## 14.6.15

### OPN Research - June 2015

"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 deficiency of $x$.

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

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

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 $3\sqrt[3]{2} \approx 3.779763$.)

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

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

$$\sigma(m^2) = \frac{2pm^2 - 2m^2 - p + 2}{p} = 2m^2 - 1 - 2\left(\frac{m^2 - 1}{p}\right)$$
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
$$\frac{m^2 - 1}{p} \geq 2.$$
$$\sigma(m^2) \leq 2m^2 - 5.$$