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

Here is my:

QUESTION

If $q^k n^2$ is an odd perfect number with special prime $q$, is $\sigma(q^k)$ coprime to $\sigma(n^2)$?

The function $\sigma(x)=\sigma_1(x)$ is called the sum of divisors of $x$.

MY ATTEMPT FOR SOME SPECIFIC VALUES OF $q$, $k$, AND $n$

For example, there may be an odd perfect number with $q^k = 17$ and $(n/3)^2$ coprime to $3$. Then $\sigma(n^2) = 17(n^2/9)$ is coprime to $\sigma(17) = 18$.

How about the general case? My hunch is that the following conjecture ought to hold:

Conjecture: If $q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(\sigma(q^k),\sigma(n^2))>1.$$

Initially, I thought that a proof of this "Conjecture" was in the following paper by Dandapat et al., but after an in-depth reading, it appears that I was mistaken.

I have therefore tagged this as a reference-request for a proof of this Conjecture.

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2 Answers 2

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This post shows that $\gcd(\sigma(q^k),\sigma(n^2))=1$ would lead to $k=1$.

Let $N=q^k n^2$ be an odd perfect number with $q$ being the special prime, $q \nmid n$ and $k \equiv 1 \pmod 4$. Then from $\sigma(N)=2N$ and multiplicativity of $\sigma$ we get $\sigma(q^k)\sigma(n^2)=2q^kn^2$. Now clearly $q^k$ is coprime with $\sigma(q^k)=1+q+\dots+q^k$, hence $q^k \mid \sigma(n^2)$. We also have $2 \mid \sigma(q^k)$ because $q,k$ are odd, thus we can write \begin{equation} \left[\frac{\sigma(q^k)}{2}\right]\cdot \left[\frac{\sigma(n^2)}{q^k}\right]=n^2. \end{equation} Now assuming $\sigma(q^k)$ and $\sigma(n^2)$ are coprime, then also the two integers on the left side are coprime. That means both are perfect squares, and so in particular $$ 2m^2=\sigma(q^k)=1+q+\dots+q^k $$ for some integer $m$. However this equation has no solution in integers for $q,k \geq 2$, so if there is an odd perfect number with this property, we must have $k=1$ (since $q \geq 5$ anyway).

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  • $\begingroup$ Thank you for very much for your time and attention, @Sil! In the paper titled Improving the Chen and Chen result for odd perfect numbers (Lemma 8, page 7), Broughan et al. show that if $$\frac{\sigma(n^2)}{q^k}$$ is a square, then $k=1$. I think my question now harkens back as to whether Dandapat et al. did indeed have a proof for $\gcd(\sigma(q^k),\sigma(n^2)) > 1$, as your proof only shows that indeed, $k = 1$ follows from $\gcd(\sigma(q^k),\sigma(n^2))=1$. $\endgroup$ Commented Sep 14, 2021 at 2:39
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    $\begingroup$ @ArnieBebita-Dris You are right, I did not realize that $k=1$ is still in the play, updated my post. $\endgroup$
    – Sil
    Commented Sep 14, 2021 at 6:34
  • $\begingroup$ You might be interested to peruse this new MO question of mine, which is related to the present post, @Sil. $\endgroup$ Commented Sep 14, 2021 at 8:33
  • $\begingroup$ You might be interested to peruse this new MSE question of mine, which is related to the present post, @Sil. $\endgroup$ Commented Nov 23, 2021 at 9:51
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Not a complete answer, I merely wanted to record some recent thoughts that occurred to me, that are directly related to this question.


Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.

Suppose to the contrary that $$\gcd(\sigma(q^k),\sigma(n^2))=1.$$

Since $q^k n^2$ is (odd) perfect, we obtain $$\sigma(q^k)\sigma(n^2) = \sigma(q^k n^2) = 2q^k n^2.$$ Applying the $\sigma$-function to both sides of the last equation, we obtain $$\sigma(\sigma(q^k))\sigma(\sigma(n^2)) = \sigma\bigg(\sigma(q^k)\sigma(n^2)\bigg) = \sigma(2 q^k n^2) = \sigma(2)\sigma(q^k)\sigma(n^2) = 3\sigma(q^k)\sigma(n^2)$$ so that $$I(\sigma(q^k))I(\sigma(n^2)) = \frac{\sigma(\sigma(q^k))}{\sigma(q^k)}\frac{\sigma(\sigma(n^2))}{\sigma(n^2)} = 3.$$

Note that, trivially, we have $\sigma(q^k) > 1$ and $\sigma(n^2) > 1$, so that $$I(\sigma(q^k)) > 1$$ and $$I(\sigma(n^2)) > 1$$ both hold. These inequalities imply that $$I(\sigma(q^k)) > 1 \implies I(\sigma(n^2)) < 3$$ and $$I(\sigma(n^2)) > 1 \implies I(\sigma(q^k)) < 3$$ so that we obtain $$1 < I(\sigma(q^k)) < 3$$ and $$1 < I(\sigma(n^2)) < 3.$$

Suppose to the contrary that $$I(\sigma(q^k)) = 2$$ holds. Then we obtain $$I(\sigma(n^2)) = \frac{3}{I(\sigma(q^k))} = \frac{3}{2} = I(2).$$ Since $2$ is solitary, this last equation forces $$\sigma(n^2) = 2,$$ which contradicts the fact that $\sigma(n^2)$ is odd. (It also contradicts $n^2 > {10}^{750}$.)

