33rd Swedish 1993 Problem 6 For reals a, b define the function f(x) = 1/(ax+b). For which a, b are there distinct reals x 1 , x 2 , x 3 such that f(x 1 ) = x 2 , f(x 2 ) = x 3 , f(x 3 ) = x 1 .
a + b 2 = 0 Solution We find that f(f(f(x))) = (a + b 2 + abx)/(b 3 +2ab + (a 2 +ab 2 )x). Now f(f(f(x))) = x is equivalent to (a2+ab2)x2 + (b3+ab)x - (a+b2) = 0. But this quadratic has three distinct solutions x 1 , x 2 , x 3 so it must vanish identically. The constant term vanishes if a + b 2 = 0, but then the other coefficients also vanish. So we must have a = -b 2 . If a = -b 2 , then it is easy to check that f(x) = 1/(b - b 2 x), f(f(x)) = (bx-1)/b 2 x, f(f(f(x))) = x, so x 1 , x 2 , x 3 certainly exist. Thanks to Suat Namli 33rd Swedish 1993 © John Scholes jscholes@kalva.demon.co.uk 24 February 2004 Last corrected/updated 24 Feb 04