Here are the latest updates regarding my research on odd perfect numbers:

Lemma 1. If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then Sorli's conjecture (i.e., $k = 1$) implies that

$\gcd(n^2, \sigma(n^2)) = 2n^2 - \sigma(n^2)$.

Lemma 1 was proved by Jaycob Coleman in the following two MSE posts:

and

We also have:

Lemma 2. Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form. If $k = 1$, then $2n^2 - \sigma(n^2) = \frac{\sigma(n^2)}{q} = \frac{n^2}{\frac{q+1}{2}}$.

The proof of Lemma 2 follows readily from the definition of perfect numbers.

Observe that, since $\sigma(n^2)$ is always odd, $2n^2 - \sigma(n^2)$ is likewise odd. Furthermore, as $q \equiv 1 \pmod 4$, $q + 1$ is even (but not divisible by $4$). We conclude, by Lemma 1, that $\gcd(n^2, \sigma(n^2)) \neq q + 1$.

We now claim the truth of the following statement:

Proposition 1. Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and let $k = 1$. If $\gcd(n^2, \sigma(n^2)) > q + 1$, then $q < n\sqrt{2}$.

Proof. By Lemmas 2 and 1,

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

Since $\gcd(n^2, \sigma(n^2)) > q + 1$, we get $q < q + 1 < n\sqrt{2}$.

We can also prove the following result:

Proposition 2. Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and let $k = 1$. If $\frac{q + 1}{2}$ is prime, then $\gcd(n^2, \sigma(n^2)) \geq \frac{q + 1}{2}$.

Proof. If $\frac{q + 1}{2}$ is prime, then

${\left(\frac{q + 1}{2}\right)}^2 \mid n^2$.

This implies that

$q{\left(\frac{q + 1}{2}\right)} \mid \sigma(n^2)$.

By Lemmas 2 and 1, these two divisibility constraints both imply that

$\gcd(n^2, \sigma(n^2)) \geq \frac{q + 1}{2}$.

It remains to rule out the case $\gcd(n^2, \sigma(n^2)) < q + 1$ in order to prove that $k = 1$ implies $q < n\sqrt{2}$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form. This can be very difficult to do, as it has been conjectured that the Euler prime $q$ might have to be the largest prime factor of the odd perfect number $N$, as discussed by Nielsen [Zeta-Flux] here. Dris has given a sufficient condition [i.e., $n < q$] for Sorli's conjecture, which further supports the conjecture that $q$ must be the largest prime divisor of $N$.

The interested reader is hereby referred to this MO post for another unsuccessful attempt of mine at improving Acquaah and Konyagin's estimate of $q < n\sqrt{3}$ to $q < n\sqrt{2}$.

Let me know via e-mail if you have any comments, questions or clarifications regarding this post. You can find my e-mail in my arXiv papers. Please send to the gmail account.

**Added [05/23/2015 12:00 NN Manila time]:**

It turns out that we can prove the following (stronger) claim:

Proposition 3. Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form. If $k = 1$, then we have:

(a) $\gcd(n^2, \sigma(n^2)) > 2(q + 1)$ implies $q < \sigma(q) < n$.

(b) $\gcd(n^2, \sigma(n^2)) < 2(q + 1)$ implies $n < q < \sigma(q)$.

Proof. The proof uses Lemmas 1 and 2, and the fact that the biconditional

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

holds.

**Added [05/24/2015 14:30 PM Manila time]:**

**If $N = q^k n^2$ is an odd perfect number given in Eulerian form, then**

$\sigma(n^2) = {q^k}\cdot{\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)}$

and

$n^2 = {\frac{\sigma(q^k)}{2}}\cdot{\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)}$.

Now, since $\gcd(q^k, \sigma(q^k)/2) = 1$, it follows that

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

Note that the

*deficiency*$D$ of $n^2$ (written simply as $D(n^2)$) is given by
$D(n^2) = 2n^2 - \sigma(n^2)$

so that

$\sigma(n^2) = 2n^2 - D(n^2)$.

In other words,

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