25th BMO 1989 Problem 1 Find the smallest positive integer a such that ax 2 - bx + c = 0 has two distinct roots in the interval 0 < x < 1 for some integers b, c.
5 Solution Let the roots be α, β. So a(x - α)(x - β) = ax 2 - bx + c. Putting x = 1 gives a(1 - α)(1 - β) = a - b + c. Multiplying by c = aαβ gives a 2 α(1 - α)β(1 - β) = c(a - b + c). Since 0  0. Also by AM/GM we have α(1 - α)  c(a - b + c) > 0. But c(a - b + c) is an integer, so a 2 /16 > 1. Hence a ≥ 5. We have b 2 > 4ac, so b ≥ 5 also. We find that 5x 2 - 5x + 1 has the roots (5 ± √5)/10, which are distinct and lie between 0 and 1. Comment. A more straightforward (but more tedious) approach is to consider separately the cases a = 1, a = 2 etc. The question first appeared in Swedish 79/5. 25th BMO 1989 © John Scholes jscholes@kalva.demon.co.uk 12 February 2004 Last corrected/updated 12 Feb 04