IMO 2002 SL A1

Origin: CZE | Category: Algebra

Problem

Find all functions f from the reals to the reals such that f(f(x) + y) = 2x + f(f(y) −x) for all real x, y.

Solution

We observe first that f is surjective. Indeed, setting y = −f(x) gives f(f(−f(x)) −x) = f(0) −2x, where the right-hand expression can take any real value. In particular, there exists x0 for which f(x0) = 0. Now setting x = x0 in the functional equation yields f(y) = 2x0 + f(f(y) −x0), so we obtain f(z) = z −x0 for z = f(y) −x0. Since f is surjective, z takes all real values. Hence for all z, f(z) = z + c for some constant c, and this is indeed a solution.