# On Odd Perfect Numbers

Hi all. My question today will be regarding what I consider to be a "stumbling block" while trying to research odd perfect numbers.

Let $N = {q^k}{n^2}$ be an odd perfect number with Euler prime $q$. Since $\gcd(q, n) = 1$, we know that $q \neq n$.

In 2008, I proved that $q^k < n^2$. This implies that, if $n < q$, then Sorli's conjecture that $k = {\nu}_q(N) = 1$ would follow.

I currently know that $I(q) \leq 6/5 < \sqrt{5/3} < I(n)$, where $I(x) = \sigma(x)/x$ is the abundancy index of $x$. In particular, this means that $$\frac{\sigma(q)}{\sigma(n)} < \frac{q}{n}$$.

Thus, if $q < n$, then $\sigma(q) < \sigma(n)$. (The contrapositive of this last implication is $\sigma(n) < \sigma(q)$ implies that $n < q$.)

Now, since $\sigma(q) = q + 1$, I believe we have three cases to consider:

Case 1: $q < \sigma(q) < n < \sigma(n)$

Case 2: $n < q < \sigma(q) \leq \sigma(n)$

Case 3: $n < \sigma(n) \leq q < \sigma(q)$

I also know that $$\frac{\sigma(q)}{n} \neq \frac{\sigma(n)}{q}.$$

My problem is: How do I dispose of Case 2? The motivation is that I want to establish an equivalence between the inequalities $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$ and $q < n$. This way, all it takes to prove Sorli's conjecture will be to show that $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$.

My idea is to show that $$\frac{\sigma(q)}{\sigma(n)} \leq 1 < \frac{q}{n}$$ cannot occur by considering two separate cases under Case 2:

Case 2A: $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$

Case 2B: $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$

I was wondering if anybody out there would have some (better) ideas on how to improve on the bounds for $\frac{n}{q}$ in Case 2A and for $\frac{\sigma(q)}{\sigma(n)}$ in Case 2B. In particular, a sharp upper bound for $I(n) = \sigma(n)/n$ would be nice!

Edit: (June 9, 2012) If $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$, then $n < q$ and $\sigma(n) < \sigma(q)$, as before. (Edit: Since $I(q) < I(n)$, then $q < n$ implies $\sigma(q) < \sigma(n)$. However, while $\sigma(n) < \sigma(q)$ does imply $n < q$, it can happen that $n < q$ AND $\sigma(q) < \sigma(n)$.) Therefore, we have two cases to consider. Under the first case:

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

Consequently, $n < \sigma(n) < q < \sigma(q)$, and we have the bounds:

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

and

$$\sqrt{\frac{5}{3}} < \frac{\sigma(q)}{n}.$$

Under the second case (i.e. $n < q < \sigma(q) \leq \sigma(n)$):

$$1 < \frac{\sigma(q)}{q} \leq \frac{\sigma(n)}{q} < \frac{\sigma(q)}{n} \leq \frac{\sigma(n)}{n} < 2,$$

and

$$1 < \frac{\sigma(n)}{q} < \sqrt{2}$$ $$\sqrt[4]{\frac{5}{3}} < \frac{\sigma(q)}{n} < 2$$

However, if $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$, then we only know that $\sigma(q) < \sigma(n)$ (from before), and no conclusion can be made about comparing $q$ to $n$, as this scenario falls under two cases:

Case 1: $q < \sigma(q) < n < \sigma(n)$

Case 2: $n < q < \sigma(q) < \sigma(n)$

Under Case 1: $$\frac{\sigma(q)}{n} < 1$$ $$\sqrt{\frac{5}{3}} < \frac{\sigma(n)}{q}$$

Under Case 2: $$1 < \frac{\sigma(q)}{n} < \frac{\sigma(n)}{q} < 2$$ $$1 < \frac{\sigma(q)}{n} < \sqrt{2}$$ $$\sqrt[4]{\frac{5}{3}} < \frac{\sigma(n)}{q} < 2$$

Again, the problem is that the upper bound $I(n) = \frac{\sigma(n)}{n} < 2$ for the abundancy index of the component/divisor $n$ is rather crude, as it only uses the fact that $n$ is a factor of a perfect number and is therefore deficient. I was wondering if anybody out there has some better ideas and/or techniques for improving this particular bound, as such will have a direct bearing on the resulting lower bound for $\frac{n}{q}$ under the case $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$, as in my own previous answer to this question. I hope I have put in sufficient detail for this question. Please do let me know if you need further clarifications/information. Thanks!

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I think the votes to close are premature. Arnie: This would be a better question if you focused in on a particular case. Forget about cases 1 and 3, and maybe even 2b. Give it a little bit of motivation as to where your question comes from, and then ask if anyone knows an improvement of a specific bound. –  Cam McLeman Feb 23 '12 at 15:15
Forgive me if this is an unfair comment, but what I think when I see this question is here is a large amount of high school algebra followed by a request that someone else supply some serious mathematics. –  Gerry Myerson Feb 24 '12 at 11:13
Update: Professor Carl Pomerance pointed me to "a new paper of Acquaah and Konyagin who almost prove that $n < q$ cannot occur". I will post the title of their paper soon. –  Jose Arnaldo Dris Mar 31 '12 at 6:32
The joint paper by Peter Acquaah and Sergei Konyagin is titled On Prime Factors of Odd Perfect Numbers". The abstract says: We prove that a prime factor $q$ of an odd perfect number $x$ satisfies the inequality $q < {(3x)}^{1/3}$." –  Jose Arnaldo Dris Mar 31 '12 at 7:33
I was about to point out a paper on the Odd Perfect Number problem which appeared only yesterday on the ArXiV, but then noticed that Arnie is the author - DOH! In case anyone else is interested, the paper is at arxiv.org/abs/1206.1548. –  John R Ramsden Jun 9 '12 at 9:23

This is a partial answer to my question, in light of a slight improvement on the lower bound for $\frac{n}{q}$ in Case 2A.

Under Case 2A, we have $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$. Therefore:

$$\frac{q}{n}I(q) < \frac{n}{q}I(n)$$

which implies that:

$${\left(\frac{n}{q}\right)}^2 > \frac{I(q)}{I(n)}.$$

Since $1 < I(q) < I(n) < 2$, we have:

$$\frac{n}{q} > \frac{\sqrt{2}}{2} \approx 0.7071.$$

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To quote Yemon Choi, who commented on an earlier question you asked (and also answered yourself): This question, and this answer, seem dangerously close to using MO as a kind of public blog or attempted polymath.... –  Woett Mar 11 '12 at 0:20
In case $I(n) < \sqrt{2}$, then the resulting bound here is: $$\frac{n}{q} > \sqrt[4]{\frac{1}{2}} \approx 0.84089641.$$ –  Jose Arnaldo Dris Jun 9 '12 at 16:34