"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]

In particular, note that

$$p\sigma(m^2) - 1 = p^{2a+1} - 1 = (p - 1)\sigma(p^{2a}) = (p - 1)q = (p - 1)(2m^2 - 1),$$

from which we obtain

$$\sigma(m^2) = 2m^2 - 1 - \left(\frac{2(m^2 - 1)}{p}\right),$$

so that we have

$$D(m^2) = 1 + \frac{2(m^2 - 1)}{p}$$

where $D(x) = 2x - \sigma(x)$ is the

$$D(m^2) = 1 + \frac{2(m^2 - 1)}{p}$$

we have the condition

$$p \mid (m^2 - 1),$$

since $\gcd(2, p) = 1$.

Additionally, from

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

we get

$$p\left(\sigma(m^2) - 1\right) = (p - 1)(q - 1).$$

Since $\gcd(p, p - 1) = 1$, we have the condition

$$p \mid (q - 1).$$

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:

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

Note that:

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

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

Note that $I(q) < I(m^2)$, since if $I(m^2) < I(q)$, then

$$I(p^{2a}) < I(q) < 1 + {10}^{-500}$$

and

$$I(m^2) < I(q) < 1 + {10}^{-500}.$$

By the Arithmetic Mean-Geometric Mean Inequality:

$$\sqrt[3]{2} = \sqrt[3]{I(q)I(p^{2a})I(m^2)} < \frac{I(q) + I(p^{2a}) + I(m^2)}{3} < 1 + {10}^{-500},$$

which is a contradiction.

Consequently,

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

Recall that, under the assumption $p = 3$, then

$$3\sqrt[3]{2} < I(q) + I(p^{2a}) + I(m^2) < 2 + 2\cdot{10}^{-500} + \frac{4}{3} \approx \frac{10}{3}.$$

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

This is a contradiction.

Consequently, $p \neq 3$.

(In fact, it is possible to do significantly better than this, since

$$I(p^2) \leq I(p^{2a}) < I(q) < 1 + {10}^{-500}$$

implies that the prime $p$ is large.)

Since

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

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

and

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

we have

$$p\sigma(m^2) - 1 = (p - 1)q = (p - 1)(2m^2 - 1) = 2pm^2 - 2m^2 - p + 1$$

which implies that

$$\sigma(m^2) = \frac{2pm^2 - 2m^2 - p + 2}{p} = 2m^2 - 1 - 2\left(\frac{m^2 - 1}{p}\right).$$

This implies that $p \mid (m^2 - 1)$, from which we get $p < m^2$.

Consequently,

$$I(m^2) = 2 - \frac{1}{m^2} - \frac{2}{p}\cdot\left(1 - \frac{1}{m^2}\right) = 2 - \frac{1}{m^2} - \frac{2}{p} + \frac{2}{p}\cdot\frac{1}{m^2} < 2 - \frac{1}{m^2} + \frac{2}{p}\cdot\frac{1}{m^2}.$$

We now obtain a lower bound for $p$. Since

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

then we have $p > {10}^{500}$ (and therefore that $m^2 > {10}^{500}$).

Therefore, we have the estimate

$$I(m^2) < 2 - \frac{1}{m^2} + {{10}^{-499}}\cdot{{10}^{-500}} < 2.$$

We then have

$$3\sqrt[3]{2} < I(p^{2a}) + I(q) + I(m^2) < 2 + 2\cdot{10}^{-500} + 2$$

whence we do not arrive at a contradiction.

Now, we try to obtain a lower bound for $I(m^2)$.

Since

$$I(m^2) = 2 - \frac{1}{m^2} - \frac{2}{p}\cdot\left(1 - \frac{1}{m^2}\right)$$

and $p > {10}^{500}$, we get

$$I(m^2) > 2 - \frac{1}{m^2} - {10}^{-499}\left(1 - \frac{1}{m^2}\right).$$

We now have the following chain of inequalities

$$2 - \frac{1}{m^2} - {10}^{-499}\left(1 - \frac{1}{m^2}\right) < I(m^2) < 2 - \frac{1}{m^2} + {{10}^{-499}}\cdot{{10}^{-500}}$$

which leads me to conjecture that

$$\sigma(m^2) = 2m^2 - 1$$

However, from the equations

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

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

and

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

I could only get

$$\sigma(\sigma(\sigma(m^2))) = 2m^2.$$

I will provide a disproof for my earlier conjecture here.

Since the equation

$$\sigma(m^2) = \frac{2pm^2 - 2m^2 - p + 2}{p} = 2m^2 - 1 - 2\left(\frac{m^2 - 1}{p}\right)$$

implies that $p \mid (m^2 - 1)$, $p \leq (m^2 - 1)$. Suppose that $p = m^2 - 1$. This contradicts the fact that $p$ is odd. Therefore, since $m^2 - 1$ is even, we have that

$$\frac{m^2 - 1}{p} \geq 2.$$

In particular,

$$\sigma(m^2) \leq 2m^2 - 5.$$

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

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]

In particular, note that

$$p\sigma(m^2) - 1 = p^{2a+1} - 1 = (p - 1)\sigma(p^{2a}) = (p - 1)q = (p - 1)(2m^2 - 1),$$

from which we obtain

$$\sigma(m^2) = 2m^2 - 1 - \left(\frac{2(m^2 - 1)}{p}\right),$$

so that we have

$$D(m^2) = 1 + \frac{2(m^2 - 1)}{p}$$

where $D(x) = 2x - \sigma(x)$ is the

**of $x$.***deficiency*__Divisibility Constraints__**From the equation**

$$D(m^2) = 1 + \frac{2(m^2 - 1)}{p}$$

we have the condition

$$p \mid (m^2 - 1),$$

since $\gcd(2, p) = 1$.

