\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \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,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \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}{\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 Repdigits as Sums of Four Fibonacci or Lucas Numbers } \vskip 1cm Benedict Vasco Normenyo\\ Institut de Math\'ematiques et de Sciences Physiques \\ Dangbo \\ B\'enin \\ \href{mailto:bvnormenyo@imsp-uac.org}{\tt bvnormenyo@imsp-uac.org} \\ \ \\ Florian Luca \\ School of Mathematics \\ University of the Witwatersrand \\ Private Bag X3 \\ Wits 2050 \\ South Africa \\ and \\ Department of Mathematics \\ Faculty of Sciences \\ University of Ostrava \\ 30 Dubna 22 \\ 701 03 Ostrava 1 \\ Czech Republic \\ \href{mailto:florian.luca@wits.ac.za}{\tt florian.luca@wits.ac.za} \\ \ \\ Alain Togb\'e \\ Department of Mathematics, Statistics and Computer Science \\ Purdue University Northwest \\ 1401 S, U.S. 421 \\ Westville, IN 46391 \\ USA \\ \href{mailto:atogbe@pnw.edu}{\tt atogbe@pnw.edu} \end{center} \vskip .2 in \begin{abstract} In this paper, we determine all base-10 repdigits expressible as sums of four Fibonacci or Lucas numbers. \end{abstract} \section{Introduction}\label{sec1} The Fibonacci sequence $(F_n)_n$ and the Lucas sequences $(L_n)_n$ are given, respectively, by $$ F_0=0, F_1=1, F_{n+2}=F_{n+1}+F_n\;\;\text{for} \;\;n\geq 0 $$ and $$ L_0=2, L_1=1, L_{n+2}=L_{n+1}+L_n\;\;\text{for} \;\;n\geq 0. $$ Luca \cite{L2012} answered the question of which repdigits can be written as sums of three Fibonacci numbers by following a general method (see \cite{L2000}). Luca \cite{L2012} showed that all nonnegative integer solutions $(m_1,m_2,m_3,n)$ of the equation $$ N=F_{m_1}+F_{m_2}+F_{m_3}=d\left(\frac{10^n-1}{9}\right) ~{\rm with}~ d\in \{1, \ldots ,9\} $$ have $$ N\in \{0,1,2,3,4,5,6,7,8,9,11,22,44,55,66,77,99,111,555,666,11111\}. $$ Luca, Normenyo, and Togbe \cite{NLT20171, NLT20172} obtained analogous results for Pell numbers and Lucas numbers. Luca \cite{L2012} conjectured that the method he employed could be used to compute all solutions of the equation \begin{eqnarray*} d\left(\frac{10^n-1}{9}\right)=F_{m_1}+F_{m_2}+F_{m_3}+F_{m_4} \end{eqnarray*} with $d\in \{1, \ldots ,9\}$ and $m_1\geq m_2\geq m_3\geq m_4$. Luca et al. \cite{NLT20173} investigated this idea for Pell numbers. Luca et al. \cite{NLT20173} showed that, all nonnegative integer solutions $(m_1,m_2,m_3,n)$ of the equation \begin{eqnarray*} N=P_{m_1}+P_{m_2}+P_{m_3}+P_{m_4}=d\left(\frac{10^n-1}{9}\right) \;\;{\rm with}\;\;d\in \{1,\ldots ,9\} \end{eqnarray*} have $$ N\in \{0,1,2,3,4,5,6,7,8,9,11,22,33,44,55,77,88,99,111,222,444,888,999\}. $$ In this paper, we compute all repdigits that can be expressed as sums of four Fibonacci or Lucas numbers. We prove Theorem \ref{mainthm1} and Theorem \ref{mainthm2} below. \begin{theorem}\rm\label{mainthm1} All nonnegative integer solutions $(m_1,m_2,m_3,m_4,n)$ of the equation \begin{eqnarray}\label{eq1} N=F_{m_1}+F_{m_2}+F_{m_3}+F_{m_4}=d\left(\frac{10^n-1}{9}\right) ~{\rm with}~ d\in \{1, \ldots ,9\} \end{eqnarray} have \begin{eqnarray*} N\in \{0,1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,99,111,222,333,555,666,777,\\999,1111,2222,11111,66666\}. \end{eqnarray*} \end{theorem} \begin{theorem}\rm\label{mainthm2} All nonnegative integer solutions $(m_1,m_2,m_3,m_4,n)$ of the equation \begin{eqnarray}\label{eq2} N=L_{m_1}+L_{m_2}+L_{m_3}+L_{m_4}=d\left(\frac{10^n-1}{9}\right) ~{\rm with}~ d\in \{1, \ldots,9\} \end{eqnarray} have \begin{eqnarray*} N\in \{4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,111,222,333,555,666,999,2222,\\4444,11111,88888\}. \end{eqnarray*} \end{theorem} Here is the organization of this paper. In the next section, we recall the useful results to prove our two main results. We use them in Section \ref{sec3} to prove Theorem \ref{mainthm1}. In Section \ref{sec4}, for the sake of completeness, we apply the same method for the entire proof of Theorem \ref{mainthm2}. \section{Preliminaries}\label{sec2} In this section, we recall some results that are useful for the proof of Theorem \ref{mainthm1} and Theorem \ref{mainthm2}. Let $\mathbb{K}$ be an algebraic number field of degree $D$ over $\mathbb{Q}$, let $\alpha _1,\ldots,\alpha _n\in \mathbb{K}\setminus \{0\}$ and let $b_1,\ldots,b_n\in \mathbb{Z}$. Set $$ B=\max \{|b_1|,\ldots ,|b_n|\} $$ and $$ \Lambda =\alpha _1^{b_1}\cdots \alpha _n^{b_n}-1. $$ Let $A_1,\ldots,A_n$ be real numbers with $$ \max\{Dh(\alpha _i),|\log \alpha _i|,0.16\}\leq A_i,\qquad 1\le i\le n, $$ where $h(\eta)$ is the logarithmic height of an algebraic number $\eta$ which is given by the formula $$ h(\eta)=\frac{1}{d(\eta)}\left(\log |a_0| +\sum _{i=1}^{d(\eta)} \log \left(\max \{|\eta ^{(i)}|,1\}\right)\right), $$ where $d(\eta)$ is the degree of $\eta$ over $\mathbb{Q}$ and $$ f(X)=a_0\prod _{i=1}^{d(\eta)} \left(X-\eta ^{(i)}\right)\in \mathbb{Z}[X] $$ the minimal polynomial of $\eta$ of degree $d(\eta)$ over $\mathbb{Z}$. \begin{lemma} \label{lem3} (\cite[Theorem 9.4]{BMS2006}) Assume that $\Lambda\neq 0$. We then have \begin{equation}\label{eq3} \log \left |\Lambda\right |> -3\times 30^{n+4}\times (n+1)^{5.5}D^2(1+\log D)(1+\log nB)A_1\cdots A_n. \end{equation} Furthermore, if $\mathbb{K}$ is real, we have \begin{equation}\label{eq4} \log \left |\Lambda\right |> -1.4\times 30^{n+3}\times n^{4.5}D^2(1+\log D)(1+\log B)A_1\cdots A_n. \end{equation} \end{lemma} We now discuss a computational method for reducing upper bounds for solutions of Diophantine equations. Let $\vartheta_1,\vartheta_2,\beta\in \mathbb{R}$ be given, and let $x_1,x_2\in \mathbb{Z}$ be unknowns. Let \begin{equation}\label{eq5} \Lambda =\beta +x_1\vartheta_1 +x_2\vartheta_2. \end{equation} Let $c$, $\delta$ be positive constants. Set $X=\max\{|x_1|,|x_2|\}$. Let $X_0$ be a (large) positive constant. Assume that \begin{equation}\label{eq6} |\Lambda|0$. We have the following results. \begin{lemma} \label{lem4} (\cite[Lemma 3.1]{W1989}) (i) If \eqref{eq6} and \eqref{eq7} hold for $x_1$, $x_2$ with $X\geq X^*$, then $(-x_2,x_1)=(p_k,q_k)$ for an index $k$ that satisfies \begin{equation}\label{eq8} k\leq -1+\frac{\log \left(1+X_0\sqrt{5}\right)}{\log \left(\frac{1+\sqrt{5}}{2}\right)}:=Y_0. \end{equation} Moreover, the partial quotient $a_{k+1}$ satisfies \begin{equation}\label{eq9} a_{k+1}>-2+\frac{|\vartheta_2|\exp(\delta q_k)}{cq_k}. \end{equation} (ii) If for some $k$ with $q_k\geq X^*$, we have \begin{equation}\label{eq10} a_{k+1}>\frac{|\vartheta_2|\exp(\delta q_k)}{cq_k}, \end{equation} then \eqref{eq6} holds for $(-x_2,x_1)=(p_k,q_k)$. \end{lemma} \begin{lemma} \label{lem5} (\cite[Lemma 3.2]{W1989}) Let $$ A=\max_{0\le k\le Y_0} a_{k+1}. $$ If \eqref{eq6} and \eqref{eq7} hold for $x_1$, $x_2$ and $\beta =0$, then \begin{equation}\label{eq11} Y<\frac{1}{\delta}\log \left(\frac{c(A+2)X_0}{|\vartheta_2|}\right). \end{equation} \end{lemma} When $\beta \vartheta_1 \vartheta_2 \neq 0$ in \eqref{eq5}, put $\vartheta=-{\vartheta_1}/{\vartheta_2}$ and $\psi={\beta}/{\vartheta_2}$. Then we have $$ \frac{\Lambda}{\vartheta_2}=\psi-x_1\vartheta+x_2. $$ Let ${p}/{q}$ be a convergent of $\vartheta$ with $q>X_0$. For a real number $x$ we define $\| x\|=\min\{|x-n|, n\in {\mathbb Z}\}$ be the distance from $x$ to the nearest integer. We have the following result. \begin{lemma} \label{lem6} (\cite[Lemma 3.3]{W1989}) Suppose that $$ \parallel q \psi\parallel>\frac{ 2X_0}{q}. $$ Then, the solutions of \eqref{eq6} and \eqref{eq7} satisfy $$ Y<\frac{1}{\delta}\log\left(\frac{q^2c}{|\vartheta_2|X_0} \right). $$ \end{lemma} \section{Proof of Theorem 1} \label{sec3} It is well known that the Fibonacci numbers are given by $$ F_m=\frac{1}{\sqrt{5}}\left(\alpha^m-\beta ^m \right) \;\; \text{for} \;\; m\geq 0, \;\; \text{where}\;\; (\alpha,\beta)=\left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\right). $$ In equation (1) we suppose that $m_1\geq m_2\geq m_3\geq m_4$. A search with Maple in the range $0\leq m_1\leq 599$ yielded only the solutions shown in the statement of Theorem 1. Let us suppose that solutions of equation (1) exist for $m_1\geq 600$. For $m_1\geq 600$, we have that $$ F_{600}\leq F_{m_1}\leq F_{m_1}+F_{m_2}+F_{m_3}+F_{m_4}=d\left(\frac{10^n-1}{9}\right)\leq 10^n-1, $$ and so $$ 125\leq \frac{\log (1+F_{600})}{\log 10}\leq n. $$ That is, $n\geq 125$. Now, \begin{align*} 10^{n-1} & \leq d\left( \frac{10^n-1}{9}\right)=F_{m_1}+F_{m_2}+F_{m_3}+F_{m_4} \\ &\leq 4F_{m_1} \\ &\leq \frac{4}{\sqrt{5}}\left(\alpha^{m_1}+|\beta|^{m_1}\right)\\ &<\frac{8}{\sqrt{5}}\alpha^{m_1}\\ & < \alpha ^{m_1+2.7}, \end{align*} since $\frac{8}{\sqrt{5}}<\alpha ^{2.7}$. This means that $10^{n-1}< \alpha ^{m_1+2.7}$, and thus $$ 4.78(n-1)<(n-1)\frac{\log 10}{\log \alpha}1.4\times 30^6\times 3^{4.5}\times D^2\times (1+\log D)\times A_1\times A_2. $$ Next, we compute $A_3$. We find that, $$ \alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}< \sqrt{5}, $$ and $$ \alpha_3^{-1}=\frac{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}{d\sqrt{5}} \leq \frac{36}{\sqrt{5}}\alpha ^{m_1-m_4}. $$ Hence, $|\log \alpha _3|<3+(m_1-m_4)\log \alpha$. Also, we have that \begin{align*} h(\alpha_3)&\leq h(d\sqrt{5})+h(9)+h(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)\\ &\leq h(9\sqrt{5})+h(9)+\log 2+h(\alpha ^{m_3-m_4}(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1))\\ & \leq h(9)+h(\sqrt{5})+h(9)+2\log 2+h(\alpha ^{m_3-m_4}) +h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\ & \leq h(\sqrt{5})+2h(9)+3\log 2+h(\alpha ^{m_3-m_4}) +h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\ & \leq h(\sqrt{5})+2h(9)+3\log 2+(m_3-m_4)h(\alpha )+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)\\ &= \frac{1}{2}\log 5+2h(9)+3\log 2+\frac{1}{2}(m_1-m_4)\log \alpha. \end{align*} Hence, $2h(\alpha_3)\leq 15+(m_1-m_4)\log \alpha.$ Therefore, we get $$ \max\{2h(\alpha _3),|\log \alpha _3|,0.16\}\leq 15+(m_1-m_4)\log \alpha =:A_3. $$ By applying Lemma \ref{lem3} to $\Gamma_1$ given by \eqref{eqG1}, and using \eqref{eq15} we have that $$ \exp (-(15+(m_1-m_4)\log \alpha )C_1(1+\log m_1))< \alpha ^{4-m_1}. $$ Thus, \begin{equation}\label{eq17} m_1\log \alpha<4\log \alpha+(15+(m_1-m_4)\log \alpha )C_1(1+\log m_1). \end{equation} {\bf Step 2}: We have that \begin{align}\label{eq18} \frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}\right)-\frac{d\times 10^n}{9} =-\frac{d}{9}+\frac{1}{\sqrt{5}}(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}-\alpha ^{m_4}). \end{align} Consequently, we get $$ \left |\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}\right)-\frac{d\times 10^n}{9}\right | \leq\frac{d}{9}+\frac{1}{\sqrt{5}}(|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4}+\alpha ^{m_4}), $$ and so \begin{equation}\label{eq19} \left |\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}\right)-\frac{d\times 10^n}{9}\right |<\frac{\alpha^{m_4+5}}{\sqrt{5}}. \end{equation} We multiply both sides of inequality \eqref{eq19} by $\frac{\sqrt{5}\alpha^{-m_1}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}}$ to get $$ \left |1-\alpha ^{-m_3}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |<\frac{\alpha ^{m_4-m_1+5}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}}, $$ which gives us \begin{equation}\label{eq20} \left |1-\alpha ^{-m_3}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |< \alpha ^{m_4-m_1+5}. \end{equation} Put \begin{equation}\label{eqG2} \Gamma_2:=1-\alpha ^{-m_3}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right). \end{equation} Suppose that $\Gamma_2=0.$ Then, we get $$ \alpha^{m_1}+\alpha^{m_2}+\alpha^{m_3}=\frac{10^n\times d\sqrt{5} }{9}. $$ Taking the conjugate of this in $\mathbb{Q}(\sqrt{5})$, we get $$ \beta^{m_1}+\beta^{m_2}+\beta^{m_3}=-\frac{10^n\times d\sqrt{5} }{9}. $$ Consequently, we obtain $$ \frac{10^{125}\times \sqrt{5} }{9}\leq \frac{10^n\times d\sqrt{5} }{9} =|\beta^{m_1}+\beta^{m_2}+\beta^{m_3}| \leq |\beta|^{m_1}+|\beta|^{m_2}+|\beta|^{m_3} < 3, $$ which means that $\frac{10^{125}\times \sqrt{5} }{9}<3$. This is false. We conclude that $\Gamma _2\neq 0$. To apply Lemma \ref{lem3} to $\Gamma_2$ given by \eqref{eqG2}, we set $$ \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)},\;\; b_1=-m_3,\;\;b_2=n,\;\;b_3=1, $$ where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$ and $b_1,b_2,b_3\in \mathbb{Z}$. We have $B=\max \{m_3,n,1\}\leq m_1$. We proceed to compute $A_3$ by first observing that $$ \alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}< \sqrt{5} $$ and $$ \alpha_3^{-1}=\frac{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}{d\sqrt{5}}\leq \frac{27}{\sqrt{5}}\alpha ^{m_1-m_3}. $$ Hence, $|\log \alpha _3|<3+(m_1-m_3)\log \alpha$. Additionally, we get \begin{align*} h(\alpha_3)& \leq h(d\sqrt{5})+h(9)+\log 2+h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\ & \leq \frac{1}{2}\log 5+2h(9)+2\log 2+h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\ &\leq \frac{1}{2}\log 5+2h(9)+2\log 2+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)\\ &= \frac{1}{2}\log 5+2h(9)+2\log 2+\frac{1}{2}(m_1-m_3)\log \alpha. \end{align*} Hence, $2h(\alpha_3)\leq 14+(m_1-m_3)\log \alpha.