Let $\mathbb{N}$ denote the set of all positive integers.Function $f:\mathbb{N}\cup{0}\rightarrow\mathbb{N}\cup{0}$ satisfies :for any two distinct positive integer $a,b$, we have $$f(a)+f(b)-f(a+b)=2019$$(1)Find $f(0)$ (2)Let $a_1,a_2,...,a_{100}$ be 100 positive integers (they are pairwise distinct), find $f(a_1)+f(a_2)+...+f(a_{100})-f(a_1+a_2+...+a_{100})$
Problem
Source: 2019 Taiwan APMO Preliminary
Tags: algebra, functional equation
hsiangshen
23.08.2020 15:01
(1)$2019$(2)$199881$
In this problem I'm not quite sure how to translate "兩兩相異" Here I translate it as "they are both arbitrary". If you have a better translation, please tell me. The attachment is the original version of the problem.
Attachments:
Keith50
27.08.2020 10:37
@above "兩兩相異" actually means pairwise distinct
Keith50
27.08.2020 10:54
2) This is just shifting, let $g(n)=f(n)-2019$, then $g$ is additive, so \[f(a_1)+f(a_2)+\cdots +f(a_{100})-f(a_1+a_2+\cdots +a_{100})=g(a_1)+g(a_2)+\cdots +g(a_{100})-g(a_1+a_2+\cdots +a_{100})+99\times 2019= 199881.\]
hsiangshen
27.08.2020 14:39
Keith50 wrote: @above "兩兩相異" actually means pairwise distinct Thank you! Edited...