<s>[INST] Is this logical deduction regarding some modular restrictions on odd perfect numbers valid? - Part II Let $p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.  Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
Here is the Abstract of Dris and San Diego's article titled "Some modular considerations regarding odd perfect numbers – Part II", published in NNTDM:

In this article, we consider the various possibilities for $p$ and $k$ modulo $16$, and show conditions under which the respective congruence classes for $\sigma(m^2)$ (modulo $8$) are attained, if $p^k m^2$ is an odd perfect number with special prime $p$.  We prove that:



(1) $\sigma(m^2) \equiv 1 \pmod 8$ holds only if $p + k \equiv 2 \pmod {16}$.




(2) $\sigma(m^2) \equiv 3 \pmod 8$ holds only if $p - k \equiv 4 \pmod {16}$.




(3) $\sigma(m^2) \equiv 5 \pmod 8$ holds only if $p + k \equiv {10} \pmod {16}$.




(4) $\sigma(m^2) \equiv 7 \pmod 8$ holds only if $p - k \equiv 4 \pmod {16}$.


From this answer by mathlove and this earlier MSE question, we get the biconditional
$$\sigma(m^2) \equiv 3 \pmod 4 \iff p \equiv {k+4} \pmod {16}.$$
Also, we have the biconditional
$$\sigma(m^2) \equiv 1 \pmod 4 \iff p \equiv k \pmod 8,$$
per the characterization theorem of Chen and Luo.
Assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
If $p=5$, then this evidently implies that $\sigma(m^2) \equiv 3 \pmod 4$.
Here is our:

QUESTION: If $p=13$, then since $13 = p \equiv {k+4=5} \pmod 8$ holds, but the conjunction
$$(p = 13) \land (k=1)$$
obviously does not satisfy $p \equiv {k+4} \pmod {16}$ nor $p \equiv k \pmod 8$, does it follow that the implication $k = 1 \implies p \neq 13$ holds, since otherwise
$$\lnot\Bigg(p \equiv {k+4} \pmod {16}\Bigg) \iff \lnot\Bigg(\sigma(m^2) \equiv 3 \pmod 4\Bigg)$$
and
$$\lnot\Bigg(p \equiv k \pmod 8\Bigg) \iff \lnot\Bigg(\sigma(m^2) \equiv 1 \pmod 4\Bigg),$$
contradicting the fact that $\sigma(m^2)$ must be odd?

 [/INST] This is not a comprehensive proof, it is a mere observation that there are other prime components that do not satisfy the condition you stated. $13^1$ is not the only prime component that violates this rule. Let us take $p^kn^2$ to be an odd perfect number and k=1. 
We know that $\sigma (p^k)/2\times gcd(n^2,\sigma (n^2))=n^2$.
We also know that $n^2\equiv 1\quad mod\quad 8$. Therefore if $\sigma (p^k)/2\equiv 3\quad mod\quad 4$ then $gcd(n^2,\sigma (n^2))$ must also be congruent to 3 mod 4. Therefore we can conclude that $gcd(n^2,\sigma (n^2))\equiv \sigma (p^k)/2$ if $gcd(n^2,\sigma (n^2))\equiv 3\quad mod\quad 4$
Now, $\sigma (p^1)/2\equiv 3\quad mod\quad 4$ if p is a prime of the form 5+8t where t is any positive integer. I will list a few examples below.

*

*[$\sigma (5^1)/2=3]\equiv 3\quad mod\quad 4,\quad$ $5=5+8(0)$

*[$\sigma (13^1)/2=7]\equiv 3\quad mod\quad 4,\quad$ $13=5+8(1)$

*[$\sigma (29^1)/2=15]\equiv 3\quad mod\quad 4,\quad$ $29=5+8(3)$

*[$\sigma (37^1)/2=19]\equiv 3\quad mod\quad 4,\quad$ $37=5+8(4)$ 
Now, a pattern emerges where we notice that $\sigma (n^2)\equiv gcd(n^2,\sigma (n^2))\equiv \sigma (p^k)/2\equiv 3\quad mod\quad 4\iff p\equiv k+4\quad (mod\quad 16)$ does not hold whenever prime p in the list above is of the form $5+8t$, where t is an odd integer. A more comprehensive list of values of p that do not satisfy the condition above is listed below

*

*$p^k$=(5+8(1))^1=13^1

*$p^k$=(5+8(3))^1=29^1

*$p^k$=(5+8(7))^1=61^1

*$p^k$=(5+8(13))^1=109^1

*$p^k$=(5+8(19))^1=157^1

*$p^k$=(5+8(21))^1=173^1

*$p^k$=(5+8(33))^1=269^1

The reader can test and satisfy himself or herself that the prime components in the above list violate the following condition 
$\sigma (n^2)\equiv gcd(n^2,\sigma (n^2))\equiv \sigma (p^k)/2\equiv 3\quad mod\quad 4\iff p\equiv k+4\quad (mod\quad 16)$
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