Find all functions $ f: \mathbb{N} \to \mathbb{Z} $ satisfying $$ n \mid f\left(m\right) \Longleftrightarrow m \mid \sum\limits_{d \mid n}{f\left(d\right)} $$holds for all positive integers $ m,n $
Problem
Source: 2018 Taiwan TST Round 1
Tags: function, algebra, functional equation
01.12.2019 23:51
12.05.2020 13:27
here's my solution: claim(1):$$ n \mid f\left(m\right) \Longleftrightarrow m \mid f(n) $$proof: let $d$ a proper divisor of $n$ $m \mid \sum\limits_{d \mid n}{f\left(d\right)} \Longleftrightarrow n|f(m) \Longleftrightarrow d \mid f(m) \Longleftrightarrow m \mid \sum\limits_{d' \mid d}{f\left(d'\right)} $ thus $ n|f(m) \Longleftrightarrow m \mid \sum\limits_{d' \mid \frac{n}{d}}{f\left(d'\right)} $ and do this infinitely many times we will get $ n \mid f\left(m\right) \Longleftrightarrow m \mid f(n) $ $\blacksquare$ now claearly $1 \mid f(n) $ thus $n \mid f(1)$ for any $n$ thus $f(1)=0$ claim(2):$f(nk) \mid f(n)$ or $f(nk)=0$ proof: $m \mid f(nk) \Longleftrightarrow nk \mid f(m) \implies n \mid f(m) \Longleftrightarrow m \mid f(n)$ $\blacksquare$ now suppose that there exists a prime $p$ such that $f(p) \neq 0$ but from claim(2) there exists $\alpha$ such that $p^{\alpha}$ doesn't divides any $f(n)$ but from claim(1) we have $p^{\alpha} \mid f(f(p^{\alpha}))$ a contradiction and we win