Example 3 If $p$ and $p+2$ are both prime numbers, then these two prime numbers are called "twin primes". Consider the following two sequences.

Fibonacci sequence: $1,1,2,3,5,8, \cdots$ (the sequence satisfying $F_{1}=1, F_{2}=1$, $F_{n+2}=F_{n+1}+F_{n}, n=1,2, \cdots$).

Twin prime sequence: $3,5,7,11,13,17,19, \cdots$ (the sequence formed by writing all twin prime pairs in ascending order).

Question: Which positive integers appear in both of the sequences above?
This is a problem related to the properties of the Fibonacci sequence. A certain term $F_{n}$ appears in the twin prime sequence if and only if $F_{n}-2$ and $F_{n}$ are both prime, or $F_{n}$ and $F_{n}+2$ are both prime. Therefore, to negate that $F_{n}$ appears in the twin prime sequence, we need to prove that $F_{n}$ is composite, or that $F_{n}-2$ and $F_{n}+2$ are both composite.

Notice that, $F_{4}=3, F_{5}=5, F_{7}=13$ all appear in the twin prime sequence. Below, we prove that when $n \geqslant 8$, $F_{n}$ does not appear in the twin prime sequence.

Using the recursive formula of the Fibonacci sequence, we have:
$$\begin{aligned}
F_{n+m+1} & =F_{1} F_{n+m-1}+F_{2} F_{n+m} \\
& =F_{1} F_{n+m-1}+F_{2}\left(F_{n-m-1}+F_{n+m-2}\right) \\
& =F_{2} F_{n+m-2}+F_{3} F_{n+m-1} \\
& =\cdots \\
& =F_{n-1} F_{m+1}+F_{n} F_{m+2}
\end{aligned}$$

In (2), let $m=n-1$, then we have $F_{2 n}=F_{n}\left(F_{n+1}+F_{n-1}\right)$. Since $F_{2}=1$, when $n \geqslant 3$, $F_{2 n}$ is composite.

On the other hand, it is well-known that: $F_{n}^{2}=F_{n-1} F_{n+1}+(-1)^{n}, n=1,2, \cdots$ (this conclusion is easily proven by mathematical induction). Combining with (2), we have
$$\begin{aligned}
F_{4 n+1} & =F_{2 n}^{2}+F_{2 n+1}^{2} \\
& =F_{2 n}^{2}+\left(F_{2 n-1}+F_{2 n}\right)^{2} \\
& =2 F_{2 n}^{2}+\left(2 F_{2 n}+F_{2 n-1}\right) F_{2 n-1} \\
& =2 F_{2 n+1} F_{2 n-1}+\left(2 F_{2 n}+F_{2 n-1}\right) F_{2 n-1}+2
\end{aligned}$$

Therefore,
$$\begin{aligned}
F_{4 n+1}-2 & =F_{2 n-1}\left(2 F_{2 n+1}+2 F_{2 n}+F_{2 n-1}\right) \\
& =F_{2 n-1}\left(F_{2 n+1}+F_{2 n+2}+F_{2 n+1}\right) \\
& =F_{2 n-1}\left(F_{2 n+3}+F_{2 n+1}\right)
\end{aligned}$$

Thus, when $n \geqslant 2$, $F_{4 n+1}-2$ is composite.
Furthermore, we also have
$$\begin{aligned}
F_{4 n+1} & =F_{2 n}^{2}+F_{2 n+1}^{2}=\left(F_{2 n+2}-F_{2 n+1}\right)^{2}+F_{2 n+1}^{2} \\
& =2 F_{2 n+1}^{2}+F_{2 n+2}\left(F_{2 n+2}-2 F_{2 n+1}\right) \\
& =2 F_{2 n+1}^{2}-F_{2 n+2} F_{2 n-1} \\
& =2 F_{2 n} F_{2 n+2}-F_{2 n+2} F_{2 n-1}-2 \\
& =F_{2 n+2}\left(2 F_{2 n}-F_{2 n-1}\right)-2 \\
& =F_{2 n+2}\left(F_{2 n}+F_{2 n-2}\right)-2
\end{aligned}$$

Therefore, when $n \geqslant 2$, $F_{4 n+1}+2$ is also composite.
Similarly, it can be proven that when $n \geqslant 2$, the numbers $F_{4 n+3}-2$ and $F_{4 n+3}+2$ are also composite. In summary, only the numbers $3,5,13$ appear in both of the above sequences.