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\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 |&gt; -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 |&gt; -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|&lt;c\cdot \exp(-\delta\cdot Y),
 \end{equation}
 \begin{equation}\label{eq7}
 X\leq X_0. 
 \end{equation}
When $\beta =0$ in \eqref{eq5}, we get 
$$
\Lambda =x_1\vartheta_1 +x_2\vartheta_2.
$$
Put $\vartheta=-{\vartheta_1}/{\vartheta_2}$. 
Let the continued fraction expansion of $\vartheta$ be given by 
$$
[a_0,a_1,a_2,\ldots],
$$ 
and let the $k$th convergent of $\vartheta$ be ${p_k}/{q_k}$ for $k=0,1,2,\ldots$.  We may assume without loss of generality that $|\vartheta_1|&lt;|\vartheta_2|$ and that $x_1&gt;0$. 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}&gt;-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}&gt;\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&lt;\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 $q&gt;X_0$. 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 
$$
\parallel q \psi\parallel&gt;\frac{ 2X_0}{q}.
$$ 
Then, the solutions of \eqref{eq6} and \eqref{eq7} satisfy 
$$
Y&lt;\frac{1}{\delta}\log\left(\frac{q^2c}{|\vartheta_2|X_0} \right).  
$$
\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} &amp; \leq  d\left( \frac{10^n-1}{9}\right)=F_{m_1}+F_{m_2}+F_{m_3}+F_{m_4} \\
&amp;\leq  4F_{m_1} \\
&amp;\leq  \frac{4}{\sqrt{5}}\left(\alpha^{m_1}+|\beta|^{m_1}\right)\\
&amp;&lt;\frac{8}{\sqrt{5}}\alpha^{m_1}\\ 
&amp; &lt;  \alpha ^{m_1+2.7},
\end{align*}  

since $\frac{8}{\sqrt{5}}&lt;\alpha ^{2.7}$.
This means that $10^{n-1}&lt; \alpha ^{m_1+2.7}$, and thus  
$$
4.78(n-1)&lt;(n-1)\frac{\log 10}{\log \alpha}&lt;m_1+2.7.
$$ 
Consequently,  
$$
n&lt;4.78n-7.48&lt;m_1,
$$ 
as $n\geq 125$. Therefore, $125\leq n&lt;m_1$.

We can  put equation \eqref{mainthm1}  in the form 
\begin{equation}\label{eq12}
\frac{\alpha ^{m_1}-\beta ^{m_1}}{\sqrt{5}}+\frac{\alpha ^{m_2}-\beta ^{m_2}}{\sqrt{5}}+\frac{\alpha ^{m_3}-\beta ^{m_3}}{\sqrt{5}}+\frac{\alpha ^{m_4}-\beta ^{m_4}}{\sqrt{5}}=\frac{d\times 10^n}{9}-\frac{d}{9}.
\end{equation}

We examine \eqref{eq12} in four different ways as follows. 


{\bf Step 1}: We express \eqref{eq12} in the form
\begin{equation}\label{eq13}
\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)-\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}\right).
\end{equation}
It follows that
\begin{align*}
&amp;\left |\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)-\frac{d\times 10^n}{9}\right |\\ &amp; \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; \leq \frac{d}{9}+\frac{1}{\sqrt{5}}\left(|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4}\right),
\end{align*}
leading to  
\begin{equation}\label{eq14}
\left |\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)-\frac{d\times 10^n}{9}\right |&lt;\frac{\alpha^{4}}{\sqrt{5}}.
\end{equation}
Multiplying both sides of inequality  \eqref{eq14} by $\frac{\sqrt{5}\alpha^{-m_1}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}}$, 
we obtain
$$
\left |1-\alpha ^{-m_4}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)\right |
&lt;\frac{\alpha ^{4-m_1}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}},
$$
and so
\begin{equation}\label{eq15}
\left |1-\alpha ^{-m_4}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)\right |&lt; \alpha ^{4-m_1}.
\end{equation}
Put 
\begin{equation}\label{eqG1}
\Gamma_1:=1-\alpha ^{-m_4}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right).
\end{equation}
Suppose that $\Gamma_1=0.$ Then we have that
$$
\alpha^{m_1}+\alpha^{m_2}+\alpha^{m_3}+\alpha^{m_4}=\frac{10^n\times d\sqrt{5} }{9}.
$$
Conjugating in $\mathbb{Q}\left(\sqrt{5}\right)$ yields 
$$
\beta^{m_1}+\beta^{m_2}+\beta^{m_3}+\beta^{m_4}=-\frac{10^n\times d\sqrt{5} }{9}.
$$
Thus,
\begin{align*}
\frac{10^{125}\times \sqrt{5} }{9} &amp; \leq   \frac{10^n\times d\sqrt{5} }{9}
=|\beta^{m_1}+\beta^{m_2}+\beta^{m_3}\beta^{m_4}|\\
&amp;\leq |\beta|^{m_1}+|\beta|^{m_2}+|\beta|^{m_3}+|\beta|^{m_4}\\
&amp;&lt; 4.
\end{align*} 
This implies  that $\frac{10^{125}\times \sqrt{5} }{9}&lt;4$, which  is false. 
Hence, it follows that $\Gamma _1\neq 0$. 

