Preamble: The present inquiry is an offshoot of What are the remaining cases to consider for this problem, specifically all the possible premises for $i(q)$?.
MOTIVATION
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:
$$i(q) \text{ is squarefree } \implies \frac{\sigma(q^k)}{2} \text{ is not squarefree.} \tag{1}$$
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
$$i(q) \text{ is a square } \implies \frac{\sigma(q^k)}{2} \text{ is a square } \implies \frac{\sigma(q^k)}{2} \text{ is not squarefree.} \tag{2}$$
This MSE answer proves the following Conjecture:
If $q^k n^2$ is an odd perfect number with special prime $q$ and $q = k$, then $\sigma(q^k)/2$ is not squarefree.
These findings highly suggest that $\sigma(q^k)/2$ is not squarefree.
My question is as follows:
Do you see a way of proving that $\sigma(q^k)/2$ is not squarefree?
MY ATTEMPT
Suppose to the contrary that $\sigma(q^k)/2$ is squarefree. Since
$$i(q) = \frac{n^2}{\sigma(q^k)/2}$$
and $i(q)$ is an (odd) integer, then $\sigma(q^k)/2 \mid n^2$. Now, the assumption that $\sigma(q^k)/2$ is squarefree would imply that $\sigma(q^k)/2 \mid n$.
But we can rewrite
$$\frac{n^2}{\sigma(q^k)/2}=\frac{\sigma(n^2)}{q^k}$$
as
$$\frac{\sigma(n^2)}{n}=\frac{q^k n}{\sigma(q^k)/2}$$
which means that $\sigma(q^k)/2 \mid n$ is equivalent to $n \mid \sigma(n^2)$, since $q^k$ and $\sigma(q^k)/2$ are coprime.
Now, let
$$G = \gcd(\sigma(q^k),\sigma(n^2)) = \sigma(q^k)/2$$
$$H = i(q) = \gcd(n^2,\sigma(n^2)) = \frac{n^2}{\sigma(q^k)/2}$$
$$I = \gcd(n,\sigma(n^2)) = n$$
$$J = \frac{n}{\gcd(\sigma(q^k)/2,n)} = \frac{n}{\sigma(q^k)/2}$$
Since
$$H = G \times J^2$$
and because of the following (which hold under the assumption that $G=\sigma(q^k)/2$ is squarefree):
(1) $J = 1$ if and only if $H$ is squarefree. (Note that, under the assumption that $\sigma(q^k)/2$ is squarefree, we get that $H$ is not squarefree. Therefore, $\sigma(q^k)/2$ is squarefree implies that $J > 1$.)
(2) $G = 1$ if and only if $H$ is a square. (Note that $G = \sigma(q^k)/2 \geq \frac{q^k + 1}{2} \geq 3$, so that $H$ is not a square, if $G = \sigma(q^k)/2$ is squarefree. This confirms the findings in this MO answer to a closely related question.)
(3) The remaining case is when $G>1$ and $J>1$.
