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In the paper titled Improving the Chen and Chen result for odd perfect numbers (Lemma 8, page 7), Broughan et al. show that if $$\frac{\sigma(n^2)}{q^k}$$ is a square, where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$ and $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then $k=1$.

Is it then possible to derive the implication $$D(n^2) = 2n^2 - \sigma(n^2) \text{ is a square } \implies k=1?$$

I could only derive the equation $$\frac{\sigma(n^2)}{q^k} = \frac{D(n^2)}{s(q^k)}$$ where $s(q^k) = \sigma(q^k) - q^k = \sigma(q^{k-1})$.

Added August 25 2018

I also currently know that $$D(n^2) \mid n^2 \iff k=1.$$

That is, the Descartes-Frenicle-Sorli conjecture is true for the odd perfect number $q^k n^2$ if and only if the non-Euler factor $n^2$ is deficient-perfect.

Added July 07 2019

If $$\frac{\sigma(n^2)}{q^k}=\frac{D(n^2)}{s(q^k)}$$ is a square, then by Broughan et al.'s result, we have $k=1$, which implies that $$s(q^k) = \sigma(q^k) - q^k = \sigma(q) - q = (q+1) - q = 1$$ is a square. Consequently, if $\sigma(n^2)/q^k$ is a square, then $D(n^2)$ is a square too.

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