HMMT 二月 2007 · 冲刺赛 · 第 24 题

英文解析

24. [ 12 ] Let x , y , n be positive integers with n > 1. How many ordered triples ( x, y, n ) of solutions aren 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 25 44
    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 apositive integer. Assume otherwise for contradiction’s sake, and suppose on the grounds of well orderingthat k is the least exponent for which a solution exists. Clearly x and y must both be even or bothn − 1 n − 1
    odd. If both are odd, we have ( x − y )( x + .... + y ) . The right factor of this expression containsan odd number of odd terms whose sum is an odd number greater than 1, impossible. Similarly if xn n k − nand y are even, write x = 2 u and y = 2 v . The equation becomes u − v = 2 . If k − n is greaterthan 0 , then our choice k could not have been minimal. Otherwise, k − n = 0, so that two consecutivepositive 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 tothe 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).
    10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth
    4 2 2 2 3 2 2