Example 20 (IMO-29 Preliminary Question) The Fibonacci numbers are defined as
$$
a_{0}=0, a_{1}=a_{2}=1, a_{n+1}=a_{n}+a_{n-1} \quad(n \geqslant 1) .
$$

Find the greatest common divisor of the 1960th and 1988th terms.
Given $a_{1}=a_{2}=1$ and
$$
a_{n+1}=a_{n}+a_{n-1} \quad(n>1)
$$

we can calculate the first 28 terms of the sequence step by step:
$1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597$, $2584,4181,6765,10946,17711,28657,46368,75025,121393,196418$, 317811.
For the Fibonacci sequence, we prove
$$
\left(a_{m}, a_{n}\right)=a_{(m, n)} \text {. }
$$

First, we use mathematical induction to prove
$$
\left(a_{n}, a_{n-1}\right)=1 \text {. }
$$

When $n=1$, $\left(a_{1}, a_{0}\right)=1$,
When $n=2$, $\left(a_{2}, a_{1}\right)=(1,1)=1$.
So, when $n=1,2$, (3) holds.
Assume that when $n=k$, (3) holds, i.e., $\left(a_{k}, a_{k-1}\right)=1$.
Then by (1), $\left(a_{k+1}, a_{k}\right)=\left(a_{k}, a_{k-1}\right)=1$.
Therefore, (3) holds for all natural numbers $n$.
Assume $m \geqslant n$, by (1) we can derive
$$
a_{m}=a_{n} a_{m-n+1}+a_{n-1} a_{m-n} \text {. }
$$

From (4), we can see that the greatest common divisor (gcd) of $a_{m-n}$ and $a_{n}$ must be a divisor of $a_{m}$, and the gcd of $a_{m}$ and $a_{n}$ must be a divisor of $a_{n-1} a_{m-n}$.
By (3), $a_{n}$ and $a_{n-1}$ are coprime, so the gcd of $a_{m}$ and $a_{n}$ is also a divisor of $a_{m-n}$, thus
$$
\left(a_{m}, a_{n}\right)=\left(a_{m-n}, a_{n}\right) \text {. }
$$

Let $m=q n+r, 0 \leqslant r < n$, continuing the above process, we get $\left(a_{m}, a_{n}\right)=a_{(m, n)}$.

Since $(1988,1960)=28$, then
$$
\left(a_{1988}, a_{1960}\right)=a_{(1988,1960)}=a_{28}=317811 .
$$