Determine all functions f: R -> R such that f(x - f(y) ) = f( f(y) ) + x f(y) + f(x) - 1 for all x, y in R. [R is the reals.]

**Solution**

*Solution communicated by Ong Shien Jin*

Let c = f(0) and A be the image f(R). If a is in A, then it is straightforward to find f(a): putting a = f(y) and x = a, we get f(a - a) = f(a) + a^{2} + f(a) - 1, so f(a) = (1 + c)/2 - a^{2}/2 (*).

The next step is to show that A - A = R. Note first that c cannot be zero, for if it were, then putting y = 0, we get: f(x - c) = f(c) + xc + f(x) - 1 (**) and hence f(0) = f(c) = 1. Contradiction. But (**) also shows that f(x - c) - f(x) = xc + (f(c) - 1). Here x is free to vary over R, so xc + (f(c) - 1) can take any value in R.

Thus given any x in R, we may find a, b in A such that x = a - b. Hence f(x) = f(a - b) = f(b) + ab + f(a) - 1. So, using (*): f(x) = c - b^{2}/2 + ab - a^{2}/2 = c - x^{2}/2.

In particular, this is true for x in A. Comparing with (*) we deduce that c = 1. So for all x in R we must have f(x) = 1 - x^{2}/2. Finally, it is easy to check that this satisfies the original relation and hence is the unique solution.

(C) John Scholes

jscholes@kalva.demon.co.uk

7 Sep 1999

Last corrected/updated 19 Aug 03