HMMT 二月 2024 · 团队赛 · 第 6 题
HMMT February 2024 — Team Round — Problem 6
题目详情
- [45] Let Q be the set of rational numbers. Given a rational number a ̸ = 0 , find, with proof, all functions f : Q → Q satisfying the equation f ( f ( x ) + ay ) = af ( y ) + x for all x, y ∈ Q .
解析
- [45] Let Q be the set of rational numbers. Given a rational number a ̸ = 0 , find, with proof, all functions f : Q → Q satisfying the equation f ( f ( x ) + ay ) = af ( y ) + x for all x, y ∈ Q . Proposed by: Yichen Xiao f ( x ) = x Answer: f ( x ) = − x f ( x ) = x + c for all rational numbers c iff a = 2 Solution: Let P ( x, y ) denote the functional equation. From P ( x, 0) , we have f ( f ( x )) = x + af (0) . Thus, the tripling trick gives f ( x + af (0)) = f ( f ( f ( x ))) = f ( x ) + af (0) . Now, here is the main idea: P ( f ( x ) , y ) gives ( ) f f ( f ( x )) + ay = af ( y ) + f ( x ) f ( x + af (0) + ay ) = f ( x ) + af ( y ) f ( x + ay ) = f ( x ) + af ( y ) − af (0) . In particular, plugging in x = 0 into this equation gives f ( ay ) = af ( y ) + (1 − a ) f (0) , so inserting it back to the same equation gives f ( x + ay ) = f ( x ) + f ( ay ) − f (0) , for all rational numbers x, y . In particular, the function g ( x ) = f ( x ) − f (0) is additive, so f is linear. Let f ( x ) = bx + c . By substituting it in, we have P ( x, y ) iff f ( ay + bx + c ) = a ( by + c ) + x 2 aby + b x + bc + c = aby + ac + x 2 ( b − 1) x + ( b + 1 − a ) c = 0 . 2 Since x is arbitrary, we can state that b − 1 = 0 and ( b + 1 − a ) c = 0 , thus b = ± 1 . As a ̸ = 0 , we know b + 1 − a = 0 only if b = 1 and a = 2 . When a ̸ = 2 or b ̸ = 1 , we know the only solutions are b = ± 1 , c = 0 , while for a = 2 , b = 1 , the equation is automatically satisfied, so the final answer is f ( x ) = x f ( x ) = − x f ( x ) = x + c for all rational number c iff a = 2 Solution 2: We will only prove that f is linear. Then, proceed as in the end of Solution 1. We know f ( f ( x )) = af (0)+ x , so as af (0)+ x can take any rational number when x takes every rational number, the range of f is Q , and so f is surjective. If f ( x ) = f ( x ) , we have x = f ( f ( x )) − af (0) = 1 2 1 1 f ( f ( x )) − af (0) = x , so x = x , implying f being injective. Thus, f is bijective. 2 2 1 2 From P ( x, 0) , we still get f ( f ( x )) = x + af (0) . − 1 − 1 Thus, from P ( f (0)) , y/a ) , we can get f ( y ) = af ( y/a ) + f (0) . Plugging again P ( f ( x ) , y/a ) , we − 1 have f ( f ( f ( x )) + y ) = f ( x + y + af (0)) = af ( y/a ) + f ( x ) = f ( x ) + f ( y ) − f ( x ) . − 1 Thus, we know f ( x + y ) = f ( x − af (0)) + f ( y ) − f (0) = f ( x ) + f ( y ) − f (0) . Hence, the function g ( x ) = f ( x ) − f (0) is additive, so g ( x ) = kx for some rational number k . Thus, f is a linear function, and we can proceed as in above solution.