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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?

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  • $\begingroup$ I noticed that, even without assuming at the outset that $G=\sigma(q^k)/2$ is squarefree, we obtain $$J = \frac{H}{I} = \frac{I}{G}$$ so that $$\gcd(\sigma(q^k),\sigma(n^2)) = G \mid I = \gcd(n,\sigma(n^2))$$ and $$G \times H = I^2 \implies G \mid I^2$$ both hold. Does this finding mean that, in fact, $\sigma(q^k)/2$ must be squarefree? $\endgroup$ Commented Feb 28, 2023 at 11:50
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    $\begingroup$ Using the idea of this answer, one can prove that if $k=(2a-1)q+2a-2$ with $a\geqslant 1$, then $\sigma(q^k)/2$ is not squarefree. $\endgroup$
    – mathlove
    Commented Mar 3, 2023 at 10:44
  • $\begingroup$ That's interesting, @mathlove! Can you expand your last comment into an actual answer, and fill in the details as needs be? Thanks! $\endgroup$ Commented Mar 3, 2023 at 11:33

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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 $P^2\mid s^2\mid q^{k+1}-1$.$\quad\blacksquare$

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    $\begingroup$ Thank you for this answer, @mathlove! I will wait for a week, and if there are no other answers, I will accept your answer. =) $\endgroup$ Commented Mar 3, 2023 at 14:13

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