In order to apply Lemma \ref{lem3} to $\Gamma_1$, we set 
 $$
  \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)},
 $$
 $$
 b_1=-m_4,\;\;b_2=n,\;\;b_3=1,
 $$ 
where $\alpha_1,\alpha_2,\alpha_3\in \mathbb{Q}\left(\sqrt{5}\right)$   and $b_1,b_2,b_3\in \mathbb{Z}$. 
Thus, the degree $D$ of $\mathbb{Q}\left(\sqrt{5}\right)$ is $2$ and $B=\max \{m_4,n,1\}\leq m_1$. 
The minimal polynomial of $\alpha$ over $\mathbb{Z}$ is  $x^2-x-1$, and so $d(\alpha)=2$ and $a_0(\alpha)=1$.
It follows that
$$
h(\alpha)=\frac{1}{2}\log \alpha.
$$
Also, the minimal polynomial of $\sqrt{5}$ over $\mathbb{Z}$ is  $x^2-5$. Thus,
$$
h\left(\sqrt{5}\right)=\frac{1}{2}\log 5.
$$
We have 
$$
\max \{2h(\alpha _1),|\log \alpha _1|,0.16\}=\log \alpha &lt;0.49=:A_1,
$$ 
$$
\max \{2h(\alpha _2),|\log \alpha _2|,0.16\}=2\log 10&lt;4.61=:A_2.
$$ 
Set 
$$
C_1=2.3\times 10^{12}&gt;1.4\times 30^6\times 3^{4.5}\times D^2\times (1+\log D)\times A_1\times A_2.
$$
Next, we compute $A_3$.
We find that, 
$$
\alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}&lt; \sqrt{5},
$$
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, $|\log \alpha _3|&lt;3+(m_1-m_4)\log \alpha$.  Also, we have that  
\begin{align*}
h(\alpha_3)&amp;\leq h(d\sqrt{5})+h(9)+h(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)\\
&amp;\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))\\
&amp; \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))\\
&amp; \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})\\
&amp; \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)\\
&amp;=  \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 
$$
\exp (-(15+(m_1-m_4)\log \alpha )C_1(1+\log m_1))&lt; \alpha ^{4-m_1}.
$$ 
Thus,
\begin{equation}\label{eq17}
m_1\log \alpha&lt;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 |&lt;\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
$$
\left |1-\alpha ^{-m_3}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |&lt;\frac{\alpha ^{m_4-m_1+5}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}},
$$
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 |&lt; \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 
$$
\frac{10^{125}\times \sqrt{5} }{9}\leq \frac{10^n\times d\sqrt{5} }{9}
=|\beta^{m_1}+\beta^{m_2}+\beta^{m_3}|
\leq |\beta|^{m_1}+|\beta|^{m_2}+|\beta|^{m_3}
&lt; 3,
$$
which means that $\frac{10^{125}\times \sqrt{5} }{9}&lt;3$. 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
$$
\alpha_3=\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}&lt; \sqrt{5}
$$
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, $|\log \alpha _3|&lt;3+(m_1-m_3)\log \alpha$.  Additionally, we get  
\begin{align*}
h(\alpha_3)&amp; \leq  h(d\sqrt{5})+h(9)+\log 2+h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\
&amp; \leq  \frac{1}{2}\log 5+2h(9)+2\log 2+h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\
&amp;\leq  \frac{1}{2}\log 5+2h(9)+2\log 2+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)\\
&amp;=  \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 
$$
\exp (-(14+(m_1-m_3)\log \alpha )C_1(1+\log m_1))&lt; \alpha ^{m_4-m_1+5}.
$$ 
Thus, we get
\begin{equation}\label{eq22}
(m_1-m_4)\log \alpha&lt;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 |&amp;\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})\\
&amp;&lt;  1+\frac{4}{\sqrt{5}}+\frac{2\alpha ^{m_3}}{\sqrt{5}}\\
&amp;\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 |&lt;  \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 
$$
\left |1-\alpha ^{-m_2}10^n\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_2}+1)}\right)\right |&lt;  \frac{\alpha^{m_3-m_1+5}}{1+\alpha ^{m_2-m_1}},
$$
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 |&lt;\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
$$
\frac{10^{125}\times \sqrt{5} }{9}\leq \frac{10^n\times d\sqrt{5} }{9}=|\beta^{m_1}+\beta^{m_2}|
\leq |\beta|^{m_1}+|\beta|^{m_2}
&lt;2,
$$
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 $|\log \alpha _3|&lt;3+(m_1-m_2)\log \alpha$.  Furthermore,  
\begin{align*}
h(\alpha_3)&amp;\leq  h(d\sqrt{5})+h(9)+\log 2+h(\alpha^{m_1-m_2})\\
&amp;\leq h(\sqrt{5})+2h(9)+\log 2+(m_1-m_2)h(\alpha)\\
&amp;= \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
$$
\max\{2h(\alpha _3),|\log \alpha _3|,0.16\}&lt;12+(m_1-m_2)\log \alpha =:A_3.
$$
Applying Lemma \ref{lem3} to $\Gamma _3$ given by  \eqref{eqG3}, and  using  \eqref{eq25} we produce  
$$
\exp (-(12+(m_1-m_2)\log \alpha)C_1(1+\log m_1))&lt; \alpha^{m_3-m_1+5},
$$
from which we obtain 
\begin{equation}\label{eq27}
(m_1-m_3)\log \alpha &lt;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| &amp; \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)\\
&amp;&lt;  1+\frac{4}{\sqrt{5}}+\frac{3}{\sqrt{5}}\alpha ^{m_2}\\
&amp;\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 |&lt; \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 |&lt; \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,
$$
\frac{\sqrt{5}\times 10^{125}}{9}\leq \frac{d\sqrt{5}\times 10^n}{9}=|\beta|^{m_1}&lt;1,
$$
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 $|\log \alpha _3|&lt;1.4$.  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 $2h(\alpha _3)&lt;10.4$.
We see that 
$$
\max \{2h(\alpha _3),|\log \alpha _3|,0.16\}&lt;10.4=:A_3.
$$
By applying Lemma \ref{lem3} to $\Gamma_4$ given by \eqref{eqG4} and using \eqref{eq30}, we obtain 
$$
\exp(-10.4C_1(1+\log m_1))&lt;\left |1-\alpha ^{-m_1}10^n\left(\frac{d\sqrt{5}}{9}\right)\right |&lt; \alpha^{m_2-m_1+5}.
$$
This means that 
\begin{equation}\label{eq32}
(m_1-m_2)\log \alpha&lt;5\log \alpha+10.4C_1(1+\log m_1)&lt;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 &amp;&lt;5\log \alpha+(12+10.5C_1(1+\log m_1))C_1(1+\log m_1)\\
\nonumber &amp;&lt;5\log \alpha +10.6C_1^2(1+\log m_1)^2\\
\nonumber&amp;&lt;10.7C_1^2(1+\log m_1)^2.
\end{align*}
That is
\begin{equation}\label{eq33}
(m_1-m_3)\log \alpha&lt;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 &amp; &lt;  5\log \alpha+\left(14+10.7C_1^2(1+\log m_1)^2 \right)C_1(1+\log m_1)\\
&amp; &lt;  5\log \alpha +10.8C_1^3(1+\log m_1)^3\\
&amp; &lt;  10.9C_1^3(1+\log m_1)^3.
\end{align*}
That is
\begin{equation}\label{eq34}
(m_1-m_4)\log \alpha&lt;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 &amp; &lt; 4 \log \alpha+(15+10.9C_1^3(1+\log m_1)^3)C_1(1+\log m_1)\\
&amp; &lt;  4\log \alpha+11.0C_1^4(1+\log m_1)^4\\
&amp;&lt;11.1C_1^4(1+\log m_1)^4\\
&amp;&lt;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&lt;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  $m_1&lt; 2.3\times 10^{59}$. 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}&amp;=\frac{\alpha ^{m_1}}{\sqrt{5}}\left(1-\alpha^{-m_1}10^n\left(\frac{d\sqrt{5}}{9}\right)\right)\\
&amp;=\frac{\alpha^{m_1}}{\sqrt{5}}\left(1-e^{\Lambda_1}\right)\\
&amp;=-\frac{d}{9}+\frac{\beta ^{m_1}}{\sqrt{5}}-F_{m_2}-F_{m_3}-F_{m_4}\\
&amp;\leq -\frac{1}{9}+\frac{|\beta| ^{600}}{\sqrt{5}}\\
&amp;&lt;0,
\end{align*}
as  $m_1\geq 600$. Thus,  $\Lambda _1&gt;0$ and so from \eqref{eq30} we obtain
$$
0&lt;\Lambda _1&lt;e^{\Lambda _1}-1=\left|1-\alpha^{-m_1}10^n\left(\frac{d\sqrt{5}}{9}\right)\right|&lt;\alpha^{m_2-m_1+5}.
$$
This means that
\begin{align*}
\log\left(\frac{d\sqrt{5}}{9}\right)+m_1(-\log \alpha)+n\log 10&amp;&lt;\alpha^5\alpha^{-(m_1-m_2)}\\
&amp;&lt;\alpha^{5.1}\exp(-0.48(m_1-m_2)),
\end{align*}
which leads  to
\begin{equation}\label{eq37}
|\Lambda_1|&lt;\alpha^{5.1}\exp(-0.48(m_1-m_2)),
\end{equation}
with $X=\max \{m_1,n\}= m_1\leq 2.3\times 10^{59}$. 
We also have that 
$$
\frac{\Lambda_1}{\log 10}=\frac{\log(d\sqrt{5}/9)}{\log 10}-m_1\frac{\log \alpha}{\log 10}+n.
$$
Thus, we take
$$
c=\alpha^{5.1},\;\;\delta=0.48,\;\;X_0=2.3\times 10^{59},\;\;\psi=\frac{\log(d\sqrt{5}/9)}{\log 10},\;\;\ Y=m_1-m_2,
$$ 
$$
\vartheta=\frac{\log \alpha}{\log 10},\;\; \vartheta_1=-\log \alpha,\;\;\vartheta_2=\log 10,\;\;\beta=\log(d\sqrt{5}/9).
$$
The smallest value of $q$ such that $q&gt;X_0$ 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)&amp;=-\frac{d}{9}+\frac{\beta^{m_1}}{\sqrt{5}}+\frac{\beta^{m_2}}{\sqrt{5}}-F_3-F_4\\
&amp;\leq  -\frac{1}{9}+\frac{|\beta|^{600}}{\sqrt{5}}+\frac{|\beta|^{290}}{\sqrt{5}}\\
&amp;&lt;0,
\end{align*}
making use of $m_1\geq 600$ and $m_2\geq 290$.
Hence,  $\Lambda _2&gt;0$, and so from \eqref{eq25} we see that 
$$
0&lt;\Lambda _2&lt;e^{\Lambda _2}-1=\left |1-\alpha ^{-m_2}10^n\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right)\right |&lt;\alpha^{m_3-m_1+5}.
$$
Thus, we have
\begin{align*}
\log\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right)+m_2(-\log \alpha)+n\log 10&amp;&lt;\alpha^{m_3-m_1+5}\\
&amp;&lt;\alpha^{5.1}\exp(-0.48(m_1-m_3)),
\end{align*} 
which gives us
\begin{equation}\label{eq39}
|\Lambda_2|&lt;\alpha^{5.1}\exp(-0.48(m_1-m_3)),
\end{equation}
where $X=\max\{m_2,n\}\leq m_1\leq 2.3\times 10^{59}$. 
We also have that 
$$
\frac{\Lambda_2}{\log 10}=\frac{1}{\log 10}\log\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right)-m_1\frac{\log \alpha}{\log 10}+n.
$$
Thus, we consider
$$
c=\alpha^{5.1},\;\;\delta=0.48,\;\;X_0=2.3\times 10^{59},\;\;\psi=\frac{1}{\log 10}\log\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right),
$$ 
$$
Y=m_1-m_3,\;\; \vartheta=\frac{\log \alpha}{\log 10},\;\;\vartheta_1=-\log \alpha,\;\;\vartheta_2=\log 10, \;\;
\beta=\log\left(\frac{d\sqrt{5} }{9(\alpha ^{m_1-m_2}+1)}\right).
$$
We find that $q=q_{132}$ satisfies the hypothesis of Lemma \ref{lem6}  for $d=1,\ldots,9$ and $0\leq m_1-m_2\leq 310$. Applying Lemma \ref{lem6}, we get $m_1-m_3\leq 328$. Hence,  $m_3\geq 272$.

