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

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:

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!

• The answer is yes because there are no odd perfect numbers. Oct 6 at 21:49
• @markvs: Are you 100% sure that there are no odd perfect numbers? =) Oct 7 at 2:24
• Yes, $100\%$ sure. Oct 7 at 2:26
• Do you have a proof for that assertion, @markvs? Oct 7 at 2:27
• 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. 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 $$3 \leq \frac{\sigma(q^k)}{2} = \gcd(\sigma(q^k),\sigma(n^2)).$$