Preamble: This post is an attempt to resolve the problem in this earlier question.

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 $\gcd(q,n)=1$.


Here is the problem that I am considering:

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


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 $$\frac{D(n^2)}{s(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.


Dris and San Diego (Theorem 5.6, page 22) proved that $$D(n^2) > \frac{q+1}{2}.$$

Since it is not true that $$\frac{D(n^2)}{s(q^k)} \leq D(n^2) \leq \frac{q+1}{2} \leq \frac{\sigma(q^k)}{2},$$ can we then say that $$\frac{D(n^2)}{s(q^k)} > \frac{\sigma(q^k)}{2}?$$

If so, this would imply that $$\frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} > \frac{\sigma(q^k)}{2}$$ which would mean that $$n > \frac{\sigma(q^k)}{2}.$$

In particular, we have that $$\frac{2n}{\sigma(q^k)} > 1$$ which implies that $$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} > n > \frac{\sigma(q^k)}{2}.$$

This appears to prove that $\sigma(n^2)/q^k \neq n$.

Here is my:

QUESTION: Does this proof hold water? If not, how can it be mended so as to produce a valid proof?

  • $\begingroup$ FYI, one has $\dfrac{D(n^2)}{s(q^k)} > \dfrac{\sigma(q^k)}{2}\iff \sigma(q^k)\lt 2n$. $\endgroup$
    – mathlove
    Commented Aug 1, 2021 at 11:37
  • $\begingroup$ Agreed, @mathlove! Please write out your last comment into a full answer so that I can upvote then accept. =) $\endgroup$ Commented Aug 1, 2021 at 11:44

1 Answer 1


Too long to comment :

One has $$\dfrac{D(n^2)}{s(q^k)} > \dfrac{\sigma(q^k)}{2}\iff \sigma(q^k)\lt 2n$$ since $$\begin{align}\frac{D(n^2)}{s(q^k)} \gt \frac{\sigma(q^k)}{2} &\iff \sigma(q^k)s(q^k)-2D(n^2)\lt 0 \\\\&\iff \sigma(q^k)(\sigma(q^k)-q^k)-2(2n^2-\sigma(n^2))\lt 0 \\\\&\iff \sigma(q^k)(\sigma(q^k)-q^k)-2\bigg(2n^2-\frac{2q^kn^2}{\sigma(q^k)}\bigg)\lt 0 \\\\&\iff \sigma(q^k)(\sigma(q^k)-q^k)-\frac{4n^2(\sigma(q^k)-q^k)}{\sigma(q^k)}\lt 0 \\\\&\iff (\sigma(q^k)-q^k)\bigg(\sigma(q^k)-\frac{4n^2}{\sigma(q^k)}\bigg)\lt 0 \\\\&\iff (\sigma(q^k)-q^k)\cdot \frac{(\sigma(q^k)+2n)(\sigma(q^k)-2n)}{\sigma(q^k)}\lt 0 \\\\&\iff \underbrace{\frac{(\sigma(q^k)-q^k)(\sigma(q^k)+2n)}{\sigma(q^k)}}_{\text{positive}}(\sigma(q^k)-2n)\lt 0 \\\\&\iff \sigma(q^k)\lt 2n\end{align}$$


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