Let $n=\prod\_{p|n}p^{\alpha\_p}$. While considering arithmetic functions of the projection $$\textrm{rad}\_k(n)=\prod\_{p|n}p^{\textrm{min}(\alpha\_p,k)},$$ a basic question has arisen that is more difficult to answer than I had supposed. 


Let $a,f:\mathbb{N}\to\mathbb{C}$ and define $$Af(n)=\sum\_{d|n}a(d)f(d)$$


so that $$A^2f(n)=\sum\_{d|n}a(d)\sum\_{e|d}a(e)f(e)$$
and so on. The question is simply this:



> 
> If $A^{k+1}=A$ for some $k\in\mathbb{N}$ and $a(1)=1$, does it follow that $a(n)$ is multiplicative? 
> 
> 
> 


Multiplicative solutions exist for every $k$; these are the functions $$a(n)=\delta\_k(n)z^{\Omega(n)}$$ where $z$ is a primitive $k$th root of unity and $\delta\_k(n)=0,1$ depending on whether $n$ is divisible by a $k$th power or not. 
There are $\phi(k)$ 
primitive roots, and an affirmative answer to the question would mean that 
accounts for all solutions. 


Interchanging the order of summation, the statement says that $$a(m)=a(m)\sum\_{d\_1|n}a(d\_1m)\cdots\sum\_{d\_k|d\_{k-1}}a(d\_km)$$ for all $m,n$ from which one can infer that $$a^{k+1}(n)=a(n)$$ for all $n$ and $$a(m)a(n)=a(mn)$$ whenever $a(m)$, $a(n)$ and $a(mn)$ are all non-zero. 


From the stated property of $A$ we have that $A^k$ is a projection and when $a(n)$ is multiplicative that $$A^kf(n)=f(\textrm{rad}\_{k-1}(n)).$$


Is every solution as such?



Projective summation over divisors: must these functions be multiplicative?
It's not the case when $k=2$ even: other solutions are, for any integer $j\ge2$,
$$
a(n) = \begin{cases}
1, &\text{if } n=1, \\
-1, &\text{if } n=j, \\
0, &\text{otherwise.}
\end{cases}
$$
In particular, this function is not multiplicative when $j=6$. Indeed, there seem to be lots of solutions that involve setting $a(p)=0$ for all primes $p$ and $a(j)\in\{-1,0\}$ for integers $j$ with $\Omega(j)=2$.


Similar remarks seem to hold for higher $k$, with $-1$ replaced by a $k$th root of unity.


