Finding all completely multiplicative arithmetic function such that $m+n\mid f(m)+f(n)$

Question

My attempt:

Since $f$ is completely multiplicative we have $f(1)=1$. $2n\mid 2f(n)$ for every n, so $f(n)=kn$ for some n. $n+1\mid f(n)+1$ for every n,so for p prime, $p+1\mid kp+1\rightarrow p+1\mid k-1$, so $k=l(p+1)+1$ for some $l$. And I'm stuck.

I also tried plugging in some basic values: $4\mid 1+f(3),5\mid f(2)+f(3), 5\mid f(2)^2+1$ etc, but I can't seem to find a pattern.

What should I do? Any hints?

Thanks!

Using an Approach0 search, there's proof check: $f$ maps positive integers to positive integers. $f(m)f(n)=f(mn)$ and $m+n$ divides $f(m)+f(n)$. The answer provides the correct solution of $f(n)=n^k$ for odd integers $k$, but it doesn't prove these are the only solutions, so I'm not voting to close as a duplicate. Note there are quite a few AoPS threads with this problem, with at least a few having apparently correct solutions. In particular, ... — John Omielan, Jul 15 '24 at 21:05
(cont.) there's Functional Number Theory (Turkey IMO TST 2016 P5), Function N to N, Function, A curious functional equation, Standard Number Theoretical Functional Equation, INMO 2018 -- Problem #6, ... — John Omielan, Jul 15 '24 at 21:08
(cont.) etc. As the titles indicate, this problem was used in at least $2$ high school Olympiads, i.e., Turkey IMO TST $2016$ P$5$ and INMO $2018$ P$6$. FYI, for the more general case which requires only that $m+n\mid f(m)+f(n)$, i.e., it doesn't need to have the $f(n)$ be multiplicative, much less completely multiplicative, there's All functions $g:\mathbb{N}\to\mathbb{Z}$ such that $m+n~|~g(m)+g(n)$. — John Omielan, Jul 15 '24 at 21:11

Finding all completely multiplicative arithmetic function such that $m+n\mid f(m)+f(n)$

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