If $q^k n^2$ is an odd perfect number with special prime $q$, is it true that $\sigma(n^2)/q^k \mid n^2$?

Question

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Question

If $q^k n^2$ is an odd perfect number with special prime $q$, is it true that $\sigma(n^2)/q^k \mid n^2$?

Motivation

Note that is easy to prove that $$\gcd(n^2, \sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$ Indeed, this follows directly from the observation that $\gcd(q^k,\sigma(q^k))=1$.

In particular, $$\dfrac{n^2}{\frac{\sigma(n^2)}{q^k}}=\dfrac{\sigma(q^k)}{2},$$ which does appear to show that $\sigma(n^2)/q^k \mid n^2$, since $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$

Does this proof suffice?

I would be interested in seeing other (alternative) proofs for the same result.

Answer 1

Your proof looks correct to me.

It seems that yours is the "simplest" proof for the claim.