In what follows, we let $\sigma=\sigma_1$ denote the classical sum-of-divisors function. Denote the deficiency function of the positive integer $x$ by $D(x)=2(x)-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)-x$.
Suppose that $N$ is an odd perfect number and $q^\alpha$ is a prime power with $q^\alpha \parallel N$.
(Euler proved that $N$ can be written in the Eulerian form $p^k n^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,n)=1$.)
Define the index $m = \sigma(N/q^\alpha)/q^\alpha$. Chen and Chen (2014) proved that, if $q = p$, then $m$ cannot take any of the $30$ forms
$$q_1, {q_1}^2, {q_1}^3, {q_1}^4, {q_1}^5, {q_1}^6, {q_1}^7, {q_1}^8, q_1 q_2, {q_1}^2 q_2, {q_1}^3 q_2, {q_1}^4 q_2, {q_1}^5 q_2, {q_1}^2 {q_2}^2, {q_1}^3 {q_2}^2, {q_1}^4 {q_2}^2, q_1 q_2 q_3, {q_1}^2 q_2 q_3, {q_1}^3 q_2 q_3,$$ $${q_1}^4 q_2 q_3, {q_1}^2 {q_2}^2 q_3, {q_1}^2 {q_2}^2 {q_3}^2, q_1 q_2 q_3 q_4, {q_1}^2 q_2 q_3 q_4, {q_1}^3 q_2 q_3 q_4, {q_1}^2 {q_2}^2 q_3 q_4, q_1 q_2 q_3 q_4 q_5, {q_1}^2 q_2 q_3 q_4 q_5, q_1 q_2 q_3 q_4 q_5 q_6, q_1 q_2 q_3 q_4 q_5 q_6 q_7$$
where $q_1, q_2, q_3, q_4, q_5, q_6, q_7$ are distinct odd primes. (A similar result is proved if $q \neq p$.)
This seems to suggest the following question:
INITIAL QUESTION
Can it be unconditionally proven that the index $m$ of an odd perfect number cannot be squarefree?
MOTIVATION / MY ATTEMPT
Let $N = p^k n^2$ be an odd perfect number with special/Eulerian prime $p$ (i.e. $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,n) = 1$).
Then we have $$\sigma(p^k)\sigma(n^2) = \sigma(p^k n^2) = \sigma(N) = 2N = 2 p^k n^2$$ so that $$\frac{\sigma(p^k)}{2}\cdot\frac{\sigma(n^2)}{p^k} = n^2.$$
Suppose that $k = 1$. We obtain $$D(n^2) = \frac{\sigma(n^2)}{p} = \frac{n^2}{(p + 1)/2},$$ where we have set $k = 1$ in $$\frac{D(n^2)}{s(p^k)} = \frac{\sigma(n^2)}{p^k} = \frac{n^2}{\sigma(p^k)/2}.$$
But Acquaah and Konyagin proved in (IJNT, 2012) that the implication $$k = 1 \implies p < n\sqrt{3}$$ holds.
Together with the implication $$k > 1 \implies p < n$$ proved by Dris in (JIS, 2012), we obtain the unconditional estimate $p < n\sqrt{3}$.
For simplicity in the computations, we will use the (weaker) estimate $p < 2n$ instead.
First, note that $p \neq 2m - 1$, since $$m = \frac{p+1}{2} \iff m^2 - p^k = m^2 - p = \left(\frac{p+1}{2}\right)^2 - p = \left(\frac{p-1}{2}\right)^2 \text{ is a square}$$ will contradict "$m^2 - p^k$ is not a square". (See: [1] and [2].)
This means that $$k = 1 \implies p < 2n-1 \implies p+1 < 2n \implies \sigma(p) < 2n \implies \sigma(p^k) < 2n$$ $$\implies D(n^2) = \frac{D(n^2)}{s(p^k)} = \frac{2n^2}{\sigma(p^k)} > n$$
Now, the condition $k = 1$ is equivalent to $D(n^2) \mid n^2$.
Since $D(n^2)$ is squarefree, we may write $$D(n^2) = \operatorname{rad(D(n^2)}) \mid n^2$$ from which it follows that $$D(n^2) = \operatorname{rad(D(n^2)}) \mid n.$$
But then this implies that $$n < D(n^2) \leq n,$$ which is a contradiction.
Hence, we conclude that if $k=1$, then the index $m = \sigma(N/q^\alpha)/q^\alpha$ where $q=p$ ($p$ is the special/Eulerian prime) is not squarefree.
FOLLOW-UP QUESTIONS
(1) Do you see an obvious (alternative) way of proving my assertion above?
(2) Can you remove the condition $k=1$ and still prove that the index is not squarefree for $q=p$?
(3) If the answer to Question (2) is YES, can you extend the proof to include the case $q \neq p$?