Suppose to the contrary that $$I(\sigma(n^2)) = 2$$ holds. Then we obtain $$I(\sigma(q^k)) = \frac{3}{I(\sigma(n^2))} = \frac{3}{2} = I(2).$$ Since $2$ is solitary, this last equation forces $$\sigma(q^k) = 2,$$ which contradicts the fact that $$\sigma(q^k) \geq q^k + 1 \geq 6$$ since $q \geq 5$ and $k \geq 1$.

Hence, neither $\sigma(q^k)$ nor $\sigma(n^2)$ could be perfect, if $\gcd(\sigma(q^k),\sigma(n^2))=1$.


Further considerations are in the following questions:

If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(n^2)$ be abundant? MSE question 4223529 Edit: (September 11, 2021 - 1:44 PM Manila time) - Finally proved that $\sigma(n^2)$ must be deficient.

If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(q^k)$ be deficient? MSE question 3831043


Added (August 14, 2021 - 15:49 PM Manila time): Here we will be proving the following claim:

CLAIM: Let $q^k n^2$ be an odd perfect number with special prime $q$, satisfying $\gcd(\sigma(q^k),\sigma(n^2))=1$.

  • $$\Bigg(\sigma(n^2) \text{ is abundant}\Bigg) \implies \Bigg(I(\sigma(q^k)) < I(\sigma(n^2))\Bigg) \implies \Bigg(\sigma(q^k) \text{ is deficient}\Bigg)$$
  • $$\Bigg(\sigma(q^k) \text{ is abundant}\Bigg) \implies \Bigg(I(\sigma(n^2)) < I(\sigma(q^k))\Bigg) \implies \Bigg(\sigma(n^2) \text{ is deficient}\Bigg).$$

Proof: Recall that both $\sigma(q^k)$ and $\sigma(n^2)$ are not perfect, if $\gcd(\sigma(q^k),\sigma(n^2)) = 1$.

Suppose to the contrary that $I(\sigma(q^k)) = I(\sigma(n^2))$.

Then this implies that $$(I(\sigma(q^k)))^2 = I(\sigma(q^k))I(\sigma(n^2)) = 3,$$ which implies $I(\sigma(q^k)) = \sqrt{3}$. This contradicts the fact that $I(\sigma(q^k))$ is rational.

Hence, we know that either $I(\sigma(q^k)) < I(\sigma(n^2))$, or $I(\sigma(n^2)) < I(\sigma(q^k))$, must hold.

Suppose that $I(\sigma(q^k)) < I(\sigma(n^2))$.

This implies that $(I(\sigma(q^k)))^2 < I(\sigma(q^k))I(\sigma(n^2)) = 3$, from which it follows that $I(\sigma(q^k)) < \sqrt{3} < 2$. Hence, $\sigma(q^k)$ is deficient, if $I(\sigma(q^k)) < I(\sigma(n^2))$. By the contrapositive, since $\sigma(q^k)$ is not perfect if $\gcd(\sigma(q^k),\sigma(n^2))=1$, then if $\sigma(q^k)$ is abundant, we have that $I(\sigma(n^2)) < I(\sigma(q^k))$.

Suppose that $I(\sigma(n^2)) < I(\sigma(q^k))$.

This implies that $(I(\sigma(n^2)))^2 < I(\sigma(q^k))I(\sigma(n^2)) = 3$, from which it follows that $I(\sigma(n^2)) < \sqrt{3} < 2$. Hence, $\sigma(n^2)$ is deficient, if $I(\sigma(n^2)) < I(\sigma(q^k))$. By the contrapositive, since $\sigma(n^2)$ is not perfect if $\gcd(\sigma(q^k),\sigma(n^2))=1$, then if $\sigma(n^2)$ is abundant, we have that $I(\sigma(q^k)) < I(\sigma(n^2))$.

QED.


SANITY CHECK: The Claim proves that $\sigma(q^k)$ and $\sigma(n^2)$ cannot be both abundant, which is true, because otherwise $$4 = 2^2 < I(\sigma(q^k))I(\sigma(n^2)) = 3,$$ which is a contradiction.

Hence, we conclude that if $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $\gcd(\sigma(q^k),\sigma(n^2))=1$, then either one of the following cases hold:

  • $\sigma(q^k)$ is deficient and $\sigma(n^2)$ is deficient
  • $\sigma(q^k)$ is abundant and $\sigma(n^2)$ is deficient
  • $\sigma(q^k)$ is deficient and $\sigma(n^2)$ is abundant - This is ruled out by the proof for a deficient $\sigma(n^2)$, as mentioned below.

Alas, this is where I get stuck!


UPDATED ON September 11, 2021 - 2:01 PM (Manila time) In this answer to a closely related MSE question, it is proved that, unconditionally, $\sigma(n^2)$ must be deficient.

This implies that $I(\sigma(n^2)) < 2$.

Since $\gcd(\sigma(q^k),\sigma(n^2))=1$ by assumption, then we have $$I(\sigma(q^k))I(\sigma(n^2))=3$$ which, together with the upper bound $I(\sigma(n^2)) < 2$, implies that $$I(\sigma(q^k))I(\sigma(n^2))=3<2I(\sigma(q^k)).$$ Finally, we obtain $$I(\sigma(q^k)) > \frac{3}{2}.$$

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