# If $q^k n^2$ is an odd perfect number with Euler prime $q$, are the following statements known to hold in general? [closed]

(Note: This has been cross-posted to MSE. However, I feel that it is more likely to receive a good answer here, because I believe that it is a research-level question. For the mathematicians who voted to put this question on hold, I invite you to peruse this preprint to give you an inkling of where I'm driving at. Thanks!)

Let $$\sigma$$ be the classical sum-of-divisors function.

A number is said to be perfect if $$\sigma(N)=2N$$.

If $$q^k n^2$$ is an odd perfect number with Euler prime $$q$$ (i.e., $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$), are the following statements known to hold in general?

$$\bf{(a)}$$ $$\gcd(n^2, \sigma(n^2))$$ is large.

$$\bf{(b)}$$ The deficiency $$D(n^2) = 2n^2 - \sigma(n^2)$$ is large.

$$\bf{(c)}$$ The index $$i(q^k) = \sigma(N/q^k)/q^k$$ is large.

Using the trivial relationships:

$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)}$$

then if the Descartes-Frenicle-Sorli conjecture that $$k = 1$$ is true, it is possible to show that a lower bound for the quantities $$\bf{(a)}$$, $$\bf{(b)}$$ and $$\bf{(c)}$$ is given by $$n/\sqrt{3}$$. (Here, I have used Acquaah and Konyagin's estimate $$q < n\sqrt{3}$$. The inequality $$q^k < n^2$$ then gives the desired large numerical bound if we use known lower bounds for the odd perfect number $$N = q^k n^2$$, latest of which are by Ochem and Rao.)

What happens when $$k > 1$$? I do know that $$\sigma(n^2)/q^k \geq 315$$ by using a result of Broughan, Delbourgo, and Zhou.

Is it possible to do better than this, apart from attempting a proof of (obviously) $$q^k < n$$?

• Why are people voting to close this question? =( – Arnie Bebita-Dris May 26 '15 at 0:28

If $k > 1$, we can do better if the following inequalities hold:

$$q < q^k < n.$$

The resulting lower bound is

$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)} > \frac{8}{5}\cdot\frac{n^2}{q^k} > \frac{8}{5}\cdot{n}.$$

Since $k > 1$ implies $q < n$, it remains to consider the case

$$q < n < q^k.$$

(Update (February 2016): In a recent preprint, Brown claims a complete proof for $q < n$, and a partial proof for the inequality $q^k < n$. If completed, Brown's proof then rules out this last remaining case.)