In what follows, we denote the classical sum of divisors of the positive integer $x$ by $$\sigma(x)=\sigma_1(x)=\sum_{d \mid x}{d},$$ and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number (hereinafter abbreviated as OPN). Euler proved that a hypothetical OPN $N$ must have the form $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is currently unknown whether any OPNs exist. It is widely believed that there are no OPNs.
Let $N = q^k n^2$ be a hypothetical OPN.
Then, since $N$ is perfect and $\gcd(q,n)=1$, and using the fact that the divisor sum $\sigma$ is multiplicative, we have the equation $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2 q^k n^2.$$
We can rewrite this equation as $$n^2 = \frac{\sigma(q^k)}{2}\cdot\frac{\sigma(n^2)}{q^k}$$ and $$q^k = \frac{\sigma(q^k)}{2}\cdot\frac{\sigma(n^2)}{n^2}.$$
Notice that $$\frac{8}{5} < I(n^2) = \frac{\sigma(n^2)}{n^2} < 2,$$ so that $I(n^2)$ is not an integer. This means that $$\frac{\sigma(q^k)}{2} \nmid \gcd(q^k,n^2) = 1,$$ which is expected, since $$3 \leq \frac{\sigma(q^k)}{2}.$$
Hence, this approach does not appear to allow us to write $$n^2 - q^k$$ in factored form.
Note that $\gcd(q,n)=1$, and that $k \equiv 1 \pmod 4$.
We do know from On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part VI, however, that we can write $$n^2 - q^k = 2^r t$$ where $2^r \neq t, r \geq 2$, and $\gcd(2,t)=1$.
Here is my inquiry:
QUESTION: Will it be possible to compute a factored expression for $n^2 - q^k$, if $q^k n^2$ is an odd perfect number with special prime $q$?
PostScript: Note that this question is related to this other inquiry.
