Is ${n^2}/D(n^2) \in \mathbb{N}$, if $q^k n^2$ is an odd perfect number?

Let $x \in \mathbb{N}$, the set of positive integers. The sum of the divisors of $x$ is denoted by $\sigma(x)$. Denote the deficiency of $x$ by $D(x):=2x-\sigma(x)$, and the sum of the aliquot parts of $x$ by $s(x):=\sigma(x)-x$. Finally, denote the abundancy index of $x$ by $I(x):=\sigma(x)/x$.

If $m$ is odd and $\sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that an odd perfect number (if one exists) must have the form $m=q^k n^2$ where $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Here is my question:

Is ${n^2}/D(n^2) \in \mathbb{N}$, if $q^k n^2$ is an odd perfect number?

MY ATTEMPT

From the fundamental equality $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}$$ one can derive $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))$$ so that we ultimately have $$\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}=\gcd(n^2,\sigma(n^2)).$$

Thus, we have $$y:=\frac{n^2}{D(n^2)}=\frac{{n^2}}{s(q^k)\gcd(n^2,\sigma(n^2))}=\frac{{n^2}D(q^k)}{2s(q^k)s(n^2)}.$$

In particular, we obtain $$\frac{n^2}{D(n^2)}=\frac{{n^2}}{s(q^k)\gcd(n^2,\sigma(n^2))}=\frac{\sigma(q^k)}{2s(q^k)}.$$

Using the reasoning in this blog post, we obtain $$\frac{q}{2} + \frac{1}{2{q^{k-1}}} - \frac{1}{2q^k} < y \leq \frac{q}{2} + \frac{1}{2{q^{k-1}}}.$$

Equality holds if and only if $k=1$.

That is, it appears that $y \in \mathbb{N}$ if and only if $k=1$ (i.e., the Descartes-Frenicle-Sorli conjecture holds).

So the question "Is ${n^2}/D(n^2) \in \mathbb{N}$?" is equivalent to asking whether $k=1$, for $m=q^k n^2$ an odd perfect number with special / Euler prime $q$.

• Unless n is prime is your first line necessarily true? – Steve B Jul 20 '18 at 3:02
• @SteveB, the first line $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}$$ is necessarily true since $m=q^k n^2$ is perfect implies that $\sigma(m)=2m$. Now use the fact that $\sigma$ is (weakly) multiplicative and that $\gcd(q,n)=1$. You will also need to use the fact that $\gcd(q^k, \sigma(q^k))=1$. – Jose Arnaldo Bebita-Dris Jul 20 '18 at 7:00
• @SteveB, by the way, it is known (by work of Nielsen, 2015) that $\omega(n) \geq 9$, where $\omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite. – Jose Arnaldo Bebita-Dris Jul 20 '18 at 13:57

We have $$\frac{n^2}{D(n^2)}\in\mathbb N\iff k=1$$
If $$\frac{n^2}{D(n^2)}\in\mathbb N$$, then we have $$\frac{2n^2}{D(n^2)}=\frac{2n^2}{2n^2-\sigma(n^2)}=\frac{2n^2}{2n^2-\frac{2n^2q^k}{\sigma(q^k)}}=\frac{2n^2}{2n^2-\frac{2n^2q^k(q-1)}{q^{k+1}-1}}=q+\frac{q-1}{q^k-1}\in\mathbb N$$ from which we have to have $$\frac{q-1}{q^k-1}\in\mathbb N$$ from which we have to have $$\frac{q-1}{q^k-1}\ge 1\implies q-1\ge q^k-1\implies q^{k-1}\le 1$$ from which $$k=1$$ follows.
If $$k=1$$, then $$\frac{n^2}{D(n^2)}=\frac{q^{2}-1}{2(q-1)}=\frac{q+1}{2}\in\mathbb N\quad\square$$