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


It is [known](https://mathoverflow.net/questions/280897) that
$$\gcd(\sigma(q^k),\sigma(n^2)) = \frac{(\gcd(n,\sigma(n^2)))^2}{\gcd(n^2,\sigma(n^2))}$$
and therefore that
$$\gcd(\sigma(q^k),\sigma(n^2)) = \gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} \geq 3$$
if and only if $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$.


Lastly, it is [known](https://math.stackexchange.com/a/4249583/28816) that
$$\gcd(\sigma(q^k),\sigma(n^2))=1$$
implies $k=1$.


Here is my:



> 
> **QUESTION:** Under what conditions is it true that
> $$\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))?$$
> 
> 
> 


I know that the **GCD** equation is true, for example, when $\sigma(n^2) = q^k n$ (and therefore, $\sigma(q^k) = 2n$). Are there other conditions under which the **GCD** equation is true?



On odd perfect numbers and a GCD - Part III
Let $p^s Q^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv s \equiv 1 \pmod 4$ and $\gcd(p,Q)=1$.


I did some more digging on when the equations


$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$
$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$
$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$


simultaneously hold. Note that we have the identity


$$\gcd(\sigma(Q^2), \sigma(p^s)) \gcd(Q^2, \sigma(Q^2)) = \left(\gcd(Q, \sigma(Q^2))\right)^2.$$


**Hence, when exactly one of the three equations above holds, then the other two equations follow.**




---


In particular, note that
$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$


is equivalent to
$$\frac{Q^2}{\sigma(p^s)/2} = \frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2}$$


which, in turn, is equivalent to


$$Q = \gcd(\sigma(p^s)/2, Q).$$


This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.




---


Furthermore, in particular, note that
$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$


is equivalent to


$$\left(\frac{Q}{\sigma(p^s)/2)}\right)\cdot\gcd(\sigma(p^s)/2, Q) = \frac{Q^2}{\sigma(p^s)/2}$$


which, in turn, is equivalent to


$$\gcd(\sigma(p^s)/2, Q) = Q.$$


This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.




---


Lastly, in particular, note that
$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$


is equivalent to
$$\frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2} = \left(\frac{Q}{\sigma(p^s)/2}\right)\cdot\gcd(\sigma(p^s)/2, Q)$$


which, in turn, is equivalent to


$$\gcd(\sigma(p^s)/2, Q) = Q.$$


This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.




---


Thus, if we set
$$G = \gcd(\sigma(Q^2), \sigma(p^s))$$
$$H = \gcd(Q^2, \sigma(Q^2))$$
$$I = \gcd(Q, \sigma(Q^2))$$


then we get the biconditional



> 
> $$G = H = I \iff Q \mid \sigma(p^s)/2.$$
> 
> 
> 


Of course, as a sanity check, when $\sigma(p^s) = 2Q$, then we obtain the conjunction


$$Q \mid \sigma(p^s)/2$$


and


$$\sigma(p^s)/2 \mid Q,$$


which by [Conjunction Elimination](https://en.wikipedia.org/wiki/Conjunction_elimination) yields


$$Q \mid \sigma(p^s)/2$$


and hence, that


$$G = H = I.$$


