# 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 $q$?

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 $$k=1$$.

• 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. Commented Sep 11, 2021 at 9:36

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 $$k=1$$.

• 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. Commented Jul 13, 2021 at 15:12