Taking $1\leq d\leq 9$, $0\leq m_2-m_3\leq m_1-m_3\leq 328$, we let
\begin{equation}\label{eqLF3}
\Lambda_3=-m_3\log \alpha +n\log 10 +\log \left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right).
\end{equation}
Equation \eqref{eq18} ensures that
\begin{align*}
\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha^{m_2-m_1}+\alpha^{m_3-m_1}\right)\left(1-e^{\Lambda_3}\right)&amp;=-\frac{d}{9}+\frac{1}{\sqrt{5}}\left(\beta^{m_1}+\beta^{m_2}+\beta^{m_3}\right)-F_4\\
&amp;\leq  -\frac{1}{9}+\frac{1}{\sqrt{5}}\left(|\beta|^{600}+|\beta|^{290}+|\beta|^{272}\right)\\
&amp;&lt;0,
\end{align*}
where we use $m_1\geq 600$, $m_2\geq 290$, and $m_3\geq 272$.
Hence,  $\Lambda _3&gt;0$, and so from \eqref{eq20} we see that 
$$
0&lt;\Lambda _3&lt;e^{\Lambda _3}-1=\left |1-\alpha ^{-m_3}10^n\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |&lt;\alpha^{m_4-m_1+5}.
$$
Hence,  we have  
\begin{align*}
\log\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)+m_3(-\log \alpha)+n\log 10&amp;&lt;\alpha^{m_4-m_1+5}\\
&amp; &lt;\alpha^{5.1}\exp(-0.48(m_1-m_4)),
\end{align*} 
leading to 
\begin{equation}\label{eq41}
|\Lambda_3|&lt;\alpha^{5.1}\exp(-0.48(m_1-m_4)),
\end{equation}
where $X=\max\{m_3,n\}\leq m_1\leq 2.3\times 10^{59}$. 
Furthermore, we obtain
$$
\frac{\Lambda_3}{\log 10}=\frac{1}{\log 10}\log\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)-m_3\frac{\log \alpha}{\log 10}+n.
$$
Thus, we take
$$
c=\alpha^{5.1},\;\;\delta=0.48,\;\;X_0=2.3\times 10^{59},\;\;\psi=\frac{1}{\log 10}\log\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right),
$$
$$
Y=m_1-m_4,\;\;\ \vartheta=\frac{\log \alpha}{\log 10},\;\;\vartheta_1=-\log \alpha,\;\;\vartheta_2=\log 10,
$$
$$
\beta=\log\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right).
$$
We find that $q=q_{135}$ satisfies the hypothesis of Lemma \ref{lem6}  for $1\leq d\leq 9$, $0\leq m_2-m_3\leq m_1-m_3\leq 328$. 
Applying Lemma \ref{lem6}, we get $m_1-m_4\leq 335$, and hence $m_4\geq 265 $.
  
