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

I discovered an interesting identity involving divisors of odd perfect numbers given in the Eulerian form $N = q^k n^2$ today (July 13, 2021). (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 $\gcd(q,n)=1$.)

The identity is:

Proposition: If $N = q^k n^2$ is an odd perfect number with special prime $q$, then $$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$

Proof:

Our starting point is the following blog post, where it is proved that

$$\gcd(n^2, \sigma(n^2)) = 2(1 - q)n^2 + q\sigma(n^2).$$

However, note that we have

$$\frac{\sigma(n^2)}{q^k} = \gcd(n^2, \sigma(n^2)).$$

These equations are equivalent to

$$2(1 - q)n^2 + q\sigma(n^2) = \frac{\sigma(n^2)}{q^k}.$$

Factoring out $qn^2$ on the LHS, we obtain

$$qn^2 \Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q^k}.$$

Multiplying both sides of the last equation by $q^{k-1}$, we get

$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$

(Note that the RHS of the last equation is an odd integer.)

This concludes our proof.

QED.

In particular, we have

$$\frac{N}{\sigma(n^2)/q} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)}.$$

But we also know from the following MSE post that

$$I(n^2) - \frac{2(q - 1)}{q} = \frac{2(q - 1)}{q\bigg(q^{k+1} - 1\bigg)}.$$

This means that we obtain

$$\frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\bigg(q^{k+1} - 1\bigg)}{2(q - 1)} = \frac{q\sigma(q^k)}{2}.$$

But $\sigma(q^k) \equiv k + 1 \pmod 4$, since $q \equiv 1 \pmod 4$, and since $k \equiv 1 \pmod 4$, then $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$

This finding implies that $\sigma(n^2)/q$ divides $N = q^k n^2$.

Here is my:

QUESTION: Does $\sigma(n^2)/q \mid q^k n^2$ imply that $\sigma(n^2)/q \mid n^2$? If not, under what condition(s) does this implication hold?

Note that a proof for $\sigma(n^2)/q \mid n^2$ would imply the Descartes-Frenicle-Sorli Conjecture that $k=1$.

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  • $\begingroup$ In fact, since $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2,$$ then we obtain $$\bigg(\frac{\sigma(q^k)}{2}\bigg)\cdot{\sigma(n^2)}=N=q^k n^2,$$ so that $$\sigma(n^2) \mid N = q^k n^2.$$ This implies that $$\frac{\sigma(n^2)}{q} \mid q^{k-1} n^2,$$ since the constraint $\gcd(q^k,\sigma(q^k))=1$ holds. $\endgroup$ Commented Sep 11, 2021 at 9:36

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This is a partial answer: I merely wanted to collect some more thoughts that recently occurred to me about this problem, after I have posted the question.


We have $$\dfrac{N}{\sigma(n^2)/q} = \dfrac{q\sigma(q^k)}{2}$$ $$\dfrac{q^k n^2}{\sigma(n^2)/q} = \dfrac{q\sigma(q^k)}{2}$$ $$\dfrac{n^2}{\sigma(n^2)/q} = \dfrac{q}{2}\cdot{I(q^k)}.$$

Note that both $q/2$ and $1 < I(q^k) < 5/4$ are non-integers.

This does not necessarily mean, however, that $$\dfrac{n^2}{\sigma(n^2)/q} = \dfrac{q}{2}\cdot{I(q^k)}$$ is a non-integer, as the RHS is equal to $(q+1)/2$ when $k=1$, which is an integer because $q \equiv 1 \pmod 4$.


I therefore conclude that a necessary and sufficient condition for $\sigma(n^2)/q \mid q^k n^2$ to imply $\sigma(n^2)/q \mid n^2$ is $k=1$.

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    $\begingroup$ May I know why this particular answer was downvoted? Any form of feedback, hopefully constructive, would go a long way in improving future answers/posts. $\endgroup$ Commented Jul 13, 2021 at 15:12

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