<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] Yes, it does.
Moreover, we can say that
$$k=1\implies p\not\equiv 13\pmod{16}$$
Proof :
Suppose that $p\equiv 13\pmod{16}$. Then, since we have
$$p \not\equiv {5} \pmod {16}\implies \sigma(m^2) \not\equiv 3 \pmod 4$$
and
$$p\not\equiv 1,9\pmod{16}\implies p\not\equiv 1\pmod 8\implies\sigma(m^2) \not\equiv 1 \pmod 4$$
we have $\sigma(m^2)\equiv 0,2\pmod 4$ which contradicts that $\sigma(m^2)$ is odd. $\ \blacksquare$
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