 Taking $1\leq d\leq 9$, $0\leq m_3-m_4\leq m_2-m_4\leq m_1-m_4\leq 335$, we let
\begin{equation}\label{eLF4}
\Lambda_4=-m_4\log \alpha +n\log 10 +\log \left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right).
\end{equation}
Using  equation \eqref{eq13}, we have that
\begin{align*}
\frac{\alpha^{m_1}}{\sqrt{5}}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)\left(1-e^{\Lambda_4}\right) 
&amp;=-\frac{d}{9}+\frac{1}{\sqrt{5}}\left(\beta^{m_1}+\beta^{m_2}+\beta^{m_3}+\beta^{m_4}\right)\\
&amp;\leq -\frac{1}{9}+\frac{1}{\sqrt{5}}\left(|\beta|^{600}+|\beta|^{290}+|\beta|^{272}+|\beta|^{}\right)\\
&amp; &lt;0
\end{align*}
making use of $m_1\geq 600$, $m_2\geq 290$, $m_3\geq 272$ and $m_4\geq 265$.
Hence,  $\Lambda _4&gt;0$, and so from \eqref{eq15} we see that 
$$
0&lt;\Lambda _4&lt;e^{\Lambda _4}-1=\left |1-\alpha ^{-m_4}10^n\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)\right |&lt;\alpha^{4-m_1}.
$$
Hence,  we have  
$$
\log\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)+m_4(-\log \alpha)+n\log 10
&lt;\alpha^{4-m_1},
$$
which implies that  
\begin{equation}\label{eq43}
|\Lambda_4|&lt;\alpha^{4.1}\exp(-0.48m_1),
\end{equation}
where $X=\max \{m_4,n\}\leq m_1&lt; 2.3\times 10^{59}$. 
In addition, 
$$
\frac{\Lambda_4}{\log 10}=\frac{1}{\log 10}\log\left( \frac{d\sqrt{5} }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)-m_4\frac{\log \alpha}{\log 10}+n.
$$
Thus, 
$$
c=\alpha^{4.1},\;\;\delta=0.48,\;\;X_0=2.3\times 10^{59}, \;\;Y=m_1
$$
$$
\psi=\frac{1}{\log 10}\log\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right),\;\; \vartheta=\frac{\log \alpha}{\log 10},
$$
$$
\vartheta_1=-\log \alpha,\;\vartheta_2=\log 10,\;\beta=\log\left( \frac{d\sqrt{5}  }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right).
$$
We find that $q=q_{141}$ satisfies the hypothesis of Lemma \ref{lem6}  for $1\leq d\leq 9$, $0\leq m_3-m_4\leq m_2-m_4\leq m_1-m_4\leq 335$.  Applying Lemma \ref{lem6}, we get $m_1\leq 392$, which contradicts the assumption that $m_1\geq 600$. This   proves the result.





\section{Proof of Theorem 2} \label{sec4}


Although this is similar to the proof of the previous theorem, we give the details for the convenience of the reader. We use the fact that
$$
L_m=\alpha^m+\beta ^m\quad {\rm\; holds\;\; for\;\; all} \quad m\geq 0, 
\; \text{ where }\;
(\alpha,\beta)=\left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\right).
$$
In equation (2), 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 2.

