I posted this question on [MSE](https://math.stackexchange.com/questions/1221761) two days ago, but did not receive any responses. I have cross-posted it on MO, hoping it gets more attention here and that it *is* appropriate for this site.


A positive integer $N$ is said to be *perfect* if $\sigma(N) = 2N$, where $\sigma(x)$ is the sum of the divisors of $x$.


An *odd perfect number* $N$ is said to be given in *Eulerian form* if $N = {q^k}{n^2}$, where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.


[Broughan, Delbourgo and Zhou (2013)](http://www.math.waikato.ac.nz/~kab/papers/chenchenformatted.pdf) defines the *perfect number index of $N$ at prime $r$* to be the integer
$$m := \frac{\sigma(N/{r^{\alpha}})}{r^{\alpha}},$$
where $r^{\alpha} || N$. They also show that $m \ge 315$. ([Chen and Chen (2014)](http://journals.impan.gov.pl/cm/Inf/136-1-4.html) extend these results in "On the index of an odd perfect number".)


Since $\sigma$ is weakly multiplicative, we have
$$\sigma(q^k)\sigma(n^2)=\sigma(N)=2N=2{q^k}{n^2}.$$
Because $\gcd(q^k,\sigma(q^k))=1$, this means that
$$\sigma(n^2)/q^k = 2{n^2}/\sigma(q^k) = s \in \mathbb{N},$$
where $s \ge 315$.


In particular, we have the simultaneous equations
$$\sigma(n^2) = s{q^k}$$
and
$$2{n^2} = s{\sigma(q^k)}.$$


We obtain
$$2{n^2} - \sigma(n^2) = s{\sigma(q^k)} - s{q^k} = s{\sigma(q^{k-1})}.$$


**UPDATE - September 27 2016**
In fact, we know that $s = \gcd(n^2,\sigma(n^2))$.


It follows that
$$\sigma(q^{k-1}) \mid (2{n^2} - \sigma(n^2)).$$


Now, here is my question:



> 
> What are the divisors of $\sigma(q^{k-1})$?
> 
> 
> 


Notice that $4 \mid (k-1)$.


By simple congruence considerations:
$$\sigma(q^{k-1}) \equiv \sum\_{i=0}^{k-1}{1} \equiv 1 + (k-1) \equiv k \equiv 1 \pmod 4.$$


Additionally, I checked using [WolframAlpha](http://www.wolframalpha.com/input/?i=1+%2B+q+%2B+q%5E2+%2B+q%5E3+%2B+q%5E4+%2B+q%5E5+%2B+q%5E6+%2B+q%5E7+%2B+q%5E8&dataset=) and found only the following factorization (for $k = 9$):
$$\sigma(q^8) = 1 + q + q^2 + q^3 + q^4 + q^5 + q^6 + q^7 + q^8 = (q^2 + q + 1)(q^6 + q^3 + 1) = \sigma(q^2)(q^6 + q^3 + 1)$$



On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$, Part II
In general there is a factorization $\sigma(q^{k-1}) = 1+q+\ldots+q^{k-1} = \frac{q^k-1}{q-1} = \prod\_{d\mid k, d>1} \Phi\_d(q)$. Here $\Phi\_d$ is the $d$'th cyclotomic polynomial, defined as the minimal polynomial of a primitive $d$'th root of unity. Your example is a special case of this since $\Phi\_3(x)=1+x+x^2$ and $\Phi\_9(x)=x^6+x^3+1$.


In certain cases it possible to factor $\Phi\_d(q)$ further. See "[aurifeuillean factorization](https://en.wikipedia.org/wiki/Aurifeuillean_factorization)" on Wikipedia. 


