+143 Let S be the set of all nonzero real numbers. Let f : S to S be a function such that f(x) + f(y) = f(xyf(x + y)) for all x, y in S such that x + y \(\neq\) 0. Let n be the number of possible values of f(4), and let s be the sum of all possible values of f(4). Find n * s.
Taking x = y = 0, you get 2f(0) = f(0), so f(0) = 0.
Taking y = 0, you get f(x) + f(0) = f(0) = 0, so f(x) = 0.
So the only function is f(x) = 0, and n*s = 1*0 = 0.
+143 The answer is to be in the form of a fraction so 0 is unreasonable.
