Let $N=q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, the aliquot sum of $x$ by $s(x)=\sigma(x)-x$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Since $N=q^k n^2$ is perfect and $\gcd(q,n)=1$, we obtain $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2.$$ Now using the fact that $\gcd(q^k,\sigma(q^k))=1$, we see that $q^k$ must divide $\sigma(n^2)$, so that $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$ Now, using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B},$$ where $B \neq 0$, $D \neq 0$, and $D-B \neq 0$, then we obtain $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}.$$
We now prove the following:
CLAIM: If $N=q^k n^2$ is an odd perfect number with special prime $q$, then the series of inequalities $$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$ holds in general.
PROOF: We prove each inequality one by one, left to right.
First, assume to the contrary that $$2n^2 - \sigma(n^2)=D(n^2)=\frac{D(n^2)}{s(q)} < \frac{2n^2}{\sigma(q)} = \frac{2n^2}{q+1}.$$ This inequality is equivalent to $$\frac{2q}{q+1}=2 - \frac{2}{q+1} < \frac{\sigma(n^2)}{n^2} = I(n^2),$$ contradicting $I(n^2) \leq 2q/(q+1)$.
Next, suppose to the contrary that $$\frac{2n^2}{q+1}=\frac{2n^2}{\sigma(q)} < \frac{\sigma(n^2)}{q}.$$ This inequality is equivalent to $$\frac{2q}{q+1} < \frac{\sigma(n^2)}{n^2} = I(n^2),$$ contradicting $I(n^2) \leq 2q/(q+1)$.
Lastly, assume to the contrary that $$\frac{\sigma(n^2)}{q} < \frac{2s(n^2)}{D(q)} = \frac{2(\sigma(n^2) - n^2)}{q - 1}.$$ This inequality is equivalent to $$(q - 1)\sigma(n^2) < 2q\sigma(n^2) - 2qn^2 \iff 2qn^2 < (q+1)\sigma(n^2) \iff \frac{2q}{q+1} < I(n^2),$$ contradicting $I(n^2) \leq 2q/(q+1)$.
Now here is my:
QUESTION: Note that equality holds in $$\frac{D(n^2)}{s(q)} \geq \frac{2n^2}{\sigma(q)} \geq \frac{\sigma(n^2)}{q} \geq \frac{2s(n^2)}{D(q)}$$ if and only if $k=1$. So do we have $$k=1 \iff \bigg(\frac{D(n^2)}{s(q)}=\frac{2n^2}{\sigma(q)}=\frac{\sigma(n^2)}{q}=\frac{2s(n^2)}{D(q)}\bigg) \iff \bigg(\frac{D(n^2)}{s(q^k)}=\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2s(n^2)}{D(q^k)}\bigg),$$ by treating $k$ as a placeholder for $1$? If so, do we then have a proof for $k=1$, since $$\bigg(\frac{D(n^2)}{s(q^k)}=\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2s(n^2)}{D(q^k)}\bigg)$$ holds unconditionally?
(This is a partial answer, mainly for my own benefit.)
Let $q^k n^2$ be an odd perfect number with special prime $q$.
Assume to the contrary that $k \neq 1$.
Then we obtain the following chain of inequalities and equations $$\dfrac{D(n^2)}{s(q)} > \dfrac{2n^2}{\sigma(q)} > \dfrac{\sigma(n^2)}{q} > \dfrac{2s(n^2)}{D(q)} > \dfrac{2s(n^2)}{D(q^k)} = \dfrac{\sigma(n^2)}{q^k} = \dfrac{2n^2}{\sigma(q^k)} = \dfrac{D(n^2)}{s(q^k)},$$ which does not result in a contradiction.