HMMT 二月 2009 · 冲刺赛 · 第 3 题

HMMT February 2009 — Guts Round — Problem 3

[ 5 ] Find all pairs of integer solutions ( n, m ) to n m 3 2 2 = 3 − 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND 0 1 2009

解析

[ 5 ] Find all pairs of integer solutions ( n, m ) to n m 3 2 2 = 3 − 1 . Answer: (0 , 0) and (1 , 1) x y Solution: We find all solutions of 2 = 3 − 1 for positive integers x and y . If x = 1, we obtain the solution x = 1 , y = 1, which corresponds to ( n, m ) = (0 , 0) in the original problem. If x > 1, consider y the equation modulo 4. The left hand side is 0, and the right hand side is ( − 1) − 1, so y is even. Thus x z z z we can write y = 2 z for some positive integer z , and so 2 = (3 − 1)(3 + 1). Thus each of 3 − 1 and z 3 + 1 is a power of 2, but they differ by 2, so they must equal 2 and 4 respectively. Therefore, the only other solution is x = 3 and y = 2, which corresponds to ( n, m ) = (1 , 1) in the original problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND 0 1 2009