The title says it all.
What is the complete (polynomial) factorization of $\sigma(p^k)$, where $p$ is prime with $p \equiv k \equiv 1 \pmod 4$?
Here, $\sigma = \sigma_{1}$ is the classical sum-of-divisors function.
My Attempt
$$\sigma(p^k) = 1 + p + \ldots + p^{k-1} + p^k = \left(1+p\right)\left(1 + p^2 + \ldots + p^{k-1}\right) = \sigma(p)\left(1 + p^2 + \ldots + p^{k-1}\right)$$
Note that $\sigma(p^k) = \sigma(p)$ if $k=1$.
Now, the finite geometric series $$1 + p^2 + \ldots + p^{k-1}$$ has $n=\dfrac{k-1}{2}+1=\dfrac{k+1}{2}$ terms, and common ratio $r=p^2$. Thus, we have $$1 + p^2 + \ldots + p^{k-1} = \dfrac{\left(p^2\right)^{(k+1)/2} - 1}{p^2 - 1} = \dfrac{p^{k+1}-1}{p^2 - 1} = \dfrac{\left(p^{(k+1)/2}+1\right)\left(p^{(k+1)/2}-1\right)}{p^2 - 1}.$$
Consequently, we obtain $$\sigma(p^k) = \sigma(p)\left(\dfrac{\left(p^{(k+1)/2}+1\right)\left(p^{(k+1)/2}-1\right)}{p^2 - 1}\right).$$
Question
Is $\sigma(p^k)$ when given as $$\sigma(p^k) = \sigma(p)\left(\dfrac{\left(p^{(k+1)/2}+1\right)\left(p^{(k+1)/2}-1\right)}{p^2 - 1}\right)$$ considered completely factored?
By complete factorization, I mean: $$\bf{e.g.} \hspace{0.5in} x^4 - 1 = (x-1)(x+1)(x^2 + 1).$$
If $k=1$ then we just have the irreducible polynomial $\sigma(p) = 1 + p$, and your second factor cancels out to just $1$. So we will assume that $k > 1$.
Have a look at cyclotomic polynomials: in general, the polynomial $\sigma(p^k) = 1 + p + \cdots + p^k = (p^{k+1}-1)/(p-1)$ will factor over $\mathbb Q[p]$ exactly as $\prod_{d\mid k+1, d>1} \Phi_d(p)$, with the $d=1$ term having been cancelled by the $p-1$ denominator. In the particular case that $k\equiv 1 \pmod 4$, you will always get the factor $\Phi_2(p) = 1+p = \sigma(p)$, and if $(k+1)/2$ happens to be prime then your other factors correspond roughly to the irreducible factors
$$\Phi_{k+1}(p) = 1 - p + p^2 - \cdots + p^{(k-1)/2},$$ $$\Phi_{(k+1)/2}(p) = 1 + p + p^2 + \cdots + p^{(k-1)/2}.$$
However in the typical case where $(k+1)/2$ is not prime, then the polynomial is certainly not completely factored by your attempt. The first time this happens (subject to the modular restriction on $k$) is when $k=17$:
$$\sigma(p^{17}) = \Phi_2(p) \Phi_3(p) \Phi_6(p) \Phi_9(p) \Phi_{18}(p) \\ = (p+1)(p^2+p+1)(p^2-p+1)(p^6+p^3+1)(p^6-p^3+1).$$
By the way, there seems to be no point whatsoever in specifying that $p\equiv 1\pmod 4$, since you are only treating $p$ as an indeterminate.