HMMT 二月 2007 · TEAM1 赛 · 第 11 题

英文解析

11. [ 30 ] Find all functions f : Q → Q such thatf ( x ) f ( y ) = f ( x ) + f ( y ) − f ( xy )
    1 + f ( x + y ) = f ( xy ) + f ( x ) f ( y )
    for all rational numbers x, y .
    Answer: f ( x ) = 1 ∀ x , and f ( x ) = 1 − x ∀ x .
    Solution. Considering the first equation, either side of the second equation is equal to f ( x ) + f ( y ).
    Now write g ( x ) = 1 − f ( x ), so thatg ( xy ) = 1 − f ( xy ) = 1 − f ( x ) − f ( y ) + f ( x ) f ( y ) = (1 − f ( x ))(1 − f ( y )) = g ( x ) g ( y )
    g ( x + y ) = 1 − f ( x + y ) = 1 − f ( x ) + 1 − f ( y ) = g ( x ) + g ( y ) ,
    By induction, g ( nx ) = ng ( x ) for all integers n , so that g ( p/q ) = ( p/q ) g (1) for integers p and q with qnonzero; i.e., g ( x ) = xg (1). As g is multiplicative, g (1) = g (1) , so the only possibilities are g (1) = 12
    and g (1) = 0. These give g ( x ) = x and g ( x ) = 0, or f ( x ) = 1 − x and f ( x ) = 1, respectively. Oneeasily checks that these functions are satisfactory.