0 < \omega_2 \leq e^{\omega_2} - 1 < \frac{3}{\alpha^k}.$$ Replacing $\omega_2$ in the above inequality by its formula \eqref{eq41} and dividing both sides of the resulting inequality by $\log\alpha$, we get \begin{equation} \label{eq42} 0 < d\left(\frac{\log2}{\log\alpha}\right) - k - \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log\alpha} < \frac{7}{\alpha^k}. \end{equation} We now put $$\gamma := \frac{\log2}{\log\alpha},\quad \mu := - \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log\alpha}, \quad A := 7 \quad\hbox{and}\quad B := \alpha.$$ Clearly $\gamma$ is an irrational number. We also put $N := 12\cdot 10^{34}$ , which is an upper bound on $d$ by Lemma \ref{lem2}. We therefore apply Lemma \ref{lem1} to inequality \eqref{eq42} for all choices of $u \in \lbrace1,\ldots,184\rbrace, v \in \lbrace1,\ldots,184\rbrace$ except when \begin{align*} (u,v) \in \varpi_1 &= \lbrace(4, 2),(13, 1),(17, 4),(17, 5),(18, 7),(19, 4),(19, 5),(22, 2),(23, 1),(23, 2),(24, 1), \\ & (24, 5),(25, 1),(28, 2),(30, 4),(31, 1),(32, 7),(33, 1),(33, 4),(33, 5),(36, 1),(36, 2), \\ & (37, 2),(38, 2),(39, 1),(43, 4),(44, 1),(45, 4),(48, 1),(48, 5),(50, 1),(51, 9),(52, 1), \\ & (52, 2),(53, 4),(54, 2),(55, 2),(56, 2),(57, 2),(59, 2),(61, 1),(62, 4),(64, 1),(66, 1), \\ & (67, 1),(69, 2),(70, 5), (75, 1),(75, 2),(83, 5),(87, 1),(87, 7),(89, 2),(90, 2),(95, 1), \\ & (95, 7),(97, 1),(98, 1),(99, 5),(100, 1),(102, 2),(107, 2),(110, 4),(111, 2),(112, 1), \\ & (112, 2),(113, 1),(113, 5),(113, 6),(114, 1),(118, 1),(122, 1),(122, 4),(122, 6), \\ & (128, 2),(129, 1),(129, 2), (130, 1),(130, 4),(130, 5),(132, 2),(133, 2),(136, 4), \\ & (138, 2),(139, 1),(139, 2),(141, 1),(141, 2),(142, 4),(147, 1),(148, 1),(158, 7)\rbrace \end{align*} and get that $$k < \frac{\log(Aq/\xi)}{\log B},$$ where $q > 6N$ is a denominator of a convergent of the continued fraction of $\gamma$ such that $\xi = \Vert \mu q\Vert - N\Vert \gamma q\Vert > 0$. Indeed, using {\tt Sagemath}, we have that $$q = q_{82} = 12054118444825786260212254516320106583249.$$ We find that if $(k,l,t,d)$ is a possible solution of the equation \eqref{eq2} with $\omega_2 > 0$ and $ (u,v) \notin \varpi_1$, then $k < 196$. This is false because our assumption is that $k > 250$. Let us now work with the cases when $(u,v) \in \varpi_1 $. We cannot study these cases as before because when applying Lemma \ref{lem1} to the expression \eqref{eq42}, the corresponding quantity $\Vert\mu q\Vert$ appearing in Lemma \ref{lem1} is zero. In these cases, the parameters $\gamma$ and $\mu$ are linearly dependent, which yields that the corresponding value of $\xi$ from Lemma \ref{lem1} is always negative and therefore the reduction method is not useful for reducing the bound on $k$ in these instances. However, one can see that if $(u,v) = (2,4)$, then the resulting inequality from \eqref{eq42} has the shape $$0 < \vert x \gamma - y \vert < \frac{7}{\alpha^{k}},$$ with $\gamma$ being an irrational number and $x,y \in \mathbb{Z}$. So, one can use to the known properties of the convergents of the continued fractions to obtain a nontrivial lower bound for $$\vert x \gamma - y \vert.