Problem 3

Determine the maximum value of

m^2+n^2

, where

m

and

n

are integers satisfying

m, n \in \{1, 2, \ldots, 1981\}

and

(n^2-mn-m^2)^2 = 1

Step 2 of 6: Find the recursive symmetry

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In plain words

This is exactly the Fibonacci recurrence hiding inside the algebra: moving forward by $(a,b)\mapsto(b,a+b)$ preserves being a solution while flipping the sign of $n^2-mn-m^2$ .

(m+k)^2-(m+k)m-m^2 = -\big(m^2-km-k^2\big)

Detailed analysis

Expanding the left side and simplifying shows that if $(k,m)$ satisfies the defining equation with value $\varepsilon=\pm1$ , then $(m,m+k)$ satisfies it with value $-\varepsilon$ . So $(a,b)$ is a solution exactly when $(b,a+b)$ is a solution, with the sign flipped.