Additionally, from

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

we get

$$p\left(\sigma(m^2) - 1\right) = (p - 1)(q - 1).$$

Since $\gcd(p, p - 1) = 1$, we have the condition

$$p \mid (q - 1).$$

__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 to the contrary 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:

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

Note that:

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

Since $p$ and $q$ are primes, then we have

$$I(p^{2a}) < I(q)$$since $p^{2a} < q$.

Note that $I(q) < I(m^2)$, since if $I(m^2) < I(q)$, then

$$I(p^{2a}) < I(q) < 1 + {10}^{-500}$$

and

$$I(m^2) < I(q) < 1 + {10}^{-500}.$$

By the Arithmetic Mean-Geometric Mean Inequality:

$$\sqrt[3]{2} = \sqrt[3]{I(q)I(p^{2a})I(m^2)} < \frac{I(q) + I(p^{2a}) + I(m^2)}{3} < 1 + {10}^{-500},$$

which is a contradiction.

Consequently,

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

Recall that, under the assumption $p = 3$, then

$$3\sqrt[3]{2} < I(q) + I(p^{2a}) + I(m^2) < 2 + 2\cdot{10}^{-500} + \frac{4}{3} \approx \frac{10}{3}.$$

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

This is a contradiction.

Consequently, $p \neq 3$.

(In fact, it is possible to do significantly better than this, since

$$I(p^2) \leq I(p^{2a}) < I(q) < 1 + {10}^{-500}$$

implies that the prime $p$ is large.)

__Note:__The rest of this blog post was pulled from an answer to this MO question.**Assume that $3 \mid m$.**

Since

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

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

and

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

we have

$$p\sigma(m^2) - 1 = (p - 1)q = (p - 1)(2m^2 - 1) = 2pm^2 - 2m^2 - p + 1$$

which implies that

$$\sigma(m^2) = \frac{2pm^2 - 2m^2 - p + 2}{p} = 2m^2 - 1 - 2\left(\frac{m^2 - 1}{p}\right).$$

This implies that $p \mid (m^2 - 1)$, from which we get $p < m^2$.

Consequently,

$$I(m^2) = 2 - \frac{1}{m^2} - \frac{2}{p}\cdot\left(1 - \frac{1}{m^2}\right) = 2 - \frac{1}{m^2} - \frac{2}{p} + \frac{2}{p}\cdot\frac{1}{m^2} < 2 - \frac{1}{m^2} + \frac{2}{p}\cdot\frac{1}{m^2}.$$

We now obtain a lower bound for $p$. Since

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

then we have $p > {10}^{500}$ (and therefore that $m^2 > {10}^{500}$).

Therefore, we have the estimate

$$I(m^2) < 2 - \frac{1}{m^2} + {{10}^{-499}}\cdot{{10}^{-500}} < 2.$$

We then have

$$3\sqrt[3]{2} < I(p^{2a}) + I(q) + I(m^2) < 2 + 2\cdot{10}^{-500} + 2$$

whence we do not arrive at a contradiction.

Now, we try to obtain a lower bound for $I(m^2)$.

Since

$$I(m^2) = 2 - \frac{1}{m^2} - \frac{2}{p}\cdot\left(1 - \frac{1}{m^2}\right)$$

and $p > {10}^{500}$, we get

$$I(m^2) > 2 - \frac{1}{m^2} - {10}^{-499}\left(1 - \frac{1}{m^2}\right).$$

We now have the following chain of inequalities

$$2 - \frac{1}{m^2} - {10}^{-499}\left(1 - \frac{1}{m^2}\right) < I(m^2) < 2 - \frac{1}{m^2} + {{10}^{-499}}\cdot{{10}^{-500}}$$

which leads me to conjecture that

**Conjecture**$$\sigma(m^2) = 2m^2 - 1$$

However, from the equations

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

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

and

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

I could only get

$$\sigma(\sigma(\sigma(m^2))) = 2m^2.$$

**Added February 6 2016**I will provide a disproof for my earlier conjecture here.

Since the equation

$$\sigma(m^2) = \frac{2pm^2 - 2m^2 - p + 2}{p} = 2m^2 - 1 - 2\left(\frac{m^2 - 1}{p}\right)$$

implies that $p \mid (m^2 - 1)$, $p \leq (m^2 - 1)$. Suppose that $p = m^2 - 1$. This contradicts the fact that $p$ is odd. Therefore, since $m^2 - 1$ is even, we have that

$$\frac{m^2 - 1}{p} \geq 2.$$

In particular,

$$\sigma(m^2) \leq 2m^2 - 5.$$

**QED**