$ As a result, we find that $$ \max\{2h(\alpha _3),|\log \alpha _3|,0.16\}\leq 14+(m_1-m_3)\log \alpha =:A_3. $$ By applying Lemma \ref{lem3} to $\Gamma_2$ given by \eqref{eqG2} and using \eqref{eq20}, we deduce that $$ \exp (-(14+(m_1-m_3)\log \alpha )C_1(1+\log m_1))< \alpha ^{m_4-m_1+5}. $$ Thus, we get \begin{equation}\label{eq22} (m_1-m_4)\log \alpha<5\log \alpha+(14+(m_1-m_3)\log \alpha )C_1(1+\log m_1). \end{equation} {\bf Step 3}: We begin with \eqref{eq12} written in the form \begin{align}\label{eq23} \frac{\alpha ^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}\right)-\frac{d\times 10^n}{9}=-\frac{d}{9}+\frac{1}{\sqrt{5}}(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}-\alpha^{m_3}-\alpha^{m_4}). \end{align} Equation \eqref{eq23} leads us to \begin{align*} \left |\frac{\alpha ^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}\right)-\frac{d\times 10^n}{9}\right |&\leq \frac{d}{9}+\frac{1}{\sqrt{5}}(|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4}+\alpha^{m_3}+\alpha^{m_4})\\ &< 1+\frac{4}{\sqrt{5}}+\frac{2\alpha ^{m_3}}{\sqrt{5}}\\ &\leq \frac{1}{\sqrt{5}}\left(\sqrt{5}+6\right)\alpha ^{m_3}, \end{align*} from which we obtain \begin{equation}\label{eq24} \left |\frac{\alpha ^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}\right)-\frac{d\times 10^n}{9}\right |< \frac{\alpha^{m_3+5}}{\sqrt{5}}. \end{equation} Multiplying both sides of inequality \eqref{eq24} by $\frac{\sqrt{5}\alpha^{-m_1}}{1+\alpha ^{m_2-m_1}}$ gives us $$ \left |1-\alpha ^{-m_2}10^n\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right)\right |< \frac{\alpha^{m_3-m_1+5}}{1+\alpha ^{m_2-m_1}}, $$ which yields \begin{equation}\label{eq25} \left |1-\alpha ^{-m_2}10^n\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right)\right |<\alpha^{m_3-m_1+5}. \end{equation} Put \begin{equation}\label{eqG3} \Gamma_3:= 1-\alpha ^{-m_2}10^n\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right). \end{equation} Suppose that $\Gamma_3=0.$ Then $$ \alpha^{m_1}+\alpha^{m_2}=\frac{10^n\times d\sqrt{5} }{9}, $$ giving us $$ \beta^{m_1}+\beta^{m_2}=-\frac{10^n\times d\sqrt{5} }{9} $$ by conjugating in $\mathbb{Q}\left(\sqrt{5}\right)$. It follows that $$ \frac{10^{125}\times \sqrt{5} }{9}\leq \frac{10^n\times d\sqrt{5} }{9}=|\beta^{m_1}+\beta^{m_2}| \leq |\beta|^{m_1}+|\beta|^{m_2} <2, $$ which is false. Hence, $\Gamma _3\neq 0$. Using the notations in Lemma \ref{lem3}, we put $$ \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}, \;\; b_1=-m_2,\;\;b_2=n,\;\;b_3=1, $$ where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$ and $b_1,b_2,b_3\in \mathbb{Z}$. We have $B=\max \{m_2,n,1\}\leq m_1$. Now, we deduce that $$ \alpha _3=\frac{d\sqrt{5} }{9(\alpha^{m_1-m_2}+1)}\leq \sqrt{5}\;\; {\rm and}\;\; \alpha _3^{-1}=\frac{9(\alpha^{m_1-m_2}+1)}{d\sqrt{5} }\leq \frac{18}{\sqrt{5}}\alpha^{m_1-m_2}. $$ So $|\log \alpha _3|<3+(m_1-m_2)\log \alpha$. Furthermore, \begin{align*} h(\alpha_3)&\leq h(d\sqrt{5})+h(9)+\log 2+h(\alpha^{m_1-m_2})\\ &\leq h(\sqrt{5})+2h(9)+\log 2+(m_1-m_2)h(\alpha)\\ &= \frac{1}{2}\log 5+2h(9)+\log 2+\frac{1}{2}(m_1-m_2)\log \alpha. \end{align*} Thus, $2h(\alpha_3)\leq 12+(m_1-m_2)\log \alpha$ and so $$ \max\{2h(\alpha _3),|\log \alpha _3|,0.16\}<12+(m_1-m_2)\log \alpha =:A_3. $$ Applying Lemma \ref{lem3} to $\Gamma _3$ given by \eqref{eqG3}, and using \eqref{eq25} we produce $$ \exp (-(12+(m_1-m_2)\log \alpha)C_1(1+\log m_1))< \alpha^{m_3-m_1+5}, $$ from which we obtain \begin{equation}\label{eq27} (m_1-m_3)\log \alpha <5\log \alpha + (12+(m_1-m_2)\log \alpha)C_1(1+\log m_1). \end{equation} {\bf Step 4}: Using equation \eqref{eq12} in the form \begin{equation}\label{eq28} \frac{\alpha ^{m_1}}{\sqrt{5}}-\frac{d\times 10^n}{9} = -\frac{d}{9}+\frac{1}{\sqrt{5}}\left(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}-\alpha ^{m_2}-\alpha ^{m_3}-\alpha ^{m_4}\right), \end{equation} we get \begin{align*} \left|\frac{\alpha ^{m_1}}{\sqrt{5}}-\frac{d\times 