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(Preamble #1: In what follows, we take $\sigma=\sigma_{1}$ to be the sum of the divisors, and denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.)

(Preamble #2: My sincerest apologies for the somewhat very long post -- I just had to put in all the details into one place for ease of quick reference later.)

PROBLEM

If $q^k n^2$ is an odd perfect number with Euler prime $q$, which of the following relationships between $q^2$ and $n$ hold?

(A) $$q^2 < \sigma(q^2) < n < \sigma(n)$$ (B) $$q^2 < n < \sigma(q^2) < \sigma(n)$$ (C) $$n < \sigma(n) < q^2 < \sigma(q^2)$$ (D) $$n < q^2 < \sigma(n) < \sigma(q^2)$$

PRELIMINARIES

(0) Of course, first of all, note that $q^2 \neq n$ since $\gcd(q,n)=1$.

(1) Note that $$\frac{q^2}{n}+\frac{n}{q^2} < \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2\cdot\bigg(\frac{q^2}{n}+\frac{n}{q^2}\bigg).$$ Consequently, we know that $$\frac{q^2}{n}+\frac{n}{q^2} \text{ is bounded from above } \iff \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} \text{ is bounded from above.}$$ In general, since the function $f(z) := z + (1/z)$ is not bounded from above, this means that we do not expect $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ to be bounded from above.

(2) $n < q^2 \implies k = 1$ [Dris, 2012]

(3) $$1 < I(q^2) \leq \frac{q^2 + q + 1}{q^2} = 1+\frac{1}{q}+\bigg(\frac{1}{q}\bigg)^2 \leq 1+\frac{1}{5}+\bigg(\frac{1}{5}\bigg)^2 = \frac{25+5+1}{5} = \frac{31}{25}$$ since $q$ being the Euler prime implies that $q$ is prime with $q \equiv 1 \pmod 4$, therefore $q \geq 5$.

Now, $1 < I(n) < 2$ since $n > 1$ is deficient, $n$ being a proper factor of the (odd) perfect number $q^k n^2$.

Consequently, $$1 < I({q^2}n) = I(q^2)I(n) = \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} < 2\cdot\frac{31}{25} = \frac{62}{25}.$$

(4) It follows from (1) and (3) that the following hold:

(a) $\sigma(n) \neq \sigma(q^2)$

Proof: Assume that $\sigma(n) = \sigma(q^2)$. Then we have (by (3)) $$I(q^2) = \frac{\sigma(q^2)}{q^2} = \frac{\sigma(n)}{q^2} < \frac{31}{25}$$ and $$I(n) = \frac{\sigma(n)}{n} = \frac{\sigma(q^2)}{n} < 2.$$ It follows that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2+\frac{31}{25}=\frac{81}{25}.$$ This implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).

(b) $\sigma(q^2) \neq n$.

Suppose to the contrary that $\sigma(q^2)=n$. Then we have (from (3)) $$1\cdot\frac{\sigma(n)}{q^2} = \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} < \frac{62}{25}.$$ It follows that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}=1+\frac{\sigma(n)}{q^2}<1+\frac{62}{25}=\frac{87}{25}.$$ This implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).

The proof for the following is similar to that of (b):

(c) $\sigma(n) \neq q^2$

MAIN RESULTS

The proofs of the succeeding three lemmas are trivial.

Lemma 1. The inequality $$I(q^2) + I(n) < \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is true if and only if the biconditional $$q^2 < n \iff \sigma(q^2) < \sigma(n)$$ is true.

Lemma 2. The inequality $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < I(q^2) + I(n)$$ is true if and only if the biconditional $$q^2 < n \iff \sigma(n) < \sigma(q^2)$$ is true.

Lemma 3. The equation $$I(q^2) + I(n) = \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is true if and only if either $q^2 = n$ or $\sigma(q^2)=\sigma(n)$ hold.

We now prove our first main result:

Theorem 1. If $q^k n^2$ is an odd perfect number, then the inequality $$I(q^2) + I(n) < \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ holds.

