IMO 1997 SL 22

Origin: UKR

Problem

(a) Do there exist functions f : R \toR and g : R \toR such that f(g(x)) = x2 and g(f(x)) = x3 for all x \inR? (b) Do there exist functions f : R \toR and g : R \toR such that f(g(x)) = x2 and g(f(x)) = x4 for all x \inR?

Solution

(a) Suppose that f and g are such functions. From g(f(x)) = x3 we have f(x1) ̸= f(x2) whenever x1 ̸= x2. In particular, f(−1), f(0), and f(1) are three distinct numbers. However, since f(x)2 = f(g(f(x))) = f(x3), each of the numbers f(−1), f(0), f(1) is equal to its square, and so must be either 0 or 1. This contradiction shows that no such f, g exist. (b) The answer is yes. We begin with constructing functions F, G : (1, \infty) \to(1, \infty) with the property F(G(x)) = x2 and G(F(x)) = x4 for x >

1. Define the functions ϕ, \psi by F(22t) = 22ϕ(t) and G(22t) = 22\psi(t). These functions determine F and G on the entire interval (1, \infty), and satisfy ϕ(\psi(t)) = t+ 1 and \psi(ϕ(t)) = t+ 2. It is easy to find examples of ϕ and \psi: for example, ϕ(t) = 1 2t + 1, \psi(t) = 2t. Thus we also arrive at an example for F, G: F(x) = 22 2 log2 log2 x+1 = 22\sqrt log2 x, G(x) = 222 log2 log2 x = 2log2 2 x. It remains only to extend these functions to the whole of R. This can be done as follows: Ff(x) = ⎧ ⎨ ⎩ F(x) for x > 1, 1/F(1/x) for 0 < x < 1, x for x \in{0, 1}; Fg(x) = ⎧ ⎨ ⎩ G(x) for x > 1, 1/G(1/x) for 0 < x < 1, x for x \in{0, 1}; and then f(x) = Ff(|x|), g(x) = Fg(|x|) for x \inR. It is directly verified that these functions have the required property.