22nd Australian 2001 Problem B1 Find all real valued functions f(x) on the reals such that f( (x - y) 2 ) = x 2 - 2y f(x) + f(y) 2 . Solution
: f(x) = x or f(x) = x + 1. Put x = y = 0, then f(0) = f(0) 2 , so f(0) = 0 or 1. Suppose first that f(0) = 1. Putting y = 0, f(x 2 ) = x 2 + 1, so f(x) = x + 1 for x >= 0. Putting x = 0, f(y 2 ) = -2y + f(y) 2 , so f(y) 2 = y 2 + 1 + 2y = (y + 1) 2 , so f(y) = y + 1 or - y - 1. Putting x = y, 1 = x 2 - 2xf(x) + f(x) 2 = (x - f(x)) 2 , so f(x) = x + 1 or x - 1. Hence f(x) = x + 1 for all x. Checking, this gives f( (x-y) 2 ) = (x-y) 2 + 1 = x 2 - 2xy + y 2 + 1, and x 2 - 2yf(x) + f(y) 2 = x 2 - 2xy - 2y + y 2 + 2y + 1 = x 2 - 2xy + y 2 + 1. So f(x) = x + 1 is a solution. Now consider the other possibility, f(0) = 0. Putting y = 0, we get f(x 2 ) = x 2 , so f(x) = x for x >= 0. Putting x = y gives 0 = x 2 - 2xf(x) + x 2 , so f(x) = x for all x. Checking: f( (x-y) 2 ) = (x-y) 2 = x 2 - 2xy + y 2 , and x 2 - 2yf(x) + f(y) 2 = x 2 - 2yx + y 2 . So f(x) = x is a solution. 22nd Aus 2001 © John Scholes jscholes@kalva.demon.co.uk 25 July 2002