### IMO 2002

Problem B2

Find all real-valued functions f on the reals such that (f(x) + f(y)) (f(u) + f(v)) = f(xu - yv) + f(xv + yu) for all x, y, u, v.

Solution

Answer: there are three possible functions: (1) f(x) = 0 for all x; (2) f(x) = 1/2 for all x; or (3) f(x) = x2.

Put x = y = 0, u = v, then 4 f(0) f(u) = 2 f(0). So either f(u) = 1/2 for all u, or f(0) = 0. f(u) = 1/2 for all u is certainly a solution. So assume f(0) = 0.

Putting y = v = 0, f(x) f(u) = f(xu) (*). In particular, taking x = u = 1, f(1)2 = f(1). So f(1) = 0 or 1. Suppose f(1) = 0. Putting x = y = 1, v = 0, we get 0 = 2f(u), so f(x) = 0 or all x. That is certainly a solution. So assume f(1) = 1.

Putting x = 0, u = v = 1 we get 2 f(y) = f(y) + f(-y), so f(-y) = f(y). So we need only consider f(x) for x positive. We show next that f(r) = r2 for r rational. The first step is to show that f(n) = n2 for n an integer. We use induction on n. It is true for n = 0 and 1. Suppose it is true for n-1 and n. Then putting x = n, y = u = v = 1, we get 2f(n) + 2 = f(n-1) + f(n+1), so f(n+1) = 2n2 + 2 - (n-1)2 = (n+1)2 and it is true for n+1. Now (*) implies that f(n) f(m/n) = f(m), so f(m/n) = m2/n2 for integers m, n. So we have established f(r) = r2 for all rational r.

From (*) above, we have f(x2) = f(x)2 ≥ 0, so f(x) is always non-negative for positive x and hence for all x. Putting u = y, v = x, we get ( f(x) + f(y) )2 = f(x2 + y2), so f(x2 + y2) = f(x)2 + 2f(x)f(y) + f(y)2 ≥ f(x)2 = f(x2). For any u > v > 0, we may put u = x2 + y2, v = x2 and hence f(u) ≥ f(v). In other words, f is an increasing function.

So for any x we may take a sequence of rationals rn all less than x we converge to x and another sequence of rationals sn all greater than x which converge to x. Then rn2 = f(rn) ≤ f(x) ≤ f(sn) = sn2 for all x and hence f(x) = x2.