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 5 of 6: Find the largest Fibonacci pair below 1981
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F16=987,F17=1597,F18=2584>1981F_{16}=987,\quad F_{17}=1597,\quad F_{18}=2584>1981
Detailed analysis

Listing Fibonacci numbers, 987987 and 15971597 are both at most 19811981 while the next one, 25842584, exceeds 19811981. So (m,n)=(987,1597)(m,n)=(987,1597) is the admissible Fibonacci pair with the largest entries.