But $G$ is squarefree, together with the following identity
$$G \times H = I^2$$
implies that
$$G \mid I.$$
Throughout this paper, we implicitly rely on the simple equality
$$\sigma(n^2) = \frac{2q^k n^2}{\sigma(q^k)}. \tag{3}$$
Unfortunately, this seems to introduce fractions. To avoid that, we can use prime factorizations, as follows. Write the prime factorization of $n$ as
$$n = {p_1}^{a_1} \cdots {p_m}^{a_m},$$
for some unique odd primes $3 \leq p_1 < \ldots < p_m$, and for some positive integer exponents $a_1, \ldots, a_m$. Since both sides of $(3)$ are integers, and since $q \equiv k \equiv 1 \pmod 4$ with $q$ prime, we know that
$$\sigma(q^k) = 2 {p_1}^{b_1} \cdots {p_m}^{b_m}$$
for some nonnegative integers $0 \leq b_i \leq 2a_i$. Thus, we have
$$\sigma(n^2) = q^k {p_1}^{2a_1 – b_1} \cdots {p_m}^{2a_m – b_m}.$$
With this information, we immediately see that
$$G := \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\left(2 {p_1}^{b_1} \cdots {p_m}^{b_m},q^k {p_1}^{2a_1 – b_1} \cdots {p_m}^{2a_m – b_m}\right)$$
$$= {p_1}^{\min(b_1,2a_1 – b_1)} \cdots {p_m}^{\min(b_m,2a_m – b_m)},$$
$$H := \gcd\left(n^2,\sigma(n^2)\right) = \gcd\left({p_1}^{2a_1} \cdots {p_m}^{2a_m}, q^k {p_1}^{2a_1 – b_1} \cdots {p_m}^{2a_m – b_m}\right)$$
$$= {p_1}^{2a_1 – b_1} \cdots {p_m}^{2a_m – b_m},$$
and
$$I := \gcd\left(n,\sigma(n^2)\right) = \gcd\left({p_1}^{a_1} \cdots {p_m}^{a_m}, q^k {p_1}^{2a_1 – b_1} \cdots {p_m}^{2a_m – b_m}\right)$$
$$= {p_1}^{\min(a_1,2a_1 – b_1)} \cdots {p_m}^{\min(a_m,2a_m – b_m)}.$$
Lastly, I know that
$$\gcd(G,J)={p_1}^{\min\left(\min(b_1,2a_1 – b_1),2a_1 – b_1 – \min(a_1,2a_1 – b_1)\right)} \cdots {p_m}^{\min\left(\min(b_m,2a_m – b_m),2a_m – b_m – \min(a_m,2a_m – b_m)\right)}$$
$$={p_1}^{\min\left(2a_1 – b_1,b_1,2a_1 – b_1 – \min(a_1,2a_1 – b_1)\right)} \cdots {p_m}^{\min\left(2a_m – b_m,b_m,2a_m – b_m – \min(a_m,2a_m – b_m)\right)}$$
$$={p_1}^{\min\left(b_1,\min(2a_1 – b_1,2a_1 – b_1 – \min(a_1,2a_1 – b_1))\right)} \cdots {p_m}^{\min\left(b_m,\min(2a_m – b_m,2a_m – b_m – \min(a_m,2a_m – b_m))\right)}$$
$$={p_1}^{\min\left(b_1,2a_1 – b_1 – \min(a_1,2a_1 – b_1)\right)} \cdots {p_m}^{\min\left(b_m,2a_m – b_m – \min(a_m,2a_m – b_m)\right)}.$$
Alas, this is where I get stuck! (I currently do not see a way of arriving at a contradiction from assuming that $\sigma(q^k)/2$ is squarefree.)
Added from a comment on 02/28/2023 – 8:03 PM – Manila time I noticed that, even without assuming at the outset that $G=\sigma(q^k)/2$ is squarefree, we obtain
$$J = \frac{H}{I} = \frac{I}{G},$$
and
$$G \times H = I^2 \implies G \mid I^2$$
so that
$$\gcd(\sigma(q^k),\sigma(n^2)) = G \mid I = \gcd(n,\sigma(n^2))$$
and
$$G \mid I^2$$
both hold. Does this finding mean that, in fact, $\sigma(q^k)/2$ must be squarefree?
Best Answer
Using the idea of this answer, one can prove the following claim :
Claim : If $k=(2a-1)q+2a-2$ with $a\geqslant 1$, then $\sigma(q^k)/2$ is not squarefree.
Proof :
Letting $s:=q+1$, we have $$\begin{align}q^{k+1}-1&=(s-1)^{k+1}-1 \\\\&\equiv (-1)^{k+1}+(k+1)s(-1)^{k}-1\pmod{s^2} \\\\&\equiv 1+(2aq-q+2a-1)s(-1)-1\pmod{s^2} \\\\&\equiv -(2as-2a-s+1+2a-1)s\pmod{s^2} \\\\&\equiv 0\pmod{s^2}\end{align}$$ Since $s=q+1\equiv 2\pmod 4$ with $s\geqslant 6$, there is an odd prime $P$ which divides $s$. The claim follows from $P^2\mid s^2\mid q^{k+1}-1$.$\quad\blacksquare$