Let $k$ be a positive integer satisfying $k \equiv 1 \pmod 4$. Let $x \in \mathbb{N}$. Let $q$ be a prime number.
If $$\sigma(x) = \sum_{d \mid x}{d}$$ is the classical sum-of-divisors function, then how do I simplify the following expression?
$$2 q^{\frac{k-1}{2}} n^2 - \sigma(q^{\frac{k-1}{2}})\sigma(n^2)$$
When $k=1$, this is just $$2n^2 - \sigma(n^2) = D(n^2)$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$.
How about when $k>1$? Does the expression have a nice form?
MY ATTEMPT
Since $k>1$ and $k \equiv 1 \pmod 4$, then $k \geq 5$, which implies that $$\sigma(q^{\frac{k-1}{2}}) = q^{\frac{k-1}{2}} + \ldots + q + 1 = q^{\frac{k-1}{2}} + \sigma(q^{\frac{k-3}{2}})$$ so that we obtain $$2 q^{\frac{k-1}{2}} n^2 - \sigma(q^{\frac{k-1}{2}})\sigma(n^2) = 2 q^{\frac{k-1}{2}} n^2 - q^{\frac{k-1}{2}} \sigma(n^2) - \sigma(q^{\frac{k-3}{2}}) \sigma(n^2)$$ $$= q^{\frac{k-1}{2}} D(n^2) - \sigma(q^{\frac{k-3}{2}}) \sigma(n^2)$$
Is my derivation correct?
Note that when $\gcd(q,n)=1$, then this is just $$2 q^{\frac{k-1}{2}} n^2 - \sigma(q^{\frac{k-1}{2}})\sigma(n^2) = 2 q^{\frac{k-1}{2}} n^2 - \sigma\left(q^{\frac{k-1}{2}} n^2\right) = D\bigg(q^{\frac{k-1}{2}} n^2\bigg).$$