2.6.17

Building on work from previous MSE question 2306650 (Re: Odd Perfect Numbers)

(Note:  This post builds on work from this previous 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 $I_{+} \leq 3 - \epsilon$, then we will have $L(q) < 3 - \epsilon$, so that (using WolframAlpha) we get
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
if $0 < \epsilon < 3 - 2\sqrt{2}$.

Here is the problem:

PROBLEM STATEMENT

Does the converse hold?  That is, if $N=q^k n^2$ is an odd perfect number, does
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon},$$
with $0 < \epsilon < 3 - 2\sqrt{2}$, imply that $I(q^k) + I(n^2) \leq 3 - \epsilon$?

MY ATTEMPT

Assume that $$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon},$$
with $0 < \epsilon < 3 - 2\sqrt{2}$, and suppose to the contrary that
$$I(q^k) + I(n^2) > 3 - \epsilon.$$

This means that $U(q) > 3 - \epsilon$, so that (by using WolframAlpha again) we obtain (for $0 < \epsilon < 3 - 2\sqrt{2}$) either
$$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon}$$
or
$$0 < q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}.$$

Here is my question:  Do the resulting inequalities
$$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon}$$
or
$$0 < q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}},$$
for $0 < \epsilon < 3 - 2\sqrt{2}$, contradict the assumption
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}?$$

The upper bound

$$q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}$$
is inconsistent with the lower bound
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q$$
because of this WolframAlpha computation, where I have set $\epsilon=x$.

However, the lower bound
$$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon}$$
is not inconsistent with the upper bound
$$q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
because the inequality
$$\frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon} < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
is satisfied for $\epsilon > 0$, per this WolframAlpha computation, where again I have set $\epsilon = x$.

We therefore conclude that, if $N=q^k n^2$ is an odd perfect number with Euler prime $q$, then given $0 < \epsilon < 3 - 2\sqrt{2}$, we have the biconditional
$$q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
if and only if $I(q^k) + I(n^2) \leq 3 - \epsilon$.

The case $\epsilon = 0.01$ verifies this to be the case, as in this MSE question.