# Determining whether $\sigma(q^k)/2$ is squarefree, where $q^k n^2$ is an odd perfect number with special prime $q$

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?

• 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? Commented Feb 28, 2023 at 11:50
• 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. Commented Mar 3, 2023 at 10:44
• That's interesting, @mathlove! Can you expand your last comment into an actual answer, and fill in the details as needs be? Thanks! Commented Mar 3, 2023 at 11:33

Claim : If $$k=(2a-1)q+2a-2$$ with $$a\geqslant 1$$, then $$\sigma(q^k)/2$$ is not squarefree.
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 $$P^2\mid s^2\mid q^{k+1}-1$$.$$\quad\blacksquare$$