$$ This clearly gives us an upper bound for $k$. For example, when $(u,v)= (2,4)$, $$- \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log\alpha} = 2- 2\frac{\log2}{\log\alpha}$$ and from \eqref{eq42}, we get that \begin{equation} \label{eq43} 0 < (d-2)\gamma - (k-2) < \frac{7}{\alpha^{k}}. \end{equation} Let $[a_0 ,a_1 ,a_2 ,a_3 ,a_4,a_5,a_6,a_7,\ldots] = [1,2,3,1,2,3,2,4\ldots]$ be the continued fraction expansion of $\gamma$, and let denote $p_n /q_n$ its $n$th convergent. Recall also that $d < 12\cdot10^{34}$ by Lemma \ref{lem2}. Furthermore, $a_N := \max\lbrace{a_i; i = 0,1,\ldots,44}\rbrace = a_{17} = 134$. So, from the known properties of continued fractions, we obtain that \begin{equation} \label{eq44} \vert (d-2)\gamma - (k-2)\vert > \frac{1}{(a_N + 2)d}. \end{equation} Comparing estimates \eqref{eq43} and \eqref{eq44}, we get that \begin{equation} \label{eq45} \alpha^{k} < 7\cdot 136\cdot d < 12\cdot 10^{37}, \end{equation} leading to $k < 184$. Using the above argument, we obtain $k<184$ in the case when $(u,v) \in \varpi_1$ except $(2,4)$. We omit the details in order to avoid unnecessary repetitions. This completes the analysis of the cases when $(u,v) \in \varpi_1$. Consequently, $k< 196$ always holds. Suppose now that $\omega_2 < 0$. First, note that $\frac{7}{\alpha^k} < \frac{1}{2}$ since $k > 250$. Then, from \eqref{eq24}, we have that $$\vert1 - e^{\omega_2}\vert < \frac{1}{2},$$ thus $$\frac{1}{2} < e^{\omega_2} < \frac{3}{2}.$$ Therefore $$e^{\vert\omega_2\vert} < 2.$$ Since $\omega_2 < 0$, we have $$0 < \vert\omega_2\vert \leq e^{\vert\omega_2\vert} -1 = e^{\vert\omega_2\vert}\vert e^{-\vert\omega_2\vert} -1\vert = e^{\vert\omega_2\vert}\vert e^{\omega_2} -1\vert < \frac{6}{\alpha^{k}}.$$ Then we obtain $$0 < -d\log2 + k\log\alpha + \log(1 + \alpha^{-u}+ \alpha^{-v}) < \frac{6}{\alpha^{k}}.$$ By the same arguments used for proving \eqref{eq24}, we obtain \begin{equation} \label{eq46} 0 < k\left(\frac{\log\alpha}{\log2}\right) - d + \frac{\log(1 + \alpha^{-u}+ \alpha^{-v})}{\log2} < \frac{9}{\alpha^k}. \end{equation} We now put $$\gamma := \frac{\log\alpha}{\log2}, \quad \mu := \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log2}, \quad A := 9 \quad\hbox{and}\quad B := \alpha.$$ Clearly $\gamma$ is an irrational number. We also put $N := 12\cdot 10^{34}$ , which is an upper bound on $d$ by Lemma \ref{lem2}. We therefore apply Lemma \ref{lem2} to inequality \eqref{eq46} for all choices of $u \in \lbrace1,\ldots,184\rbrace, v \in \lbrace1,\ldots ,184\rbrace$ except when \begin{align*} (u,v) \in \varpi_2 &= \lbrace (2, 1),(3, 3),(4, 2),(5, 2),(7, 7), (8, 2),(9, 5),(9, 6),(10, 2),(11, 4),(12, 1),(13, 1), \\ & (14, 2),(17, 4),(19, 5),(19, 6),(20, 9),(22, 1),(23, 1),(24, 1),(24, 5),(25, 1), (25, 2), \\ & (28, 2),(30, 1),(31, 4),(32, 7),(33, 1),(36, 3),(37, 2), (38, 1),(40, 2),(44, 1),(47, 2), \\ & (48, 1),(48, 6),(50, 1),(50, 2), (51, 9),(53, 1),(54, 2),(56, 2),(57, 2),(58, 5),(58, 6), \\ & (58, 8), (62, 1),(62, 4),(64, 1),(66, 1),(69, 2),(70, 6),(75, 1),(75, 2), (76, 1), \\ & (76, 4),(77, 2),(77, 3),(78, 2),(79, 2),(83, 6),(87, 1), (87, 7),(89, 1),(89, 2),(91, 