# Does $\sigma(n^2)/q \mid q^k n^2$ imply $\sigma(n^2)/q \mid n^2$, if $q^k n^2$ is an odd perfect number with special prime $q$? - Part II

(Preamble: This inquiry is an offshoot of this MSE question.)

Denote the classical sum of divisors of the positive integer $$x$$ by $$\sigma(x)=\sigma_1(x)$$. Denote the abundancy index of $$x$$ as $$I(x)=\sigma(x)/x$$.

Recall that an odd perfect number $$N = q^k n^2$$ is said to be given in Eulerian form if $$q$$ is the special prime satisfying $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$.

Also, elsewhere we showed that $$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$

In the hyperlinked post, we showed that we have the equation $$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q},$$ from which it follows that $$\frac{N}{\sigma(n^2)/q} = \frac{1}{I(n^2) - \frac{2(q - 1)}{q}} = \frac{q\left(q^{k+1} - 1 \right)}{2(q - 1)} = \frac{q\sigma(q^k)}{2},$$ since $$I(n^2) = \frac{2}{I(q^k)} = \frac{2q^k (q - 1)}{q^{k+1} - 1}.$$

This shows that the divisibility condition $$\sigma(n^2)/q \mid N = q^k n^2$$ holds.

Under the assumption $$k=1$$, since $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = \gcd\left(q,\frac{\sigma(n^2)}{q^k}\right) = \gcd\left(q,\gcd(n^2,\sigma(n^2))\right) = \gcd(\gcd(q,n^2),\sigma(n^2)) = \gcd(1,\sigma(n^2)) = 1,$$ then it follows from $$\sigma(n^2)/q \mid q^k n^2$$ and $$\gcd(q,\sigma(n^2)/q)=1$$ that $$\sigma(n^2)/q \mid n^2$$, which implies that $$k=1$$.

Under the assumption $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = 1$$ we obtain $$\sigma(n^2)/q \mid n^2$$, which implies that $$k=1$$.

Hence, we actually have the biconditional $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = 1 \iff k = 1. \tag{*}$$

Here is my question:

Is my proof of the biconditional depicted in the section marked with a (*) correct? If not, how can it be mended so as to produce a valid argument?

This answer presents an alternative proof for the argument in the original post, and shows that the argument is indeed valid.

Taking off from Devansh Singh's preprint in ResearchGate, we have $$\frac{\sigma(n^2)}{q} = \gcd\left(q^{k-1} n^2, \sigma(q^{k-1})\sigma(n^2) - q^{k-1} n^2\right).$$

Therefore, $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = \gcd\left(q,\gcd\Bigg(q^{k-1} n^2, \sigma(q^{k-1})\sigma(n^2) - q^{k-1} n^2\Bigg)\right).$$

Suppose that $$k \neq 1$$.

It follows that $$\gcd\left(q,\gcd\Bigg(q^{k-1} n^2, \sigma(q^{k-1})\sigma(n^2) - q^{k-1} n^2\Bigg)\right) = q \geq 5,$$ since $$\gcd(q, q^{k-1} n^2) = q$$ under the assumption $$k \neq 1$$, and also $$q \mid \sigma(n^2)$$ and $$q \mid q^{k-1}$$. Hence we have that $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) \neq 1.$$

On the other hand, suppose that $$k=1$$.

It follows that $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right)=\gcd\left(q,\gcd\Bigg(q^{k-1} n^2, \sigma(q^{k-1})\sigma(n^2) - q^{k-1} n^2\Bigg)\right)=\gcd\left(q,\gcd\Bigg(n^2, \sigma(n^2) - n^2\Bigg)\right)=\gcd\left(\gcd(q,n^2),\sigma(n^2) - n^2\right)=\gcd\left(1,\sigma(n^2)-n^2\right)=1.$$

QED

• To the quick downvoter: Would you mind sharing feedback regarding your vote? As it is, I am totally clueless. Any form of feedback, hopefully constructive, would go a long way towards improving future answers/posts. Oct 18, 2022 at 5:25