Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
The problem under consideration is as in the title:
Is the following expression bounded when $p \equiv 1 \pmod 4$ is prime and $t \equiv 1 \pmod 4$? $$\dfrac{\sigma\bigg(p^{\dfrac{t-1}{2}}\bigg)\sigma\bigg(p^{\dfrac{t+1}{2}}\bigg)}{\sigma(p^t)} - 1$$
MY ATTEMPT
First, the following expression $$\sigma\bigg(p^{\dfrac{t-1}{2}}\bigg)\sigma\bigg(p^{\dfrac{t+1}{2}}\bigg) - \sigma(p^t)$$ can be rewritten as $$\dfrac{p\bigg(p^{\dfrac{t-1}{2}} - 1\bigg)\bigg(p^{\dfrac{t+1}{2}} - 1\bigg)}{\bigg(p-1\bigg)^2}.$$ Dividing through by $\sigma(p^t)$, we get $$\dfrac{\sigma\bigg(p^{\dfrac{t-1}{2}}\bigg)\sigma\bigg(p^{\dfrac{t+1}{2}}\bigg)}{\sigma(p^t)} - 1 = \dfrac{p\bigg(p^{\dfrac{t-1}{2}} - 1\bigg)\bigg(p^{\dfrac{t+1}{2}} - 1\bigg)}{\bigg(p^{t+1} - 1\bigg)\bigg(p-1\bigg)}.$$
Since $p \equiv 1 \pmod 4$ is prime and $t \equiv 1 \pmod 4$ is an integer, then it is evident that $$f(p,t) := \dfrac{p\bigg(p^{\dfrac{t-1}{2}} - 1\bigg)\bigg(p^{\dfrac{t+1}{2}} - 1\bigg)}{\bigg(p^{t+1} - 1\bigg)\bigg(p-1\bigg)} > 0.$$
It remains to find an upper bound. I claim that:
CLAIM $f(p,t) < 1$
Proof Attempt Suppose to the contrary that $f(p,t) \geq 1$.
Then we obtain
$$p^{t+2} - p^{t+1} - p + 1 = \bigg(p^{t+1} - 1\bigg)\bigg(p-1\bigg) \leq p\bigg(p^{\dfrac{t-1}{2}} - 1\bigg)\bigg(p^{\dfrac{t+1}{2}} - 1\bigg) = p^{t+1} - p^{\dfrac{t+3}{2}} - p^{\dfrac{t+1}{2}} + p,$$ which leads to a contradiction.
Hence, we have $$0 < \dfrac{p\bigg(p^{\dfrac{t-1}{2}} - 1\bigg)\bigg(p^{\dfrac{t+1}{2}} - 1\bigg)}{\bigg(p^{t+1} - 1\bigg)\bigg(p-1\bigg)} < 1$$ and the expression in question is bounded.
Here are my:
QUESTIONS
(1) Is my proof attempt valid? If not, how can it be mended to produce a correct demonstration?
(2) Would it be possible to improve on the lower and upper bounds obtained for $f(p,t)$?
(1)
Yes. (We have $0\color{red}{\le}f(p,t)$ if $t\ge 1$.)
(2)
We have $$\frac{\partial f(p,t)}{\partial t}=\frac{(p + 1) p^{(t + 1)/2}(p^{(t+1)/2}-1)^2 \log p}{2 (p - 1) (p^{t + 1} - 1)^2}\gt 0$$ and $$\lim_{t\to\infty}f(p,t)=\lim_{t\to\infty}\frac{p\bigg(1-p^{-(t-1)/2}\bigg)\bigg(1-p^{-(t+1)/2}\bigg)}{\bigg(p-p^{-t}\bigg)(p-1)}=\frac{1}{p-1}$$
So, if $t\ge 1$, then we obtain $$0=f(p,1)\le f(p,t)\lt \frac{1}{p-1}$$ to conclude $$\color{red}{0\le f(p,t)\lt \frac 14}$$