Let $\mathbb{N}$ denote the set of natural numbers (i.e., positive integers).

A number $N \in \mathbb{N}$ is said to be *perfect* if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum of divisors. For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect. (Note that $6$ is even.) Denote the *abundancy index* of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.

Euler proved that an odd perfect number $N$, if any exists, must take the form $N=q^k n^2$, where $q$ is prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Suppose that $k=q$. Since $q$ is prime and $q \equiv 1 \pmod 4$, this implies that $k \geq 5$. (In particular, $k \neq 1$, so that the Descartes-Frenicle-Sorli conjecture is false in this case.)

$$I(q^k)=I(q^q)=\frac{q^{q+1}-1}{{q^q}(q-1)} \leq \frac{3906}{3125} = 1.24992$$

which corresponds to the lower bound

$$I(n^2)=\frac{2}{I(q^k)} \geq \frac{3125}{1953} \approx 1.6001\ldots.$$

Consider the product

$$\bigg(I(q^q) - \frac{3906}{3125}\bigg)\bigg(I(n^2) - \frac{3906}{3125}\bigg).$$

This product is nonpositive. Therefore,

$$I(q^q)I(n^2) + \bigg(\frac{3906}{3125}\bigg)^2 \leq \frac{3906}{3125}\cdot\bigg(I(q^q) + I(n^2)\bigg).$$

Since $N=q^k n^2$ is perfect with $q=k$, then $I(q^k)I(n^2)=I(q^q)I(n^2)=2$, so that

$$I(q^q) + I(n^2) \geq \frac{3906}{3125} + \frac{3125}{1953} = \frac{17394043}{6103125} \approx 2.850022406554\ldots.$$

$$I(q^k) + I(n^2) \leq \frac{3q^2 + 2q + 1}{q(q+1)} = 3 - \frac{q-1}{q(q+1)}$$

with equality occurring if and only if $k=1$.

In our case, since $k = q \geq 5$, we obtain

$$\frac{17394043}{6103125} \leq I(q^q) + I(n^2) = I(q^k) + I(n^2) < 3 - \frac{q-1}{q(q+1)}$$

$$q > \frac{3125}{781} \approx 4.00128\ldots.$$

Here is my question:

**Why does the bound **

$$I(q^q) + I(n^2) \geq \frac{3906}{3125} + \frac{3125}{1953} = \frac{17394043}{6103125} \approx 2.850022406554\ldots$$

*not imply* that $q > 5$?

I am thinking along the lines that:

**(1)** $57/20 < I(q^k) + I(n^2) < 3$ is best-possible.

**(2)** Improving the upper bound $3$ would result in a finite upper bound for the Euler prime $q$.

**(3)** Therefore, improving the lower bound $57/20$ would result in a lower bound for $q$ better than the currently known $q \geq 5$.

**REFERENCES**

**POSTED ANSWER**

I am guessing that it has got something to do with the *interaction* between the conditions $k=1$ and $q=5$.

When $k=1$, we have the bounds

$$I(q^k)=I(q)=1+\frac{1}{q} \leq \frac{6}{5}$$

and

$$I(n^2)=\frac{2}{I(q)} \geq \frac{5}{3}.$$

When $q=5$, we have the bounds
$$I(n^2) \leq 2 - \frac{5}{3q} = \frac{5}{3}$$

and

$$I(q^k) \geq \frac{6}{5}.$$

Note that, when $k=1$, we have the lower bound

$$I(q^k) + I(n^2) \geq \frac{43}{15} = 2.8\overline{666} > 2.85$$

Note further that, when $q=5$, we have the upper bound

$$I(q^k) + I(n^2) \leq \frac{43}{15}.$$

**ADDITIONAL NOTES**

It is easy to show that

$$\bigg(q = 5\bigg) \land \bigg(k = 1\bigg) \Longrightarrow I(q^k) + I(n^2) = \frac{43}{15}.$$

I am currently in the process of writing up a proof for the reverse implication

$$I(q^k) + I(n^2) = \frac{43}{15} \Longrightarrow \bigg(\left(q = 5\right) \land \left(k = 1\right)\bigg).$$