(Note: This has been cross-posted to MO.)

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

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 $4 \mid (k-1)$.

  • $\begingroup$ By simple congruence considerations, $$\sigma(q^{k-1}) \equiv \sum_{i=0}^{k-1}{1} \equiv 1 + (k - 1) \equiv k \equiv 1 \pmod 4.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 6 '15 at 0:29
  • $\begingroup$ Additionally, I checked using WolframAlpha and found only the following factorization (for $k = 9$): $$\sigma(q^8) = 1 + q + q^2 + q^3 + q^4 + q^5 + q^6 + q^7 + q^8 = (q^2 + q + 1)(q^6 + q^3 + 1) = \sigma(q^2)(q^6 + q^3 + 1).$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 6 '15 at 0:54

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