### IMO 1992

Problem A2

Find all functions f defined on the set of all real numbers with real values, such that f(x2 + f(y)) = y + f(x)2 for all x, y.

Solution

The first step is to establish that f(0) = 0. Putting x = y = 0, and f(0) = t, we get f(t) = t2. Also, f(x2+t) = f(x)2, and f(f(x)) = x + t2. We now evaluate f(t2+f(1)2) two ways. First, it is f(f(1)2 + f(t)) = t + f(f(1))2 = t + (1 + t2)2 = 1 + t + 2t2 + t4. Second, it is f(t2 + f(1 + t)) = 1 + t + f(t)2 = 1 + t + t4. So t = 0, as required.

It follows immediately that f(f(x)) = x, and f(x2) = f(x)2. Given any y, let z = f(y). Then y = f(z), so f(x2 + y) = z + f(x)2 = f(y) + f(x)2. Now given any positive x, take z so that x = z2. Then f(x + y) = f(z2 + y) = f(y) + f(z)2 = f(y) + f(z2) = 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 < x, then let z = x - y and f(z) = f(x - y) = f(x) - f(y) = y - x. In either case we get some z > 0 with f(z) = -z < 0. But now take w so that w2 = z, then f(z) = f(w2) = f(w)2 >= 0. Contradiction. So we must have f(x) = x.

Solutions are also available in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.