Let us suppose that solutions of equation (2) exist for  $m_1\geq 600$. We observe that
$$
L_{600}\leq L_{m_1}\leq L_{m_1}+L_{m_2}+L_{m_3}+L_{m_4}=d\left(\frac{10^n-1}{9}\right)\leq 10^n-1.
$$ 
This leads us to
$$
125\leq \frac{\log (1+L_{600})}{\log 10}\leq n,
$$
and so  $n\geq 125$. 
We further observe that   
$$
10^{n-1}  \leq  d\left( \frac{10^n-1}{9}\right)=L_{m_1}+L_{m_2}+L_{m_3}+L_{m_4} \leq  4\left(\alpha^{m_1}+|\beta|^{m_1}\right)
&lt;\alpha ^{m_1+4.33}.
$$
Hence, we obtain  
$$
4.78(n-1)&lt;(n-1)\frac{\log 10}{\log \alpha}&lt;m_1+4.33,
$$ 
which gives us  
$$
n&lt;4.78n-9.11&lt;m_1,
$$ 
for $n\geq 125$. Therefore, $125\leq n&lt;m_1$.

We can  put equation \eqref{mainthm2}  in the form 
\begin{equation}\label{eq44}
\alpha ^{m_1}+\beta ^{m_1}+\alpha ^{m_2}+\beta ^{m_2}+\alpha ^{m_3}+\beta ^{m_3}+\alpha ^{m_4}+\beta ^{m_4}-\frac{d\times 10^n}{9}=-\frac{d}{9}.
\end{equation}

Equation\eqref{eq44} is treated in four different ways in the  steps that follow. 


{\bf Step 1}: We express \eqref{eq44} in the form
\begin{equation}\label{eq45}
\alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)-\frac{d\times 10^n}{9} 
=-\frac{d}{9}-\left(\beta ^{m_1}+\beta ^{m_2}+\beta ^{m_3}+\beta ^{m_4}\right).
\end{equation}
It follows that
$$
\left |\alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)-\frac{d\times 10^n}{9}\right |
 \leq \frac{d}{9}+\left(|\beta| ^{m_1}+|\beta| ^{m_2}+|\beta| ^{m_3}+|\beta| ^{m_4}\right),
$$
leading to  
\begin{equation}\label{eq46}
\left |\alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)-\frac{d\times 10^n}{9}\right |&lt;\alpha^{3.35}.
\end{equation}
Multiplication of both sides of  \eqref{eq46} by $\frac{\alpha^{-m_1}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}}$ 
gives us
$$
\left |1-\alpha ^{-m_4}10^n\left(\frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)\right |
&lt;\frac{\alpha ^{3.35-m_1}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}},
$$
from which we get
\begin{equation}\label{eq47}
\left |1-\alpha ^{-m_4}10^n\left(\frac{d }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)\right |&lt; \alpha ^{3.35-m_1}.
\end{equation}
Put 
\begin{equation}\label{eqG11}
\Gamma_1:=1-\alpha ^{-m_4}10^n\left(\frac{d }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right).
\end{equation}
Suppose that $\Gamma_1=0.$ Then, we have that
$$
\alpha^{m_1}+\alpha^{m_2}+\alpha^{m_3}+\alpha^{m_4}=\frac{10^n\times d }{9}.
$$
Conjugating in $\mathbb{Q}\left(\sqrt{5}\right)$ yields 
$$
\beta^{m_1}+\beta^{m_2}+\beta^{m_3}+\beta^{m_4}=\frac{10^n\times d }{9}.
$$
Thus,
$$
\frac{10^{125} }{9}\leq \frac{10^n\times d}{9}=|\beta^{m_1}+\beta^{m_2}+\beta^{m_3}\beta^{m_4}|
\leq |\beta|^{m_1}+|\beta|^{m_2}+|\beta|^{m_3}+|\beta|^{m_4}&lt; 4.
$$
This implies  that $\frac{10^{125}}{9}&lt;4$, which  is false. 
Hence, it follows that $\Gamma _1\neq 0$. 

