On odd perfect numbers and a GCD - Part III

Question

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 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 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?

Answer 1

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.

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

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

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

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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 yields

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

and hence, that

$$G = H = I.$$