Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
Note that both $\sigma$ and $I$ are multiplicative functions.
A number $m$ is said to be perfect if $\sigma(m)=2m$. Equivalently, $I(m)=2$.
Euler proved that an odd perfect number, if one exists, must have the form
$$m = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $q$ is prime, we have
$$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} – 1}{q^k (q – 1)} < \frac{q^{k+1}}{q^k (q – 1)} = \frac{q}{q – 1}$$
from which it follows that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Note that we then have the lower bound
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5}$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.
Here is my initial question:
Can we improve the lower bound for $I(n^2)$ to
$$I(n^2) \geq \frac{5}{3}$$
using the following argument?
$$\bigg(\frac{2q}{q+1} \geq I(n^2) > \frac{5}{3}\bigg) \implies q > 5 \implies q \geq 13 \implies \bigg(I(n^2) > \frac{2(q-1)}{q} \geq \frac{24}{13} > \frac{5}{3}\bigg)$$
Thus, we have the biconditional
$$I(n^2) > \frac{5}{3} \iff q > 5.$$
Next, we have the implication
$$I(n^2) = \frac{5}{3} \implies q = 5.$$
It then suffices to prove the implication
$$q = 5 \implies I(n^2) = \frac{5}{3}$$
to finally show that
$$I(n^2) \geq \frac{5}{3},$$
since $q \geq 5$ holds.
But note that, if $q=5$, then
$$\frac{5}{3} = I(n^2) = \frac{2}{I(5^k)} = \frac{2\cdot{5^k}(5-1)}{5^{k+1}-1}$$
which implies that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
Still, notice that we have
$$k=1 \implies I(q^k) = I(q) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2\cdot{5}}{6} = \frac{5}{3},$$
which is what we set out to prove.
Here is my final question:
Would it be possible to remove the reliance of the proof on the Descartes-Frenicle-Sorli Conjecture?
Best Answer
I think that the answer for your initial question is yes. I've found no errors in the argument.
I think that the answer for your final question is no since under the condition that $q=5$, we see that $I(n^2)\ge \dfrac 53$ is equivalent to $k=1$ as follows :
$$\begin{align}I(n^2)\ge\frac 53&\iff \frac{8\cdot 5^k}{5^{k+1}-1}\ge\frac 53 \\\\&\iff 24\cdot 5^k\ge 5(5^{k+1}-1) \\\\&\iff 5^k\le 5 \\\\&\iff k\le 1 \\\\&\iff k=1\end{align}$$