A question about $\gcd$ and divisibility

Question

Let $\sigma$ be the classical sum-of-divisors function.

Suppose that I have the following equations:

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

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

and assume that $\sigma(n^2)/q^k$ and ${2n^2}/\sigma(q^k)$ are both (positive) integers. Lastly, suppose that we have $k \equiv 1 \pmod 4$, and that $\gcd(q,n) = 1$. (Edit: I forgot to indicate that $q$ must be prime.)

My question is this:

What is $\gcd(n^2, \sigma(n^2))$?

Update: I have transferred the transcript of my attempt to an actual answer to this MSE question, in response to a suggestion from mixedmath in a comment to another one of my questions.

Answer 1

Here is my attempt at an answer (although, admittedly, I am a bit confused by that $\sigma(q^{k-1})$):

Under the given conditions, both equations imply that $$\sigma(q^{k-1}) \mid \left(2n^2 - \sigma(n^2)\right).$$

Consequently, we can rewrite the equations as $$\sigma(n^2) = {q^k}\cdot{\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)}$$ and $$n^2 = {\frac{\sigma(q^k)}{2}}\cdot{\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)}.$$

Now, since $\gcd(q^k, \sigma(q^k)/2) = 1$, it now follows that $$\gcd(n^2, \sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}.$$

I hope this is correct.