If $k \equiv 1 \pmod 4$ and $q > 3$ (where $q$ is prime), does it follow that ${q^k}\sigma(q^k) \equiv q\sigma(q) \equiv 2 \pmod {q - 1}$?
Observe that $q\sigma(q) \mid {q^k}\sigma(q^k)$ when $k \equiv 1 \pmod 4$.
I got $q\sigma(q) \equiv 2 \pmod {q - 1}$ (for $q > 3$) from this Wolfram link.
Edit: After taking to anon, I would like to add that, for the problem I am considering, I actually have $q \equiv k \equiv 1 \pmod 4$. anon's answer gave $k \equiv 1 \pmod {q - 1}$ as a condition to check for the validity of the conjecture above.