If $q \equiv k \equiv 1 \pmod 4$, is it necessarily true that $\gcd\bigg(\sigma(q^k),\sigma(q^{(k-1)/2})\bigg)=1$?

Question

Let $\sigma$ denote the classical sum-of-divisors function. In what follows, we let $q$ be a prime number.

Here is my question:

If $q \equiv k \equiv 1 \pmod 4$, is it necessarily true that $\gcd\bigg(\sigma(q^k),\sigma(q^{(k-1)/2})\bigg)=1$?

MY ATTEMPT

I pattern my approach after the answer to this closely related MSE question.

$$\sigma(q^{\frac{k-1}{2}})=\frac{q^{\frac{k+1}{2}} - 1}{q - 1}$$ $$\sigma(q^k)=\frac{q^{k+1} - 1}{q - 1}$$

Therefore, $$\gcd\bigg(\sigma(q^{\frac{k-1}{2}}),\sigma(q^k)\bigg)=\bigg(\frac{1}{q-1}\bigg)\gcd\bigg(q^{\frac{k+1}{2}} - 1, q^{k+1} - 1\bigg)=\frac{q^{\gcd\bigg(\frac{k+1}{2}, \hspace{0.05in} k+1\bigg)} - 1}{q - 1}$$ $$=\frac{q^{\frac{k+1}{2}} - 1}{q - 1} = \sigma(q^{\frac{k-1}{2}}).$$

But $\sigma(q^{(k-1)/2}) = 1$ is true if and only if $$q^{(k-1)/2} = 1 \iff (k-1)/2 = 0 \iff k = 1.$$

Is this proof correct?

Answer 1

The work shown here follows the same pattern as that of the closely related MSE question that was referenced. Based on that, plus my own checking, what you've shown in your proof attempt all looks fine to me.

One small thing to note is that since $q^{k + 1} - 1 = \left(q^{(k+1)/2} - 1\right)\left(q^{(k+1)/2} + 1\right)$, this is another way to see that $\gcd\left(q^{k + 1} - 1, q^{(k+1)/2} - 1\right) = q^{(k+1)/2} - 1$.