Denote the sum of divisors of $x \in \mathbb{N}$ by $\sigma(x)$.
Here is my question:
What is $\sigma(q^k)$ in terms of $k$ if $q \equiv 5 \pmod 8$?
I know that if $q \equiv 1 \pmod 8$, then $$\sigma(q^k) = \sum_{i=0}^{k}{q^i} \equiv 1 + \sum_{i=1}^{k}{q^i} \equiv 1 + \sum_{i=1}^{k}{1^i} \equiv k + 1 \pmod 8,$$ but I am having trouble trying to wrap my head around the corresponding derivation for the case when $q \equiv 5 \pmod 8$.
MY ATTEMPT
I first computed the residues of powers of $5$ modulo $8$: $$5^0 \equiv 1 \pmod 8$$ $$5^1 \equiv 5 \pmod 8$$ $$5^2 \equiv 1 \pmod 8$$ $$5^3 \equiv 5 \pmod 8$$ etc.
Therefore, I get: $$\sigma(q^k) \equiv \begin{cases} 1 \pmod 8, \text{ if } k \equiv 0 \pmod 8 \\ 1 + 5 \equiv 6 \pmod 8, \text{ if } k \equiv 1 \pmod 8 \\ 6 + 1 \equiv 7 \pmod 8, \text{ if } k \equiv 2 \pmod 8 \\ 7 + 5 \equiv 4 \pmod 8, \text{ if } k \equiv 3 \pmod 8 \\ 4 + 1 \equiv 5 \pmod 8, \text{ if } k \equiv 4 \pmod 8 \\ 5 + 5 \equiv 2 \pmod 8, \text{ if } k \equiv 5 \pmod 8 \\ 2 + 1 \equiv 3 \pmod 8, \text{ if } k \equiv 6 \pmod 8 \\ 3 + 5 \equiv 0 \pmod 8, \text{ if } k \equiv 7 \pmod 8 \\ \end{cases} $$
Is my derivation correct?