I am interested in the following problem:

Let $f:\mathbb N\to \mathbb N$ be a function and let $k$ be some fixed constant natural number more than $1$. If$$f(kx)=kf(x)$$holds for all $x\in\mathbb{N}$, what can we say about the function $f(x)$? Do there exist any non trivial solution (i.e., some solution other than identical mapping) $?$

For $k=3,$ I found kinda similar problem about existence of such a function here.

This problem also motivates us to think about existence of function $f:\mathbb N\to\mathbb N$ for which $$f\circ f\circ\underbrace{\cdots}_{k~\text{compositions}}\circ f(n) = mn$$ for some fixed pair $(m,k)\in\mathbb N^2$. Does such a function exists for all values of $m$ and $k$?

The above problem is kinda similar problem as the first one as we can take one more compositions both sides in the above equation to get $mf(n)=f(mn)$. But here identical mapping won't be a solution of course! As of now, I have no idea how we can proceed for construction for such functions or do they even exists.

Any help regarding this will be highly appreciated!


For the first functional equation, there exists infinitely many pathological solutions. Simply set $f(x)$ to be whatever you want for $k \nmid x$. Then, you follow the requirement and set $f(kx)=kf(x)$ for $k \nmid x$. Then, you set $f(k^2x)=kf(kx)=k^2f(x)$ for $k \nmid x$ and so on. Clearly, any positive integer can be written as $k^nx$ where $x$ is a positive integer and $k$ is a non-negative integer. You can clearly see that this construction obeys the given rule.

For the second functional equation, simply take the set of all numbers not divisible by $m$ and divide it into $k$-tuples. Let $(x_1,x_2,\ldots, x_k)$ be such a $k$-tuple. You simply define $f(m^tx_i)=m^tx_{i+1}$ when $i<k$ and define $f(m^tx_k)=m^{t+1}x_1$. Do the same for all $k$-tuples. You can clearly see that this satisfies our requirements. You Thus, our answer is yes.


Partial answer: Let $k$ be prime. Then all $n\in\mathbb{N}$ may uniquely be written as $n=k^j m$ with $j\geq 0$ and $m\in \mathbb{N}_k:=\{l\in\mathbb{N} \mid k\not\mid l\}$. Then, given any function $g\colon \mathbb{N}_k\to\mathbb{N}$, the function $f$, defined by $f(k^j m):=k^j g(m)$, satisfies $f(k n)=k f(n)$ for all $n$.


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