14. Determine the number of ordered pairs of integers $(m, n)$ for which $m n \geq 0$ and $m^{3}+n^{3}+99 m n=33^{3}$.
(2 marks)
求滿足 $m n \geq 0$ 及 $m^{3}+n^{3}+99 m n=33^{3}$ 的整數序偶 $(m, n)$ 的數目。
(2 分)
14. 35
14. Using the identity
$$
a^{3}+b^{3}+c^{3}-3 a b c=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right],
$$
we have
$$
\begin{aligned}
m^{3}+n^{3}+99 m n-33^{3} & =m^{3}+n^{3}+(-33)^{3}-3 m n(-33) \\
& =\frac{1}{2}(m+n-33)\left[(m-n)^{2}+(m+33)^{2}+(n+33)^{2}\right]
\end{aligned}
$$

For this expression to be equal to 0 , we either have $m+n=33$ or $m=n=-33$. The latter gives one solution $(-33,-33)$ while the former gives the 34 solutions $(0,33),(1,32), \ldots,(33,0)$. Hence the answer is 35 .