IMO 1986 · Problem 1

Let \(d\) be any positive integer not equal to \(2\), \(5\), or \(13\). Show that one can find distinct \(a, b\) in the set \(\{2, 5, 13, d\}\) such that \(ab-1\) is not a perfect square.

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

Solution

Suppose, for a contradiction, that every product of two distinct members of \(\{2,5,13,d\}\), minus \(1\), is a perfect square. Since \(d\) is distinct from \(2,5,13\), there are nonnegative integers \(x,y,z\) such that \[ 2d-1=x^2,\qquad 5d-1=y^2,\qquad 13d-1=z^2. \]

Every integer square is congruent to \(0\) or \(1\pmod4\). If \(d\) were even, then \(x^2=2d-1\equiv3\pmod4\), which is impossible. Thus \(d\) is odd.

Consequently, \(y^2=5d-1\) and \(z^2=13d-1\) are even, so \(y\) and \(z\) are even. Write \(y=2u\) and \(z=2v\), where \(u,v\) are integers. Subtracting their defining equations gives \[ 4(v^2-u^2)=8d, \] hence \[ v^2-u^2=2d\equiv2\pmod4. \] But the difference of two squares, each congruent to \(0\) or \(1\pmod4\), can only be congruent to \(0,1,\) or \(3\pmod4\). This is a contradiction.

Therefore at least one of \(2d-1,5d-1,13d-1\) is not a perfect square, and its corresponding pair satisfies the requirement. \(\square\)

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