# On odd perfect numbers and a GCD

(Note: This question is closely related to this other one in MSE.)

Let $$N = q^k n^2$$ be an odd perfect number.

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 $$5$$ by a result of Dris and Luca.

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 $$G:=\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg({n^2}/{i(q)}, i(q)\bigg)$$.

(Edited Sept 12 2017) Here are my questions:

Original Question

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 $$G = i(q)$$?

Some authors have already considered the possibility that $$i(q)$$ may be a square. This would imply that $$G$$ is also a square. Would $$G$$ a square mean that $$G = i(q)$$ holds?

(End Edit)

Thank you.

• $G = i(q)$ if $i(q) \mid n$. Usually, this is the case when $i(q)$ is prime (since $i(q) \mid n^2$ must hold). Alas, the statement $i(q)$ is prime has been disproved in Chen and Chen (2012). Commented Sep 12, 2017 at 0:51
• If the downvoter could kindly explain his/her vote, I would consider it as constructive feedback. As it is, I am totally clueless. Commented Sep 13, 2017 at 14:29
• Commented Feb 9, 2020 at 15:06

It turns out that $$G \text{ is a square } \iff i(q) \text{ is a square.}$$

The proof is essentially contained in this answer to a closely related MSE question.

Thus, we have the implication $$G \text{ is a square } \implies k=1$$ by a result of Broughan, Delbourgo, and Zhou (Improving the Chen and Chen Result for Odd Perfect Numbers).

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 OP, 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 $$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(\sigma(n^2), n).$$

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 $$\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2)) = i(q).$$