1981 IMO Problems/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 $( n^2 - mn - m^2 )^2 = 1$.


We first observe that since $\gcd(m,n)=1$, $m$ and $n$ are relatively prime. In addition, we note that $n \ge m$, since if we had $n < m$, then $n^2 -nm -m^2 = n(n-m) - m^2$ would be the sum of two negative integers and therefore less than $-1$. We now observe

\[(m+k)^2 -(m+k)m - m^2 = -(m^2 - km - k^2)\],

i.e., $(m,n) = (a,b)$ is a solution iff. $(b, a+b)$ is also a solution. Therefore, for a solution $(m, n)$, we can perform the Euclidean algorithm to reduce it eventually to a solution $(1,n)$. It is easy to verify that if $n$ is a positive integer, it must be either 2 or 1. Thus by trivial induction, all the positive integer solutions are of the form $(F_{n}, F_{n+1})$, where the $F_i$ are the Fibonacci numbers. Simple calculation reveals $987$ and $1597$ to be the greatest Fibonacci numbers less than $1981$, giving $987^2 + 1597^2=3524578$ as the maximal value.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1981 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
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