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Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.

A number $M$ is said to be perfect if $\sigma(M)=2M$. For example, $6$ and $28$ are perfect since $$\sigma(6) = 1 + 2 + 3 + 6 = 2\cdot{6}$$ and $$\sigma(28) = 1 + 2 + 4 + 7 + 14 + 28 = 2\cdot{28}.$$

It is currently unknown if there are infinitely many even perfect numbers. It is also an open problem whether any odd perfect numbers exist. It is widely believed that there are no odd perfect numbers.

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

QUESTION

Here is my question:

Can the odd perfect number $N$ be of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n}?$$

I would certainly appreciate it if anybody could point me to papers/articles/publications in the literature where this particular inquiry is covered.

CONTEXT

Slowak (1999) proved that the odd perfect number $N$ must be of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{d},$$ where $d > 1$.

Dris (2017) showed further that $d$ must have the form $$\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$.

Jose Arnaldo Bebita Dris
  • 8,471
  • What is a difference between he claim in your question an the claim from cited Slowak’s result? Is it required that $(q,n)=1$, whereas not necessarily $(d,n)=?$ – Alex Ravsky Aug 20 '20 at 04:49
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    Thank you for your comment, @AlexRavsky. Slowak does not necessarily claim that $d=n$. Yes, it is necessarily true that $\gcd(q,n)=1$, while it is also necessarily true that $$\gcd(d,n) > 1$$ since $\frac{\sigma(q^k)}{2} \mid n^2$ and $$\sigma(q^k) < \frac{2n^2}{3}.$$ But if I may just ask, why are you considering $\gcd(d,n)$? – Jose Arnaldo Bebita Dris Aug 20 '20 at 05:17
  • Opps, $(d,n)=?$ was a misprint, sorry. There should be $\operatorname{gcd}(d,q)=1$ instead. – Alex Ravsky Aug 20 '20 at 05:23
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    I meant $$\sigma(q^k) \leq \frac{2n^2}{3}$$ in my last inequality. Yes, $\gcd(d,q)=1$ follows readily from $\gcd(q,n)=1$. The divisibility constraint $\gcd(q,n)=1$ necessarily holds since prime powers are deficient. – Jose Arnaldo Bebita Dris Aug 20 '20 at 06:34

1 Answers1

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(In what follows, the quantities $G$, $H$, and $J$ are defined in this publication.)

Suppose that $N = q^k n^2$ is an odd perfect number given in the form $$N = \left(\frac{{q^k}\sigma(q^k)}{2}\right)\cdot{n}.$$

In other words, suppose that $\sigma(q^k)/2 = n$. (Note that this means that $\sigma(q^k)/2$ is not squarefree since otherwise, this will contradict a result of Steuerwald from the year 1937, that $n$ must contain a square factor.)

Note that this assumption implies $$1=J=\frac{n}{\sigma(q^k)/2}$$ since $$H=\gcd\left(n^2,\sigma(n^2)\right)=\frac{n^2}{\sigma(q^k)/2}={\sigma(q^k)/2}\times\Bigg(\frac{n}{\sigma(q^k)/2}\Bigg)^2$$

In particular, $H$ is squarefree (since $H = G \times J^2$ and $J=1$). We infer that $G=\sigma(q^k)/2$ is likewise squarefree. This contradicts our earlier findings.

Hence, $H$ is not squarefree, which is equivalent to $J \neq 1$. In particular, this means that $\sigma(q^k)/2 \neq n$.

QED

Jose Arnaldo Bebita Dris
  • 8,471