On odd perfect numbers and a GCD - Part V

(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?

Solution 1:

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 $\gcd(n,\sigma(n^2))$.