### IMO 1986

**Problem A1**
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.

**Solution**

Consider residues mod 16. A perfect square must be 0, 1, 4 or 9 (mod 16). d must be 1, 5, 9, or 13 for 2d - 1 to have one of these values. However, if d is 1 or 13, then 13d - 1 is not one of these values. If d is 5 or 9, then 5d - 1 is not one of these values. So we cannot have all three of 2d - 1, 5d - 1, 13d - 1 perfect squares.

*Alternative solution from Marco Dalai*

Suppose 2d-1, 5d-1, 13d-1 are all squares. Squares mod 4 must be 0 or 1, considering 2d-1, so d must be odd. Put d = 2k+1. Then 10k+4 = b^{2}. So b must be even, so k must be even. Put k = 2h, then 5k+1 is a square. Similarly, 52h+12 is a square, so 13h+3 is a square. Hence (13h+3)-(5h+1) = 8h+2 is a difference of two squares, which is impossible (a difference of two squares must be 0, 1, or 3 mod 4).

Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

27th IMO 1986

© John Scholes

jscholes@kalva.demon.co.uk

26 Oct 1998