IMO 2004 SL A6

Origin: RUS | Category: Algebra

Problem

Find all functions f : R \toR satisfying the equation f
x2 + y2 + 2f(xy)  = (f(x + y))2 for all x, y \inR.

Solution

Let us make the substitution z = x + y, t = xy. Given z, t \inR, x, y are real if and only if 4t \leqz2. Define g(x) = 2(f(x) −x). Now the given functional equation transforms into f
z2 + g(t)  = (f(z))2 for all t, z \inR with z2 \geq4t. (1) Let us set c = g(0) = 2f(0). Substituting t = 0 into (1) gives us f(z2 + c) = (f(z))2 for all z \inR. (2) If c < 0, then taking z such that z2 + c = 0, we obtain from (2) that f(z)2 = c/2, which is impossible; hence c \geq0. We also observe that

x > c implies f(x) \geq0. (3) If g is a constant function, we easily find that c = 0 and therefore f(x) = x, which is indeed a solution. Suppose g is nonconstant, and let a, b \inR be such that g(a)−g(b) = d > 0. For some sufficiently large K and each u, v \geqK with v2 −u2 = d the equality u2 + g(a) = v2 + g(b) by (1) and (3) implies f(u) = f(v). This further leads to g(u) −g(v) = 2(v −u) = d u+ \sqrt u2+d. Therefore every value from some suitably chosen segment [\delta, 2\delta] can be expressed as g(u)−g(v), with u and v bounded from above by some M. Consider any x, y with y > x \geq2 \sqrt M and \delta < y2 −x2 < 2\delta. By the above considerations, there exist u, v \leqM such that g(u) −g(v) = y2 −x2, i.e., x2 + g(u) = y2 + g(v). Since x2 \geq4u and y2 \geq4v, (1) leads to f(x)2 = f(y)2. Moreover, if we assume w.l.o.g. that 4M \geqc2, we conclude from (3) that f(x) = f(y). Since this holds for any x, y \geq2 \sqrt M with y2 −x2 \in[\delta, 2\delta], it follows that f(x) is eventually constant, say f(x) = k for x \geqN = 2 \sqrt M. Setting x > N in (2) we obtain k2 = k, so k = 0 or k = 1. By (2) we have f(−z) = \pmf(z), and thus |f(z)| \leq1 for all z \leq−N. Hence g(u) = 2f(u) −2u \geq−2 −2u for u \leq−N, which implies that g is unbounded. Hence for each z there exists t such that z2 + g(t) > N, and consequently f(z)2 = f(z2 + g(t)) = k = k2. Therefore f(z) = \pmk for each z. If k = 0, then f(x) \equiv0, which is clearly a solution. Assume k = 1. Then c = 2f(0) = 2 (because c \geq0), which together with (3) implies f(x) = 1 for all x \geq2. Suppose that f(t) = −1 for some t < 2. Then t −g(t) = 3t + 2 > 4t. If also t −g(t) \geq0, then for some z \inR we have z2 = t−g(t) > 4t, which by (1) leads to f(z)2 = f(z2+g(t)) = f(t) = −1, which is impossible. Hence t −g(t) < 0, giving us t < −2/3. On the other hand, if X is any subset of (−\infty, −2/3), the function f defined by f(x) = −1 for x \inX and f(x) = 1 satisfies the requirements of the problem. To sum up, the solutions are f(x) = x, f(x) = 0 and all functions of the form f(x) = .1, x ̸\inX, −1, x \inX, where X \subset(−\infty, −2/3).