On the biconditional $I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5)$, where $q^k n^2$ is an odd perfect number

Question

MOTIVATION

Let $N$ be an odd perfect number given in the so-called Eulerian form $$N = q^k n^2,$$ i.e., $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In what follows, I let $$I(x)=\frac{\sigma(x)}{x}$$ denote the abundancy index of the positive integer $x$. ($\sigma(x)$ is the sum of divisors of $x$.)

CLAIM

If $q^k n^2$ is an odd perfect number given in Eulerian form, then $$I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5).$$

PROOF OF CLAIM

Let $q^k n^2$ be an odd perfect number given in Eulerian form. Suppose that $$I(n^2) = 2 - \frac{5}{3q}.$$ Since $q^k n^2$ is perfect, then we have $$I(q^k n^2) = I(q^k)I(n^2) = 2$$ where we have used the fact that $I$ is multiplicative. Hence, $$I(n^2) = I(q^k)I(n^2) - \frac{5}{3q} \implies \frac{5}{3q} = I(n^2)\bigg(I(q^k) - 1\bigg) \geq I(n^2)\bigg(1+\frac{1}{q}-1\bigg).$$ This implies that $$I(n^2) \leq \frac{5}{3}.$$

Assume to the contrary that $$I(n^2) = 2 - \frac{5}{3q} < \frac{5}{3}.$$ Then we have $$\frac{6q - 5}{3q} < \frac{5}{3} \implies 18q - 15 < 15q \implies 3q < 15 \implies q < 5,$$ contradicting $q \geq 5$.

Added to the Proof of Claim (Dec 15 2019) Hence, $$I(n^2) = 2 - \frac{5}{3q} \implies I(n^2) = \frac{5}{3} \implies (k=1 \land q=5)$$ while the proof of the direction $$(k=1 \land q=5) \implies I(n^2) = \frac{5}{3} \implies I(n^2) = 2 - \frac{5}{3q}$$ is trivial.

QUESTION

It can be proved (page 17) that $$I(n^2) \leq 2 - \frac{5}{3q}$$ holds in general for an odd perfect number $q^k n^2$ given in Eulerian form.

My question is: Can a biconditional similar to the one proved here be derived for the case $$I(n^2) < 2 - \frac{5}{3q}?$$

Answer 1

Let $q^k n^2$ be an odd perfect number given in Eulerian form.

Since $$I(n^2) \leq 2 - \frac{5}{3q}$$ holds in general, and since the biconditional $$\bigg(I(n^2) = 2 - \frac{5}{3q}\bigg) \iff \bigg(k=1 \land q=5\bigg)$$ holds, the simplest biconditional that I could come up with is $$\bigg(I(n^2) < 2 - \frac{5}{3q}\bigg) \iff \bigg(k>1 \lor q>5\bigg).$$

Of course, by Material Implication, we also have the biconditionals $$\bigg(k>1 \lor q>5\bigg) \iff \bigg(k=1 \implies q>5\bigg) \iff \bigg(q=5 \implies k>1\bigg).$$

Answer 2

Let $q^k n^2$ be an odd perfect number given in Eulerian form.

Since the equation $$I(n^2) = 2 - \frac{5}{3q}$$ is true if and only if the conjunction $$k=1 \land q=5$$ holds, then $I(n^2) = 2 - {5/(3q)}$ is true if and only if $$I(n^2) = 2 - \frac{5}{3\cdot{5}} = 2 - \frac{1}{3} = \frac{5}{3}.$$

It follows that $$I(n^2) < 2 - \frac{5}{3q} \iff I(n^2) \neq \frac{5}{3}.$$

Suppose that $I(n^2) > 5/3$. We get $$2 - \frac{5}{3q} > I(n^2) > \frac{5}{3} \implies q > 5.$$ (Another way to get the same conclusion is to use the inequality $$I(n^2) \leq \frac{2q}{q+1},$$ which holds generally.)

Now, assume that $I(n^2) < 5/3$. We obtain $$\frac{2(q-1)}{q} < I(n^2) < \frac{5}{3} \implies q < 6 \implies q = 5.$$ Thus, $$I(5^k) = I(q^k) = \frac{2}{I(n^2)} > \frac{6}{5} \implies k > 1.$$

We summarize our results below:

If $I(n^2) > 5/3$, then $q > 5$.

If $I(n^2) < 5/3$, then $q = 5$ and $k > 1$.

Since Cohen and Sorli (2012, page 4) have proved that $q=5$ and $k=5$ is untenable, then we have $$I(n^2) < \frac{5}{3} \implies \bigg(q=5 \land k>1\bigg) \implies \bigg(q=5 \land k \geq 9\bigg) \implies I(n^2) < \frac{2}{I(5^9)} = \frac{1953125}{1220703} \approx 1.60000016384,$$ whence there is no contradiction, since it is known (unconditionally) that $I(n^2) > 8/5$ holds.