Proof: Suppose to the contrary that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} \leq I(q^2)+I(n).$$

Since $I(q^2) \leq 31/25$ and $I(n) < 2$, this implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2 + \frac{31}{25} = \frac{81}{25}$$ so that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).

By using Lemma 1, we are able to obtain the following Corollary to Theorem 1.

Corollary 2. If $q^k n^2$ is an odd perfect number, then the biconditionals $$q^2 < n \iff \sigma(q^2) < \sigma(n) \iff \frac{\sigma(q^2)}{n} < \frac{\sigma(n)}{q^2}.$$

Proof: Trivial.

Note that Corollary 2 proves that the list of inequalities given in the Problem Statement exhausts all possible cases.

We now claim that:

Theorem 3. If $q^k n^2$ is an odd perfect number, then $q^2 < \sigma(n)$ and $n < \sigma(q^2)$ cannot be both true.

Proof:

Suppose that both $$1 < \frac{\sigma(q^2)}{n}$$ and $$1 < \frac{\sigma(n)}{q^2}$$ hold. Then it follows that both $$\frac{\sigma(n)}{q^2} < \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} = I({q^2}n) < \frac{62}{25}$$ and $$\frac{\sigma(q^2)}{n} < \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} = I({q^2}n) < \frac{62}{25}$$ hold. This means that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2\cdot\frac{62}{25} = \frac{124}{25}$$ which implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).

Note that Theorem 3 immediately rules out Case (B) and Case (D) above.

INQUIRIES

We are left to consider the following two (2) remaining cases:

(A) $$q^2 < \sigma(q^2) < n < \sigma(n)$$ (C) $$n < \sigma(n) < q^2 < \sigma(q^2)$$

Question #1 By this answer, we know (?) that the implication $n < q^{k+1} \implies k \neq 1$ is true if and only if $q^2 < n$. Can this result be improved?

Question #2 Following this answer, it is conjectured that $k \neq 1$. This would follow if we could rule out $q = q^k < \sigma(q) = \sigma(q^k) < n < \sigma(n)$. (That is, if we could prove that $\sigma(q^k) < n$ is false.) To what extent can a proof along this thread of thought be pursued, given Brown's arguments for a partial proof of $q^k < n$ in A Partial Proof of a Conjecture of Dris and the considerations in this MSE post?

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This post (and other related posts) seem to prove the following assertion:

CONJECTURE If there is an odd perfect number, then there are infinitely many of them.

Proof: Suppose that there are only a finite number of odd perfect numbers.

This means that there exists an effectively computable constant $\mathcal{C}$ such that $$q^k n^2 < \mathcal{C}.$$

Since $q^k < n^2$ [Dris, 2012], then it follows that $$q^{2k} < {q^k}{n^2} < \mathcal{C}.$$ This implies that $$q^2 \leq q^{2k} < \mathcal{C},$$ so that $$q < \mathcal{C}^{1/2}$$ which means that the Euler prime is bounded from above.

Note that we also have $$n^2 < {q^k}{n^2} < \mathcal{C}$$ so that $n < \mathcal{C}^{1/2}$, that is to say, the square root of the non-Euler part (which is $n$) is bounded from above.

Going back to the considerations in this post, we either have $q^2 < n$ or $n < q^2$. (In both cases, note that we have $q < n$ by [Brown, 2016], [Starni, 2017], and [Dris, 2017].)

In the first case, we have $$q^2 < n < \mathcal{C}^{1/2}$$ so that $$\frac{q^2}{n}+\frac{n}{q^2} < 1 + \frac{\mathcal{C}^{1/2}}{25}.$$ This contradicts (1) in the question.

In the second case, we have $$q < n < q^2$$ so that $n > {10}^{500}$, and $$\frac{q^2}{n}+\frac{n}{q^2} < 1 + \frac{\mathcal{C}}{{10}^{500}}.$$ Again, this contradicts (1) in the question, and the CONJECTURE in this answer is proved.

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