## 2.6.17

### If $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

(Note:  This blog post was copied verbatim from this MSE question.)

Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$.  If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number.  Denote the abundancy index $I$ of $N$ as $I(N)=\sigma(N)/N$.

Euler proved that every odd perfect number $N$ has to have the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In this paper, Dris proved that we have the bounds
$$L(q) = 3 - \frac{q-2}{q(q-1)} < I_{+} \leq 3 - \frac{q-1}{q(q+1)} = U(q)$$
for the sum of the abundancy indices $I_{+}=I(q^k)+I(n^2)$.  In particular, if $I_{+} \leq 2.99$, then we will have $L(q) < 2.99$, so that (using WolframAlpha) we get
$$q < \frac{101 + \sqrt{9401}}{2} \approx 98.9794,$$
(since $q$ satisfies $q \geq 5$) from which we obtain $q \leq 97$ (since $q$ is prime).

Here is the problem:

PROBLEM STATEMENT

Does the converse hold?  That is, if $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

MY ATTEMPT

Since $N=q^k n^2$ is perfect and $\gcd(q,n)=1$, then $I(q^k)I(n^2)=2$, from which we have
$$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2}{1 + \frac{1}{q}}.$$

By assumption, $q \leq 97$, so that we get
$$\frac{1}{q} \geq \frac{1}{97} \Longrightarrow 1 + \frac{1}{q} \geq \frac{98}{97} \Longrightarrow \frac{2}{1 + \frac{1}{q}} \leq \frac{97}{49} \approx 1.97959\ldots.$$

However, this does not match up with the known upper bound
$$I(q^k) < \frac{5}{4} = 1.25.$$

ANOTHER APPROACH

Assume that $q \leq 97$, and suppose to the contrary that
$$I(q^k) + I(n^2) > 2.99.$$

This means that $U(q) > 2.99$, so that (by using WolframAlpha again) we obtain
$$q > \frac{99 + \sqrt{9401}}{2} \approx 97.9794,$$
contradicting $q \leq 97$.

Here is my question:  Is this proof in the ANOTHER APPROACH section correct?