Rasputin Numbers

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