## 26.5.23

### Some New Results on Odd Perfect Numbers - Part IV

Continuing from this earlier blog post: (In particular, if $m$ is almost perfect, then $m$ must be a square (since $m$ is odd). Consequently, if the equation $\sigma(p^k)=2m$ holds (which is true if and only if $G=H=I$), then $\sigma(p^k)/2$ is also a square. This implies that $k=1$, since $$p^k + 1 \leq \sigma(p^k)=2m<2p$$ holds under the assumption $m<p$.)

Summarizing our results so far, we have the chain of implications:
$$m < p \Rightarrow \Bigg((m < p^k) \land (k = 1) \land (D(m) = 1)\Bigg).$$
But this blog post proves that the condition $D(m) = 1$ is equivalent to $m < p$.

Since $\sigma(m^2)/p^k$ is also a square if $\sigma(p^k)/2$ is a square, then under the assumption that $m < p$, we obtain $\sigma(m^2)/p^k = \sigma(p^k)/2 = m$ is a square. (Note that the difference $m^2 - p^k$ is also a square.)

$$\sigma(p^k) = p^k + 1$$
since $k=1$ follows from the assumption $m < p$.  Since $m^2 - p^k$ is a square if and only if
$$m^2 - p^k = (m - 1)^2,$$
it then follows from the assumption $m < p$ that $2m - 1 = p^k$, whence we obtain
$$2m = p^k + 1 = \sigma(p^k).$$

But note that
$$m < p \Rightarrow \Bigg((k = 1) \land (m < p^k) \land (\sigma(p^k) = 2m)\Bigg).$$

Since $m^2 - p^k$ is not a square, then we obtain
$$m^2 - p^k = 2^r t$$
where $r \geq 2$ and $2^r \neq t$.

Now, consider the following scenario in this MathOverflow question, under which the inequality $m < p^k$ is sure to hold:

$$\text{Case (A): } m > \max(2^r, t)$$

Case (A) implies that

$$\left(m - \max(2^r, t)\right)\left(m + \min(2^r, t)\right) > 0$$
$$p^k = m^2 - 2^r t = m^2 - \min(2^r, t)\max(2^r, t)$$
$$> m\left(\max(2^r, t) - \min(2^r, t)\right) = m\left|2^r - t\right|,$$

from which we get

$$p^k > m\left|2^r - t\right|.$$

Note that this implies that $m < p^k$ since $\left|2^r - t\right| \geq 1$ must necessarily hold.

Next, notice that the other (remaining) cases are

$$\text{Case (B): } \min(2^r, t) < m < \max(2^r, t)$$

$$\text{Case (C): } m < \min(2^r, t).$$

The disposal of Case (C) is very easy and is left as an exercise for the interested reader.

We then have the biconditional

$$\left(m < \max(2^r, t)\right) \iff \left(p^k < m\left|2^r - t\right|\right).$$

$$p^k < m \Rightarrow \text{ Case (B) } \Rightarrow \left|2^r - t\right| \neq 1.$$

By the contrapositive, we obtain
$$\left|2^r - t\right| = 1 \Rightarrow m < p^k.$$

However, under the assumption $m < p$, we know that $\sigma(p^k) = 2m$.  It follows that $p^k < 2m$.

Consequently, under Case (A), we derive
$$m \leq m\left|2^r - t\right| < p^k < 2m$$
from which it follows that
$$1 \leq \left|2^r - t\right| < 2$$
which forces $\left|2^r - t\right| = 1$.

We conclude that
$$m < p^k \iff \left|2^r - t\right| = 1.$$

Conclusion:  We now claim that the inequality $p < m$ indeed holds. Indeed, assume to the contrary that $m < p$ holds. Then we obtain both $k = 1$ and $\sigma(p^k)=2m$ as valid inferences from this assumption.  In particular,
$$2m=\sigma(p^k)=p+1$$
$$2m - 1 = p$$

But then again, as a by-product of Acquaah and Konyagin's results published in the International Journal of Number Theory in 2012, we have the upper bound
$$p < m\sqrt{3}.$$

Consequently,
$$2m - 1 = p < m\sqrt{3}$$
$$m(2 - \sqrt{3}) < 1.$$

This contradicts $m \geq 4$.

Hence, we infer that the estimate $p < m$ must be true.