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.

**Solution**

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 < 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 w^{2} = z, then f(z) = f(w^{2}) = 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.

© John Scholes

jscholes@kalva.demon.co.uk

13 Nov 1998