IMO 2000 SL A3

Origin: BLR | Category: Algebra

Problem

Find all pairs of functions f : R \toR, g : R \toR such that f(x + g(y)) = xf(y) −yf(x) + g(x) for all x, y \inR.

Solution

Let us first solve the problem under the assumption that g(\alpha) = 0 for some \alpha. Setting y = \alpha in the given equation yields g(x) = (\alpha+1)f(x)−xf(\alpha). Then the given equation becomes f(x+g(y)) = (\alpha+1−y)f(x)+(f(y)−f(\alpha))x, so setting y = \alpha + 1 we get f(x + n) = mx, where n = g(\alpha + 1) and m = f(\alpha + 1) −f(\alpha). Hence f is a linear function, and consequently g is also linear. If we now substitute f(x) = ax + b and g(x) = cx + d in the given equation and compare the coefficients, we easily find that

f(x) = cx −c2 1 + c and g(x) = cx −c2, c \inR \ {−1}. Now we prove the existence of \alpha such that g(\alpha) = 0. If f(0) = 0 then putting y = 0 in the given equation we obtain f(x + g(0)) = g(x), so we can take \alpha = −g(0). Now assume that f(0) = b ̸= 0. By replacing x by g(x) in the given equation we obtain f(g(x) + g(y)) = g(x)f(y) −yf(g(x)) + g(g(x)) and, analogously, f(g(x) + g(y)) = g(y)f(x) −xf(g(y)) + g(g(y)). The given functional equation for x = 0 gives f(g(y)) = a −by, where a = g(0). In particular, g is injective and f is surjective, so there exists c \inR such that f(c) = 0. Now the above two relations yield g(x)f(y) −ay + g(g(x)) = g(y)f(x) −ax + g(g(y)). (1) Plugging y = c in (1) we get g(g(x)) = g(c)f(x) −ax + g(g(c)) + ac = kf(x) −ax + d. Now (1) becomes g(x)f(y) + kf(x) = g(y)f(x) + kf(y). For y = 0 we have g(x)b + kf(x) = af(x) + kb, whence g(x) = a −k b f(x) + k. Note that g(0) = a ̸= k = g(c), since g is injective. From the surjectivity of f it follows that g is surjective as well, so it takes the value 0.