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This question is related to this post1 and this post2. Hereinafter, we will let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$, and $\gcd(y,z)$ will denote the greatest common divisor of the (positive) integers $y$ and $z$.

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

From the hyperlinked posts, we have $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))}=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\sigma(n^2)/q^k}=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot{\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{n^2}}.$$

Since the LHS is an integer, the RHS is also an integer. But $\sigma(q^k)/2$ is an integer. Therefore, $$\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{n^2} = \frac{\gcd\bigg(n^2,(\sigma(n^2))^2\bigg)}{n^2} = \gcd\Bigg(1, \bigg(\frac{\sigma(n^2)}{n}\bigg)^2\Bigg)$$ is an integer, which is just equal to $1$, and this further means that $n \mid \sigma(n^2)$.

Hence, we have the following proposition:

THEOREM A: If $q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}.$$

In particular, this Theorem implies the following Corollary:

COROLLARY: If $q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(\sigma(q^k),\sigma(n^2)) \geq 3.$$

Here are my:

QUESTIONS: Is this argument logically sound? If it is not, how can it be mended so as to produce a valid proof?

EDIT: (September 26, 2021 - 2:16 PM Manila time) From the comments, I realized that what I actually have is the following modification to THEOREM A (note the additional divisibility condition on $\sigma(n^2)$):

THEOREM B: If $q^k n^2$ is an odd perfect number with special prime $q$, satisfying $n \mid \sigma(n^2)$, then $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}.$$

Jose Arnaldo Bebita Dris
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  • I don't see why you can say that $\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer from $\gcd(\sigma(q^k),\sigma(n^2))=\dfrac{\sigma(q^k)}{2}\cdot{\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}}$. Something like $75=\dfrac{54}{2}\cdot\dfrac{5^2}{3^2}$ might happen where $\dfrac{5^2}{3^2}$ is not an integer. I think that to prove that $\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer, you have to prove that $\dfrac{\sigma(q^k)}{2}$ divides $\gcd(\sigma(q^k),\sigma(n^2))$. – mathlove Sep 25 '21 at 17:45
  • That is exactly my point, @mathlove! We do have $$\frac{\sigma(q^k)}{2} \mid \sigma(q^k) \mid \gcd(\sigma(q^k),\sigma(n^2)).$$ (The first divisibility constraint makes use of $q \equiv k \equiv 1 \pmod 4$, so that $\sigma(q^k)/2$ is an integer, and at the same time, $A/2 \mid A$, where $A=\sigma(q^k)$. The second divisibility constraint just makes use of the definition of GCD, so that $A$ divides $\gcd(A, B)$, where $B=\sigma(n^2)$.) – Jose Arnaldo Bebita Dris Sep 25 '21 at 23:22
  • $A$ does not always divide $\gcd(A,B)$. For example, take $A=7,B=2$ for which $\gcd(A,B)=1$ holds. – mathlove Sep 26 '21 at 04:52
  • Yes, I realized my mistake while scouring the web for the appropriate GCD property to back my claim, for which I found none, @mathlove. If you could write out your last two comments as an actual answer, then I will upvote and then accept. Thanks! =) – Jose Arnaldo Bebita Dris Sep 26 '21 at 06:15
  • Do you have a proof for your assertion that "to prove that $\frac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer, then one has to prove that $\frac{\sigma(q^k)}{2}$ divides $\gcd(\sigma(q^k),\sigma(n^2))$"? Can you please add that proof to your answer below, @mathlove? Thanks a lot! =) – Jose Arnaldo Bebita Dris Sep 29 '21 at 09:47
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    You have $\dfrac{A}{B}=C$ where $A=\gcd(\sigma(q^k),\sigma(n^2)),B=\dfrac{\sigma(q^k)}{2}$ and $C=\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$. So, $C$ is an integer $\iff\dfrac AB$ is an integer $\iff B\mid A$. – mathlove Sep 29 '21 at 09:59
  • Thank you very much for your time and attention, @mathlove! Now, I understand. =) – Jose Arnaldo Bebita Dris Sep 29 '21 at 14:10

1 Answers1

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I don't see why you can say that $\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer from $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}\cdot\frac{(\gcd(n,\sigma(n^2)))^2}{n^2}.$$ Something like $75=\dfrac{54}{2}\cdot\dfrac{5^2}{3^2}$ might happen where $\dfrac{5^2}{3^2}$ is not an integer. I think that to prove that $\dfrac{(\gcd(n,\sigma(n^2)))^2}{n^2}$ is an integer, you have to prove that $\dfrac{\sigma(q^k)}2$ divides $\gcd(\sigma(q^k),\sigma(n^2))$.

It is true that $\dfrac{\sigma(q^k)}{2}\mid \sigma(q^k)$, but I don't see why you can say that $$\sigma(q^k)\mid \gcd(\sigma(q^k),\sigma(n^2))$$

$A$ does not always divide $\gcd(A,B)$. For example, take $A=7,B=2$ for which $\gcd(A,B)=1$ holds.

mathlove
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