(Preamble: This post is an offshoot of this MSE question and this MO question.)
My primary aim in this post is to compute a (hopefully factorable) expression for the quantity $n^2 - q^k$, if $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$, $\gcd(q,n)=1$, $n^2 - q^k = 2^r t$, with $r \geq 2$, and $\gcd(2,t)=1$. (The last GCD constraint holds because $n^2 - q^k$ is not a square.)
Letting $$i(q):=\text{ index of } N \text{ at the special prime } q,$$ as defined by Broughan et. al, then we have $$i(q):=\frac{\sigma(N/q^k)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))$$ whereupon we obtain $$n^2 = i(q)\cdot\frac{\sigma(q^k)}{2}$$ and $$q^k = \frac{\sigma(n^2)}{i(q)},$$ giving the difference $$n^2 - q^k = \frac{\bigg(i(q)\bigg)^2 \cdot\sigma(q^k) - 2\sigma(n^2)}{2i(q)}.$$
Let us now try to evaluate and simplify the numerator of $n^2 - q^k$, using the following linear combination formula for $i(q)=\gcd(n^2,\sigma(n^2))$ from this answer: $$i(q) = q\sigma(n^2) - 2(q - 1)n^2$$ $$\bigg(i(q)\bigg)^2 \cdot\sigma(q^k) - 2\sigma(n^2) = \bigg(q\sigma(n^2) - 2(q - 1)n^2\bigg)^2 \cdot\sigma(q^k) - 2\sigma(n^2).$$
I then tried to factor the resulting expression using WolframAlpha but was unsuccessful.
Alas, this is where I get stuck!
