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12.9.17

On computing $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ when $q^k n^2$ is an odd perfect number with Euler prime $q$

In this blog post, we compute
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right)$$
when $N = q^k n^2$ is an odd perfect number with Euler prime $q$.

From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$

In particular, we know that the index $i(q)$ is an integer greater than $5$ [Dris, Luca (2016)] (a copy is available in arXiv).

We now attempt to compute an expression for $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ in terms of $i(q)$.

First, since we have 
$$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$
we obtain
$$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$
and
$$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$
so that we get
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$

Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) =  1$ and $i(q)$ is odd, we get
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

Hence, we conclude that
$$G := \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

I posed the following question in MathOverflow:

I seem to recall that somebody (was it Pomerance [?] et. al) proved that
$$G \neq 1.$$
Does anybody here happen to know a reference?  Additionally, does $G \neq 1$ imply that $G = i(q)$?



UPDATED (August 1, 2020 -- 5:41 PM Manila time)

Here is a conditional proof that
$$G = \gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

As derived in the original post, we have
$$G = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

This is equivalent to
$$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$

But we also have
$$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(n, \sigma(n^2)).$$

Consequently, we obtain
$$G = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$

In particular, we get
$$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2))$$
if and only if
$$\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2)),$$
which would imply that
$$n = \frac{\sigma(q^k)}{2}.$$
(See this closely related MSE question for the details.)