AMC10 2007 A

AMC10 2007 A · Q23

AMC10 2007 A · Q23. It mainly tests Quadratic equations, Divisibility & factors.

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

有多少对正整数有序对$(m, n)$，满足$m > n$，且它们的平方差为$96$？

(A) 3 3

(B) 4 4

(D) 9 9

(E) 12 12

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $x$ and $y$ be, respectively, the larger and smaller of the integers. Then $96=x^2-y^2=(x+y)(x-y)$. Because $96$ is even, $x$ and $y$ are both even or are both odd. In either case $x+y$ and $x-y$ are both even. Hence there are four possibilities for $(x+y,x-y)$, which are $(48,2)$, $(24,4)$, $(16,6)$, and $(12,8)$. The four corresponding values of $(x,y)$ are $(25,23)$, $(14,10)$, $(11,5)$, and $(10,2)$.

答案（B）：设 $x$ 和 $y$ 分别为这两个整数中较大和较小的一个。则 $96=x^2-y^2=(x+y)(x-y)$。因为 $96$ 是偶数，$x$ 和 $y$ 要么同为偶数，要么同为奇数。在任一情况下，$x+y$ 和 $x-y$ 都为偶数。因此 $(x+y,x-y)$ 有四种可能，分别是 $(48,2)$、$(24,4)$、$(16,6)$、$(12,8)$。相应的四组 $(x,y)$ 分别为 $(25,23)$、$(14,10)$、$(11,5)$ 和 $(10,2)$。

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