The ratio between Fibonacci numbers soon settles down to a number close to $1.618$. This number is called the Golden Ratio.
You get an extra digit every time the Fibonacci numbers have increased by a factor 10.
$1.618^4=6.854$ and $1.618^5=11.09$
Once the ratio settles down, you get at least one extra digit every five numbers. Sometimes the extra digit arrives sooner, and you only get four numbers with so many digits.
$\textbf{Exercise}$ Let $(a,b)=1$ and $c>0$. Prove that there is an integer $x$ such that $(a+bx,c)=1$.
$\textbf{Solution.}$ Let $p_{1},p_{2},\cdots,p_{m}$ be the primes which appear in the prime factorization of $b$. Then since $(a,b)=1$, we have $(a,p_{i})=1$ for all $i$. If the prime factorization of $c$ contains only primes from the set $\{p_{1},p_{2},\ldots,p_{m}\}$, then our required integer $x=0$ since $(a,p_i)=1$ for each $i$ implies $(a,c)=1$. Now suppose that $c$ contains extra primes $q_{1},q_{2},\ldots,q_{n}$. That is $c$ is of the form $$c=\prod p_{i}q_j \quad \text{or} \quad c =\prod_{i=1}^{n} q_{i}$$ then we want to find a integer $x$ such that $(a+bx,p_i)=1$ for all $i$ and $(a+bx,q_j)=1$ for all $j$. It is clear that $(a+bx,p_i)=(a,p_i)=1$. So basically we want to find an integer $x$ such that $(a+bx,q_j)=1$ for all $j$. We know that $(q_{j}+1,q_j)=1$ always so we need $x$ such that $bx+a=q_{j}+1\equiv 1\pmod{q_j}$ for all $j$, that is $bx\equiv 1-a\pmod{q_j}$ for all $j$. Since $(b,q_j)=1$ for all $j$, therefore $b$ has an inverse and so $x=(1-a)b^{-1}\pmod{q_j}$. Now the system of congruences
\begin{align*}
x &\equiv (1-a)b^{-1}\pmod{q_1}\\
x &\equiv (1-a)b^{-1}\pmod{q_2}\\
& \cdot\\
& \cdot\\
x&\equiv (1-a)b^{-1}\pmod{q_j}
\end{align*}
has a solution by Chinese Remainder Theorem and that solution is our required $x$.
$\textbf{Remark.}$ If we assume Dirichlet's Theorem then this problem can be solved as follows: Since $(a,b)=1$, The set $\{a+bx \ : x \in \mathbb{Z}\}$ contains infinitely many primes. Since $c$ is fixed so its finite so has only finitely many prime factors. Let $P$ be the largest prime factor of $c$. Now choose $x$ large enough so that $a+bx$ is a prime which is greater than $P$. Then for that $x$, $(a+bx,c)=1$.
Best Answer
The very first observation is that if for some prime number $p$ and positive integer $n$ we have $p|a_n$ then $a_m$ is not divisible by $p$ for all $m>n$ (*) (the modulo will be $0,-1,1,-1,1,\dots$). We will return to this observation later in this post.
Let $P(x)=x^2-x-1$
For any odd prime number $p$, consider the sequence $b_n = (a_n \mod p)$ for all $n \ge 0$.
Note that for any number $a$, $P(a) \equiv P(1-a) (\mod p)$ hence the modulo of $P(a)$ mod $p$ has at most $\frac{p+1}{2}$ different values .
So by pigeonhole principle, there are two distinct numbers $1 \le m<n \le \frac{p+1}{2}+1$ such that $b_m=b_n$, i.e.
From the position $m$, the sequence $(b_k)$ repeats with the period $n-m$ (1).
We will show next there is not any $k$ that lies between $m,n$ such that $b_k=0$(2).
Indeed, if its true, that leads to $b_{k+m-n}=0$, which means $a_k$ and $a_{k+m-n}$ are divisible by $p$. This is a contradiction to our observation (*).
From (1) and (2), we conclude that $b_k \ne 0$ for all $k \ge m$. Besides $m \le \frac{p+1}{2}$, thus in general, $b_k \ne 0 $ with all $k \ge \frac{p+1}{2}$.
In other words, $a_k$ is not divisible by $p$ for all $k\ge \frac{p+1}{2}$. Hence the conclusion.