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

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.

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 $$$$\left[\frac{\sigma(q^k)}{2}\right]\cdot \left[\frac{\sigma(n^2)}{q^k}\right]=n^2.$$$$ Now assuming $$\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).

• 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 $\gcd(\sigma(q^k),\sigma(n^2))=1$. Commented Sep 14, 2021 at 2:39
• @ArnieBebita-Dris You are right, I did not realize that $k=1$ is still in the play, updated my post.
– Sil
Commented Sep 14, 2021 at 6:34
• You might be interested to peruse this new MO question of mine, which is related to the present post, @Sil. Commented Sep 14, 2021 at 8:33
• You might be interested to peruse this new MSE question of mine, which is related to the present post, @Sil. Commented Nov 23, 2021 at 9:51

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 $$I(\sigma(q^k)) > \frac{3}{2}.$$