IMO 1977 LL HUN24

Origin: HUN

Problem

Determine all real functions f(x) that are defined and contin- uous on the interval (−1, 1) and that satisfy the functional equation f(x + y) = f(x) + f(y) 1 −f(x)f(y) (x, y, x + y \in(−1, 1)).

Solution

Setting x = y = 0 gives us f(0) = 0. Let us put g(x) = arctanf(x). The given functional equation becomes tan g(x + y) = tan(g(x) + g(y)); hence g(x + y) = g(x) + g(y) + k(x, y)\pi, where k(x, y) is an integer function. But k(x, y) is continuous and k(0, 0) = 0, therefore k(x, y) = 0. Thus we obtain the classical Cauchy’s functional equation g(x + y) = g(x) + g(y) on the interval (−1, 1), all of whose continuous solutions are of the form g(x) = ax for some real a. Moreover, g(x) \in(−\pi, \pi) implies |a| \leq\pi/2. Therefore f(x) = tan ax for some |a| \leq\pi/2, and this is indeed a solution to the given equation.