put $ x = y = 0$,then we have $ f(0) = - 4$ put $ x = 0$ ,then we have $ c = 0$, put $ x = 1, y = - 1$, we have $ f(0) = f( - 1) + f(1) - b + 4 ,f( - 1) = b - 9$ put $ x = y = - 1$, $ f( - 2) = 2f( - 1) - 2a + b + 4 = - 2a + 3b - 14$ put $ x = y = - 2$, $ f( - 4) = 2f( - 2) - 16a + 4b + 4, - 4 = - 20a + 10b - 24, - 2 = 2a - b .....(1)$ put $ x = y = 1$, we have, $ f(2) = 2f(1) + 2a + b + 4$ $ 14 = 2a + b .......(2)$ use $ (1), (2)$, we have $ a = 3, b = 8$ so $ f(x + y) = f(x) + f(y) + 3xy(x + y) + 8xy + 4$ put y=1, we have $ f(x + 1) - f(x) = 3x^2 + 11x + 5$ $ \sum_{x = 1}^n[f(x + 1) - f(x)] = \sum_{x = 1}^n[3x^2 + 11x + 5]$ hence, $ f(x) = x^3 + 4x^2 - 4$ (b)$ f(x) > = mx^2 + (5m + 1)x + 4m$ $ x^3 + 4x^2 - 4 > = mx^2 + 5mx + 4m + x$ $ (x + 4)(x + 1)(x - 1) > = m(x + 1)(x + 4)$ $ x > = m + 1$ for all non-negative integer x hence, the greatest possible value of m is $ - 1$