Let $\sigma(x)=\sigma\_1(x)$ be the *classical sum of divisors* of the positive integer $x$.


It is known that
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))}$$
if $q^k n^2$ is an odd perfect number with special prime $q$.


Hence, **if it is known that $n \mid \sigma(n^2)$**, then it follows that
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}$$
since we can compute
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot\Bigg(\gcd\bigg(1,\frac{\sigma(n^2)}{n}\bigg)\Bigg)^2.$$


But since $n^2 \nmid \sigma(n^2)$, then $n \mid \sigma(n^2)$ implies that
$$n\cdot{\frac{\sigma(q^k)}{2}}=\gcd(n^2,\sigma(n^2))\cdot\gcd(\sigma(q^k),\sigma(n^2))=\Bigg(\gcd(n,\sigma(n^2))\Bigg)^2=n^2$$
from which we obtain
$$\frac{\sigma(q^k)}{2}=n=\frac{\sigma(n^2)}{q^k}.$$




---


**Edit: October 5, 2021 - 1:56 PM Manila time** I have just found a gap in the proof. If $n \mid \sigma(n^2)$, then it does not follow from $n^2 \nmid \sigma(n^2)$ that $\gcd(n^2, \sigma(n^2)) = n$. (In fact, since $\sigma(n^2) = cn$ for some $c > (8/5)n$, then $c$, ~~which is just a proper divisor of $n$~~, must be large.) We then have
$$\gcd(n^2, \sigma(n^2)) = cn$$
which contradicts
$$\gcd(n^2, \sigma(n^2)) = n$$
to several orders of magnitude.




---


(We can then derive the estimates
$$\frac{8n}{5} < q^k < 2n$$
by considering either the resulting abundancy index of $q^k$ or that of $n^2$.)
Note that we then have
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}=n=\gcd(n^2,\sigma(n^2))$$
under the assumption that $n \mid \sigma(n^2)$.


Here is my question:



> 
> Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?
> 
> 
> 


**MY ATTEMPT**


I tried checking for examples of numbers $n$ satisfying the divisibility constraint
$$n \mid \sigma(n^2)$$
using a Pari-GP script, via [Sage Cell Server](https://sagecell.sagemath.org/?z=eJxLyy_SqLA10lEwNAADHYXMNA0N3_wUjeLM9NxEjYo4I02dCk0FW1sFA02dgqLMvBKNCp20xOQSkEZNIAAAZksS2Q==&lang=gp&interacts=eJyLjgUAARUAuQ==):



```
for(x=2, 1000000, if((Mod(sigma(x^2),x) == 0),print(x,factor(x))))

```

Here is the output:



```
39[3, 1; 13, 1]
793[13, 1; 61, 1]
2379[3, 1; 13, 1; 61, 1]
7137[3, 2; 13, 1; 61, 1]
13167[3, 2; 7, 1; 11, 1; 19, 1]
76921[13, 1; 61, 1; 97, 1]
78507[3, 2; 11, 1; 13, 1; 61, 1]
230763[3, 1; 13, 1; 61, 1; 97, 1]
238887[3, 2; 11, 1; 19, 1; 127, 1]
549549[3, 2; 7, 1; 11, 1; 13, 1; 61, 1]
692289[3, 2; 13, 1; 61, 1; 97, 1]
863577[3, 2; 11, 2; 13, 1; 61, 1]

```

Note that all of the known examples are **odd**.


**Alas, this is where I get stuck!**



Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?
Let $p^k m^2$ be an odd perfect number with special prime $p$.


It follows that
$$\frac{\sigma(m^2)}{p^k}\cdot\frac{\sigma(p^k)}{2}=m^2.$$


Let $t\_1 = \sigma(m^2)/p^k$, $t\_2 = \sigma(p^k)/2$. It follows that $m^2 = t\_1 t\_2$.


