### 35th IMO 1994 shortlist

Problem A4

h and k are reals. Find all real-valued functions f defined on the positive reals such that f(x) f(y) = yh f(x/2) + xk f(y/2) for all x, y.

Solution

Answer: f(x) = 0 is always a solution. If h = k, then there is an additional solution f(x) = 2(x/2)h.

Consider first the case h = k. Putting x = y, f(x)2 = 2xhf(x/2), so f(x/2) = x-hf(x)2/2. Hence f(x) f(y) = (y/x)h f(x)2/2 + (x/y)h f(y)2 or ( (y/x)h/2f(x) - (x/y)h/2f(y) )2 = 0. So f(x)/xh = f(y)/yh. This is true for all x, y, so we have f(x) = A xh for some constant A. Substituting back into the original equation gives A2 = A/2h + A/2h, so A = 0 or A = 21-h. Checking, it is clear that A = 0, corresponding to f(x) = 0, is a solution. If A = 21-h, then f(x) = 2 (x/2)h. The lhs of the original equation then becomes 4 (x/2)h(y/2)h, and the rhs becomes yh 2 (x/4)h + xh 2 (y/4)h = 4 (x/2)h(y/2)h = lhs. So it is indeed a solution.

Now consider h and k unequal. We have f(x) f(y) = yh f(x/2) + xk f(y/2) and f(y) f(x) = xh f(y/2) + yk f(x/2). Subtracting, 0 = (yh - yk) f(x/2) - (xh - xk) f(y/2). So f(x/2) = A (xh - xk) for some constant A for all x except possibly x = 1.

Substituting back in the original equation, we get A2( (2x)h - (2x)k)( (2y)h - (2y)k) = A(xh - xk) yh + A(yh - yk) xk, except possibly for x or y = 1/2 or 1. Comparing coefficients, 4hA2 = A, 4kA2 = -A. So A = 0, and hence f(x) = 0 for all x except possibly 1/2 and 1. So, in particular f(2) = 0. Putting x = y = 2 in the original equation gives: 0 = (2h + 2k) f(1), so f(1) = 0 also. Similarly, putting x = y = 1 in the original equation gives f(1/2) = 0. So f(x) = 0 for all x.