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?


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:

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!

  • $\begingroup$ The answer is yes because there are no odd perfect numbers. $\endgroup$
    – markvs
    Oct 6 at 21:49
  • $\begingroup$ @markvs: Are you 100% sure that there are no odd perfect numbers? =) $\endgroup$ Oct 7 at 2:24
  • $\begingroup$ Yes, $100\%$ sure. $\endgroup$
    – markvs
    Oct 7 at 2:26
  • $\begingroup$ Do you have a proof for that assertion, @markvs? $\endgroup$ Oct 7 at 2:27
  • $\begingroup$ This assertion cannot be formally proved at the time but so is the opposite. I also cannot prove by myself that $E=mc^2$, the Fermat's last theorem, and many other things. It does not mean that I am not $100\%$ sure these are true. $\endgroup$
    – markvs
    Oct 7 at 2:34

This is only a partial answer.

Since $q^k n^2$ is (odd) perfect with special prime $q$, then $$\sigma(n^2) = \frac{2n^2 q^k}{\sigma(q^k)}$$ which implies that $\sigma(q^k) \mid 2n^2$, because $\gcd(q^k,\sigma(q^k))=1$. But since $q \equiv k \equiv 1 \pmod 4$, then $\sigma(q^k)/2$ is an integer, so that $$\frac{\sigma(q^k)}{2} \mid n^2.$$

Now, suppose that $\sigma(q^k)/2$ is squarefree.

This implies that $$\frac{\sigma(q^k)}{2} \mid n,$$ so that $$\frac{\sigma(n^2)}{n}=\frac{2n q^k}{\sigma(q^k)}$$ would be an integer, granted $\sigma(q^k)/2$ is indeed squarefree.

Again, by assumption, we have that $\sigma(q^k)/2$ is squarefree. By the considerations in the previous section, this implies that $n \mid \sigma(n^2)$.

But as shown in the previous section, $n \mid \sigma(n^2)$ would imply that $$\frac{\sigma(q^k)}{2} \mid n,$$ which implies that $\sigma(q^k)/2$ is squarefree, since $\sigma(q^k)/2 \mid n^2$.

Hence, we actually have the biconditional $$n \mid \sigma(n^2) \iff \frac{\sigma(q^k)}{2} \mid n \iff \frac{\sigma(q^k)}{2} \text{ is squarefree.}$$

Therefore, the problem of whether $n$ divides $\sigma(n^2)$ reduces to determining if $\sigma(q^k)/2$ is squarefree.

In particular, note that we have shown that $$\frac{\sigma(q^k)}{2} = \gcd(\sigma(q^k),\sigma(n^2))$$ if $\sigma(q^k)/2$ is squarefree, so that $$3 \leq \frac{\sigma(q^k)}{2} = \gcd(\sigma(q^k),\sigma(n^2)).$$


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