IMO 1994 SL A3

Origin: GBR | Category: Algebra

Problem

Let S be the set of real numbers greater than −1. Find all functions f : S \toS such that f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all x and y in S, and f(x)/x is strictly increasing for −1 < x < 0 and for 0 < x.

Solution

The last condition implies that f(x) = x has at most one solution in (−1, 0) and at most one solution in (0, \infty). Suppose that for u \in(−1, 0), f(u) = u. Then putting x = y = u in the given functional equation yields f(u2 + 2u) = u2 + 2u. Since u \in(−1, 0) ⇒u2 + 2u \in(−1, 0), we deduce that u2 + 2u = u, i.e., u = −1 or u = 0, which is impossible. Similarly, if f(v) = v for v \in(0, \infty), we are led to the same contradiction. However, for all x \inS, f(x + (1 + x)f(x)) = x + (1 + x)f(x), so we must have x + (1 + x)f(x) = 0. Therefore f(x) = − x 1+x for all x \inS. It is directly verified that this function satisfies all the conditions.