"Maybe this is the case that needs to be eliminated:

$$N = {q}{p^{2a}}{m^2}$$

where

$$\sigma(m^2) = p^{2a},$$

$$\sigma(p^{2a}) = q,$$

and

$$q + 1 = 2{m^2}.$$

It would seem that an odd perfect $N$ of this form would contradict [the] conjecture that $q < n$. $\ldots$ maybe this is the only problem case."

[From an e-mail communication of Professor Carl Pomerance dated June 1, 2015]

By using the fact that the abundancy index $I(x) = \sigma(x)/x$ satisfies $1 < I(x)$ for all $x > 1$, then we have the inequalities

$$m^2 < p^{2a} < q.$$

In particular, note that we have

$$m < p^a,$$

and also that $q$ is the largest prime divisor of $N$.

We now establish the following claims:

$I(q) = I(p^{2a}) = I(m^2)$ is false.

Suppose to the contrary that

$$I(q) = I(p^{2a}) = I(m^2)$$ is true.

Then

$$2 = I(q)I(p^{2a})I(m^2) = (I(q))^3 = (I(p^{2a}))^3 = (I(m^2))^3.$$

This implies that

$$\sqrt[3]{2} = I(q) = I(p^{2a}) = I(m^2).$$

The number $\sqrt[3]{2}$ is irrational, while all of the abundancy indices $I(q)$, $I(p^{2a})$, and $I(m^2)$ are rational.

$I(q) \neq I(m^2)$ is true.

Suppose to the contrary that $I(q) = I(m^2)$.

This then contradicts the fact that the prime $q$ is solitary.

The proof of the following claim is similar to that of

$I(p^{2a}) \neq I(m^2)$ is true.

Use the fact that the prime-power $p^{2a}$ is solitary.

$I(q) \neq I(p^{2a})$ is true.

Note that $q$ is prime and $p^{2a}$ is a prime power.

We then have:

$$\sigma(m^2) = 3^{2a}$$

$$\sigma(3^{2a}) = \frac{3^{2a + 1} - 1}{2} = q$$

$$2m^2 = q + 1.$$

This then implies that

$$3\sigma(m^2) - 1 = 2q = 2(2m^2 - 1).$$

This finally gives:

$$\frac{\sigma(m^2)}{m^2} + \frac{1}{3m^2} = \frac{4}{3}.$$

Consequently:

$$I(m^2) < \frac{4}{3}.$$

We are now able to improve the following bounds:

$$\frac{10}{9} < I(m^2) = \frac{2{p^{2a}}}{q + 1} < 2$$

$$1 < I(q) = \frac{2m^2}{\sigma(p^{2a})} < 1 + {10}^{-500}$$

$$1 < I(p^{2a}) = \frac{q}{\sigma(m^2)} < \frac{3}{2}$$

In particular, we obtain the estimates

$$\frac{\sqrt{10}}{3} < I(m)$$

$$1 < I(q)$$

$$\frac{\sqrt{5}}{2} < I(p^a).$$

By the Arithmetic Mean-Geometric Mean Inequality:

$$\sqrt[3]{2} = \sqrt[3]{I(q)I(p^{2a})I(m^2)} < \frac{I(q) + I(p^{2a}) + I(m^2)}{3}.$$

This then gives the (trivial [?]) lower bound

$$3\sqrt[3]{2} < I(q) + I(p^{2a}) + I(m^2).$$

(Note the rational approximation $3\sqrt[3]{2} \approx 3.779763$.)

But we have the (nontrivial) upper bounds

$$I(m^2) < \frac{4}{3}$$

$$I(q) < 1 + {10}^{-500}$$

$$I(p^{2a}) < \frac{3}{2}$$

from which we obtain the upper bound

$$I(q) + I(p^{2a}) + I(m^2) < \frac{4}{3} + \frac{3}{2} + 1 + {10}^{-500} \approx 3.8\bar{3}.$$

$$\sqrt{\frac{10}{9}} < I(m) < \frac{4}{3}$$

$$1 < I(q) < 1 + {10}^{-500}$$

$$\sqrt{\frac{5}{4}} < I(p^a) < \frac{3}{2}.$$

$$N = {q}{p^{2a}}{m^2}$$

where

$$\sigma(m^2) = p^{2a},$$

$$\sigma(p^{2a}) = q,$$

and

$$q + 1 = 2{m^2}.$$

It would seem that an odd perfect $N$ of this form would contradict [the] conjecture that $q < n$. $\ldots$ maybe this is the only problem case."

