Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$. (That is, $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$.) Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$.
Define $$D(n^2) := 2n^2 - \sigma(n^2)$$ to be the deficiency of the non-Euler part $n^2$.
CLAIM
$\gcd(n^2, D(n^2)) \neq 1$.
MY ATTEMPT
From this preprint, we have the relationships
$$\gcd(n^2, \sigma(n^2)) = \dfrac{D(n^2)}{\sigma(q^{k-1})} = \dfrac{\sigma(n^2)}{q^k}.$$
We also have the lower bound
$$\dfrac{\sigma(n^2)}{q^k} \geq 3.$$
Since $\gcd(a,b) = \gcd(a, ax+by)$ for $x, y \in \mathbb{Z}$, we have
$$\gcd(n^2, \sigma(n^2)) = \gcd(n^2, 2n^2 - \sigma(n^2)) = \gcd(n^2, D(n^2)) \geq 3.$$
Here is my question:
QUESTION
Is this proof correct?
Added February 13 2017
Note that, for the Descartes spoof $d = 198585576189 = KM$ (where $K$ is a square and $M$ is the quasi-Euler prime), then $$D(K) = D({3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}) = 819 = {3^2}\cdot{7}\cdot{13}$$ which divides $K = {3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}$. (In other words, $K$ is an odd deficient-perfect number.)
In particular, note that $$\gcd(K, D(K)) = {3^2}\cdot{7}\cdot{13} = D(K) \neq 1.$$