10^n}{9}\right| & \leq \frac{d}{9}+\frac{1}{\sqrt{5}}\left(|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4}+\alpha ^{m_2}+\alpha ^{m_3}+\alpha ^{m_4}\right)\\ &< 1+\frac{4}{\sqrt{5}}+\frac{3}{\sqrt{5}}\alpha ^{m_2}\\ &\leq \frac{1}{\sqrt{5}}(\sqrt{5}+7)\alpha ^{m_2}, \end{align*} which means that \begin{equation}\label{eq29} \left |\frac{\alpha ^{m_1}}{\sqrt{5}}-\frac{d\times 10^n}{9}\right |< \frac{\alpha^{m_2+5}}{\sqrt{5}}. \end{equation} Multiplying both sides of \eqref{eq29} by $\sqrt{5}\alpha ^{-m_1}$ yields \begin{equation}\label{eq30} \left |1-\alpha ^{-m_1}10^n\left(\frac{d\sqrt{5} }{9}\right)\right |< \alpha^{m_2-m_1+5}. \end{equation} Put \begin{equation}\label{eqG4} \Gamma_4:=1-\alpha ^{-m_1}10^n\left(\frac{d\sqrt{5} }{9}\right). \end{equation} Suppose that $\Gamma_4=0$. Then $$ \alpha^{m_1}=\frac{d\sqrt{5}\times 10^n}{9}, $$ which implies that $$ \beta^{m_1}=-\frac{d\sqrt{5}\times 10^n}{9}. $$ Consequently, $$ \frac{\sqrt{5}\times 10^{125}}{9}\leq \frac{d\sqrt{5}\times 10^n}{9}=|\beta|^{m_1}<1, $$ which is impossible. Hence, $\Gamma_4 \neq 0$. In order to apply Lemma \ref{lem3} to $\Gamma_4$ given by \eqref{eqG4}, we take $$ \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{\sqrt{5} d}{9},\;\;b_1=-m_1,\;\;b_2=n,\;\;b_3=1, $$ where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$ and $b_1,b_2,b_3\in \mathbb{Z}$. To compute $A _3$, we observe that $$ \alpha _3=\frac{d\sqrt{5} }{9}\leq \sqrt{5}\quad {\rm and} \quad \alpha _3^{-1}=\frac{9}{d\sqrt{5} }\leq \frac{9}{\sqrt{5}}, $$ so $|\log \alpha _3|<1.4$. In addition, we have $$ h(\alpha _3)\leq h(d\sqrt{5})+h(9)\leq \frac{1}{2}\log 5+2h(9). $$ As a result, we have that $2h(\alpha _3)<10.4$. We see that $$ \max \{2h(\alpha _3),|\log \alpha _3|,0.16\}<10.4=:A_3. $$ By applying Lemma \ref{lem3} to $\Gamma_4$ given by \eqref{eqG4} and using \eqref{eq30}, we obtain $$ \exp(-10.4C_1(1+\log m_1))<\left |1-\alpha ^{-m_1}10^n\left(\frac{d\sqrt{5}}{9}\right)\right |< \alpha^{m_2-m_1+5}. $$ This means that \begin{equation}\label{eq32} (m_1-m_2)\log \alpha<5\log \alpha+10.4C_1(1+\log m_1)<10.5C_1(1+\log m_1). \end{equation} Putting together \eqref{eq32} and \eqref{eq27} yields \begin{align*} (m_1-m_3)\log \alpha &<5\log \alpha+(12+10.5C_1(1+\log m_1))C_1(1+\log m_1)\\ \nonumber &<5\log \alpha +10.6C_1^2(1+\log m_1)^2\\ \nonumber&<10.7C_1^2(1+\log m_1)^2. \end{align*} That is \begin{equation}\label{eq33} (m_1-m_3)\log \alpha<10.7C_1^2(1+\log m_1)^2. \end{equation} Combining \eqref{eq33} and \eqref{eq22}, we obtain \begin{align*} (m_1-m_4)\log \alpha & < 5\log \alpha+\left(14+10.7C_1^2(1+\log m_1)^2 \right)C_1(1+\log m_1)\\ & < 5\log \alpha +10.8C_1^3(1+\log m_1)^3\\ & < 10.9C_1^3(1+\log m_1)^3. \end{align*} That is \begin{equation}\label{eq34} (m_1-m_4)\log \alpha<10.9C_1^3(1+\log m_1)^3. \end{equation} We now combine \eqref{eq34} and \eqref{eq17} to obtain \begin{align*} m_1\log \alpha & < 4 \log \alpha+(15+10.9C_1^3(1+\log m_1)^3)C_1(1+\log m_1)\\ & < 4\log \alpha+11.0C_1^4(1+\log m_1)^4\\ &<11.1C_1^4(1+\log m_1)^4\\ &<11.1\left(2.3\times 10^{12}\right)^4(1+\log m_1)^4. \end{align*} That is \begin{equation}\label{eq35} m_1\log \alpha<11.1\left(2.3\times 10^{12}\right)^4(1+\log m_1)^4. \end{equation} Inequality \eqref{eq35} gives rise to the inequality $m_1< 2.3\times 10^{59}$. Now, we lower the bound. Let \begin{equation}\label{eqLF1} \Lambda_1 =-m_1\log \alpha +n\log 10+\log\left(\frac{d\sqrt{5} }{9}\right). \end{equation} Equation \eqref{eq28} leads us to \begin{align*} \frac{\alpha^{m_1}}{\sqrt{5}}-\frac{d\times 10^n}{9}&=\frac{\alpha ^{m_1}}{\sqrt{5}}\left(1-\alpha^{-m_1}10^n\left(\frac{d\sqrt{5}}{9}\right)\right)\\ &=\frac{\alpha^{m_1}}{\sqrt{5}}\left(1-e^{\Lambda_1}\right)\\ &=-\frac{d}{9}+\frac{\beta ^{m_1}}{\sqrt{5}}-F_{m_2}-F_{m_3}-F_{m_4}\\ &\leq -\frac{1}{9}+\frac{|\beta| ^{600}}{\sqrt{5}}\\ &<0, \end{align*} as $m_1\geq 600$. Thus, $\Lambda _1>0$ and so from \eqref{eq30} we obtain $$ 0<\Lambda _1X_0$ is $q=q_{125}$. We find that $q=q_{128}$ satisfies the hypothesis of Lemma \ref{lem6} for $d=1,\dots,9$. Applying Lemma \ref{lem6}, we get $m_1-m_2\leq 310$, and hence $m_2\geq 290$. Taking $1\leq d\leq 9$ and $0\leq m_1-m_2\leq 310$, we let \begin{equation}\label{eqLF2} \Lambda_2=-m_2\log \alpha +n\log 10 +\log \left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right). \end{equation} We see from equation \eqref{eq23} that \begin{align*} \frac{\alpha^{m_1}}{\sqrt{5}}(1+\alpha^{m_2-m_1})\left(1-e^{\Lambda_2}\right)&=-\frac{d}{9}+\frac{\beta^{m_1}}{\sqrt{5}}+\frac{\beta^{m_2}}{\sqrt{5}}-F_3-F_4\\ &\leq -\frac{1}{9}+\frac{|\beta|^{600}}{\sqrt{5}}+\frac{|\beta|^{290}}{\sqrt{5}}\\ &<0, \end{align*} making use of $m_1\geq 600$ and $m_2\geq 290$. Hence, $\Lambda _2>0$, and so from \eqref{eq25} we see that $$ 0<\Lambda _20$, and so from \eqref{eq20} we see that $$ 0<\Lambda _30$, and so from \eqref{eq15} we see that $$ 0<\Lambda _41.4\times 30^6\times 3^{4.5}\times D^2\times (1+\log D)\times A_1\times A_2. $$ Next, we compute $A_3$. We find that, $$ \alpha_3=\frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}< 1, $$ and $$ \alpha_3^{-1}=\frac{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}{d} \leq 36\alpha ^{m_1-m_4}, $$ hence, $|\log \alpha _3|<4+(m_1-m_4)\log \alpha$. Also, we have that \begin{align*} h(\alpha_3)&\leq h(d)+h(9)+\log 2+h(\alpha ^{m_3-m_4}(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1))\\ & \leq 2h(9)+2\log 2+h(\alpha ^{m_3-m_4})+h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\ & \leq 2h(9)+3\log 2+h(\alpha ^{m_3-m_4})+h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\ & \leq 2h(9)+3\log 2+(m_3-m_4)h(\alpha )+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)&\\ &= 2h(9)+3\log 2+\frac{1}{2}(m_1-m_4)\log \alpha. \end{align*} Hence, $2h(\alpha_3)\leq 13+(m_1-m_4)\log \alpha.$ Therefore, we get $$ \max\{2h(\alpha _3),|\log \alpha _3|,0.16\}\leq 13+(m_1-m_4)\log \alpha =:A_3. $$ By applying Lemma \ref{lem3} to $\Gamma_1$ given by \eqref{eqG11}, and using \eqref{eq47} we have that $$ \exp (-(13+(m_1-m_4)\log \alpha )C_1(1+\log m_1))< \alpha ^{3.35-m_1}. $$ Thus, \begin{equation}\label{eq49} m_1\log \alpha<3.35\log \alpha+(13+(m_1-m_4)\log \alpha )C_1(1+\log m_1). \end{equation} {\bf Step 2}: Writing equation \eqref{eq44} as \begin{equation}\label{eq50} \alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}\right)-\frac{d\times 10^n}{9} =-\frac{d}{9}-\alpha ^{m_4} -(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}), \end{equation} we get $$ \left |\alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}\right)-\frac{d\times 10^n}{9}\right | \leq\frac{d}{9}+\alpha ^{m_4} +|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4}, $$ and so \begin{equation}\label{eq51} \left |\alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}\right)-\frac{d\times 10^n}{9}\right |<\alpha^{m_4+3.73}. \end{equation} By multiplying both sides of inequality \eqref{eq51} by $\frac{\alpha^{-m_1}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}}$ we obtain $$ \left |1-\alpha ^{-m_3}10^n\left(\frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |<\frac{\alpha ^{m_4-m_1+3.73}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}}, $$ which leads to \begin{equation}\label{eq52} \left |1-\alpha ^{-m_3}10^n\left(\frac{d}{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |< \alpha ^{m_4-m_1+3.73}. \end{equation} Put \begin{equation}\label{eqG22} \Gamma_2:=1-\alpha ^{-m_3}10^n\left(\frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right). \end{equation} Suppose that $\Gamma_2=0.$ Then, we get $$ \alpha^{m_1}+\alpha^{m_2}+\alpha^{m_3}=\frac{10^n\times d }{9}. $$ Taking the conjugate of this in $\mathbb{Q}(\sqrt{5})$, we get $$ \beta^{m_1}+\beta^{m_2}+\beta^{m_3}=\frac{10^n\times d}{9}, $$ which implies that $$ \frac{10^{125} }{9}\leq \frac{10^n\times d }{9}=|\beta^{m_1}+\beta^{m_2}+\beta^{m_3}|\leq |\beta|^{m_1}+|\beta|^{m_2}+|\beta|^{m_3}< 3. $$ Thus, $\frac{10^{125}}{9}<3$, which is false. We conclude that $\Gamma _2\neq 0$. To apply Lemma \ref{lem3} to $\Gamma_2$ given by \eqref{eqG22}, we set $$ \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d}{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)},\;\; b_1=-m_3,\;\;b_2=n,\;\;b_3=1, $$ where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$ and $b_1,b_2,b_3\in \mathbb{Z}$. Also, we obtain $B=\max \{m_3,n,1\}\leq m_1$. We proceed to compute $A_3$ by first observing that $$ \alpha_3=\frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}< 1, $$ and $$ \alpha_3^{-1}=\frac{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}{d}\leq 27\alpha ^{m_1-m_3}. $$ Hence, $|\log \alpha _3|<4+(m_1-m_3)\log \alpha$. Additionally, we get \begin{align*} h(\alpha_3)& \leq h(d)+h(9)+\log 2+h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\ & \leq 2h(9)+2\log 2+h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\ &\leq 2h(9)+2\log 2+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)\\ &= 2h(9)+2\log 2+\frac{1}{2}(m_1-m_3)\log \alpha. \end{align*} Hence, $2h(\alpha_3)\leq 12+(m_1-m_3)\log \alpha.