IMO 1979 SL 26

Origin: YUG

Problem

Prove that the functional equations f(x + y) = f(x) + f(y), and f(x + y + xy) = f(x) + f(y) + f(xy) (x, y \inR) are equivalent.

Solution

Let us assume that f(x + y) = f(x) + f(y) for all reals. In this case we trivially apply the equation to get f(x + y + xy) = f(x + y) + f(xy) = f(x)+f(y)+f(xy). Hence the equivalence is proved in the first direction. Now let us assume that f(x + y + xy) = f(x) + f(y) + f(xy) for all reals. Plugging in x = y = 0 we get f(0) = 0. Plugging in y = −1 we get f(x) = −f(−x). Plugging in y = 1 we get f(2x + 1) = 2f(x) + f(1) and hence f(2(u + v + uv) + 1) = 2f(u + v + uv) + f(1) = 2f(uv) + 2f(u) + 2f(v) + f(1) for all real u and v. On the other hand, plugging in x = u and y = 2v + 1 we get f(2(u + v + uv) + 1) = f(u + (2v +

1. • u(2v + 1)) = f(u) + 2f(v) + f(1) + f(2uv + u). Hence it follows that 2f(uv) + 2f(u) + 2f(v) + f(1) = f(u) + 2f(v) + f(1) + f(2uv + u), i.e., f(2uv + u) = 2f(uv) + f(u). (1) Plugging in v = −1/2 we get 0 = 2f(−u/2) + f(u) = −2f(u/2) + f(u). Hence, f(u) = 2f(u/2) and consequently f(2x) = 2f(x) for all reals. Now (1) reduces to f(2uv + u) = f(2uv) + f(u). Plugging in u = y and x = 2uv, we obtain f(x) + f(y) = f(x + y) for all nonzero reals x and y. Since f(0) = 0, it trivially holds that f(x + y) = f(x) + f(y) when one of x and y is 0.