HMMT 二月 2007 · TEAM1 赛 · 第 11 题

HMMT February 2007 — TEAM1 Round — Problem 11

[ 30 ] Find all functions f : Q → Q such that f ( 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 .

解析

[ 30 ] Find all functions f : Q → Q such that f ( 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 that g ( 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 q 2 nonzero; i.e., g ( x ) = xg (1). As g is multiplicative, g (1) = g (1) , so the only possibilities are g (1) = 1 and g (1) = 0. These give g ( x ) = x and g ( x ) = 0, or f ( x ) = 1 − x and f ( x ) = 1, respectively. One easily checks that these functions are satisfactory.