IMO 2007 · Problem 5

Let \(a\) and \(b\) be positive integers. Show that if \(4ab-1\) divides \((4a^2-1)^2\), then \(a=b\).

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

Solution

Put \(d=4ab-1\). Since \[ b(4a^2-1)=ad+(a-b), \] the given divisibility implies \[ d\mid b^2(4a^2-1)^2 \quad\Longrightarrow\quad d\mid(a-b)^2. \]

Suppose, for a contradiction, that \(a\ne b\). Then \[ (a-b)^2=k(4ab-1) \] for some positive integer \(k\). We will show by infinite descent that this equation has no positive integer solutions.

Fix this \(k\), and choose a positive integer solution \((x,y)\) of \[ (x-y)^2=k(4xy-1) \tag{1} \] with \(x+y\) as small as possible. Such a solution exists by our assumption. The right-hand side is positive, so \(x\ne y\); by symmetry, we may assume \(x>y\).

Rearranging (1) gives \[ x^2-(4k+2)yx+y^2+k=0. \] Define the integer \[ z=(4k+2)y-x. \] The polynomial \[ f(t)=t^2-(4k+2)yt+y^2+k \] therefore factors as \[ f(t)=(t-x)(t-z). \] Comparing constant terms gives \[ xz=y^2+k>0, \] so \(z>0\). Moreover, \[ (y-x)(y-z)=f(y)=k(1-4y^2)<0. \] Since \(y-x<0\), we obtain \(y-z>0\). Thus \[ 0<z<y<x. \]

Finally, \(f(z)=0\) gives \[ z^2-(4k+2)yz+y^2+k=0, \] or equivalently, \[ (z-y)^2=k(4zy-1). \] Hence \((z,y)\) is another positive integer solution of (1), but \[ z+y<x+y, \] contradicting the minimal choice of \((x,y)\).

Therefore \(a\ne b\) is impossible, and \(\boxed{a=b}\).

The solution was checked against the official IMO shortlist solutions.