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

Define the GCDs: $$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$$ $$I = \gcd\bigg(n,\sigma(n^2)\bigg).$$

In this preprint, Dris shows that the identity $$G \times H = I^2$$ holds.

The proof of the divisibility constraint $I \mid H$ trivially follows from the GCD property $$\bigg(a \mid b\bigg) \implies \bigg(\gcd(a,c) \mid \gcd(b,c)\bigg).$$

It then follows that $$J = \frac{I}{G} = \frac{H}{I} = \sqrt{\frac{H}{G}}$$ must be an (odd) integer.

In an answer to this question, the anonymous MSE user mathlove shows that the value of $J$ may be given as $$J = \frac{n}{\gcd\bigg(\sigma(q^k)/2,n\bigg)}.$$

Thus, we get $$I = \gcd\bigg(n,\sigma(n^2)\bigg) = \bigg(\frac{n}{\sigma(q^k)/2}\bigg)\cdot\gcd\bigg(\sigma(q^k)/2,n\bigg), \tag{*}$$ since it is known that $$H = \gcd\bigg(n^2,\sigma(n^2)\bigg) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$

Here is my:

QUESTION: Can you show that Equation (*) unconditionally implies $\sigma(q^k)/2 \mid n$ (since $I$ must be an integer)?

Jose Arnaldo Bebita Dris
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  • Since $\sigma(q^k)/2 \mid n^2$ holds in general, then $\sigma(q^k)/2 \mid n$ is true if $\sigma(q^k)/2$ is squarefree. – Jose Arnaldo Bebita Dris Apr 17 '22 at 06:25
  • Furthermore, note that the divisibility constraint $\sigma(q^k)/2 \mid n$ implies the divisibility condition $G \mid I$, which is true. – Jose Arnaldo Bebita Dris Apr 17 '22 at 07:21
  • Lastly, note that $$\frac{\gcd(n,\sigma(n^2))}{\gcd(n,\sigma(q^k)/2)} = \frac{n}{\sigma(q^k)/2},$$ so that the divisibility condition $\sigma(q^k)/2 \mid \sigma(n^2)$ implies that the divisibility constraint $\sigma(q^k)/2 \mid n$ holds. – Jose Arnaldo Bebita Dris Apr 17 '22 at 09:01
  • What is the evidence that $\frac{\gcd(n,\sigma(n^2))}{\gcd(n,\sigma(q^k)/2)} = \frac{n}{\sigma(q^k)/2}$, holds? – User4576283 Aug 08 '22 at 09:45
  • Isn't there a possibility that this equation could also hold? $\frac{\gcd(n,\sigma(n^2))}{\gcd(n,\sigma(q^k)/2)} = \frac{gcd(n^2, \sigma(n^2)) }{n}$ – User4576283 Aug 08 '22 at 09:48
  • @Mathguru: Thank you for your time and attention. In fact, $$\gcd(n^2,\sigma(n^2)) = \frac{n^2}{\sigma(q^k)/2}$$ holds, so that both $$\frac{\gcd(n,\sigma(n^2))}{\gcd(n,\sigma(q^k)/2)} = \frac{n}{\sigma(q^k)/2}$$ and $$\frac{\gcd(n,\sigma(n^2))}{\gcd(n,\sigma(q^k)/2)} = \frac{\gcd(n^2,\sigma(n^2))}{n}$$ hold. For more details, you are hereby referred to this preprint. – Jose Arnaldo Bebita Dris Aug 08 '22 at 11:17

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