Let $N = q^k n^2$ be a hypothetical odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
In a letter to Mersenne dated November $15$, $1638$, Descartes showed that $$\mathscr{D} = {{3}^2}\cdot{{7}^2}\cdot{{11}^2}\cdot{{13}^2}\cdot{22021} = 198585576189$$ would be an odd perfect number if $22021$ were prime.
"Equating" $N$ to $\mathscr{D}$, we obtain $$n_{\mathscr{D}} = 3 \cdot 7 \cdot {11} \cdot {13} = 3003$$ and $$k_{\mathscr{D}} = 1$$ $$q_{\mathscr{D}} = 22021,$$ where we "pretend" that $q_{\mathscr{D}}$ is prime. (Note that we already know that, for an actual odd perfect number $N = q^k n^2$, $n$ must contain a square factor, and hence that $n$ cannot be squarefree, by a result of Steuerwald in $1937$. But, oh well, let us see what we have got.)
This paper shows that the following conditions are equivalent for an odd perfect number $N = q^k n^2$:
$$\sigma(q^k)/2 \mid n$$ $$n \mid \sigma(n^2)$$ $$G = \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \sigma(q^k)/2$$ $$I = \gcd\left(n,\sigma(n^2)\right) = n$$
With $n_{\mathscr{D}} = 3003$, $k_{\mathscr{D}} = 1$, and $q_{\mathscr{D}} = 22021$ (where we pretend that $22021$ is prime), we obtain $$\sigma({q_{\mathscr{D}}}^{k_{\mathscr{D}}})/2 = \sigma(22021)/2 = 22022/2 = 11011 = 7 \cdot {11}^2 \cdot {13} \nmid 3 \cdot 7 \cdot {11} \cdot {13} = 3003 = n_{\mathscr{D}},$$ $$n_{\mathscr{D}} = 3003 = 3 \cdot 7 \cdot {11} \cdot {13} \nmid {3}^2 \cdot 7 \cdot {13} \cdot {19}^2 \cdot {61} = 18035199 = \sigma({3003}^2) = \sigma({n_{\mathscr{D}}}^2),$$ $$G = \gcd\left(\sigma({q_{\mathscr{D}}}^{k_{\mathscr{D}}}),\sigma({n_{\mathscr{D}}}^2)\right) = \gcd\left(7 \cdot {11}^2 \cdot {13}, {3}^2 \cdot 7 \cdot {13} \cdot {19}^2 \cdot {61}\right) = 7 \cdot {13} = 91 \neq 11011 = \sigma({q_{\mathscr{D}}}^{k_{\mathscr{D}}})/2$$ $$I = \gcd\left(n_{\mathscr{D}},\sigma({n_{\mathscr{D}}}^2)\right) = \gcd\left(3 \cdot 7 \cdot {11} \cdot {13}, {3}^2 \cdot 7 \cdot {13} \cdot {19}^2 \cdot {61}\right) = 3 \cdot 7 \cdot 13 = 273 \neq 3003 = 3 \cdot 7 \cdot {11} \cdot {13} = {n_{\mathscr{D}}}.$$
For an actual odd perfect number $N = q^k n^2$ (assuming such a number exists), since $\sigma(q^k)/2 \mid n^2$ holds in general, then we have the implication:
$$\sigma(q^k)/2 \text{ is squarefree } \implies \sigma(q^k)/2 \mid n$$
Here is my:
QUESTION: Based on the findings from the Descartes spoof $\mathscr{D}$, since $$\sigma({q_{\mathscr{D}}}^{k_{\mathscr{D}}})/2 \nmid n_{\mathscr{D}},$$ does it follow that, in fact, $\sigma(q^k)/2$ is not squarefree for an actual odd perfect number $N = q^k n^2$ with special prime $q$?
At the very least, this appears to corroborate the partial results in this earlier MSE question.
Reference:
R. Steuerwald, "Verschärfung einer notwendigen Bedingung für die Existenz einer ungeraden vollkommenen Zahl," S.-B. Math.-Nat. Abt. Bayer. Akad. Wiss., 1937, pp. 68-73.