In the notation of  Lemma \ref{lem3}, we set 
 $$
  \alpha_1=\alpha,\;\;\alpha_2=10,\;\;\alpha_3=\frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)},
 $$
 $$
 b_1=-m_4,\;\;b_2=n,\;\;b_3=1,
 $$ 
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 $B=\max \{m_4,n,1\}\leq m_1$. The minimal polynomial of $\alpha$ over $\mathbb{Z}$ is  $x^2-x-1$, and so $d(\alpha)=2$ and $a_0(\alpha)=1$.
It is known that
$$
h(\alpha)=\frac{1}{2}\log \alpha.
$$
We have 
$$
\max \{2h(\alpha _1),|\log \alpha _1|,0.16\}=\log \alpha &lt;0.49=:A_1,
$$ 
$$
\max \{2h(\alpha _2),|\log \alpha _2|,0.16\}=2\log 10&lt;4.61=:A_2.
$$ 
Set 
$$
C_1=2.4\times 10^{12}&gt;1.4\times 30^6\times 3^{4.5}\times D^2\times (1+\log D)\times A_1\times A_2.
$$
Next, we compute $A_3$.
We find that, 
$$
\alpha_3=\frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}&lt; 1,
$$
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, $|\log \alpha _3|&lt;4+(m_1-m_4)\log \alpha$.  Also, we have that  
\begin{align*}
h(\alpha_3)&amp;\leq h(d)+h(9)+\log 2+h(\alpha ^{m_3-m_4}(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1))\\
&amp; \leq  2h(9)+2\log 2+h(\alpha ^{m_3-m_4})+h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\
&amp; \leq  2h(9)+3\log 2+h(\alpha ^{m_3-m_4})+h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\
&amp; \leq  2h(9)+3\log 2+(m_3-m_4)h(\alpha )+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)&amp;\\
&amp;=  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 
$$
\exp (-(13+(m_1-m_4)\log \alpha )C_1(1+\log m_1))&lt; \alpha ^{3.35-m_1}.
$$ 
Thus,
\begin{equation}\label{eq49}
m_1\log \alpha&lt;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 |&lt;\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
$$
\left |1-\alpha ^{-m_3}10^n\left(\frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |&lt;\frac{\alpha ^{m_4-m_1+3.73}}{1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}},
$$
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 |&lt; \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 
$$
\frac{10^{125} }{9}\leq \frac{10^n\times d }{9}=|\beta^{m_1}+\beta^{m_2}+\beta^{m_3}|\leq |\beta|^{m_1}+|\beta|^{m_2}+|\beta|^{m_3}&lt; 3.
$$  
Thus, $\frac{10^{125}}{9}&lt;3$, 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
$$
\alpha_3=\frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}&lt; 1,
$$
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, $|\log \alpha _3|&lt;4+(m_1-m_3)\log \alpha$.  Additionally, we get  
\begin{align*}
h(\alpha_3)&amp; \leq  h(d)+h(9)+\log 2+h(\alpha^{m_2-m_3}(\alpha^{m_1-m_2}+1))\\
&amp; \leq  2h(9)+2\log 2+h(\alpha^{m_2-m_3})+h(\alpha^{m_1-m_2})\\
&amp;\leq  2h(9)+2\log 2+(m_2-m_3)h(\alpha)+(m_1-m_2)h(\alpha)\\
&amp;=  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 
$$
\exp (-(12+(m_1-m_3)\log \alpha )C_1(1+\log m_1))&lt; \alpha ^{m_4-m_1+3.73}.
$$ 
Thus, we get
\begin{equation}\label{eq54}
(m_1-m_4)\log \alpha&lt;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 |&lt; \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 
$$
\left |1-\alpha ^{-m_2}10^n\left( \frac{d  }{9(\alpha ^{m_1-m_2}+1)}\right)\right |&lt;  \frac{\alpha^{m_3-m_1+4.05}}{1+\alpha ^{m_2-m_1}},
$$
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 |&lt;\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
$$
\frac{10^{125}}{9}\leq \frac{10^n\times d}{9}=|\beta^{m_1}+\beta^{m_2}|\leq |\beta|^{m_1}+|\beta|^{m_2}&lt;2,
$$  
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 $|\log \alpha _3|&lt;3+(m_1-m_2)\log \alpha$.  Additionally,  we have 
\begin{align*}
h(\alpha_3)&amp;\leq h(d)+h(9)+\log 2+h(\alpha^{m_1-m_2})\\
&amp;\leq 2h(9)+\log 2+(m_1-m_2)h(\alpha)\\
&amp;= 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
$$
\max\{2h(\alpha _3),|\log \alpha _3|,0.16\}&lt;11+(m_1-m_2)\log \alpha =:A_3.
$$
Applying Lemma \ref{lem3} to $\Gamma _3$ given by  \eqref{eqG33}, and  using  \eqref{eq57} we produce  
$$
\exp (-(11+(m_1-m_2)\log \alpha)C_1(1+\log m_1))&lt; \alpha^{m_3-m_1+4.05},
$$
from which we obtain 
\begin{equation}\label{eq59}
(m_1-m_3)\log \alpha &lt;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 |&lt; \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 |&lt; \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,
$$
\frac{10^{125}}{9}\leq \frac{d\times 10^n}{9}=|\beta|^{m_1}&lt;1,
$$
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 $|\log \alpha _3|&lt;2.2$.  In addition, we have
$$
h(\alpha _3)\leq h(d)+h(9)\leq 2h(9).
$$
This gives us $2h(\alpha _3)&lt;8.79$.
And so we have 
$$
\max \{2h(\alpha _3),|\log \alpha _3|,0.16\}&lt;8.79=:A_3.
$$
By applying Lemma \ref{lem3} to $\Gamma_4$ given by \eqref{eqG44} and using \eqref{eq62}, we obtain 
$$
\exp(-8.79C_1(1+\log m_1))&lt;\left |1-\alpha ^{-m_1}10^n\left(\frac{d}{9}\right)\right |&lt; \alpha^{m_2-m_1+4.33}.
$$
This means that 
\begin{equation}\label{eq64}
(m_1-m_2)\log \alpha&lt;4.33\log \alpha+8.79C_1(1+\log m_1)&lt;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 &amp;&lt;4.05\log \alpha+(11+8.80C_1(1+\log m_1))C_1(1+\log m_1)\\
\nonumber &amp;=4.05\log \alpha +11C_1(1+\log m_1) +8.80C_1^2(1+\log m_1)^2\\
\nonumber&amp;&lt;8.81C_1^2(1+\log m_1)^2,
\end{align*}
since $4.05\log \alpha +11C_1(1+\log m_1) &lt;0.01C_1^2(1+\log m_1)^2$.
Hence,
\begin{equation}\label{eq65}
(m_1-m_3)\log \alpha&lt;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 &amp; &lt;  3.73\log \alpha+\left(12+8.81C_1^2(1+\log m_1)^2 \right)C_1(1+\log m_1)\\
&amp; = 3.73\log \alpha +12C_1(1+\log m_1) +8.81C_1^3(1+\log m_1)^3\\
&amp; &lt;  8.82C_1^3(1+\log m_1)^3.
\end{align*}
since $3.73\log \alpha +12C_1(1+\log m_1) &lt;0.01C_1^3(1+\log m_1)^3$.
Thus,
\begin{equation}\label{eq66}
(m_1-m_4)\log \alpha&lt;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 &amp; &lt; 3.35 \log \alpha+(13+8.82C_1^3(1+\log m_1)^3)C_1(1+\log m_1)\\
&amp; =  3.35\log \alpha +13C_1(1+\log m_1)+8.82C_1^4(1+\log m_1)^4\\
&amp;&lt; 8.83C_1^4(1+\log m_1)^4\\
&amp;&lt; 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&lt;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  $m_1&lt; 2.2\times 10^{59}$. 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}&amp;=\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)\\&amp;=-\frac{d}{9}-\beta ^{m_1}-L_{m_2}-L_{m_3}-L_{m_4}\\
&amp;\leq -\frac{1}{9}+|\beta| ^{600}\\
&amp;&lt;0,
\end{align*}
as  $m_1\geq 600$. Thus,  $\Lambda _1&gt;0$ and so from \eqref{eq62} we obtain
$$
0&lt;\Lambda _1&lt;e^{\Lambda _1}-1=\left|1-\alpha^{-m_1}10^n\left(\frac{d}{9}\right)\right|&lt;\alpha^{m_2-m_1+4.33}.
$$
This means that
$$
\log\left(\frac{d}{9}\right)+m_1(-\log \alpha)+n\log 10&lt;\alpha^{4.33}\alpha^{-(m_1-m_2)}
&lt;\alpha^{4.34}\exp(-0.48(m_1-m_2)),
$$
which leads  to
\begin{equation}\label{eq69}
|\Lambda_1|&lt;\alpha^{4.34}\exp(-0.48(m_1-m_2)),
\end{equation}
where $X=\max\{m_1,n\}= m_1\leq 2.2\times 10^{59}$. 
We also have that 
$$
\frac{\Lambda_1}{\log 10}=\frac{\log(d/9)}{\log 10}-m_1\frac{\log \alpha}{\log 10}+n.
$$
Hence, we set
$$
c=\alpha^{4.34},\;\;\delta=0.48,\;\;X_0=2.2\times 10^{59},\;\;\psi=\frac{\log(d/9)}{\log 10}, \;\;Y=m_1-m_2
$$ 
$$
\vartheta=\frac{\log \alpha}{\log 10},\;\; \vartheta_1=-\log \alpha,\;\;\vartheta_2=\log 10,\;\;\beta=\log(d/9).
$$
When $d=9$, $\beta=0$. Substituting $X_0=2.2\times 10^{59}$ into inequality \eqref{eq8} yields $0\leq k\leq 284$. In the notation of Lemma \ref{lem5} we find that $A=a_{138}=770$,  from the continued fraction expansion of $\frac{\log \alpha}{\log 10}$.
Applying Lemma \ref{lem5}, we get  $m_1-m_2\leq 301$. 
We now consider the case $\beta\neq 0$. The smallest value of $q$ such that $q&gt;X_0$ 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)&amp;=-\frac{d}{9}-\beta^{m_1}-\beta^{m_2}-L_3-L_4\\
&amp;\leq  -\frac{1}{9}+|\beta|^{600}+|\beta|^{291}\\
&amp;&lt;0,
\end{align*}
making use of $m_1\geq 600$ and $m_2\geq 291$.
Hence,  $\Lambda _2&gt;0$, and so from \eqref{eq57} we see that 
$$
0&lt;\Lambda _2&lt;e^{\Lambda _2}-1=\left |1-\alpha ^{-m_2}10^n\left(\frac{d }{9(\alpha ^{m_1-m_2}+1)}\right)\right |&lt;\alpha^{m_3-m_1+4.05}.
$$
Thus, we have
\begin{align*}
\log\left(\frac{d}{9(\alpha ^{m_1-m_2}+1)}\right)+m_2(-\log \alpha)+n\log 10 &amp;&lt;\alpha^{m_3-m_1+4.05}\\
&amp;&lt;\alpha^{4.06}\exp(-0.48(m_1-m_3)),
\end{align*}
which gives us
\begin{equation}\label{eq71}
|\Lambda_2|&lt;\alpha^{4.06}\exp(-0.48(m_1-m_3)),
\end{equation}
where $X=\max\{m_2,n\}\leq m_1\leq 2.2\times 10^{59}$. 
In addition, we have  
$$
\frac{\Lambda_2}{\log 10}=\frac{1}{\log 10}\log\left(\frac{d }{9(\alpha ^{m_1-m_2}+1)}\right)-m_1\frac{\log \alpha}{\log 10}+n.
$$
So, we take
$$
c=\alpha^{4.06},\;\;\delta=0.48,\;\;X_0=2.2\times 10^{59},\;\;\psi=\frac{1}{\log 10}\log\left(\frac{d }{9(\alpha ^{m_1-m_2}+1)}\right),
$$ 
$$
Y=m_1-m_3, \;\;\vartheta=\frac{\log \alpha}{\log 10},\;\;\vartheta_1=-\log \alpha,\;\;\vartheta_2=\log 10,\;\;
\beta=\log\left(\frac{d }{9(\alpha ^{m_1-m_2}+1)}\right).
$$
We find that $q=q_{132}$ satisfies the hypothesis of Lemma \ref{lem6}  for $d=1,\ldots,9$ and $0\leq m_1-m_2\leq 309$. Applying Lemma \ref{lem6}, we get $m_1-m_3\leq 327$. Hence,  $m_3\geq 273$.


