(Preamble: The method presented here to compute the GCD $g$ is patterned after the method used to compute a similar GCD in this answer to a closely related MSE question.)
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)-x$.
If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must necessarily have the form $$N = q^k n^2,$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
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 $$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \frac{s(n^2)}{D(q^k)/2}.$$
Consequently, we have $$s(n^2) = \sigma(n^2) - n^2 = \left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2).$$
For simpler algebra, let $$u = q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)$$ and let $$v = (-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2).$$
Then the equation above involving $s(n^2)$ becomes $$\sigma(n^2) - n^2 = u\cdot{n^2} + v\cdot\sigma(n^2)$$ so that $$(1-v)\sigma(n^2) = (1+u){n^2}. \tag{1}$$
Dividing both sides of Equation $(1)$ by $\gcd(n^2,\sigma(n^2))$, we obtain: $$(1-v)q^k = (1+u)\cdot(\sigma(q^k)/2). \tag{2}$$ Since $\gcd(q^k,\sigma(q^k)/2)=1$, it follows from Equation $(2)$ that $$q^k \mid 1+u$$ so that there exists an integer $g$ such that $$1+u = g\cdot{q^k}. \tag{3}$$ Substituting Equation $(3)$ in Equation $(2)$, we get $$(1-v)q^k = g\cdot{q^k}\cdot(\sigma(q^k)/2) \implies 1-v=g\cdot(\sigma(q^k)/2). \tag{4}$$ It follows that $$g=\gcd(1+u,1-v)$$ since $\gcd(q^k,\sigma(q^k)/2)=1$.
Here, we prove that, in fact, $g$ must be odd, and thereby derive an explicit formula for $g$.
Subtracting Equation $(4)$ from Equation $(3)$, we obtain: $$u+v = g\cdot(D(q^k)/2) = \left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right) + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right),$$ which implies that $$g=\gcd(1+u,1-v)=\left(q^k t - 2(q - 1)\right) + \left((-\sigma(q^k)/2)\cdot{t} + q\right)=(D(q^k)/2)\cdot{t} - (q - 2) \equiv 1 \pmod 2,$$ since $D(q^k)/2$ must be even.
Here is my:
QUESTION: Is my derivation of the formula $$g=\gcd(1+u,1-v)=(D(q^k)/2)\cdot{t} - (q - 2)$$ correct?
This answer gives an alternative derivation for the formula $$g = \gcd(1+u,1-v) = (D(q^k)/2)\cdot{t} - (q - 2),$$ and thereby attempts to answer the original question.
From this WolframAlpha computation, we get the following fully simplified expression for $1+u$:
$$1+u = \frac{q^k (tq^{k+1} - 2tq^k - 2q^2 + 6q + t - 4)}{2(q - 1)} = {q^k}\cdot\frac{\left(tq^{k+1} - 2tq^k + t - 2q^2 + 6q - 4\right)}{2(q - 1)}$$ $$= {q^k}\cdot\left(t\cdot\frac{q^{k+1} - 2q^k + 1}{2(q - 1)}-\frac{2(q^2 - 3q + 2)}{2(q - 1)}\right) = {q^k}\cdot\left(t\cdot(D(q^k)/2)-(q-2)\right).$$
Similarly, from this WolframAlpha computation, we get the following fully simplified expression for $1-v$:
$$1-v=\frac{(q^{k+1} - 1)(tq^{k+1} - 2tq^k - 2q^2 + 6q + t - 4)}{4(q - 1)^2}$$ $$=\frac{q^{k+1} - 1}{2(q - 1)}\cdot\frac{\left(tq^{k+1} - 2tq^k + t - 2q^2 + 6q - 4\right)}{2(q - 1)}$$ $$=\frac{\sigma(q^k)}{2}\cdot\left(t\cdot(D(q^k)/2) - (q - 2)\right).$$
Let $g' = t\cdot(D(q^k)/2) - (q - 2)$.
It follows that $$\gcd(1+u,1-v)=\gcd\left({q^k}\cdot{g'},(\sigma(q^k)/2)\cdot{g'}\right)={g'}\cdot\gcd(q^k,\sigma(q^k)/2)=g'.$$
Note that $$g=(D(q^k)/2)\cdot{t} - (q - 2)=t\cdot(D(q^k)/2) - (q - 2)=g'.$$