12・111 Given integers $m, n$ satisfying $m, n \in\{1,2, \cdots, 1981\}$ and
$$
\left(n^{2}-m n-m^{2}\right)^{2}=1 \text {. }
$$

Find the maximum value of $m^{2}+n^{2}$.
(22nd International Mathematical Olympiad, 1981)
［Solution］If $m=n$, from (1) we get $(m n)^{2}=1$, and since $m, n \in\{1,2, \cdots, 1981\}$, it follows that $m=n=1$.
If $m \neq n$, and we assume $n>m$.
Let $n=m+u_{k}$, then from (1) we have
$$
\left[\left(m+u_{k}\right)^{2}-m\left(m+u_{k}\right)-m^{2}\right]^{2}=1,
$$

which simplifies to
$$
\left(m^{2}-u_{k} m-u_{k}^{2}\right)^{2}=1.
$$

Now let $m=u_{k}+u_{k-1}$, similarly we have
$$
\left(u_{k}^{2}-u_{k-1} u_{k}-u_{k-1}^{2}\right)^{2}=1.
$$

If $u_{k} \neq u_{k-1}$, the above steps can continue until
$$
\left(u_{k-i-1}^{2}-u_{k-i} u_{k-i-1}-u_{k-i}^{2}\right)^{2}=1,
$$

and
$$
u_{k-i}=u_{k-i-1}=1.
$$

This results in the sequence
$$
n, m, u_{k}, u_{k-1}, \cdots, u_{k-i}(=1), u_{k-i-1}(=1).
$$

Any two consecutive terms in this sequence satisfy equation (1), and they also satisfy $u_{j}=u_{j-1}+u_{j-2}$, which is precisely the Fibonacci sequence.
It is easy to find that the Fibonacci numbers in the set $\{1,2, \cdots, 1981\}$ are
$$
\begin{array}{l}
1,1,2,3,5,8,13,21,34,55, \\
89,144,233,377,610,987,1597.
\end{array}
$$

From this, we know that when $m=987, n=1597$, the value of $m^{2}+n^{2}$ is maximized, with the maximum value being $987^{2}+1597^{2}=3524578$.