1961 AMC 12 第 33 题

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33.

方程 22x32y=552^{2x}-3^{2y}=55xxyy 均为整数,其解的个数为:

The number of solutions of 22x32y=55,2^{2x}-3^{2y}=55, in which xx and yy are integers, is:

00

11

22

33

多于三个,但为有限个

more than three, but finite

答案:B
知识点:平方差丢番图方程因数
难度评级:1550
小提示:

将左边按平方差因式分解

Factor the left side as a difference of squares

大提示:

两个正因数的乘积为 5555,且奇偶性相同

The two positive factors multiply to 5555 and have the same parity

解答:

因式分解得 (2x3y)(2x+3y)=55 (2^x-3^y)(2^x+3^y)=55\text{。}首先,x>0x>0。若 y0y\le0,则 55<4x5655<4^x\le56,而 44 的整数次幂不可能落在这个范围内。因此 y>0y>0,两个因数都是正奇整数。因数对 551111 给出 2x+1=162^{x+1}=1623y=62\cdot3^y=6,所以 (x,y)=(3,1)(x,y)=(3,1)。因数对 115555 则要求 2x=282^x=28,不可能。因此恰有一个解。

所以,正确答案是 B

Factor: (2x3y)(2x+3y)=55. (2^x-3^y)(2^x+3^y)=55. First, x>0.x>0. If y0,y\le0, then 55<4x56,55<4^x\le56, which is impossible for an integral power of 4.4. Hence y>0,y>0, so both factors are positive odd integers. The factor pair 5,5, 1111 gives 2x+1=162^{x+1}=16 and 23y=6,2\cdot3^y=6, so (x,y)=(3,1).(x,y)=(3,1). The factor pair 1,1, 5555 would require 2x=28,2^x=28, impossible. Hence there is exactly one solution.

Thus, the correct answer is B.

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