On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II

Solution 1:

On OP's request, I am converting my comment into an answer.

$p^k\lt m$ is equivalent to $$m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have$$\begin{align}p^k\lt m&\iff m^2-2^rt\lt m \\\\&\iff m^2-m-2^rt\lt 0 \\\\&\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\end{align}$$

$(7)$ is better than $2m\lt 2^r+t+1$ since $$\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$ holds.

To see that $(8)$ holds, note that $$\begin{align}(2)&\iff \sqrt{1+2^{r+2}t}\lt 2^r+t \\\\&\iff 1+2^{r+2}t\lt 2^{r+1}+2^{r+1}t+t^2 \\\\&\iff (2^r-t)^2\gt 1 \\\\&\iff |2^r-t|\gt 1\end{align}$$ which does hold.

We can say that $$\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$ since $$\begin{align}(9)&\iff \bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \\\\&\iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \\\\&\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0 \\\\&\iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t} \\\\&\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t) \\\\&\iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0 \\\\&\iff (2^r-t)^2\gt 1 \\\\&\iff |2^r-t|\gt 1\end{align}$$ which does hold.