IMO 1999 SL A5

Origin: JAP | Category: Algebra

Problem

Find all the functions f : R \toR that satisfy f(x −f(y)) = f(f(y)) + xf(y) + f(x) −1 for all x, y \inR.

Solution

Let A = {f(x) | x \inR} and f(0) = c. Plugging in x = y = 0 we get f(−c) = f(c) + c −1, hence c ̸= 0. If x \inA, then taking x = f(y) in the original functional equation we get f(x) = c+1 −x2 2 for all x \inA. We now show that A −A = {x1 −x2 | x1, x2 \inA} = R. Indeed, plugging in y = 0 into the original equation gives us f(x−c)−f(x) = cx+f(c)−1, an expression that evidently spans all the real numbers. Thus, each x can be represented as x = x1 −x2, where x1, x2 \inA. Plugging x = x1 and f(y) = x2 into the original equation gives us f(x) = f(x1−x2) = f(x1)+x1x2+f(x2)−1 = c−x2 1 + x2 +x1x2 = c−x2 2 . Hence we must have c = c+1 2 , which gives us c = 1. Thus f(x) = 1 −x2 for all x \inR. It is easily checked that this function satisfies the original functional equation.