On the conjectured inequality $q > k$, where $q^k n^2$ is an odd perfect number with special prime $q$

Question

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

Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds. (Beasley (2013), page 25 - third paragraph)

I recently came across the following question: Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$.

Note that, specializing to $a = q$ and $n = k$, we obtain $$q^{k+1} + k \geq (k + 1)q$$ $$q^{k+1} - 1 \geq (k + 1)(q - 1)$$ $$\frac{q^{k+1} - 1}{q - 1} \geq k + 1$$ $$k + 1 \leq \sigma(q^k),$$ where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.

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Edit: December 27, 2021 - 1:00 PM - Manila time

Since equality occurs in $$(n+1)a \leq a^{n+1}+n$$ if and only if $a=1$, then equality does not occur in $$k + 1 \leq \sigma(q^k),$$ since $q$ is a prime satisfying $q \equiv 1 \pmod 4$ implies that $q \geq 5$, so that what we actually have is the inequality $$k + 1 < \sigma(q^k).$$

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In particular, note that when $k=1$, we get $$k + 1 < \sigma(q^k) = q + 1,$$ which implies that $q > k$. (Sanity check: If $k=1$, then since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, we have $q \geq 5 > 1 = k$.)

Notice that it is true in general that $$q + 1 \leq \sigma(q^k).$$

Here is my:

QUESTION: Will it be possible to remove the reliance of the proof for the inequality $q > k$ on the condition $k=1$? That is, do we unconditionally have $$k + 1 < q + 1 \leq \sigma(q^k)?$$

MY ATTEMPT

Suppose to the contrary that $q < k$. Then we obtain $$q + 1 < k + 1 < \sigma(q^k)$$ which implies that $k \neq 1$. (More is actually true. A similar reasoning via the starting point $k > q$ proves the following: Since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, and since $k \equiv 1 \pmod 4$, then $k > q$ implies that $k > q \geq 5$, which implies that $k > 5$, from which it follows that $k \geq 9$.)

Alas, this is where I get stuck.

I currently do not see a way to derive a contradiction between the assumption $q < k$ and the known result $k + 1 < \sigma(q^k)$.

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Elsewhere, various authors (Cohen and Sorli, and then Robbins) have ruled out values for $q$ such that $q = k$. I then dare to conjecture that:

CONJECTURE: If $q^k n^2$ is an odd perfect number with special prime $q$, then $q \neq k$.

Alas, I have no proof for this Conjecture.

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REFERENCES

    Sorli, R. M., & Cohen, G. L. (2012). On odd perfect numbers and even 3-perfect numbers, Integers, 12A, Article #A6.

    Robbins, N. (2013). On the Eulerian factor of an odd perfect number, Universal Journal of Mathematics and Mathematical Sciences, 4 (2), 237–239.

Answer 1

Taking cue from mathlove's hint, I can get the following derivation:

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Suppose that $$\sigma(q^k) \geqslant k+5,$$ which we know to be true as obtained by mathlove in the comments.

Set $$f(k):= \sigma(q^k) - k - 5 \geqslant 0.$$

Then, since $$\frac{\partial }{\partial k}\Bigg(\sigma(q^k) - k - 5\Bigg)=\frac{\partial }{\partial k}\Bigg(\sigma(q^k) - k - 1\Bigg) = \frac{q(q^k \ln(q) - 1) + 1}{q - 1} > 0,$$ then we know that $f(k)$ is an increasing function of $k$. In particular, we obtain $$f(k) = \sigma(q^k) - k - 5 \geqslant f(1) = (q + 1) - 6 = q - 5$$ from which we get $$\sigma(q^k) \geqslant q + k,$$ which finally yields $$\sigma(q^k) \geqslant q + k \geqslant k + 5,$$ thus showing that mathlove's bound cannot be significantly improved in this manner.

Answer 2

On OP's request, I am converting my comment into an answer.

FYI, one can get $\sigma(q^k)\geqslant k+5$ which is better than $\sigma(q^k)\gt k+1$ since $$\dfrac{\partial}{\partial k}\bigg(\sigma(q^k)-k-1\bigg)\gt 0$$ and $$\sigma(q^k)-k-1\geqslant (q+1)-2\geqslant 4.$$

Answer 3

Comment converted into an answer

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A better bound for $\sigma(q^k)$ is as follows: $$\sigma(q^k) \geqslant q^k + 1 \geqslant q + k,$$ where the second inequality $$q^k + 1 \geqslant q + k$$ holds for all primes $q$ satisfying $q \equiv 1 \pmod 4$ and all positive integers $k$ such that $k \equiv 1 \pmod 4$.