HMMT 二月 2019 · 冲刺赛 · 第 16 题

HMMT February 2019 — Guts Round — Problem 16

[ 7 ] Let R be the set of real numbers. Let f : R → R be a function such that for all real numbers x and y , we have 2 2 2 f ( x ) + f ( y ) = f ( x + y ) − 2 xy. 2019 ∑ Let S = f ( n ). Determine the number of possible values of S . n = − 2019 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2019, February 16, 2019 — GUTS ROUND Organization Team Team ID#

解析

[ 7 ] Let R be the set of real numbers. Let f : R → R be a function such that for all real numbers x and y , we have 2 2 2 f ( x ) + f ( y ) = f ( x + y ) − 2 xy. 2019 ∑ Let S = f ( n ). Determine the number of possible values of S . n = − 2019 Proposed by: Yuan Yao ( ) 2020 Answer: 2039191 OR + 1 2 Letting y = − x gives 2 2 2 2 f ( x ) + f ( x ) = f (0) + 2 x for all x . When x = 0 the equation above gives f (0) = 0 or f (0) = 2. 2 2 If f (0) = 2, then f ( x ) = x + 2 for all nonegative x , so the LHS becomes x + y + 4, and RHS becomes 2 2 x + y + 4 x + 4 y + 4 for all x + y ≥ 0, which cannot be equal to LHS if x + y > 0. If f (0) = 0 then f ( x ) = x for all nonnegative x . Moreover, letting y = 0 gives 2 2 f ( x ) = f ( x ) ⇒ f ( x ) = ± x for all x . Since negative values are never used as inputs on the LHS and the output on the RHS is always squared, we may conclude that for all negative x , f ( x ) = x and f ( x ) = − x are both possible (and the values are independent). Therefore, the value of S can be written as 2019 ∑ S = f (0) + ( f (1) + f ( − 1)) + ( f (2) + f ( − 2)) + · · · + ( f (2019) + f ( − 2019)) = 2 iδ i i =1 S for δ , δ , . . . , δ ∈ { 0 , 1 } . It is not difficult to see that can take any integer value between 0 and 1 2 2019 2 2020 · 2019 = 2039190 inclusive, so there are 2039191 possible values of S . 2