Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors by $s(z):=\sigma(z)-z$. Finally, let the abundancy index of $z$ be denoted by $I(z):=\sigma(z)/z$.
If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the form $n=p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Starting from the fundamental equality $$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}$$ one can derive $$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}=\gcd(m^2,\sigma(m^2))$$ so that we ultimately have $$\frac{D(m^2)}{s(p^k)}=\frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k}=\gcd(m^2,\sigma(m^2))$$
The following is copied verbatim from this answer to a closely related question:
It turns out that it is possible to express $\gcd(m^2, \sigma(m^2))$ as an integral linear combination of $m^2$ and $\sigma(m^2)$, in terms of $p$ alone.
To begin with, write $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{\sigma(p^{k-1})}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$
Now, using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$ where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)},$$ from which we get $$\gcd(m^2,\sigma(m^2))=\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)=2m^2 - p(2m^2 - \sigma(m^2)) = 2m^2 - pD(m^2),$$ or equivalently, $$\gcd(m^2,\sigma(m^2))=2(1 - p)m^2 + p\sigma(m^2).$$
Here is my Question #1:
Is it possible to obtain the following equation? $$2m^2 - \sigma(m^2)=\gcd(m^2,\sigma(m^2))=2(1 - p)m^2 + p\sigma(m^2)$$
MY ATTEMPT
Suppose that $2m^2 - \sigma(m^2)=\gcd(m^2,\sigma(m^2))$. Then it follows that $m^2$ is deficient-perfect, which then implies that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds. But notice that the variable $k$ is not present in the last equation above, so that it is not really relevant to consider the value of $k$ here.
By assumption, we have $$2m^2 - \sigma(m^2)=\gcd(m^2,\sigma(m^2))=2(1 - p)m^2 + p\sigma(m^2),$$ from which it follows (?) that $$\bigg(2(1-p)=2\bigg) \land \bigg(-1 = p\bigg),$$ which results (?) in the contradiction $$\bigg(0 = p\bigg) \land \bigg(-1 = p\bigg).$$ Note that either conjunct is already a contradiction, as $p$ being the special / Euler prime it ought to satisfy $p \equiv 1 \pmod 4$.
We therefore conclude that $m^2$ is not deficient-perfect. This is equivalent to $k \neq 1$.
Here is my Question #2:
Is this proof correct? Is it even logically sound?
