On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$

Question

Note: This question was cross-posted from MO.

Preamble: I apologize in advance if this particular MSE post would appear to be a bit of a polymath approach, I just had to put down all the details to present my argument for this particular math problem.

A positive integer $N$ is said to be perfect if $\sigma(N)=2N$, where $\sigma(x)$ is the sum of the divisors of $x$.

An odd perfect number $N$ is said to be given in Eulerian form if $N = {q^k}{n^2}$, where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $\gcd(q,n)=1$ and prime powers are deficient, we have $q \neq n$ and $q^k \neq n$.

In an earlier version of the paper The Abundancy Index of Divisors of Odd Perfect Numbers (see here), it was conjectured that the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

is true. (Note that the proof of the inequation

$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k}$$

is trivial.)

Recently, an attempt to prove the said biconditional appears to have been completed in this preprint.

I present here the highlights of the said proof:

One direction of the biconditional is trivial:

<p>$$q^k &lt; n \Longrightarrow \frac{\sigma(q^k)}{n} &lt; \frac{\sigma(n)}{q^k}.$$</p>

<p>This is proved by noting that $I(q^k) &lt; \sqrt[3]{2} &lt; I(n)$ (where $I(x) = \sigma(x)/x$ is the <em>abundancy index</em> of $x$), from which the following chain of implications follow:</p>

<p>$$q^k &lt; n \Longrightarrow \sigma(q^k) &lt; \sigma(n)$$
$$q^k &lt; n \Longrightarrow \frac{1}{n} &lt; \frac{1}{q^k}$$
$$\{\sigma(q^k) &lt; \sigma(n)\} \land \{\frac{1}{n} &lt; \frac{1}{q^k}\} \Longrightarrow \frac{\sigma(q^k)}{n} &lt; \frac{\sigma(n)}{q^k}$$</p>

<p>Therefore:</p>

<p>$$q^k &lt; n \Longrightarrow \sigma(q^k) &lt; \sigma(n) \Longrightarrow \frac{\sigma(q^k)}{n} &lt; \frac{\sigma(n)}{q^k}.$$</p>

For the other direction:

The implication

<p>$$\frac{\sigma(q^k)}{n} &lt; \frac{\sigma(n)}{q^k} \Longrightarrow \sigma(q^k) &lt; \sigma(n)$$
can be proved, again by observing that $I(q^k) &lt; \sqrt[3]{2} &lt; I(n)$.</p>

<p>Now, to prove the last implication:</p>

<p>$$\sigma(q^k) &lt; \sigma(n) \Longrightarrow q^k &lt; n$$</p>

<p>we take an indirect approach.</p>

First we show that:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form and

<p>$$I(q^k) + I(n) &lt; \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k},$$</p>

<p>then $q^k &lt; n \Longleftrightarrow \sigma(q^k) &lt; \sigma(n)$.</p>

Similarly, we can prove that:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form and

<p>$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} &lt; I(q^k) + I(n),$$</p>

<p>then $q^k &lt; n \Longleftrightarrow \sigma(n) &lt; \sigma(q^k)$.</p>

Observe that

$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

if and only if

$$\sigma(q^k) = \sigma(n).$$

Also, observe that if we assume

$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n)$$

then the biconditional

$$q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k)$$

will contradict $I(q^k) < \sqrt[3]{2} < I(n)$.

Therefore, the following inequality must be true:

<p>$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}.$$</p>

It suffices to consider the case when

$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

which is true if and only if

$$\sigma(q^k) = \sigma(n).$$

This last equation, together with the inequality $I(q^k) < \sqrt[3]{2} < I(n)$, implies that

$$1 = \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}$$

from which it follows that $n < q^k$. Thus, $1/q^k < 1/n$, which then gives, together with the equation $\sigma(n) = \sigma(q^k)$, the inequality

$$\frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}.$$

But this last inequality, together with $I(q^k) < \sqrt[3]{2} < I(n)$, is known to imply

$$n < q^k.$$

Consequently, if

<p>$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$</p>

<p>then</p>

<p>$$n &lt; q^k \Longleftrightarrow \frac{\sigma(n)}{q^k} &lt; \frac{\sigma(q^k)}{n}$$ </p>

Now, here is the part where I am a bit unsure about logical solidity:

Lastly, note that since

$$\sigma(q^k) = \sigma(n)$$

implies $n < q^k$, the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n)$$

is vacuously true, under this case.

NOW HERE IS MY QUESTION (and this is also the main reason for this MSE post):

<p>I have been told that there is a <em>gap</em> in the last part of this proof.  As I
myself am having a hard time spotting where that particular <em>error</em> is, would
somebody be kind enough as to help by skimming through this argument and then 
(after glossing over the details) give it either a <em>PASS</em> or a <em>FAIL</em>?</p>

Answer 1

Update (January 30, 2015)

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.

While $q^k < n$ does imply $\sigma(q^k) < \sigma(n)$ (via $I(q^k) < I(n)$), $\sigma(q^k) < \sigma(n)$ by itself DOES NOT imply $$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$ Thus, the implication $$\sigma(q^k) < \sigma(n) \implies \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ remains unproven.

Similarly, while $$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ does imply $\sigma(q^k) < \sigma(n)$ (via $I(q^k) < I(n)$), $\sigma(q^k) < \sigma(n)$ by itself DOES NOT imply $q^k < n$.

Thus, the implication $\sigma(q^k) < \sigma(n) \implies q^k < n$ remains unproven.

Further Update (July 14, 2016)

Brown has released a preprint titled A Partial Proof of a Conjecture of Dris where he announced a proof for the inequality $q < n$, as well as a partial proof that $q^k < n$ holds "in many cases".

If the inequality $q^k < n$ is proved, it will follow that the biconditionals $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$ are also true.