Question

有多少个整数有序对 $(m, n)$ 满足下列方程？ $m^2 + mn + n^2 = m^2 n^2?$

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2 + mn + n^2 = m^2 n^2?$

$7$

$1$

$3$

$6$

$5$

答案：C

知识点：丢番图方程极限情形界定分类讨论

难度评级：1660

解答：

若 $m = 0$ ，方程迫使 $n^2 = 0$ ，得到 $(0, 0)$ 。否则两者都非零。不妨假设 $|m| \le |n|$ 。则 $m^2 n^2 = m^2 + mn + n^2 \le 3n^2$ ，所以 $m^2 \le 3$ ，即 $m = \pm 1$ 。当 $m = 1$ 时， $1 + n + n^2 = n^2$ ，得 $n = -1$ 。当 $m = -1$ 时，得 $n = 1$ 。加上 $(0,0), (1,-1), (-1,1)$ ，共有三个有序对。所以正确答案是 C。

If $m = 0,$ the equation forces $n^2 = 0,$ giving $(0, 0).$ Otherwise both are nonzero; assume $|m| \le |n|.$ Then $m^2 n^2 = m^2 + mn + n^2 \le 3n^2,$ so $m^2 \le 3$ and $m = \pm 1.$ Take $m = 1:$ $1 + n + n^2 = n^2$ gives $n = -1.$ Take $m = -1:$ $n = 1.$ That leaves $(0,0), (1,-1), (-1,1),$ three in all. Therefore, the answer is C.

2023 AMC 10B 第 14 题

答案：C

难度评级：1660

解答：

其他年份的第 14 题