Question

满足 $mn \ge 0$ 的整数有序对 $(m, n)$ ，且

$m^3 + n^3 + 99mn = 33^3$

的个数等于

The number of ordered pairs of integers $(m, n)$ for which $mn \ge 0$ and

$m^3 + n^3 + 99mn = 33^3$

is equal to

$2$

$3$

$33$

$35$

$99$

答案：D

知识点：因式分解丢番图方程

难度评级：2460

解答：

令 $z = -33$ ，方程变为 $m^3 + n^3 + z^3 - 3mnz = 0$ ，因式分解得 $\begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0. \end{aligned}$

第二个因子等于 $\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big]$ ，只在 $(m, n) = (-33, -33)$ 时为 $0$ ，且满足 $mn \ge 0$ 。

否则 $m + n = 33$ 。由 $mn \ge 0$ ，二者同为非负，得到 $(0, 33), (1, 32), \ldots, (33, 0)$ ，共 $34$ 对。总计 $35$ 对。

所以正确答案是 D。

Writing $z = -33,$ the equation becomes $m^3 + n^3 + z^3 - 3mnz = 0,$ which factors as $\begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0. \end{aligned}$

The second factor equals $\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big],$ which is $0$ only at $(m, n) = (-33, -33);$ this satisfies $mn \ge 0.$

Otherwise $m + n = 33.$ With $mn \ge 0$ both are nonnegative, giving $(0, 33), (1, 32), \ldots, (33, 0),$ which is $34$ pairs. Together there are $35$ solutions.

Thus, the correct answer is D.

1999 AMC 12 第 30 题

答案：D

难度评级：2460

解答：