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.)

Additionally, we obtain
$$\sigma(p^k) = p^k + 1$$
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 divisibility constraint $\sigma(p^k) \mid 2m$ that
$$2m - 1 = p^k < p^k + 1 = \sigma(p^k) \leq 2m.$$

We therefore conclude that $p < m$.  (Of course, this implies that $m^2 - p^k$ is not a square.)

Note then that we still do not know anything about $p^k < m$ nor $m < p^k$.

Summarizing, notice that we obtain

$$p < m < p^k < 2m,$$

if it so happens that $\sigma(p^k)=2m$.

Notice that, by the contrapositive, we obtain

$$(m < p) \Rightarrow (\sigma(p^k) \neq 2m).$$

( To be continued$\ldots$ )

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