IMO 1981 · Problem 3

Determine the maximum value of \(m^3+n^3\), where \(m\) and \(n\) are integers satisfying \(m,n\in\{1,2,\ldots,1981\}\) and \((n^2-mn-m^2)^2=1\).

Official text of the problem. Official problem paper (PDF)

Answer

The maximum is \[ \boxed{5\,034\,507\,976}, \] attained uniquely at \((m,n)=(987,1597)\).

Solution

Write \[ Q(m,n)=n^2-mn-m^2. \] The given equation is equivalent to \(Q(m,n)=\pm1\). We first classify all positive integer pairs satisfying this condition.

If \(n\le m\), then \[ Q(m,n)=n(n-m)-m^2\le -m^2. \] Thus \(|Q(m,n)|=1\) forces \(m=1\), and consequently \(n=1\).

For any other solution, therefore, \(n>m\). The pair \[ (m',n')=(n-m,m) \] has positive integer coordinates and satisfies \[ Q(m',n') =m^2-m(n-m)-(n-m)^2 =-Q(m,n). \] It is another solution, with strictly smaller coordinate sum: \[ m'+n'=n<m+n. \] Repeating this transformation must eventually reach \((1,1)\), since a positive integer sum cannot decrease indefinitely.

The transformation has the unique inverse \[ (a,b)\longmapsto(b,a+b). \] Moreover, \[ Q(b,a+b)=-Q(a,b), \] so starting from \((1,1)\), every pair produced by this inverse transformation is indeed a solution. Hence all positive integer solutions are precisely \[ (m,n)=(F_k,F_{k+1})\qquad(k\ge1), \] where \[ F_1=F_2=1,\qquad F_{k+2}=F_{k+1}+F_k. \]

The successive Fibonacci numbers through the first one exceeding \(1981\) are \[ \begin{gathered} 1,1,2,3,5,8,13,21,34,55,89,144,\\ 233,377,610,987,1597,2584. \end{gathered} \] Thus the last admissible pair is \((987,1597)\): the next pair is \((1597,2584)\), which violates the bound.

Along these pairs neither coordinate decreases, and the second coordinate strictly increases. Therefore \(m^3+n^3\) strictly increases, so its unique maximum occurs at \((987,1597)\). Its value is \[ 987^3+1597^3 =961\,504\,803+4\,073\,003\,173 =\boxed{5\,034\,507\,976}. \]

The solution was checked against John Scholes' IMO solution archive (Kalva).