15th Australian 1994 Problem A4 Q is the rationals and R is the reals. Find all functions f : Q → R such that f(x + y) = f(x) + f(y) + 2xy for all x, y.
x 2 + kx for any real k Solution Put g(x) = f(x) - x 2 . Then g(x+y) = f(x+y) - (x+y) 2 = f(x) + f(y) - x 2 - y 2 = g(x) + g(y). g(0) = g(0) + g(0), so g(0) = 0. Let g(1) = k. Suppose g(1/m) = h, where m is any positive integer. Then by a trivial induction g(n/m) = nh. In particular g(1) = mh = k, so g(n/m) = kn/m. In other words, g(x) = kx for any positive rational x. But g(0) = g(x) + g(-x), so g(-x) = -g(x) = -kx. So g(x) = kx for all x. Hence f(x) = x 2 + kx. It is easy to check that this satisfies the relation given for any k. Thanks to Suat Namli 15th Aus 1994 © John Scholes jscholes@kalva.demon.co.uk 5 February 2003 Last updated/corrected 5 Feb 04