On the golden ratio and even perfect numbers

Question

(Note: This post is an offshoot of this earlier MSE question.)

Here is my question in this post:

Is $I(2^{p-1}) - 1 > 1/I(2^{p-1})$ true when $I(2^{p-1}) = \sigma(2^{p-1})/2^{p-1}$ is the abundancy index of $2^{p-1}$ and $6 \neq 2^{p-1}(2^p - 1)$ is an even perfect number (with corresponding Mersenne prime $2^p - 1$)?

MY ATTEMPT

Claim For $p \geq 3$, $$\frac{7}{4} \leq I(2^{p-1}) < 2.$$

Proof: Only the left-hand inequality is not evident (as $2^{p-1}$ is deficient, being a proper divisor of the perfect number $2^{p-1}(2^p - 1)$). $$I(2^{p-1}) = \frac{\sigma(2^{p-1})}{2^{p-1}} = \frac{2^p - 1}{2^{p-1}} = 2 - \bigg(\frac{1}{2^{p-1}}\bigg).$$ But since $6 \neq 2^{p-1}(2^p - 1)$, then $p \geq 3$, which implies that $$2^{p-1} \geq 4 \implies \frac{1}{2^{p-1}} \leq \frac{1}{4} \implies 2 - \bigg(\frac{1}{2^{p-1}}\bigg) \geq 2 - \frac{1}{4} = \frac{7}{4}.$$ QED

Checking now whether this inequality is satisfied: $$I(2^{p-1}) - 1 > \frac{1}{I(2^{p-1})}$$

For $p \geq 3$: $$I(2^{p-1}) - 1 \geq \frac{3}{4} > \frac{4}{7} \geq \frac{1}{I(2^{p-1})}.$$

For general $p$: $$I(2^{p-1}) - 1 = 1 - \bigg(\frac{1}{2^{p-1}}\bigg) = \frac{2^{p-1} - 1}{2^{p-1}}$$ $$\frac{1}{I(2^{p-1})} = \frac{2^{p-1}}{2^p - 1}$$ $$\bigg(I(2^{p-1}) - 1\bigg) - \frac{1}{I(2^{p-1})} = \frac{2^{p-1} - 1}{2^{p-1}} - \frac{2^{p-1}}{2^p - 1} > 0 \text{ when } p \geq 3.$$

Note that, since $I(2^{p-1})=x$ satisfies the inequality $$I(2^{p-1}) - 1 > \frac{1}{I(2^{p-1})} \iff x - 1 > \frac{1}{x} \implies x^2 - x - 1 > 0 \text{ since } x > 1 > 0 \implies x > \frac{1 + \sqrt{5}}{2} = \varphi \approx 1.618,$$ which is trivial compared to the inequality $$I(2^{p-1})=x \geq \frac{7}{4} = 1.75.$$

Finally, notice that $6 = 2^{2 - 1} \cdot (2^2 - 1)$ was excluded in this analysis because it is squarefree.

Answer 1

You have written a proof for the follwing claim :

Claim For $p \geq 3$, $\frac{7}{4} \leq I(2^{p-1}).$

$$2^{p-1} \geq 4 \implies \frac{1}{2^{p-1}} \leq \frac{1}{4} \implies 2 - \bigg(\frac{1}{2^{p-1}}\bigg) \geq 2 - \frac{1}{4} = \frac{7}{4}.$$

I've found no errors here.

(One can also say "Since $I(2^{p-1})$ is increasing, we have $I(2^{p-1})\ge I(2^{3-1})=\frac 74$".)

After this, you have written

For $p \geq 3$:$$I(2^{p-1}) - 1 \geq \frac{3}{4} > \frac{4}{7} \geq \frac{1}{I(2^{p-1})}.$$

I've found no errors here.

In conclusion, I think that you have correctly proved the inequality.