[From an e-mail communication of Professor Carl Pomerance dated June 1, 2015]

__Preliminary Results__By using the fact that the abundancy index $I(x) = \sigma(x)/x$ satisfies $1 < I(x)$ for all $x > 1$, then we have the inequalities

$$m^2 < p^{2a} < q.$$

In particular, note that we have

$$m < p^a,$$

and also that $q$ is the largest prime divisor of $N$.

We now establish the following claims:

**Claim 1:**$I(q) = I(p^{2a}) = I(m^2)$ is false.

**Proof of Claim 1.**Suppose to the contrary that

$$I(q) = I(p^{2a}) = I(m^2)$$ is true.

Then

$$2 = I(q)I(p^{2a})I(m^2) = (I(q))^3 = (I(p^{2a}))^3 = (I(m^2))^3.$$

This implies that

$$\sqrt[3]{2} = I(q) = I(p^{2a}) = I(m^2).$$

The number $\sqrt[3]{2}$ is irrational, while all of the abundancy indices $I(q)$, $I(p^{2a})$, and $I(m^2)$ are rational.

**QED.**

**Claim 2:**$I(q) \neq I(m^2)$ is true.

**Proof of Claim 2.**Suppose to the contrary that $I(q) = I(m^2)$.

This then contradicts the fact that the prime $q$ is solitary.

**QED.**The proof of the following claim is similar to that of

*Claim 2*.**Claim 3:**$I(p^{2a}) \neq I(m^2)$ is true.

**Proof of Claim 3.**Use the fact that the prime-power $p^{2a}$ is solitary.

**Claim 4:**$I(q) \neq I(p^{2a})$ is true.

**Proof of Claim 4.**Note that $q$ is prime and $p^{2a}$ is a prime power.

__Main Results__**Assume that $p = 3$.**

We then have:

$$\sigma(m^2) = 3^{2a}$$

$$\sigma(3^{2a}) = \frac{3^{2a + 1} - 1}{2} = q$$

$$2m^2 = q + 1.$$

This then implies that

$$3\sigma(m^2) - 1 = 2q = 2(2m^2 - 1).$$

This finally gives:

$$\frac{\sigma(m^2)}{m^2} + \frac{1}{3m^2} = \frac{4}{3}.$$

Consequently:

$$I(m^2) < \frac{4}{3}.$$

We are now able to improve the following bounds:

$$\frac{10}{9} < I(m^2) = \frac{2{p^{2a}}}{q + 1} < 2$$

$$1 < I(q) = \frac{2m^2}{\sigma(p^{2a})} < 1 + {10}^{-500}$$

$$1 < I(p^{2a}) = \frac{q}{\sigma(m^2)} < \frac{3}{2}$$

to

$$\frac{10}{9} < I(m^2) < \frac{4}{3}$$

$$1 < I(q) < 1 + {10}^{-500}$$

$$\frac{5}{4} < I(p^{2a}) < \frac{3}{2}.$$

$$\frac{\sqrt{10}}{3} < I(m)$$

$$1 < I(q)$$

$$\frac{\sqrt{5}}{2} < I(p^a).$$

By the Arithmetic Mean-Geometric Mean Inequality:

$$\sqrt[3]{2} = \sqrt[3]{I(q)I(p^{2a})I(m^2)} < \frac{I(q) + I(p^{2a}) + I(m^2)}{3}.$$

This then gives the (trivial [?]) lower bound

$$3\sqrt[3]{2} < I(q) + I(p^{2a}) + I(m^2).$$

(Note the rational approximation $3\sqrt[3]{2} \approx 3.779763$.)

But we have the (nontrivial) upper bounds

$$I(m^2) < \frac{4}{3}$$

$$I(q) < 1 + {10}^{-500}$$

$$I(p^{2a}) < \frac{3}{2}$$

from which we obtain the upper bound

$$I(q) + I(p^{2a}) + I(m^2) < \frac{4}{3} + \frac{3}{2} + 1 + {10}^{-500} \approx 3.8\bar{3}.$$

__Further Results__**Using the (trivial) estimate $I(x^2) < {(I(x))}^2$ (which is true for all $x > 1$), we get the bounds**

$$\sqrt{\frac{10}{9}} < I(m) < \frac{4}{3}$$

$$1 < I(q) < 1 + {10}^{-500}$$

$$\sqrt{\frac{5}{4}} < I(p^a) < \frac{3}{2}.$$