IMO 1992 SL 6

Origin: IND

Problem

Find all functions f : R \toR such that f(x2 + f(y)) = y + f(x)2 for all x, y in R.

Solution

It is easy to see that f is injective and surjective. From f(x2 + f(y)) = f((−x)2 + f(y)) it follows that f(x)2 = (f(−x))2, which implies f(−x) = −f(x) because f is injective. Furthermore, there exists z \inR such that f(z) = 0. From f(−z) = −f(z) = 0 we deduce that z = 0. Now we have f(x2) = f(x2 + f(0)) = 0 + (f(x))2 = f(x)2, and consequently f(x) = f(\sqrtx)2 > 0 for all x > 0. It also follows that f(x) < 0 for x < 0. In other words, f preserves sign. Now setting x > 0 and y = −f(x) in the given functional equation we obtain f(x −f(x)) = f(\sqrtx 2 + f(−x)) = −x + f(\sqrtx)2 = −(x −f(x)). But since f preserves sign, this implies that f(x) = x for x > 0. Moreover, since f(−x) = −f(x), it follows that f(x) = x for all x. It is easily verified that this is indeed a solution.