MathLabs

Problem 3

Determine the maximum value of m2+n2m^2+n^2, where mm and nn are integers satisfying m,n∈{1,2,…,1981}m, n \in \{1, 2, \ldots, 1981\} and (n2−mn−m2)2=1(n^2-mn-m^2)^2 = 1.
Step 1 of 6: Basic constraints on m,nm,n
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In plain words

Before hunting for the extremal pair, pin down the shape every solution must have: coprime, with the second entry not smaller than the first.

(n2−mn−m2)2=1  ⟹  gcd⁡(m,n)=1 and n≥m(n^2-mn-m^2)^2=1 \implies \gcd(m,n)=1 \text{ and } n\ge m
Detailed analysis

Any common prime factor of m,nm,n would have its square divide n2−mn−m2n^2-mn-m^2, contradicting that this quantity is ±1\pm1; so gcd⁡(m,n)=1\gcd(m,n)=1. If n<mn<m, then n(n−m)−m2≤−n−m2≤−2n(n-m)-m^2\le -n-m^2\le -2, contradicting n2−mn−m2=±1n^2-mn-m^2=\pm1. Hence every positive solution has n≥mn\ge m.