IMO 1992 Problem A2 Find all functions f defined on the set of all real numbers with real values, such that f(x 2 + f(y)) = y + f(x) 2 for all x, y.
The first step is to establish that f(0) = 0. Putting x = y = 0, and f(0) = t, we get f(t) = t 2 . Also, f(x 2 +t) = f(x) 2 , and f(f(x)) = x + t 2 . We now evaluate f(t 2 +f(1) 2 ) two ways. First, it is f(f(1) 2 + f(t)) = t + f(f(1)) 2 = t + (1 + t 2 ) 2 = 1 + t + 2t 2 + t 4 . Second, it is f(t 2 + f(1 + t)) = 1 + t + f(t) 2 = 1 + t + t 4 . So t = 0, as required. It follows immediately that f(f(x)) = x, and f(x 2 ) = f(x) 2 . Given any y, let z = f(y). Then y = f(z), so f(x 2 + y) = z + f(x) 2 = f(y) + f(x) 2 . Now given any positive x, take z so that x = z 2 . Then f(x + y) = f(z 2 + y) = f(y) + f(z) 2 = f(y) + f(z 2 ) = f(x) + f(y). Putting y = -x, we get 0 = f(0) = f(x + -x) = f(x) + f(-x). Hence f(-x) = - f(x). It follows that f(x + y) = f(x) + f(y) and f(x - y) = f(x) - f(y) hold for all x, y. Take any x. Let f(x) = y. If y > x, then let z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z. If y  0 with f(z) = -z = 0. Contradiction. So we must have f(x) = x. s are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 33rd IMO 1992 © John Scholes jscholes@kalva.demon.co.uk 13 Nov 1998