If $N = q^k n^2$ is an odd perfect number and $n < q^{k+1}$, does it follow that $k > 1$?

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form. Let $$I(x) = \dfrac{\sigma(x)}{x}$$ be the abundancy index of $x$.

This is a partial answer to the original question, and proves the claim in the affirmative, subject to the validity of a recent proof claim by Patrick A. Brown that $q^k < n$ holds (in many cases).

First, we show the following lemmas:

Lemma 1. $$I(q^k) + I(n) < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \implies \left(q^k < n \iff \sigma(q^k) < \sigma(n)\right)$$

Proof. $$I(q^k) + I(n) < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \implies \left(n - q^k\right)\left(\sigma(q^k) - \sigma(n)\right) < 0 \implies \left(q^k < n \iff \sigma(q^k) < \sigma(n)\right)$$ QED

Lemma 2. $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < I(q^k) + I(n) \implies \left(q^k < n \iff \sigma(n) < \sigma(q^k)\right)$$

Proof. $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < I(q^k) + I(n) \implies \left(q^k - n\right)\left(\sigma(q^k) - \sigma(n)\right) < 0 \implies \left(n < q^k \iff \sigma(q^k) < \sigma(n)\right)$$ QED

Lemma 3. $$I(q^k) + I(n) = \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \iff \sigma(q^k) = \sigma(n)$$

Proof. $$I(q^k) + I(n) = \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \iff \left(q^k - n\right)\left(\sigma(q^k) - \sigma(n)\right) = 0 \iff \sigma(q^k) = \sigma(n)$$ since $q^k \neq n$.

QED

We are now ready to prove the following proposition:

Proposition 1. $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$

Proof. Since $I(q^k) < \sqrt[3]{2} < I(n)$ [Dris, 2012], then $\sigma(q^k) = \sigma(n) \implies n < q^k$. So it suffices to consider the remaining case under Lemma 2. It thus remains to consider $n < q^k < \sigma(q^k) < \sigma(n).$ But these are already ruled out, assuming Brown's proof for $q^k < n$ is successfully completed.

Therefore, we are left with the case under Lemma 1. It follows that $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}.$$ QED

Next, we prove the following proposition:

Proposition 2. $$k = 1 \implies \sigma(q^k) < n$$

Proof. We use Proposition 1 to list all possible permutations of the set $\left\{q^k, n, \sigma(q^k), \sigma(n)\right\}$.

So we have $$\bf{A}: q^k < n < \sigma(q^k) < \sigma(n)$$ $$\bf{B}: q^k < \sigma(q^k) < n < \sigma(n)$$ $$\bf{C}: n < q^k < \sigma(n) < \sigma(q^k)$$ $$\bf{D}: n < \sigma(n) \leq q^k < \sigma(q^k)$$

(Notice that $\sigma(q^k) \neq n$ since $\sigma(q^k) \equiv k+1 \equiv 2 \pmod 4$ while $n$ is odd.)

Note that Brown's result rules out cases $\bf{C}$ and $\bf{D}$.

Now, $k = 1$ rules out case $\bf{A}$.

Therefore, under the assumption $k = 1$, we are left with case $\bf{B}$.

QED

We are left with proving the following lemma.

"Lemma 4". $$n < \sigma(q^k) \iff n < q^{k+1}$$

"Proof". $n < \sigma(q^k) < q^{k+1}$ since $I(q^k) < \dfrac{5}{4} < 5 \leq q$. Hence, $n < \sigma(q^k) \implies n < q^{k+1}$.

It now remains to show that $n < q^{k+1} \implies n < \sigma(q^k)$.

"QED"

We now get the following "Theorem":

"Theorem". $n < q^{k+1} \implies k \neq 1$

Proof. Take the contrapositive of Proposition 2, and use "Lemma 4".

"QED"

In particular, by noting that $n < q^2 \implies k = 1$ is true, we obtain the following "Corollary":

"Corollary"

$q^2 < n$

Added September 08 2016 I just realized today that we actually have the following biconditional: $$q^2 < n \iff \left\{n<q^{k+1} \implies k>1\right\}$$