HMMT 二月 2007 · 冲刺赛 · 第 24 题
HMMT February 2007 — Guts Round — Problem 24
题目详情
- [ 12 ] Let x , y , n be positive integers with n > 1. How many ordered triples ( x, y, n ) of solutions are n n 100 there to the equation x − y = 2 ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND 4 2 2 2 3 2 2
解析
- [ 12 ] Let x , y , n be positive integers with n > 1. How many ordered triples ( x, y, n ) of solutions are n n 100 there to the equation x − y = 2 ? Answer: 49 . Break all possible values of n into the four cases: n = 2, n = 4, n > 4 and n odd. By ( ) 4 4 25 4 Fermat’s theorem, no solutions exist for the n = 4 case because we may write y + 2 = x . n n k We show that for n odd, no solutions exist to the more general equation x − y = 2 where k is a positive integer. Assume otherwise for contradiction’s sake, and suppose on the grounds of well ordering that k is the least exponent for which a solution exists. Clearly x and y must both be even or both n − 1 n − 1 odd. If both are odd, we have ( x − y )( x + .... + y ) . The right factor of this expression contains an odd number of odd terms whose sum is an odd number greater than 1, impossible. Similarly if x n n k − n and y are even, write x = 2 u and y = 2 v . The equation becomes u − v = 2 . If k − n is greater than 0 , then our choice k could not have been minimal. Otherwise, k − n = 0, so that two consecutive positive integers are perfect n th powers, which is also absurd. For the case that n is even and greater than 4, consider the same generalization and hypotheses. m m m m k m m a k Writing n = 2 m, we find ( x − y )( x + y ) = 2 . Then x − y = 2 < 2 . By our previous work, we see that m cannot be an odd integer greater than 1. But then m must also be even, contrary to the minimality of k . 2 2 100 a b Finally, for n = 2 we get x − y = 2 . Factoring the left hand side gives x − y = 2 and x + y = 2 , b − 1 a − 1 b − 1 a − 1 where implicit is a < b. Solving, we get x = 2 + 2 and y = 2 − 2 , for a total of 49 solutions. Namely, those corresponding to ( a, b ) = (1 , 99) , (2 , 98) , · · · , (49 , 51). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND 4 2 2 2 3 2 2