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?
Best Answer
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
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
$$G = H = I.$$