6. 89 Determine the maximum value of $m^{2}+n^{2}$. Where $m, n$ are integers, and $m, n \in\{1$, $2, \cdots, 1981\},\left(n^{2}-m n-m^{2}\right)^{2}=1$

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6. 89 Determine the maximum value of $m^{2}+n^{2}$. Where $m, n$ are integers, and $m, n \in\{1$, $2, \cdots, 1981\},\left(n^{2}-m n-m^{2}\right)^{2}=1$
［Solution］First, consider the case where $m=n$.
Since $\left(n^{2}-m n-m^{2}\right)^{2}=1$, we have $(m n)^{2}=1$, and since $m, n \in\{1,2,3, \cdots, 1981\}$, it must be that $m=n=1$. Next, consider the case where $m \neq n$.

Clearly, $n>m$. Let $n=m+u_{k}$.
Then $\left[\left(m+u_{k}\right)^{2}-m\left(m+u_{k}\right)-m^{2}\right]^{2}=1$.
Simplifying, we get $\left(m^{2}-u_{k} m-u_{k}^{2}\right)^{2}=1, \left(m>u_{k}\right)$
Let $m=u_{k}+u_{k-1}$,
Substituting into (1) gives $\left(u_{k}^{2}-u_{k-1} u_{k}-u_{k-1}^{2}\right)^{2}=1$.
If $u_{k} \neq u_{k-1}$, then continue the above transformation until
$$\left(u_{k-i}^{2}-u_{k-i-1} u_{k-i}-u_{k-i-1}^{2}\right)^{2}=1 \text {, and } u_{k-i}=u_{k-i-1}=1 \text {,}$$

stopping here. From all the transformations above, we get the sequence:
$$u_{k-i-1}, u_{k-i}, \cdots, u_{k-1}, m, n$$

Any two consecutive terms in this sequence satisfy the condition $\left(u_{j}^{2}-u_{j-1} u_{j}-u_{j-1}^{2}\right)^{2}=1$, and the sequence satisfies $u_{j}=u_{j-1}+u_{j-2}$, which is clearly the Fibonacci sequence.
The Fibonacci numbers in $\{1,2, \cdots, 1981\}$ are: $1,1,2,3,5,8,13,21,34,55$, $89,144,233,377,610,987,1597$.

To maximize $m^{2}+n^{2}$, we need to take $m=987, n=1597$.
Thus, $m^{2}+n^{2}=987^{2}+1597^{2}=3524578$ is the maximum value that satisfies the given conditions.