1), \\ & (91, 2),(92, 1),(94, 1),(95, 1), (95, 7),(96, 2),(97, 1),(98, 1),(99, 5),(99, 6),(102, 2), \\ & (102, 3), (112, 2),(113, 1),(113, 5),(113, 6),(120, 1),(122, 1),(122, 5), (122, 6), \\ & (123, 2),(129, 2),(129, 3),(130, 4),(130, 5),(134, 7), (135, 2),(138, 1),(139, 1), \\ & (141, 1),(141, 2),(142, 4),(143, 1), (147, 1),(148, 1),(149, 2) \rbrace, \end{align*} and get that $$k < \frac{\log(Aq/\xi)}{\log B},$$ where $q > 6N$ is a denominator of a convergent of the continued fraction of $\gamma$ such that $\xi = \Vert \mu q\Vert - N\Vert \gamma q\Vert > 0$. Indeed, using {\tt Sagemath}, we have hat $$q = q_{82} = 1234165504911193651820557190855668171489.$$ We find that if $(k,l,t,d)$ is a possible solution of the equation \eqref{eq2} with $\omega_2 < 0$ and $ (u,v) \notin \varpi_2$, then $k < 192$. This is false because our assumption is that $k > 250$. With the same arguments as in the case $\omega_2>0$, when $(u,v) \in \varpi_2$, we obtain that $k<192$. Thus, Theorem \ref{thm3} is proven. \section{Acknowledgments} This work was partially supported by a grant from the Simons Foundation. Part of this work was done during a very enjoyable visit of B. Edjeou at AIMS S\'en\'egal. This author thanks AIMS S\'en\'egal for the hospitality and support. We also thank Bernadette Faye and Eva Goedhart for useful discussions. \begin{thebibliography}{9} \bibitem{1} A. Baker and H. Davenport, The equations $3x^2 - 2 = y^2$ and $8x^2 - 7 = z^2$, \textit{Quart. J. Math. Oxford Ser.} \textbf{20} (1969), 129--137. \bibitem{4} E. F. Bravo and J. J. Bravo, On the Diophantine equation $F_n + F_m + F_l = 2^d$, \textit{Lith. Math. J.} \textbf{55} (2015), 301--311. \bibitem{2} J. J. Bravo and F. Luca, On the Diophantine equation $L_n + L_m = 2^a$, \textit{J. Integer Sequences} \textbf{17} (2014), Article $14.8.3$. \bibitem{3}A. Dujella and A. Peth\H{o}, A generalization of a theorem of Baker and Davenport, \textit{Quart. J. Math. Oxford Ser. (2)} \textbf{49} (1998), 291--306. \bibitem{5}A. C. G. Lomel\'i and S. H. Hern\'andez, Powers of two as sums of two Padovan numbers, \textit{INTEGERS} \textbf{18} (2018), A84. \bibitem{6} M. Laurent, M. Mignotte, and Y. Nesterenko, Formes lin\'eaires en deux logarithmes et d\'eterminants d'interpolation, \textit{J. Number Theory} \textbf{55} (1995), 285--321. \bibitem{7} F. Luca and S. Siksek, Factorials expressible as sums of two and three Fibonacci numbers, \textit{Proc. Edinb. Math. Soc.} \textbf{53} (2010), 747--763. \bibitem{8} E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers. II, \textit{Izv. Ran. Ser. Mat.} \textbf{64} (2000), 125--180; \textit{Izv. Math.} \textbf{64} (2000), 1217--1269. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39, Secondary 11J86. \noindent \emph{Keywords: } Lucas number, Diophantine equation, linear form in logarithms. \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A000032}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received August 21 2019; revised versions received August 9 2020; August 26 2020. Published in {\it Journal of Integer Sequences}, October 13 2020. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}