In what follows, set $\sigma(x)$ to be the sum of divisors of $x \in \mathbb{N}$, and let $$D(x) = 2x - \sigma(x)$$ be the deficiency of $x$, and let $$s(x) = \sigma(x) - x$$ be the sum of the aliquot divisors of $x$.
An earlier question asks:
What is the common (and simplified) value for $D(q^k)D(n^2) = 2s(q^k)s(n^2)$ when $q^k n^2$ is an odd perfect number?
In the accepted answer, MSE user mathlove writes:
$$\begin{align}D(q^k)D(n^2) &= 2s(q^k)s(n^2) \\\\&=\frac{2n^2(q^k-1)(q^{k+1}-2q^k+1)}{(q-1)(q^{k+1}-1)}\end{align}.$$
Now, I know that $$s(q^k) = \sigma(q^k) - q^k = \sigma(q^{k-1}) = \frac{q^k - 1}{q - 1}$$ and that $$\gcd(D(q^k), s(q^k)) = 1.$$
Here is my question in this post:
Does mathlove's answer imply that $D(n^2) \mid n^2$?
I ask because the divisibility condition $D(n^2) \mid n^2$ for the non-Euler part $n^2$ is known to be equivalent to the Descartes-Frenicle-Sorli Conjecture (that is, $k=1$) for odd perfect numbers.
