The short answer isLet NO, there are no more possible conditions under which $$\gcd(n,\sigma(n^2)) = \gcd(n^2,\sigma(n^2)),$$ other than$p^s Q^2$ be an odd perfect number with special prime $n = \gcd(n^2,\sigma(n^2))$$p$ satisfying $p \equiv s \equiv 1 \pmod 4$ and $\gcd(p,Q)=1$. (This follows from an unconditional proof for
I did some more digging on when the conditionequations
$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$ $$n = \gcd(n,\sigma(n^2)).$$$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$ See below$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$
simultaneously hold.) Note that we have the identity
$$\gcd(\sigma(Q^2), \sigma(p^s)) \gcd(Q^2, \sigma(Q^2)) = \left(\gcd(Q, \sigma(Q^2))\right)^2.$$
Proof:Hence, when exactly one of the three equations above holds, then the other two equations follow.
Consider $\gcd(q^k n, 1)=1$. This equation trivially holds.In particular, note that $$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$
Let $$i(q):=\gcd(n^2,\sigma(n^2)) = \dfrac{\sigma(n^2)}{q^k}.$$is equivalent to $$\frac{Q^2}{\sigma(p^s)/2} = \frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2}$$
We obtain $$1 = \gcd(q^k n, 1) = \dfrac{\gcd\bigl(n\sigma(n^2),i(q)\bigr)}{i(q)} = \dfrac{\gcd\bigl(n\sigma(n^2),\gcd(n^2,\sigma(n^2))\bigr)}{i(q)} = \dfrac{\gcd\bigl(\gcd(n\sigma(n^2),n^2),\sigma(n^2)\bigr)}{i(q)} = \dfrac{\gcd\bigl(n\gcd(\sigma(n^2),n),\sigma(n^2)\bigr)}{i(q)}.$$which, in turn, is equivalent to
We consider two cases:$$Q = \gcd(\sigma(p^s)/2, Q).$$
Case 1: This last GCD equation holds if and only if $\gcd(n,\sigma(n^2))=1$$Q \mid \sigma(p^s)/2$.
If $\gcd(n,\sigma(n^2))=1$Furthermore, then since it is known that $n^2 \nmid \sigma(n^2)$ and $\sigma(n^2) \nmid n$in particular, so that we have $$1 = \dfrac{\gcd\bigl(n\gcd(\sigma(n^2),n),\sigma(n^2)\bigr)}{i(q)}.$$ But we knownote that $$3 \leq \gcd(n^2,\sigma(n^2)) = \gcd(n,\sigma(n^2)) = 1,$$ which$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$
is a contradiction.equivalent to
Case 2: $\gcd(n,\sigma(n^2))=n$$$\left(\frac{Q}{\sigma(p^s)/2)}\right)\cdot\gcd(\sigma(p^s)/2, Q) = \frac{Q^2}{\sigma(p^s)/2}$$
We now prove that, indeedwhich, $\gcd(n,\sigma(n^2)) = n$ holds in general.turn, is equivalent to
Based from the results in this MSE question titled "On odd perfect numbers and a$$\gcd(\sigma(p^s)/2, Q) = Q.$$
This last GCD equation holds if and only if - Part V", we have the following proposition:$Q \mid \sigma(p^s)/2$.
THEOREM: If $m = q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(n^2,\sigma(n^2))=\gcd(n^i,\sigma(n^2))$$ holds, where $i \geq 2$ is an integer.
From Case 1 aboveLastly, we have the equationin particular, note that $$1 = \dfrac{\gcd\bigl(n\gcd(\sigma(n^2),n),\sigma(n^2)\bigr)}{i(q)}.$$$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$
This means that $$\gcd(I\cdot{n},\sigma(n^2)) = \gcd(n^2,\sigma(n^2)),$$is equivalent to where $I = \gcd(n,\sigma(n^2))$.$$\frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2} = \left(\frac{Q}{\sigma(p^s)/2}\right)\cdot\gcd(\sigma(p^s)/2, Q)$$
By the THEOREM abovewhich, we know that $I = n^j$ for some integerin turn, is equivalent to
$$\gcd(\sigma(p^s)/2, Q) = Q.$$
This last GCD equation holds if and only if $j \geq 1$$Q \mid \sigma(p^s)/2$.
HoweverThus, if we know from the results of this preprint thatset $$I = \frac{n}{\sigma(q^k)/2}\cdot\gcd(\sigma(q^k)/2,n).$$$$G = \gcd(\sigma(Q^2), \sigma(p^s))$$ $$H = \gcd(Q^2, \sigma(Q^2))$$ $$I = \gcd(Q, \sigma(Q^2))$$
We therefore have $$n^{j-1} = \frac{\gcd(\sigma(q^k)/2,n)}{\sigma(q^k)/2}.$$then we get the biconditional
$$G = H = I \iff Q \mid \sigma(p^s)/2.$$
SinceOf course, as a sanity check, when $j \geq 1$ is an integer$\sigma(p^s) = 2Q$, then we obtain the RHS is also an integer, so that $$\sigma(q^k)/2 \mid \gcd(\sigma(q^k)/2,n).$$conjunction
But we know that $$\gcd(\sigma(q^k)/2,n) \mid \sigma(q^k)/2$$ by the definition of GCD.$$Q \mid \sigma(p^s)/2$$
Hence, we do in fact have $$\gcd(\sigma(q^k)/2,n) = \sigma(q^k)/2$$ fromand
$$\sigma(p^s)/2 \mid Q,$$
which it follows thatby $j=1$.Conjunction Elimination yields
Consequently, $\gcd(n,\sigma(n^2))=n$$$Q \mid \sigma(p^s)/2$$
and hence, which is equivalent to $\sigma(q^k)/2 \mid n$.that
$$G = H = I.$$
