(Note: The following post was copied verbatim from this MSE question.)
On the golden ratio and odd perfect numbers
Here is my question:
Is $I(n^2) - 1 > 1/I(n^2)$ true when $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $k>1$?
My Attempt
If $k>1$, then since $q$ is the special prime, then $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In particular, we know that $q \geq 5$ and $k \geq 5$.
We know that
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} \leq \frac{5}{4}.$$
It follows that
$$I(n^2) = \frac{2}{I(q^k)} > \frac{2(q - 1)}{q} \geq \frac{8}{5}.$$
Thus,
$$I(n^2) - 1 > \frac{2(q - 1)}{q} - 1 = \frac{(2q - 2) - q}{q} = \frac{q - 2}{q} > \frac{q}{2(q - 1)} > \frac{1}{I(n^2)}$$
where the inequality
$$\frac{q - 2}{q} > \frac{q}{2(q - 1)}$$
holds provided $q > 3+\sqrt{5} \approx 5.23607$.
However, the resulting inequality for $I(n^2)$ from
$$I(n^2) - 1 > \frac{1}{I(n^2)}$$
together with the following upper bound for $I(n^2)$ (which holds when $k>1$)
$$\frac{2q}{q+1} > I(n^2)$$
only yields
$$\frac{2q}{q+1} > I(n^2) > \frac{\sqrt{5}+1}{2}$$
thereby giving
$$q > \frac{1+\sqrt{5}}{3-\sqrt{5}} = 2+\sqrt{5} \approx 4.23607.$$
Comments
Note that there is no discrepancy when $k=1$, as then we have
$$I(n^2) - 1 \geq \frac{2}{3} > \frac{3}{5} \geq \frac{1}{I(n^2)}$$
yielding the lower bound $q > 2+\sqrt{5} \approx 4.23607$ from
$$\frac{2q}{q+1}=I(n^2) > \frac{\sqrt{5}+1}{2}.$$
When $k>1$, we get
$$I(n^2) - 1 > \frac{3}{5} \not\gt \frac{5}{8} > \frac{1}{I(n^2)}.$$
