IMO 2003 · Problem 2

Find all pairs \((m,n)\) of positive integers such that \(\frac{m^2}{2mn^2-n^3+1}\) is a positive integer.

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

Answer

All such pairs are \[ \boxed{(m,n)=(2t,1),\quad (t,2t),\quad (8t^4-t,2t), \qquad t\in\mathbb Z_{>0}.} \]

Solution

If \(n=1\), the quotient is \(m/2\), so precisely the pairs \((2t,1)\) work.

Now suppose \(n>1\), and write \[ m^2=k(2mn^2-n^3+1),\qquad k\in\mathbb Z_{>0}. \] Thus \(m\) is a root of \[ f(x)=x^2-2kn^2x+k(n^3-1). \] The other root is the integer \(m'=2kn^2-m\). Moreover, \[ mm'=k(n^3-1)>0, \] so \(m'>0\).

Since \[ f(n)=n^2-kn^3-k<0, \] the number \(n\) lies strictly between the two roots. Consequently, their smaller root \(r\) is an integer satisfying \[ 0<r<n. \] As \(f(r)=0\), we have \[ r^2=k\bigl(n^2(2r-n)+1\bigr). \] If \(2r-n\le -1\), the right-hand side is negative, which is impossible. If \(2r-n\ge1\), then \[ r^2\ge n^2+1, \] contradicting \(r<n\). Therefore \(2r-n=0\). It follows that \[ r=\frac n2,\qquad k=r^2=\frac{n^2}{4}. \] In particular, \(n\) is even. The larger root is \[ 2kn^2-r=\frac{n^4-n}{2}. \] Writing \(n=2t\), the original \(m\) must therefore be either \(t\) or \(8t^4-t\).

Finally, both possibilities work. For \((m,n)=(t,2t)\), the denominator is \(1\), so the quotient is \(t^2\). For \((m,n)=(8t^4-t,2t)\), we have \[ m=t(8t^3-1),\qquad 2mn^2-n^3+1=(8t^3-1)^2, \] so the quotient is again \(t^2\). Together with the case \(n=1\), this proves the answer.

The solution was checked against Evan Chen's published IMO solution notes.