Problem 5 - Entrance Test

We are given the functional equation f(x+y) = f(x) + f(y) + xy and f(1) = 1. We need to find f(3). Let's find f(2) first: Set x = 1 and y = 1: f(1+1) = f(1) + f(1) + (1)(1) f(2) = f(1) + f(1) + 1 Since f(1) = 1: f(2) = 1 + 1 + 1 = 3. Now, let's find f(3): We can set x = 2 and y = 1: f(2+1) = f(2) + f(1) + (2)(1) f(3) = f(2) + f(1) + 2 Substitute the values we found for f(2) and f(1): f(3) = 3 + 1 + 2 f(3) = 6. Alternatively, we could set x=1, y=2: f(1+2) = f(1) + f(2) + (1)(2) f(3) = f(1) + f(2) + 2 f(3) = 1 + 3 + 2 = 6. (Consistent) Another approach could be to find the general form of f(x). If g(x) = f(x) - x^2/2, then g(x+y) = f(x+y) - (x+y)^2/2 = f(x)+f(y)+xy - (x^2+2xy+y^2)/2 = f(x)+f(y)+xy - x^2/2 - xy - y^2/2 = (f(x)-x^2/2) + (f(y)-y^2/2) = g(x)+g(y). This is Cauchy's functional equation. If g(x)=cx for rational x, then f(x) = cx + x^2/2. Given f(1) = 1: c(1) + 1^2/2 = 1 c + 1/2 = 1 c = 1/2. So, f(x) = x/2 + x^2/2. Then f(3) = 3/2 + 3^2/2 = 3/2 + 9/2 = 12/2 = 6.