Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.
I discovered an interesting identity involving divisors of odd perfect numbers given in the Eulerian form $N = q^k n^2$ today (July 13, 2021).
The identity is:
Proposition: If $N = q^k n^2$ is an odd perfect number with special prime $q$, then
$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$
Proof:
Our starting point is the following blog post, where it is proved that
$$\gcd(n^2, \sigma(n^2)) = 2(1 - q)n^2 + q\sigma(n^2).$$
However, note that we have
$$\frac{\sigma(n^2)}{q^k} = \gcd(n^2, \sigma(n^2)).$$
These equations are equivalent to
$$2(1 - q)n^2 + q\sigma(n^2) = \frac{\sigma(n^2)}{q^k}.$$
Factoring out $qn^2$ on the LHS, we obtain
$$qn^2 \Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q^k}.$$
Multiplying both sides of the last equation by $q^{k-1}$, we get
$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$
(Note that the RHS of the last equation is an odd integer.)
This concludes our proof.
QED.
In particular, we have
$$\frac{N}{\sigma(n^2)/q} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)}.$$
But we also know from the following MSE post that
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{2(q - 1)}{q\bigg(q^{k+1} - 1\bigg)}.$$
This means that we obtain
$$\frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\bigg(q^{k+1} - 1\bigg)}{2(q - 1)} = \frac{q\sigma(q^k)}{2}.$$
But $\sigma(q^k) \equiv k + 1 \pmod 4$, since $q \equiv 1 \pmod 4$, and since $k \equiv 1 \pmod 4$, then $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$
This finding implies that $\sigma(n^2)/q$ divides $N = q^k n^2$, which is almost a proof of the desired divisibility constraint $\sigma(n^2)/q \mid n^2$, as the latter constraint would imply $k = 1$.
