On the equation $s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$, if $q^k n^2$ is an odd perfect number with special prime $q$ Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In what follows, let us keep in mind the following lemma:

LEMMA: If $q^k n^2$ is an odd perfect number given in Eulerian form, then $k = 1$ holds if and only if
$$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)},$$
where $s(x)=\sigma(x)-x$ is the aliquot sum of the positive integer $x$, $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.


Now, we start with
$$\gcd(n^2,\sigma(n^2)) = \left(q^k t - 2(q - 1)\right)n^2 + \left((-\sigma(q^k)/2)\cdot{t} + q\right)\sigma(n^2),$$
like Tony Kuria Kimani did in this recent ResearchGate preprint.
But the equation
$$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \frac{s(n^2)}{D(q^k)/2} = \frac{D(n^2)}{s(q^k)}$$
holds.  Consequently, we obtain the simultaneous equations
$$s(n^2) = (D(q^k)/2)\cdot\gcd(n^2,\sigma(n^2))$$
$$= \left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$
$$D(n^2) = {s(q^k)}\cdot\gcd(n^2,\sigma(n^2)) = \left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2).$$
We now test whether the equation
$$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$$
holds; that is, whether the equation
$$\left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$
$$= \left(\frac{q-1}{2}\right)\cdot\Bigg(\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2)\Bigg) \tag{1}$$
is true.
For simpler algebra, let
$$u_1 = q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)$$
$$u_2 = \left(\frac{q-1}{2}\right)\cdot\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)$$
$$v_1 = (-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)$$
and
$$v_2 = \left(\frac{q-1}{2}\right)\cdot\left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right).$$
Then Equation $(1)$ becomes
$$\left(u_1 - u_2\right)\cdot{n^2} = \left(v_2 - v_1\right)\cdot{\sigma(n^2)}. \tag{2}$$
We now attempt to get simplified expressions for $u_1 - u_2$ and $v_2 - v_1$ using WolframAlpha.  We obtain the following:
$$u_1 - u_2 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(tq^k - 2q + 2\right)$$
$$v_2 - v_1 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$
Dividing both sides of Equation $(2)$ by $d = \gcd(n^2,\sigma(n^2))$, we obtain
$$\left(u_1 - u_2\right)\cdot\Bigg({\frac{n^2}{d}}\Bigg) = \left(v_2 - v_1\right)\cdot\Bigg(\frac{\sigma(n^2)}{d}\Bigg).$$
Recall that
$$d = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$
Consequently, we have
$$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = \left(v_2 - v_1\right)\cdot{q^k}. \tag{4}$$
Since $\gcd(q^k,\sigma(q^k)/2)=1$, then there exists an integer $G$ such that
$$v_2 - v_1 = G\cdot\left(\sigma(q^k)/2\right).$$
Substituting into Equation $(4)$, we get
$$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = G\cdot\left(\sigma(q^k)/2\right)\cdot{q^k},$$
from which we finally obtain
$$u_1 - u_2 = G\cdot{q^k}.$$
We therefore finally get
$$\gcd(u_1 - u_2, v_2 - v_1) = \gcd\Bigg(G\cdot{q^k}, G\cdot\left(\sigma(q^k)/2\right)\Bigg) = G\cdot\gcd(q^k, \sigma(q^k)/2) = G. \tag{5}$$

To recap, we have obtained
$$G = \gcd(u_1 - u_2, v_2 - v_1).$$
However, from the equations in $(3)$, we also get
$$\gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$
Hence, we finally have
$$G = \gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$
Since GCDs are always nonnegative, then $G \geq 0$, which means that we have
$$q^k - q \leq 0.$$
This implies that $k \leq 1$.  Since $k \geq 1$ ought to hold (because $k$ is a positive integer satisfying $k \equiv 1 \pmod 4$), then we now know that $k=1$.
Quite apparently, this implies that $G = 0$, whereupon we obtain $u_1 = u_2$ and $v_1 = v_2$.

Here is my question:

Can I define GCDs to be always positive (so that $G > 0$ in the previous section) and thereby get a proof for $k \neq 1$?

 A: Posting this self-answer, thanks to hints from Professor Dujella in the chat room.

First, we show that the assumption $G > 0$ leads to a contradiction.
It is clear that, under the assumption $G > 0$, $k \neq 1$ holds.  (This is because $k = 1$ implies that $G = 0$.)
Then I obtain (from Equation $(4)$ in the original post) that
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{2(v_2 - v_1)}{u_1 - u_2} = \frac{q(tq^k - 2q + 2) - t}{(q - 1)(tq^k - 2q + 2)} = \frac{q}{q - 1} - \frac{t}{(q - 1)(tq^k - 2q + 2)}. \tag{1}$$
(Note that from Equation $(1)$ in this answer, we have $t \neq 0$.)
We obtain
$$\frac{1}{q^k (q - 1)} = \frac{t}{(q - 1)(tq^k - 2q + 2)}$$
which is equivalent to
$$0 = \left(\frac{1}{q - 1}\right)\cdot\left(\frac{1}{q^k} - \frac{t}{tq^k - 2q + 2}\right) = \left(\frac{1}{q - 1}\right)\cdot\left(\frac{tq^k - 2q + 2 - tq^k}{tq^k - 2q + 2}\right) = -\frac{2(q - 1)}{(q - 1)(tq^k - 2q + 2)}.$$
Now, since a priori we know that $q \geq 5$ holds, then we may cancel $q - 1$ from both numerator and denominator of the last fraction, to get
$$0 = -\frac{2}{tq^k - 2q + 2},$$
resulting in the contradiction
$$0 = -2.$$
(Note that $tq^k - 2q + 2 \neq 0$, because otherwise we would get
$$t = \frac{2(q - 1)}{q^k},$$
which implies that $q^k \mid 2(q - 1)$ (since $t$ must be an integer), contradicting $\gcd(2,q)=\gcd(q-1,q)=1$.)
Hence, we conclude that $G = 0$ must hold.

Now, we claim that:

CLAIM: $G = 0$ if and only if $k = 1$.

Proof: The implication $k = 1 \implies G = 0$ is obvious.
For the converse, assume that $G = 0$.
Then we obtain the system of equations
$$0 = u_1 - u_2 = - \left(\frac{q^k - q}{2(q - 1)}\right)\cdot(tq^k - 2q + 2) \tag{2}$$
$$0 = v_2 - v_1 = - \left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$
Now, note that (referencing Equations $(2)$ and $(3)$ in this answer), we have
$$(2) \iff \left((q^k - q = 0) \lor (tq^k - 2q + 2 = 0)\right) \land (q \neq 1) \iff k = 1,$$
since $tq^k - 2q + 2 = 0$ contradicts the requirement that $t$ must be an integer.
$$(3) \iff \left((q^k - q = 0) \lor (q(tq^k - 2q + 2) - t = 0)\right) \land (q \neq 1) \iff \left((k = 1) \lor (t(q^{k+1} - 1) = 2q(q - 1))\right) \land (q \neq 1) \iff \Bigg((k = 1) \lor \left(1 < t = \frac{2q}{\sigma(q^k)} = \frac{2q}{q + 1} < 2\right)\Bigg) \land (q \neq 1) \iff k = 1,$$
which completes the proof.
