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 $q$

Question

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.

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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
$$3 \leq \frac{\sigma(q^k)}{2}.$$

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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.

Answer 1

This has been partially answered in MathOverflow.

In particular, since we know that $$n^2 - q^k \equiv 0 \pmod 4,$$ then we can write $$n^2 - q^k = a^2 - b^2$$ for some positive integers $$a=f(q,k,n)$$ and $$b=g(q,k,n).$$

It follows that $$n^2 - q^k = \left(f(q,k,n) - g(q,k,n)\right)\cdot\left(f(q,k,n) + g(q,k,n)\right).$$