Now define the GCDs
\begin{align\*}
G&=\gcd(\sigma(p^k),\sigma(m^2))=\gcd(\sigma(p^k)/2,\sigma(m^2)/p^k) =\gcd(\sigma(p^k)/2,H) \\
H&=\gcd(m^2,\sigma(m^2))=\gcd(m^2,\sigma(m^2)/p^k)=\gcd(m^2,H) \\
I&=\gcd(m,\sigma(m^2))=\gcd(m,\sigma(m^2)/p^k))= \gcd(m,H).
\end{align\*}


It follows that
$$G = \gcd\left(\sigma(p^k)/2,H\right) = \gcd(t\_1, t\_2)$$
$$H = \gcd(m^2,H) = \gcd\left(t\_1 t\_2, p^k t\_1\right) = t\_1 \gcd(t\_2, p^k) = t\_1$$
$$I = \gcd(m,H) = \gcd\left(\sqrt{t\_1 t\_2}, p^k t\_1\right) = \sqrt{t\_1}\gcd\left(\sqrt{t\_2}, p^k \sqrt{t\_1}\right). \tag{\*}$$


These findings follow from Equation $(\*)$:



> 
> **(1)** Since $t\_1$ is a square if and only if $t\_2$ is a square, if $t\_2 = \sigma(p^k)/2$ is a square, then we have
> $$I = \sqrt{H}\cdot\gcd\left(\sqrt{\sigma(p^k)/2},p^k \sqrt{H}\right) 
>  = \sqrt{H}\cdot\gcd\left(\sqrt{\sigma(p^k)/2}, \sqrt{H}\right)$$
> $$= \sqrt{H}\cdot\sqrt{\gcd(\left(\sigma(p^k)/2,H\right)} = \sqrt{HG},$$
> which checks with the expected result $G \times H = I^2$.
> 
> 
> 



> 
> **(2)** Note that neither $t\_1$ nor $t\_2$ could be squarefree in Equation $(\*)$, since it would make the $RHS$ irrational, while the $LHS$ must necessarily be an integer. (Another way to see it is via the following (lengthier) proof: The implication
> $$\left(t\_1 \text{ is squarefree}\right) \implies \left(t\_1 \mid I\right)$$
> holds. But $I \mid t\_1$ since $I=\gcd(m,\sigma(m^2))$ and $t\_1 = \gcd(m^2,\sigma(m^2))$. Hence, we have that $t\_1 \text{ is squarefree } \implies t\_1 = I$. Therefore, $t\_1=I=t\_2$. This implies that $J=\frac{t\_1}{t\_2}=1$. Consequently, by the contrapositive to [Theorem **4.2 (page $5$)** of this paper](https://arxiv.org/abs/2202.08116), $t\_2 = \sigma(p^k)/2$ is *not squarefree*.)
> 
> 
> **(3)** Lastly, note that if $t\_1$ and $t\_2$ are both not squares and not squarefree, then again $\sqrt{t\_1}$ and $\sqrt{t\_2}$ would be irrational, making the $RHS$ of Equation $(\*)$ irrational, contradicting the fact that the $LHS$ must necessarily be an integer.
> 
> 
> 


I believe this rough sketch for a proof argument that $\sigma(p^k)/2$ (and therefore, $\sigma(m^2)/p^k$) must necessarily be squares holds water. (I thank Professor [Pace Nielsen](https://mathoverflow.net/users/3199) for giving the appropriate hints in this [MO answer](https://mathoverflow.net/a/446925/10365).) In particular, this appears to prove that the Descartes-Frenicle-Sorli Conjecture necessarily holds for an odd perfect number. (See [this paper](http://math.colgate.edu/%7Eintegers/n39/n39.pdf) for a proof of the implication
$$\sigma(p^k)/2 = \square \implies k = 1,$$
where $x = \square$ means that $x$ is a square.)


Note that
$$\gcd(\sigma(p^k),\sigma(m^2)) = G = t\_2 = \sigma(p^k)/2$$
**and, therefore**
$$I = \gcd(m,\sigma(m^2)) = m,$$
follows directly from
$$G \times H = I^2$$
and
$$(\sigma(p^k)/2)\cdot(\sigma(m^2)/p^k)=m^2,$$
since $H=\sigma(m^2)/p^k$.


