IMO 1992 SL 2

Origin: CHN

Problem

Let R+ be the set of all nonnegative real numbers. Given two positive real numbers a and b, suppose that a mapping f : R+ \toR+ satisfies the functional equation f(f(x)) + af(x) = b(a + b)x. Prove that there exists a unique solution of this equation.

Solution

Let us define xn inductively as xn = f(xn−1), where x0 \geq0 is a fixed real number. It follows from the given equation in f that xn+2 = −axn+1 + b(a + b)xn. The general solution to this equation is of the form xn = \lambda1bn + \lambda2(−a −b)n, where \lambda1, \lambda2 \inR satisfy x0 = \lambda1 + \lambda2 and x1 = \lambda1b −\lambda2(a + b). In order to have xn \geq0 for all n we must have \lambda2 = 0. Hence x0 = \lambda1 and f(x0) = x1 = \lambda1b = bx0. Since x0 was arbitrary, we conclude that f(x) = bx is the only possible solution of the functional equation. It is easily verified that this is indeed a solution.