 Taking $1\leq d\leq 9$, $0\leq m_2-m_3\leq m_1-m_3\leq 327$, we let
\begin{equation}\label{eqLF33}
\Lambda_3=-m_3\log \alpha +n\log 10 +\log \left( \frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right).
\end{equation}
Equation \eqref{eq50} allows us to write
\begin{align*}
\alpha^{m_1}\left(1+\alpha^{m_2-m_1}+\alpha^{m_3-m_1}\right)\left(1-e^{\Lambda_3}\right)&amp;=-\frac{d}{9}-\left(\beta^{m_1}+\beta^{m_2}+\beta^{m_3}\right)-L_4\\
&amp;\leq  -\frac{1}{9}+|\beta|^{600}+|\beta|^{291}+|\beta|^{273}\\
&amp;&lt;0,
\end{align*}
where we use $m_1\geq 600$, $m_2\geq 291$, and $m_3\geq 273$.
Hence,  $\Lambda _3&gt;0$, and so from \eqref{eq52} we see that 
$$
0&lt;\Lambda _3&lt;e^{\Lambda _3}-1=\left |1-\alpha ^{-m_3}10^n\left( \frac{d  }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)\right |
&lt;\alpha^{m_4-m_1+3.73}.
$$
Hence,  we have  
\begin{align*}
\log\left( \frac{d}{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)+m_3(-\log \alpha)+n\log 10
&amp;&lt;\alpha^{m_4-m_1+3.73}\\
&amp;&lt;\alpha^{3.74}\exp(-0.48(m_1-m_4)),
\end{align*}
leading to 
\begin{equation}\label{eq73}
|\Lambda_3|&lt;\alpha^{3.74}\exp(-0.48(m_1-m_4)),
\end{equation}
where $X=\max\{m_3,n\}\leq m_1\leq 2.2\times 10^{59}$. 
Furthermore, we obtain
$$
\frac{\Lambda_3}{\log 10}=\frac{1}{\log 10}\log\left( \frac{d}{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right)-m_3\frac{\log \alpha}{\log 10}+n.
$$
Thus, we take
$$
c=\alpha^{3.74},\;\;\delta=0.48,\;\;X_0=2.2\times 10^{59},\;\; Y=m_1-m_4,
$$
$$
\psi=\frac{1}{\log 10}\log\left( \frac{d}{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right), \;\; \vartheta=\frac{\log \alpha}{\log 10},\;\;\vartheta_1=-\log \alpha,
$$
$$
\vartheta_2=\log 10,\;\; \beta=\log\left( \frac{d }{9(\alpha ^{m_1-m_3}+\alpha ^{m_2-m_3}+1)}\right).
$$
We find that $q=q_{138}$ satisfies the hypothesis of Lemma \ref{lem6}  for $1\leq d\leq 9$, $0\leq m_2-m_3\leq m_1-m_3\leq 327$. 
Applying Lemma \ref{lem6}, we get $m_1-m_4\leq 371$, and hence $m_4\geq 229$.
  
