Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. (Note that the divisor sum $\sigma$ is a multiplicative function.)
A number $P$ is said to be perfect if $\sigma(P)=2P$. If a perfect number $N$ is odd, then $N$ is called an odd perfect number. Euler proved that a hypothetical odd perfect number $N$ must have the form
$$N = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is known that
$$i(q)=\gcd(n^2,\sigma(n^2))=\frac{n^2}{\sigma(q^k)/2}=\frac{\sigma(n^2)}{q^k},$$
where $i(q)=\sigma(N/{q^k})/{q^k}$ is the index of $N$ at the (special) prime $q$, as initially defined by Broughan, Delbourgo, and Zhou, and whose results were eventually improved upon by Chen and Chen.
In a recent preprint, Dris proves that the following implication holds:
$$i(q) \text{ is squarefree } \implies \frac{\sigma(q^k)}{2} \text{ is not squarefree.}$$
We likewise obtain the biconditional
$$i(q) \text{ is a square } \iff \frac{\sigma(q^k)}{2} \text{ is a square.}$$
This implies that we have the chain of implications
$$i(q) \text{ is a square } \implies \frac{\sigma(q^k)}{2} \text{ is a square } \implies \frac{\sigma(q^k)}{2} \text{ is not squarefree.}$$
These findings highly suggest that $\sigma(q^k)/2$ is not squarefree.
My question is as follows:
What are the remaining cases to consider for this problem, specifically all the possible premises for $i(q)$?
Best Answer
FYI, here says that "Any arbitrary positive integer $n$ can be represented in a unique way as the product of a square and a square-free integer: $n=m^{2}k$. In this factorization, $m$ is the largest divisor of $n$ such that $m^2$ is a divisor of $n$". Using this, one can say that $m=1$ if and only if $n$ is squarefree, and that $k=1$ if and only if $n$ is a square. Therefore, the remaining case is when $m\gt 1$ and $k\gt 1$.