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

The index $i(q)$ of $N$ at the prime $q$ is then equal to $$i(q):=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}=\frac{D(n^2)}{s(q^k)}=\frac{s(n^2)}{D(q^k)/2}=\gcd(n^2,\sigma(n^2)),$$ where $D(x):=2x-\sigma(x)$ is the deficiency and $s(x):=\sigma(x)-x$ is the sum of aliquot divisors of $x \in \mathbb{N}$.

In this MSE question, it is shown that

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

Furthermore, in this MSE question, taking $m=n$ and noting that $i(q)=\gcd(n^2,\sigma(n^2))$, essentially it has been proven that $$\gcd\bigg(\gcd(n^2,\sigma(n^2)),\frac{n^2}{\gcd(n^2,\sigma(n^2))}\bigg)=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))},$$ so that $$G := \gcd(\sigma(q^k),\sigma(n^2)) \text{ is a square} \iff i(q) = \gcd(n^2,\sigma(n^2)) \text{ is a square.}$$

Note that $$G = \gcd(\sigma(q^k),\sigma(n^2)) = \gcd\bigg(\sigma(q^k)/2,\sigma(n^2)\bigg)$$ as $\sigma(n^2)$ is odd.

I now wish to prove the following proposition:

PROPOSITION If $q^k n^2$ is an odd perfect number with special prime $q$, then its index at the prime $q$ is not a square.

Proof Assume to the contrary that $$i(q) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}$$ is a square.

It follows that $\sigma(q^k)/2$ and $$G = \gcd(\sigma(q^k),\sigma(n^2)) = \gcd\bigg(\sigma(q^k)/2,\sigma(n^2)\bigg)$$ are both squares. Hence, $\sigma(n^2)$ is also a square, by the GCD property $$\gcd(x^2, y^2) = \bigg(\gcd(x,y)\bigg)^2.$$

But $i(q) = \sigma(n^2)/q^k$ and $\sigma(n^2)$ are both squares would imply that $q^k$ is a square, which clearly contradicts $k \equiv 1 \pmod 4$. (In fact, $i(q)$ is a square implies $k=1$ by a result of Broughan, Delbourgo, and Zhou (Improving the Chen and Chen Result for Odd Perfect Numbers).)

QED

Question

Is this proof logically sound/correct?

Jose Arnaldo Bebita Dris
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    It seems to me that you are claiming that $\sigma(n^2)$ is a square without any proof. That $\sigma(n^2)$ is a square does not follow from $\gcd(x^2, y^2) = (\gcd(x,y))^2$. What you have to prove is that if $\gcd(a^2,b)=c^2$, then $b$ is a square. However, this is not true in general. A counterexample is $(a,b,c)=(6,27,3)$. – mathlove Feb 10 '20 at 09:37
  • Thank you for your comment, @mathlove! I think that qualifies as a response to my original inquiry. I invite you to write that out as an answer, so that I may be able to accept it. – Jose Arnaldo Bebita Dris Feb 10 '20 at 09:44
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    I converted my comment into an answer. – mathlove Feb 10 '20 at 09:48
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    Gladly accepting your answer now, @mathlove! =) – Jose Arnaldo Bebita Dris Feb 10 '20 at 09:51

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It seems to me that you are claiming that $\sigma(n^2)$ is a square without any proof.

That $\sigma(n^2)$ is a square does not follow from $$\gcd(x^2,y^2)=\bigg(\gcd(x,y)\bigg)^2$$

What you have to prove is that if $\gcd(a^2,b)=c^2$, then $b$ is a square.

However, this is not true in general. A counterexample is $(a,b,c)=(6,27,3)$.

mathlove
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