0.1 Given the Fibonacci sequence defined as follows: $F_{1}=1, F_{2}=1, F_{n+2}=F_{n+1}+F_{n}$ (n 1), find $\left(F_{2017}, F_{99} F_{101}+1\right)$ (40 points)
For any positive integers $\mathrm{m}, \mathrm{n}$ (let $\mathrm{m}>\mathrm{n}$), we have
$$
\begin{aligned}
\left(F_{m}, F_{n}\right) & =\left(F_{m-1} F_{2}+F_{m-2} F_{1}, F_{n}\right) \\
& =\left(F_{m-2} F_{2}+F_{m-3} F_{2}+F_{m-2} F_{1}, F_{n}\right) \\
& =\left(F_{m-2} F_{3}+F_{m-3} F_{2}, F_{n}\right) \\
& =\left(F_{m-3} F_{4}+F_{m-4} F_{3}\right) \\
& =\ldots \\
& =\left(F_{n} F_{m+1-n}+F_{n-1} F_{m-n}, F_{n}\right) \\
& =\left(F_{m-n}, F_{n}\right)
\end{aligned}
$$

By the Euclidean algorithm, we know
$$
\left(F_{m}, F_{n}\right)=F_{(m, n)}
$$

We will prove by induction: $F_{n+1}^{2}=F_{n} F_{n+2}-(-1)^{n}$
When $n=1$, it is clearly true.
Assume the statement holds for $n-1$ ($n>1$), then
$$
\begin{aligned}
F_{n} F_{n+2}-(-1)^{n} & =F_{n}\left(F_{n+1}+F_{n}\right)-(-1)^{n} \\
& =F_{n}^{2}+F_{n} F_{n+1}-(-1)^{n} \\
& =F_{n-1} F_{n+1}+F_{n} F_{n+1}-(-1)^{n-1}-(-1)^{n} \\
& =F_{n+1}^{2}
\end{aligned}
$$

By the principle of induction, the proposition is proved.
Thus, $\left(F_{2017}, F_{100}\right)=1$
$$
\left(F_{2017}, F_{99} F_{101}+1\right)=\left(F_{2017}, F_{100}^{2}\right)=1
$$