HMMT 二月 2017 · ALGNT 赛 · 第 3 题

HMMT February 2017 — ALGNT Round — Problem 3

Let f : R → R be a function satisfying f ( x ) f ( y ) = f ( x − y ). Find all possible values of f (2017). 2017

解析

Let f : R → R be a function satisfying f ( x ) f ( y ) = f ( x − y ). Find all possible values of f (2017). Proposed by: Alexander Katz 2 Let P ( x, y ) be the given assertion. From P (0 , 0) we get f (0) = f (0) = ⇒ f (0) = 0 , 1. 2 From P ( x, x ) we get f ( x ) = f (0). Thus, if f (0) = 0, we have f ( x ) = 0 for all x , which satisfies the given constraints. Thus f (2017) = 0 is one possibility. Now suppose f (0) = 1. We then have P (0 , y ) = ⇒ f ( − y ) = f ( y ), so that P ( x, − y ) = ⇒ f ( x ) f ( y ) = ( ) x x f ( x − y ) = f ( x ) f ( − y ) = f ( x + y ). Thus f ( x − y ) = f ( x + y ), and in particular f (0) = f − = 2 2 ( ) x x f + = f ( x ). It follows that f ( x ) = 1 for all x , which also satisfies all given constraints. 2 2 Thus the two possibilities are f (2017) = 0 , 1 . 2017