$ As a result, we find that $$ \max\{2h(\alpha _3),|\log \alpha _3|,0.16\}\leq 12+(m_1-m_3)\log \alpha =:A_3. $$ By applying Lemma \ref{lem3} to $\Gamma_2$ given by \eqref{eqG22} and using \eqref{eq52}, we deduce that $$ \exp (-(12+(m_1-m_3)\log \alpha )C_1(1+\log m_1))< \alpha ^{m_4-m_1+3.73}. $$ Thus, we get \begin{equation}\label{eq54} (m_1-m_4)\log \alpha<3.73\log \alpha+(12+(m_1-m_3)\log \alpha )C_1(1+\log m_1). \end{equation} {\bf Step 3}: Writing \eqref{eq44} as \begin{equation}\label{eq55} \alpha ^{m_1}\left(1+\alpha ^{m_2-m_1}\right)-\frac{d\times 10^n}{9}=-\frac{d}{9}-(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}) -(\alpha^{m_3}+\alpha^{m_4}), \end{equation} gives us $$ \left |\alpha ^{m_1}\left(1+\alpha ^{m_2-m_1}\right)-\frac{d\times 10^n}{9}\right |\leq \frac{d}{9}+|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4} +\alpha^{m_3}+\alpha^{m_4} \leq 7\alpha ^{m_3}, $$ which leads to \begin{equation}\label{eq56} \left |\alpha ^{m_1}\left(1+\alpha ^{m_2-m_1}\right)-\frac{d\times 10^n}{9}\right |< \alpha^{m_3+4.05}. \end{equation} Multiplying both sides of \eqref{eq56} by $\frac{\alpha^{-m_1}}{1+\alpha ^{m_2-m_1}}$ gives us $$ \left |1-\alpha ^{-m_2}10^n\left( \frac{d }{9(\alpha ^{m_1-m_2}+1)}\right)\right |< \frac{\alpha^{m_3-m_1+4.05}}{1+\alpha ^{m_2-m_1}}, $$ which yields \begin{equation}\label{eq57} \left |1-\alpha ^{-m_2}10^n\left( \frac{d }{9(\alpha ^{m_1-m_2}+1)}\right)\right |<\alpha^{m_3-m_1+4.05}. \end{equation} Put \begin{equation}\label{eqG33} \Gamma_3:= 1-\alpha ^{-m_2}10^n\left( \frac{d }{9(\alpha ^{m_1-m_2}+1)}\right). \end{equation} Suppose that $\Gamma_3=0.$ Then $$ \alpha^{m_1}+\alpha^{m_2}=\frac{10^n\times d }{9}, $$ giving us $$ \beta^{m_1}+\beta^{m_2}=\frac{10^n\times d}{9} $$ by conjugating in $\mathbb{Q}\left(\sqrt{5}\right)$. We see that $$ \frac{10^{125}}{9}\leq \frac{10^n\times d}{9}=|\beta^{m_1}+\beta^{m_2}|\leq |\beta|^{m_1}+|\beta|^{m_2}<2, $$ which is false. Hence, $\Gamma _3\neq 0$. Using the notations in Lemma \ref{lem3}, we put $$ \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d }{9(\alpha ^{m_1-m_2}+1)},\;\;b_1=-m_2,\;\;b_2=n,\;\;b_3=1, $$ where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$ and $b_1,b_2,b_3\in \mathbb{Z}$. We get $B=\max \{m_2,n,1\}\leq m_1$. It is easily seen that $$ \alpha _3=\frac{d }{9(\alpha^{m_1-m_2}+1)}\leq 1\;\; {\rm and}\;\; \alpha _3^{-1}=\frac{9(\alpha^{m_1-m_2}+1)}{d}\leq 18\alpha^{m_1-m_2}. $$ So $|\log \alpha _3|<3+(m_1-m_2)\log \alpha$. Additionally, we have \begin{align*} h(\alpha_3)&\leq h(d)+h(9)+\log 2+h(\alpha^{m_1-m_2})\\ &\leq 2h(9)+\log 2+(m_1-m_2)h(\alpha)\\ &= 2h(9)+\log 2+\frac{1}{2}(m_1-m_2)\log \alpha. \end{align*} Thus, $2h(\alpha_3)\leq 11+(m_1-m_2)\log \alpha$ and so $$ \max\{2h(\alpha _3),|\log \alpha _3|,0.16\}<11+(m_1-m_2)\log \alpha =:A_3. $$ Applying Lemma \ref{lem3} to $\Gamma _3$ given by \eqref{eqG33}, and using \eqref{eq57} we produce $$ \exp (-(11+(m_1-m_2)\log \alpha)C_1(1+\log m_1))< \alpha^{m_3-m_1+4.05}, $$ from which we obtain \begin{equation}\label{eq59} (m_1-m_3)\log \alpha <4.05\log \alpha + (11+(m_1-m_2)\log \alpha)C_1(1+\log m_1). \end{equation} {\bf Step 4}: Writing equation \eqref{eq44} as \begin{equation}\label{eq60} \alpha ^{m_1}-\frac{d\times 10^n}{9} = -\frac{d}{9}-\left(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}\right) -(\alpha ^{m_2}+\alpha ^{m_3}+\alpha ^{m_4}), \end{equation} we get $$ \left|\alpha ^{m_1}-\frac{d\times 10^n}{9}\right| \leq \frac{d}{9}+|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4} +\alpha ^{m_2}+\alpha ^{m_3}+\alpha ^{m_4}\leq 8\alpha ^{m_2}, $$ which means that \begin{equation}\label{eq61} \left |\alpha ^{m_1}-\frac{d\times 10^n}{9}\right |< \alpha^{m_2+4.33}. \end{equation} Multiplying both sides of \eqref{eq61} by $\alpha ^{-m_1}$ yields \begin{equation}\label{eq62} \left |1-\alpha ^{-m_1}10^n\left(\frac{d }{9}\right)\right |< \alpha^{m_2-m_1+4.33}. \end{equation} Put \begin{equation}\label{eqG44} \Gamma_4:=1-\alpha ^{-m_1}10^n\left(\frac{d }{9}\right). \end{equation} Suppose that $\Gamma_4=0$. Then $$ \alpha^{m_1}=\frac{d\times 10^n}{9}, $$ and by conjugation $$ \beta^{m_1}=\frac{d\times 10^n}{9}. $$ Consequently, $$ \frac{10^{125}}{9}\leq \frac{d\times 10^n}{9}=|\beta|^{m_1}<1, $$ which is impossible. Hence, $\Gamma_4 \neq 0$. In order to apply Lemma \ref{lem3} to $\Gamma_4$ given by \eqref{eqG44}, we take $$ \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d}{9},\;\;b_1=-m_1,\;\;b_2=n,\;\;b_3=1, $$ where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$ and $b_1,b_2,b_3\in \mathbb{Z}$. To compute $A _3$, we observe that $$ \alpha _3=\frac{d}{9}\leq 1\quad {\rm and} \quad \alpha _3^{-1}=\frac{9}{d}\leq 9, $$ so $|\log \alpha _3|<2.2$. In addition, we have $$ h(\alpha _3)\leq h(d)+h(9)\leq 2h(9). $$ This gives us $2h(\alpha _3)<8.79$. And so we have $$ \max \{2h(\alpha _3),|\log \alpha _3|,0.16\}<8.79=:A_3. $$ By applying Lemma \ref{lem3} to $\Gamma_4$ given by \eqref{eqG44} and using \eqref{eq62}, we obtain $$ \exp(-8.79C_1(1+\log m_1))<\left |1-\alpha ^{-m_1}10^n\left(\frac{d}{9}\right)\right |< \alpha^{m_2-m_1+4.33}. $$ This means that \begin{equation}\label{eq64} (m_1-m_2)\log \alpha<4.33\log \alpha+8.79C_1(1+\log m_1)<8.80C_1(1+\log m_1). \end{equation} Putting together \eqref{eq64} and \eqref{eq59} yields \begin{align*} (m_1-m_3)\log \alpha &<4.05\log \alpha+(11+8.80C_1(1+\log m_1))C_1(1+\log m_1)\\ \nonumber &=4.05\log \alpha +11C_1(1+\log m_1) +8.80C_1^2(1+\log m_1)^2\\ \nonumber&<8.81C_1^2(1+\log m_1)^2, \end{align*} since $4.05\log \alpha +11C_1(1+\log m_1) <0.01C_1^2(1+\log m_1)^2$. Hence, \begin{equation}\label{eq65} (m_1-m_3)\log \alpha<8.81C_1^2(1+\log m_1)^2. \end{equation} Combining \eqref{eq65} and \eqref{eq54}, we obtain \begin{align*} (m_1-m_4)\log \alpha & < 3.73\log \alpha+\left(12+8.81C_1^2(1+\log m_1)^2 \right)C_1(1+\log m_1)\\ & = 3.73\log \alpha +12C_1(1+\log m_1) +8.81C_1^3(1+\log m_1)^3\\ & < 8.82C_1^3(1+\log m_1)^3. \end{align*} since $3.73\log \alpha +12C_1(1+\log m_1) <0.01C_1^3(1+\log m_1)^3$. Thus, \begin{equation}\label{eq66} (m_1-m_4)\log \alpha<8.82C_1^3(1+\log m_1)^3. \end{equation} We now combine \eqref{eq66} and \eqref{eq49} to obtain \begin{align*} m_1\log \alpha & < 3.35 \log \alpha+(13+8.82C_1^3(1+\log m_1)^3)C_1(1+\log m_1)\\ & = 3.35\log \alpha +13C_1(1+\log m_1)+8.82C_1^4(1+\log m_1)^4\\ &< 8.83C_1^4(1+\log m_1)^4\\ &< 8.83\left(2.4\times 10^{12}\right)^4(1+\log m_1)^4. \end{align*} That is \begin{equation}\label{eq67} m_1\log \alpha<8.83\left(2.4\times 10^{12}\right)^4(1+\log m_1)^4. \end{equation} Inequality \eqref{eq67} gives rise to the inequality $m_1< 2.2\times 10^{59}$. Now, we need to lower the bound. Let \begin{equation}\label{eqLF11} \Lambda_1 =-m_1\log \alpha +n\log 10+\log\left(\frac{d}{9}\right). \end{equation} Making use of equation \eqref{eq60}, we have that \begin{align*} \alpha^{m_1}-\frac{d\times 10^n}{9}&=\alpha ^{m_1}\left(1-\alpha^{-m_1}10^n\left(\frac{d}{9}\right)\right)=\alpha^{m_1}\left(1-e^{\Lambda_1}\right)\\&=-\frac{d}{9}-\beta ^{m_1}-L_{m_2}-L_{m_3}-L_{m_4}\\ &\leq -\frac{1}{9}+|\beta| ^{600}\\ &<0, \end{align*} as $m_1\geq 600$. Thus, $\Lambda _1>0$ and so from \eqref{eq62} we obtain $$ 0<\Lambda _1X_0$ is $q=q_{125}$. We find that $q=q_{127}$ satisfies the hypothesis of Lemma \ref{lem6} for $d=1,\dots,8$. Applying Lemma \ref{lem6}, we get $m_1-m_2\leq 309$. We see that $m_1-m_2\leq 309$ for $d=1,\dots,9$ and hence $m_2\geq 291$. Taking $1\leq d\leq 9$ and $0\leq m_1-m_2\leq 309$, we let \begin{equation}\label{eqLF22} \Lambda_2=-m_2\log \alpha +n\log 10 +\log \left(\frac{d}{9(\alpha ^{m_1-m_2}+1)}\right). \end{equation} We use equation \eqref{eq55} to arrive at \begin{align*} \alpha^{m_1}(1+\alpha^{m_2-m_1})\left(1-e^{\Lambda_2}\right)&=-\frac{d}{9}-\beta^{m_1}-\beta^{m_2}-L_3-L_4\\ &\leq -\frac{1}{9}+|\beta|^{600}+|\beta|^{291}\\ &<0, \end{align*} making use of $m_1\geq 600$ and $m_2\geq 291$. Hence, $\Lambda _2>0$, and so from \eqref{eq57} we see that $$ 0<\Lambda _20$, and so from \eqref{eq52} we see that $$ 0<\Lambda _30$, and so from \eqref{eq47} we see that $$ 0<\Lambda _4