30th CanMO 1998 Problem 2 Find all real x equal to √(x - 1/x) + √(1 - 1/x).
Squaring are rearranging: x 3 - x 2 - x - 2 = 2√(x 3 - x 2 - x + 1). Squaring again, x 2 (x 4 - 2x 3 - x 2 - 2x + 1) = 0. Referring to the original equation, x = 0 is not a solution, so we must have x 4 - 2x 3 - x 2 - 2x + 1 = 0. Factorising (x 2 - x - 1) 2 = 0, so x = (1 + √5)/2 or (1 - √5)/2. But referring to the original equation, we must have x ≥ 1, so the only candidate is x = (1 + √5)/2. With this value 1/x = (√5 - 1)/2, so x - 1/x = 1 and 1 - 1/x = ( (√5 - 1)/2) 2 and hence √(x - 1/x) + √(1 - 1/x) = x, as required. 30th CanMO 1998 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002