IMO 1995 SL A5

Origin: UKR | Category: Algebra

Problem

Let R be the set of real numbers. Does there exist a function f : R \toR that simultaneously satisfies the following three conditions? (a) There is a positive number M such that −M \leqf(x) \leqM for all x. (b) f(1) = 1. (c) If x ̸= 0, then f  x + 1 x2  = f(x) + f  1 x  2 .

Solution

Suppose that a function f satisfies the condition, and let c be the least upper bound of {f(x) | x \inR}. We have c \geq2, since f(2) = f(1 + 1/12) = f(1) + f(1)2 = 2. Also, since c is the least upper bound, for each k = 1, 2, . . . there is an xk \inR such that f(xk) \geqc −1/k. Then c \geqf  xk + 1 x2 k  \geqc −1 k + f  1 xk 2 =⇒ f  1 xk  \geq−1 \sqrt k . On the other hand, c \geqf  1 xk

• x2 k  = f  1 xk 
• f(xk)2 \geq−1 \sqrt k

 c −1 k 2 . It follows that \sqrt k −1 k2 \geqc  c −1 −2 k  , which cannot hold for k sufficiently large. Second solution. Assume that f exists and let n be the least integer such that f(x) \leqn 4 for all x. Since f(2) = 2, we have n \geq8. Let f(x) > n−1 4 . Then f(1/x) = f(x + 1/x2) −f(x) < 1/4, so f(1/x) > −1/2. On the other hand, this implies
n−1 2 < f(x)2 = f(1/x+x2)−f(1/x) < n 4 + 1 2, which is impossible when n \geq8.