If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).

Question

I tried Googling for the keywords "the theory of odd perfect numbers" and one of the search results that came up was this document, titled ON THE DIVISORS OF THE SUM OF A GEOMETRICAL SERIES WHOSE FIRST TERM IS UNITY AND COMMON RATIO ANY POSITIVE OR NEGATIVE INTEGER. (The author turns out to be J. J. Sylvester.)

Let me copy the first two paragraphs from that document:

A REDUCED Fermatian, $\frac{r^p - 1}{r - 1}$, is obviously only another name for the sum of a geometrical series whose first term is unity and common ratio an integer, $r$.

If $p$ is a prime number, it is easily seen that the above reduced Fermatian will not be divisible by $p$, unless $r - 1$ is so, in which case (unless $p$ is $2$) it will be divisible by $p$, but not by $p^2$.

Now, let me translate the second paragraph into modern language, fitted to the case of the divisor-sum function of the special prime power component of an odd perfect number $N = q^k n^2$, where $q$ is the special prime and satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. That is, we will be considering $$\sigma(q^k) = \frac{q^{k+1} - 1}{q - 1},$$ where $\sigma(x)=\sigma_1(x)$ is the classical divisor sum function of $x$.

(Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.)

Let us now see what we have got. Translating to modern language (in the context of $\sigma(q^k)$):

PROPOSITION: If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).

Here is my:

QUESTION: What happens to the PROPOSITION when $k=1$?

MY ATTEMPT

Suppose that the Descartes-Frenicle-Sorli Conjecture holds. Then $k=1$.

Then $k+1=2$ is prime. Also, since $q \equiv 1 \pmod 4$, then $q - 1 \equiv 0 \pmod 4$, so that $2 = (k + 1) \mid (q - 1)$. Then $\sigma(q^k)$ is divisible by $(k+1)=2$ but not by $(k+1)^2=4$, which is true. So what's with the UNLESS?

Here is a proof for $\sigma(q^k) \equiv 2 \pmod 4$:

First, since $q \equiv 1 \pmod 4$, then $$\sigma(q^k) = 1 + q + \ldots + q^k \equiv k + 1 \pmod 4.$$ Next, since $k \equiv 1 \pmod 4$, then $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$ QED

Answer 1

This is deduced due to a famous lemma often called as "LTE lemma" (Lifting the exponents). The lemma is as follows:

If $p\not=2$ is a prime number and $a,b$ are positive integers such that $p|a-b$, then for every positive integer $n$ we have : $V_p(a^n-b^n)=V_p(a-b)+V_p(n)$

The lemma also holds for $p=2$ but it's slightly different in that case. This theorem can be easily proved using induction.
Anyways using LTE lemma on your problem, since $p=k+1|q-1$, then $$V_p(\frac{q^{k+1}-1}{q-1})=V_p(q^{k+1}-1)-V_p(q-1)=V_p(q-1)+V_p(k+1)-V_p(q-1)=V_p(k+1)$$ But we know that $V_p(k+1)=V_p(p)=1$ so finally $V_p(\frac{q^{k+1}-1}{q-1})=1$ hence the sum of divisors is divisible by $p$ but not by $p^2$.

P.S. Here is almost everything on the lemma