Hong Kong competition problems

a.
We have $f(x) = x^3 + 4x^2 - 4$.

Label the equation as follows.
$$f(x + y) = f(x) + f(y) + axy(x + y) + bxy + c(x + y) + 4 \quad (1)$$

Putting $x = y = 0$ in (1), we have
$$f(0) = f(0) + f(0) + 4,$$
which implies $f(0) = -4$. Next, putting $x = 1$ and $y = 0$ in (1), we have
$$f(1) = f(1) + f(0) + c + 4.$$
This yields $c = 0$. Now, putting $x = y = 1$ in (1), we have
$$f(2) = f(1) + f(1) + 2a + b + 4.$$
As it is given that $f(1) = 1$ and $f(2) = 20$, we obtain
$$2a + b = 14. \qquad (2)$$
Also, we put $x = y = 2$ in (1). This gives
$$f(4) = f(2) + f(2) + 16a + 4b + 4.$$
This means $f(4) = 16a + 4b + 44$. Moreover, putting $x = 4$ and $y = -4$ in (1), we find that
$$f(0) = f(4) + f(-4) - 16b + 4.$$
This becomes
$$4a - 3b = -12. \qquad (3)$$
Solving (2) and (3), we have
$$a = \frac{3(2a + b) + (4a - 3b)}{10} = 3$$
and $b = 14 - 2a = 8$. Equation (1) becomes
$$f(x + y) = f(x) + f(y) + 3xy(x + y) + 8xy + 4. \qquad (4)$$
Putting $y = 1$ in (4), we get
$$f(x + 1) = f(x) + 3x(x + 1) + 8x + 5.$$
For $x \in \mathbb{Z}^{+}$, adding
$$\begin{aligned} f(x) &= f(x - 1) + 3(x - 1)x + 8(x - 1) + 5, \\ f(x - 1) &= f(x - 2) + 3(x - 2)(x - 1) + 8(x - 2) + 5, \\ &\vdots, \\ f(2) &= f(1) + 3(1)(2) + 8(1) + 5, \end{aligned}$$
we obtain
$$\begin{aligned} f(x) &= f(1) + 3 \sum_{k=1}^{x-1} k(k+1) + 8 \sum_{k=1}^{x-1} k + 5(x-1) \\ &= (x-1)x(x+1) + 4(x-1)x + 5x - 4 \\ &= x^3 + 4x^2 - 4. \end{aligned}$$

This also holds when $x = 0$ as $f(0) = -4$. Now, putting $y = -x$ in (4), we have
$$f(0) = f(x) + f(-x) - 8x^2 + 4.$$
For $x < 0$, this yields
$$f(x) = -f(-x) + 8x^2 - 8 = -((-x)^3 + 4(-x)^2 - 4) + 8x^2 - 8 = x^3 + 4x^2 - 4.$$
So the same formula holds for all $x \in \mathbb{Z}$.
Lastly, for $f(x) = x^3 + 4x^2 - 4$, we check that
$$\begin{aligned} f(x+y) &= (x+y)^3 + 4(x+y)^2 - 4 \\ &= x^3 + 3x^2y + 3xy^2 + y^3 + 4x^2 + 8xy + 4y^2 - 4 \end{aligned}$$
and
$$\begin{aligned} f(x) + f(y) + 3xy(x+y) + 8xy + 4 \\ &= x^3 + 4x^2 - 4 + y^3 + 4y^2 - 4 + 3xy(x+y) + 8xy + 4. \end{aligned}$$
It is routine to check that the two sides are identical. Thus, this is the only solution.

b.
The greatest possible value of $m$ is $-1$.

Note that
$$\begin{align*} & x^3 + 4x^2 - 4 \ge mx^2 + (5m + 1)x + 4m \\ \Leftrightarrow & x^3 + 4x^2 - x - 4 \ge mx^2 + 5mx + 4m \\ \Leftrightarrow & (x+1)(x+4)(x-1) \ge m(x+1)(x+4) \\ \Leftrightarrow & x - 1 \ge m. \end{align*}$$
By putting $x = 0$, we need $m \le -1$. When $m = -1$, this inequality holds for all $x \ge 0$. Thus, the greatest possible value of $m$ is $-1$.