Is this divisor-sum camouflage criterion for prime-order Dirichlet characters correct, and is it already known?

Let \chi be a Dirichlet character of prime order r, and define

[ A_\chi(n)=\sum_{d\mid n}\chi(d). ]

For a prime p, one trivially has

[ A_\chi(p)=1+\chi(p). ]

I have been looking at composites n that satisfy the same identity

[ A_\chi(n)=1+\chi(n). \tag{1} ]

I would like to know whether the following argument is correct, whether I am overlooking a standard theorem, and in particular whether the constructive converse below has a gap.

Assume throughout that n is unramified for \chi.

Write

[ n=\prod_{i=1}^s p_i^{e_i}, ]

and choose a primitive r-th root of unity \zeta_r. Write

[ \chi(p_i)=\zeta_r^{a_i}. ]

Then

[ A_\chi(n)

\prod_i \left( 1+\zeta_r^{a_i} +\zeta_r^{2a_i} +\cdots+ \zeta_r^{e_i a_i} \right). ]

Define the integer polynomial

[ F(x)= \prod_i \left( 1+x^{a_i}+\cdots+x^{e_i a_i} \right)

1-x^{\sum_i e_i a_i}. ]

If (1) holds, then F(\zeta_r)=0. Since r is prime, the minimal polynomial of \zeta_r is

[ \Phi_r(x)=1+x+\cdots+x^{r-1}. ]

Thus \Phi_r(x)\mid F(x). Evaluating at x=1 gives

[ r=\Phi_r(1)\mid F(1)

\prod_i(e_i+1)-2. ]

Therefore

[ \boxed{\tau(n)\equiv2\pmod r.} \tag{2} ]

So (2) is a necessary condition for a composite to satisfy the prime-like identity (1).

There also seems to be a constructive converse at the level of character states.

Set

[ m_i=e_i+1 ]

and define mixed-radix weights

[ W_1=1,\qquad W_i=\prod_{h<i}m_h. ]

Every divisor

[ d=\prod_i p_i^{j_i}, \qquad 0\le j_i<m_i, ]

corresponds uniquely to

[ t=\sum_i j_iW_i \in{0,1,\ldots,\tau(n)-1}. ]

Now prescribe

[ \chi(p_i)=\zeta_r^{W_i}. \tag{3} ]

Then

[ \chi(d)=\zeta_r^t, ]

and therefore

[ A_\chi(n)

\sum_{t=0}^{\tau(n)-1}\zeta_r^t. ]

If

[ \tau(n)=kr+2, ]

then

[ A_\chi(n)

k(1+\zeta_r+\cdots+\zeta_r^{r-1}) +1+\zeta_r

1+\zeta_r. ]

Also,

[ \sum_i e_iW_i=\tau(n)-1, ]

so

[ \chi(n)=\zeta_r^{\tau(n)-1}=\zeta_r. ]

Hence

[ A_\chi(n)=1+\chi(n). ]

Thus the congruence

[ \tau(n)\equiv2\pmod r ]

appears to be not only necessary, but constructively sharp at the character-state level.

For several compatible characters \chi_j of distinct prime orders r_j, the same mixed-radix weights can be used in every channel:

[ \chi_j(p_i)=\zeta_{r_j}^{W_i}. ]

This suggests the simultaneous necessary condition

[ \boxed{ \tau(n)\equiv2 \pmod{\operatorname{lcm}(r_1,\ldots,r_k)} } ]

is likewise constructively sharp, provided the prescribed joint character values are actually realizable by primes. I am assuming here that the chosen characters are independent enough that the joint character map onto the desired roots-of-unity states is surjective; CRT and Dirichlet's theorem would then supply primes in the corresponding reduced residue classes.

As one consequence, if n is squarefree with s=\omega(n), then

[ \tau(n)=2^s. ]

For character orders 11,23,41, simultaneous camouflage requires

[ 2^{s-1}\equiv1 \pmod{11,23,41}. ]

Since

[ \operatorname{ord}{11}(2)=10,\qquad \operatorname{ord}{23}(2)=11,\qquad \operatorname{ord}_{41}(2)=20, ]

one obtains

[ 220\mid(s-1). ]

Thus, apart from the prime case s=1, the first squarefree value not excluded by this condition is

[ s=221. ]

My questions are:

1. Is the cyclotomic argument proving [ A_\chi(n)=1+\chi(n)\Rightarrow\tau(n)\equiv2\pmod r ] correct as stated?

2. Is the mixed-radix construction a valid converse at the character-state level?

3. Under the stated independence/surjectivity condition on several characters, is the passage from prescribed character states to actual prime factors justified as I have described it?

4. Is this criterion or mixed-radix construction already standard under another name or contained in known results on twisted divisor sums, cyclotomic polynomials, or Dirichlet characters?

I am not claiming this gives a faster primality test or factoring algorithm. My interest is the structural question of which composite factorizations can imitate the identity A_\chi(p)=1+\chi(p) simultaneously across several character channels.