Question

有多少个正整数有序对 $(m, n)$ 满足 $m \gt n$ ，且它们的平方差为 $96$ ？

How many ordered pairs $(m, n)$ of positive integers, with $m \gt n,$ have the property that their squares differ by $96?$

$3$

$4$

$6$

$9$

$12$

将平方差写成 $m^2 - n^2 = (m + n)(m - n) = 96$ 。

Write $m^2 - n^2 = (m + n)(m - n) = 96$

$m + n$ 和 $m - n$ 奇偶性相同，所以二者都必须是偶数。

$m + n$ and $m - n$ have the same parity, so both must be even

因数 $m + n$ 和 $m - n$ 的奇偶性相同。因为它们的乘积是 $96$ ，所以不可能都是奇数，只能都是偶数。

偶数因数对为 $(48, 2)$ 、 $(24, 4)$ 、 $(16, 6)$ 和 $(12, 8)$ ，对应 $(m, n) = (25, 23)$ 、 $(14, 10)$ 、 $(11, 5)$ 和 $(10, 2)$ 。

所以共有 $4$ 个有序对。

所以正确答案是 B。

The factors $m + n$ and $m - n$ have the same parity. Since their product is $96,$ they cannot both be odd, so both must be even.

The even factor pairs are $(48, 2),$ $(24, 4),$ $(16, 6),$ and $(12, 8),$ giving $(m, n) = (25, 23),$ $(14, 10),$ $(11, 5),$ and $(10, 2).$

So there are $4$ ordered pairs.

Thus, the correct answer is B.