IMO 2002 SL A4

Origin: IND | Category: Algebra

Problem

Find all functions f from the reals to the reals such that (f(x) + f(z))(f(y) + f(t)) = f(xy −zt) + f(xt + yz) for all real x, y, z, t.

Solution

Putting x = z = 0 and t = y into the given equation gives 4f(0)f(y) = 2f(0) for all y. If f(0) ̸= 0, then we deduce f(y) = 1 2, i.e., f is identically equal to 1 2. Now we suppose that f(0) = 0. Setting z = t = 0 we obtain f(xy) = f(x)f(y) for all x, y \inR. (1) Thus if f(y) = 0 for some y ̸= 0, then f is identically zero. So, assume f(y) ̸= 0 whenever y ̸= 0.

Next, we observe that f is strictly increasing on the set of positive reals. Actually, it follows from (1) that f(x) = f(\sqrtx)2 \geq0 for all x \geq0, so that the given equation for t = x and z = y yields f(x2+y2) = (f(x)+f(y))2 \geq f(x2) for all x, y \geq0. Using (1) it is easy to get f(1) = 1. Now plugging t = y into the given equation, we are led to 2[f(x) + f(z)] = f(x −z) + f(x + z) for all x, z. (2) In particular, f(z) = f(−z). Further, it is easy to get by induction from (2) that f(nx) = n2f(x) for all integers n (and consequently for all rational numbers as well). Therefore f(q) = f(−q) = q2 for all q \inQ. But f is increasing for x > 0, so we must have f(x) = x2 for all x. It is easy to verify that f(x) = 0, f(x) = 1 2 and f(x) = x2 are indeed solutions.