(Note: This post is tangentially related to this earlier MSE question.)
Let $N = q^k n^2$ be an odd perfect number with special/Eulerian prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is known that
$$i(q):=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}$$
and this follows easily from the fact that $\gcd(q^k,\sigma(q^k))=1$.
Recall that GCD is associative:
$$\gcd(A,\gcd(B,C))=\gcd(\gcd(A,B),C)$$
for positive integers $A, B, C$, and that GCD is commutative:
$$\gcd(D,E)=\gcd(E,D)$$
for positive integers $D, E$.
With these initial considerations out of the way, let us prove the following propositions:
Proposition 1: If $q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(n^4,\sigma(n^2))=\gcd(n^2,\sigma(n^2)).$$
Proof: Since $\gcd(q,n)=1$ and $q$ is prime, then
$$1=\gcd(q^k,n^2)=\gcd\Bigg(\frac{\sigma(n^2)}{i(q)},n^2\Bigg)=\frac{\gcd\Bigg(\sigma(n^2),n^2 {i(q)}\Bigg)}{i(q)}=\frac{\gcd\Bigg(\sigma(n^2),{n^2}\cdot{\gcd(n^2,\sigma(n^2))}\Bigg)}{i(q)}.$$
It follows that by GCD Commutativity that
$$1=\frac{\gcd\Bigg(\sigma(n^2),\gcd(n^2{\sigma(n^2)},n^4)\Bigg)}{i(q)}$$
and then, by GCD Associativity, that
$$1=\frac{\gcd\Bigg(\gcd(\sigma(n^2),n^2{\sigma(n^2)}),n^4\Bigg)}{i(q)}.$$
Since $\sigma(n^2) \mid {n^2 \sigma(n^2)}$, then we obtain
$$1=\frac{\gcd(\sigma(n^2),n^4)}{i(q)}$$
from which we get
$$1=\frac{\gcd(n^4,\sigma(n^2))}{i(q)}$$
by GCD Commutativity.
Hence, we finally have
$$\gcd(n^2,\sigma(n^2))=i(q)=\gcd(n^4,\sigma(n^2)),$$
which is what we set out to prove for in Proposition 1.
QED.
Similarly, by considering $\gcd(q^k,n)=1$ as a starting point and following exactly the same line of reasoning as in the proof of Proposition 1, we obtain the following proposition:
Proposition 2: If $q^k n^2$ is an odd perfect number with special prime $q$, then
$$\gcd(n^3,\sigma(n^2))=\gcd(n^2,\sigma(n^2)).$$
Proof: (The proof is left as an exercise for the interested reader.)
Here are my:
QUESTIONS: Could it then be proved that
$$\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))?$$
If a(n) (alternative) proof for this last equation is not possible, could you explain why?
Best Answer
On OP's request, I am converting my comment into an answer.
FYI, you idea suggests that if $a$ is such that $1=\gcd(q^k,n^a)$, then $$\gcd(n^2,\sigma(n^2))=\gcd(n^{a+2},\sigma(n^2))$$ holds.
Since $a$ cannot be $−1$, I think that another idea is needed to deal with $\gcd(n,\sigma(n^2))$.