IMO 1984 · Problem 6

Let \(a, b, c\) and \(d\) be odd integers such that \(0<a<b<c<d\) and \(ad=bc\). Prove that if \(a+d=2^k\) and \(b+c=2^m\) for some integers \(k\) and \(m\), then \(a=1\).

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

Solution

First, using \(ad=bc\), we obtain \[ a\bigl((a+d)-(b+c)\bigr) =a^2+bc-ab-ac =(b-a)(c-a)>0. \] Thus \(2^k>2^m\), so \(k>m\). Also, \(b\ge3\) and \(c\ge5\), whence \(m\ge3\).

Reducing the equations \(a+d=2^k\) and \(b+c=2^m\) modulo \(2^m\) gives \[ d\equiv-a,\qquad c\equiv-b\pmod{2^m}. \] Since \(ad=bc\), it follows that \[ a^2\equiv b^2\pmod{2^m}, \qquad\text{so}\qquad 2^m\mid(b-a)(b+a). \]

Both \(b-a\) and \(b+a\) are even. Their difference is \(2a\), which is congruent to \(2\) modulo \(4\). Therefore exactly one of these two factors is divisible by \(4\), while the other is divisible by \(2\) but not by \(4\). Consequently, the former factor must be divisible by \(2^{m-1}\).

However, \[ 0<b-a<b<\frac{b+c}{2}=2^{m-1}, \] so \(b-a\) cannot be divisible by \(2^{m-1}\). Hence \(2^{m-1}\mid b+a\). As \[ 0<b+a<b+c=2^m, \] we must have \[ a+b=2^{m-1}. \]

Put \(T=2^{m-1}\). Then \(b=T-a\), and \(b+c=2T\) gives \(c=T+a\). Therefore \[ a2^k=a(a+d)=a^2+ad =a^2+bc =a^2+(T-a)(T+a) =T^2=2^{2m-2}. \] Thus \(a\) is a positive odd divisor of a power of \(2\), forcing \(\boxed{a=1}\).

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