 Taking $1\leq d\leq 9$, $0\leq m_3-m_4\leq m_2-m_4\leq m_1-m_4\leq 371$, we let
\begin{equation}\label{eqLF44}
\Lambda_4=-m_4\log \alpha +n\log 10 +\log \left( \frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right).
\end{equation}
Using equation \eqref{eq45}, we get
\begin{align*}
\alpha^{m_1}\left(1+\alpha ^{m_2-m_1}+\alpha ^{m_3-m_1}+\alpha ^{m_4-m_1}\right)\left(1-e^{\Lambda_4}\right) 
&amp;=-\frac{d}{9}-\left(\beta^{m_1}+\beta^{m_2}+\beta^{m_3}+\beta^{m_4}\right)\\
&amp;\leq -\frac{1}{9}+\left(|\beta|^{600}+|\beta|^{291}+|\beta|^{273}+|\beta|^{229}\right)\\
&amp; &lt;0
\end{align*}
making use of $m_1\geq 600$, $m_2\geq 291$, $m_3\geq 273$ and $m_4\geq 229$.
Hence,  $\Lambda _4&gt;0$, and so from \eqref{eq47} we see that 
$$
0&lt;\Lambda _4&lt;e^{\Lambda _4}-1=\left |1-\alpha ^{-m_4}10^n\left( \frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)\right |&lt;\alpha^{3.35-m_1}.
$$
Hence,  we have  
$$
\log\left( \frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)+m_4(-\log \alpha)+n\log 10
&lt;\alpha^{3.35-m_1},
$$
which implies that  
\begin{equation}\label{eq75}
|\Lambda_4|&lt;\alpha^{3.36}\exp(-0.48m_1),
\end{equation}
where $X=\max\{m_4,n\}\leq m_1&lt; 2.2\times 10^{59}$. 
In addition, 
$$
\frac{\Lambda_4}{\log 10}=\frac{1}{\log 10}\log\left( \frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right)-m_4\frac{\log \alpha}{\log 10}+n.
$$
Thus, 
$$
c=\alpha^{3.36},\;\;\delta=0.48,\;\;X_0=2.2\times 10^{59}, \;\; Y=m_1,
$$
$$
\psi=\frac{1}{\log 10}\log\left( \frac{d }{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right),\;\; \vartheta=\frac{\log \alpha}{\log 10},
$$
$$
\vartheta_1=-\log \alpha,\;\;\vartheta_2=\log 10,\;\;\beta=\log\left( \frac{d}{9(\alpha ^{m_1-m_4}+\alpha ^{m_2-m_4}+\alpha ^{m_3-m_4}+1)}\right).
$$
We find that $q=q_{145}$ satisfies the hypothesis of Lemma \ref{lem6}  for $1\leq d\leq 9$, $0\leq m_3-m_4\leq m_2-m_4\leq m_1-m_4\leq 371$. 
  Applying Lemma \ref{lem6}, we get $m_1\leq 403$, which contradicts the assumption that $m_1\geq 600$. This   proves the result.
  

\section{Acknowledgments}
\label{sec7}

We thank the referee for comments which improved the quality of this
paper. F. L. was supported in parts by Grants CPRR160325161141 of NRF and
the Focus Area Number Theory Grant from CoEMaSS at Wits (South Africa)
and CGA 17-02804S (Czech Republic). This paper was completed during a
visit of F. L. at Purdue University Northwest in February, 2018. This
author thanks this institution for the hospitality and the  support.



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\bibitem{L2000}F.~Luca, Distinct digits in base $b$ expansions of linear recurrences, \emph{Quaest. Math.} {\bf 23} (2000), 389--404.

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\end{thebibliography}

\bigskip
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\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11A25; Secondary 11B39, 11J86.

\noindent \emph{Keywords:} Fibonacci number, Lucas number, linear form.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032} and
\seqnum{A000045}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 26 2018;
revised versions received  May 27 2018; August 23 2018; August 29 2018.
Published in {\it Journal of Integer Sequences}, September 8 2018.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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