If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$? Let $\sigma(x)$ be the sum of the divisors of $x$.  Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$.
Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?
  $$D(q^k)D(n^2)=2s(q^k)s(n^2)$$

I only know that $k=1$ is true if and only if (one of) the following conditions hold:
(1) $\sigma(n^2)/q \mid n^2$
(2) $D(n^2) \mid n^2$
(3) $\gcd(n^2, \sigma(n^2)) = D(n^2)$
Source of Equation
From the fundamental equation
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$
we obtain
$$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \frac{D(n^2)}{s(q^k)} = \gcd(n^2, \sigma(n^2))$$
and
$$\frac{\sigma(n^2) - n^2}{\frac{2q^k - \sigma(q^k)}{2}} = \frac{2s(n^2)}{D(q^k)} = \gcd(n^2, \sigma(n^2)),$$
using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}.$$
Reference
Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers
 A: The equation is actually an identity since we get
$$\begin{align}D(q^k)D(n^2)&=(2q^k-\sigma(q^k))(2n^2-\sigma(n^2))
\\\\&=(2q^k-\sigma(q^k))\left(2n^2-\frac{2n^2q^k}{\sigma(q^k)}\right)
\\\\&=4n^2q^k-\frac{4n^2q^{2k}}{\sigma(q^k)}-2n^2\sigma(q^k)+2n^2q^k
\\\\&=4n^2q^k-2n^2\sigma(q^k)-\frac{4n^2q^{2k}}{\sigma(q^k)}+2n^2q^k
\\\\&=2\left(\sigma(q^k)-q^k\right)\left(\frac{2n^2q^k}{\sigma(q^k)}-n^2\right)
\\\\&=2\left(\sigma(q^k)-q^k\right)\left(\sigma(n^2)-n^2\right)
\\\\&=2s(q^k)s(n^2)
\end{align}$$
A: This is not an answer, just some comments that are too long to fit in the space beneath my original question.
For what it is worth, there is the simple relationship
$$D(x) := x - s(x)$$
so that we obtain
$$2s(q^k)s(n^2) = D(q^k)D(n^2) = \bigg(q^k - s(q^k)\bigg)\bigg(n^2 - s(n^2)\bigg)$$ 
$$= q^k n^2 - q^k s(n^2) - n^2 s(q^k) + s(q^k)s(n^2),$$
from which it follows that
$$s(q^k)s(n^2) + q^k s(n^2) + n^2 s(q^k) - q^k n^2 = 0$$
$$s(q^k)s(n^2) + q^k s(n^2) + n^2 s(q^k) + q^k n^2 = 2q^k n^2$$
$$\bigg(s(q^k) + q^k\bigg)s(n^2) + \bigg(s(q^k) + q^k\bigg)n^2 = 2q^k n^2$$
$$\bigg(s(q^k) + q^k\bigg)\bigg(s(n^2) + n^2\bigg) = 2q^k n^2$$
But $s(x) + x = (\sigma(x) - x) + x = \sigma(x)$.  Therefore, we have:
$$\sigma(q^k)\sigma(n^2) = 2q^k n^2.$$
However, it is not clear to me whether the preceding argument can or cannot force $k=1$.
