12. (NET) ${ }^{\mathrm{IMO} 3}$ Determine the maximum value of $m^{2}+n^{2}$ where $m$ and $n$ are integers satisfying  
$$ m, n \in\{1,2, \ldots, 100\} \quad \text { and } \quad\left(n^{2}-m n-m^{2}\right)^{2}=1 $$
12. We will solve the contest problem (in which \( m, n \in \{1,2, \ldots, 1981\} \)). For \( m=1 \), \( n \) can be either 1 or 2. If \( m>1 \), then \( n(n-m)=m^{2} \pm 1 > 0 \); hence \( n-m > 0 \). Set \( p=n-m \). Since \( m^{2}-m p-p^{2}=m^{2}-p(m+p) = -\left(n^{2}-n m-m^{2}\right) \), we see that \((m, n)\) is a solution of the equation if and only if \((p, m)\) is a solution too. Therefore, all the solutions of the equation are given as two consecutive members of the Fibonacci sequence 
\[ 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584, \ldots \]
So the required maximum is \( 987^{2}+1597^{2} \).