8th Irish 1995 Problem A5 Find all real-valued functions f(x) such that xf(x) - yf(y) = (x-y) f(x+y) for all real x, y.
f(x) = ax + b Solution We show first that f(x) = ax + b for x rational. Let f(0) = b, f(1) = a+b. Putting y = -x, we get f(x) + f(-x) = 2b. Replacing x by 2x, and y by -x, we get 2f(2x) + f(-x) = 3f(x). Hence f(2x) = 2f(x) - b. We conjecture that f(nx) = nf(x) - (n-1)b. Suppose true for n. Then nf(nx) - f(x) = (n-1)f((n+1)x), so f((n+1)x) = (n+1)f(x) - nb. So the result is true for all positive integers n. But f(-nx) = 2b - f(nx) = -nf(x) - (-nb - 1). So the result is true for all integers n. Taking x = 1/n, we get a+b = nf(1/n) - (n-1)b, so f(1/n) = a(1/n) + b. Hence f(m/n) = m(a(1/n) + b) - (m- 1)b = a(m/n) + b. Put g(x) = f(x) - b. Then xg(x) - yg(y) = xf(x) - yf(y) - (x-y)b, and (x-y)g(x+y) = (x-y)f(x+y) - (x-y)b. So xg(x) - yg(y) = (x-y)g(x+y) (*). Also g(-x) = -g(x). So putting -y for y in (*) we get xg(x) - yg(y) = (x+y)g(x-y). Hence g(x+y)/(x+y) = g(x-y)/(x-y) for all x, y. Now given any k ≠ 1, put x = (k+1)/2, y = (k-1)/2. Then x+y = k, x-y = 1, so g(k)/k = g(1) = a. Hence g(k) = ak for all k. 8th Irish 1995 © John Scholes jscholes@kalva.demon.co.uk 17 Dec 2003 Last updated/corrected 17 Dec 03