$5 \cdot 7$ Fibonacci numbers are defined as
$$
a_{0}=0, a_{1}=a_{2}=1, a_{n+1}=a_{n}+a_{n-1}(n \geqslant 1) .
$$

Find the greatest common divisor of the 1960th and 1988th terms.
(29th International Mathematical Olympiad Shortlist, 1988)
[Solution] 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 as follows:
$$
\begin{array}{r}
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597, \\
2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,
\end{array}
$$
317811.
For the Fibonacci sequence, we prove that
$$
\left(a_{m}, a_{n}\right)=a_{(m, n)}.
$$

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

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
$$
a_{m}=a_{n} a_{m-n+1}+a_{n-1} a_{m-n}.
$$

From (4), it can be seen that the greatest common divisor of $a_{m-n}$ and $a_{n}$ must be a divisor of $a_{m}$, and the greatest common divisor 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 greatest common divisor of $a_{m}$ and $a_{n}$ is also a divisor of $a_{m-n}$, thus
$$
\left(a_{m}, a_{n}\right)=\left(a_{m-n}, a_{n}\right).
$$

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