It is fairly easy to show that a perfect number $\Gamma$ cannot be written in the form $\Gamma=n^2$ for integer values of $n$. However, does this property hold true for multi-perfect numbers---that is, integers $R$ such that $\sigma(R)=kR,$ for $k>2$? This question is out of pure curiosity, as I can not find an answer online.


I now know that for even $k$ and/or an even $k$-perfect number $R$ the answer is no, as it follows from contradiction by assuming $R=n^2$ and then computing $$\sigma(R)=\prod_{i=1}^{l}\sigma(p_i^{2a_i}), $$ where $n=\prod_{i=1}^lp_i^{a_i}$ (here all $p_i$ are distinct primes and all $a_i\geq1$).

Given this result, my question is now whether the same is true for odd $k$ and, assuming they exist, an odd $k$-perfect number $Q$. That is, if both $k$ and $Q$ are odd in $\sigma(Q)=kQ,$ is there a way to show that $Q$ cannot be a perfect square? I believe this could be done by considering the number of divisors of $Q=\prod_{i=1}^rp_i^{a_i}$---given by $$d(Q)=\prod_{i=1}^r(a_i+1)$$ and showing that at least one $a_i\equiv1\pmod2,$ but I have not found any results similar to this online.

  • $\begingroup$ Your statement about the number of divisors of $Q$ is equivalent to saying all the primes $p_i$ in $Q$'s decomposition are have even powers (which is a more intuitive way of thinking about square numbers, IMO). $\endgroup$ – Alexis Olson Sep 23 '16 at 23:51
  • $\begingroup$ Indeed it is more intuitive to think of squares that way. This is why the number of divisors function above returns an odd value for perfect squares. $\endgroup$ – fruitegg Sep 24 '16 at 0:37

I have found references to a proof that any odd triperfect is a square.

See here and here. They both reference the following German paper:

H.-J. Kanold, "Über mehrfach vollkommene Zahlen. II," J. Reine Angew. Math., v. 197, 1957, pp. 82-96. MR 18, 873.

I'm in the process of translating and extracting the relevant portion and will edit that in soon.


From p. 88-89, here's what I have extracted:

Lemma 1. Let $\displaystyle n = \prod_{i=1}^k p_i^{\alpha_i}$ be an $(s-1)$-fold perfect number and also let $sn \equiv 1 \pmod 2$. Then $n > 10^{20}$.

Proof: We start from the relationship

$$sn = s \prod_{i=1}^k p_i^{\alpha_i} = \prod_{i=1}^k \sigma\bigl(p_i^{\alpha_i}\bigr).$$

From this, we see that

$$\alpha_i \equiv 0 \pmod 2 \text{ for } i = 1, \ldots, k$$

must be true.

[Remainder of proof excluded as this is all we need.]

This is what we want since $sn \equiv 1 \pmod 2$ means $n$ is odd (since $sn$ is odd implies both $s$ and $n$ are odd) and $\alpha_i \equiv 0 \pmod 2$ means that $n$ is a perfect square (since each prime power is even).

To explain this a bit more, suppose some $\alpha_j$ is odd and use the following formula (reference):

$$\sigma\bigl(p_j^{\alpha_j}\bigr) = 1 + p_j + p_j^2 + \cdots + p_j^{\alpha_j}.$$

Since $n$ is odd, we know $p_j$ must be odd and hence any power of $p_j$ is also odd. Thus we have an odd number of odd integers plus the remaining $1$. Thus the sum is even, which implies $sn$ is even. This is a contradiction. Therefore, all the $\alpha_i$ must be even and hence $n$ is a perfect square.

Edit: (Clarification of $(s-1)$-fold perfect number notation)

This is the first couple sentences from the paper:

Wir schließen uns in dieser Arbeit der Bezeichnungsweise einer früheren an Danach heißt eine naturliche Zahl $\displaystyle n=\prod_{x=1}^k p_x^{\alpha_x}$ eine $(s-1)$-fach vollkommene Zahl, wenn sie der Bedingung $\sigma(n) = s \cdot n$ genügt. Die $p_x$ bedeuten Primzahlen, $\sigma(n)$ bezeichnet die Summe aller positiven Teiler von $n$.

I have translated this as:

We agree in this work with earlier notation used before by calling a natural number $\displaystyle n=\prod_{x=1}^k p_x^{\alpha_x}$ an $(s-1)$-fold perfect number if satisfies the condition $\sigma(n) = s \cdot n$. The $p_x$ denote primes and $\sigma(n)$ denotes the sum of all positive divisors of $n$.

This means what they refer to as a "$(s-1)$-fold perfect number" is what we would call an "$s$-perfect number". This understanding also agrees with the proof.

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  • $\begingroup$ Interesting articles for sure! I'm curious to see your extraction from the German manuscript. $\endgroup$ – fruitegg Sep 24 '16 at 1:55
  • $\begingroup$ @fruitegg I've got it. It was rather simpler than I anticipated. $\endgroup$ – Alexis Olson Sep 24 '16 at 2:39
  • $\begingroup$ How is the relationship in the proof established? The author states $n$ is an $(s-1)$-fold perfect number, which implies $\sigma(n)=\prod_{i=1}^k\sigma(p_i^{a_i})=n(s-1)\neq ns$... $\endgroup$ – fruitegg Sep 24 '16 at 2:51
  • $\begingroup$ They have a rather odd definition. In the opening paragraph of the paper, they define $(s-1)$-fold perfect to mean $\sigma(n) = sn$ rather than what you assumed. $\endgroup$ – Alexis Olson Sep 24 '16 at 3:00
  • $\begingroup$ I see. Does this also generalize to any $s$-fold (or ($s-1$)-fold in Kanold's case) where $s$ is odd? That is, if $s$ is odd then all $Q$ in $\sigma(Q)=sQ$ must be perfect squares? $\endgroup$ – fruitegg Sep 24 '16 at 3:05

A paper by Chen and Luo (2012) published in the Bulletin of the Australian Mathematical Society seems to contain most of the details that you need. A preprint is available in arXiv.

In particular, Chen and Luo's theorem on the explicit structure for odd $k$-perfect numbers $n$ (for any $k \geq 2$) imply that $n$ cannot be a square. (At the same time, $n$ cannot also be squarefree. With regard to this latter consideration, there is a related MO question here.)

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    $\begingroup$ I am guessing some information may have been lost in translation. For instance, I am a bit doubtful about that $(s-1)$-fold perfect number. You have to keep in mind that the parity of $s$ and $s-1$ are different. The explicit structure theorem for all odd $k$-perfect numbers (for all $k \geq 2$) by Chen and Luo is in page 4 (Theorem 2.2) of their preprint. (continued...) $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 25 '16 at 1:06
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    $\begingroup$ (... continued) There, they implicitly state that the exponents of the primes in the factorization of what they call as the Euler part of an odd $k$-perfect number, are all odd. Since the non-Euler part is $M^2$ (a square!), this means that odd multiperfect numbers cannot be squares. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 25 '16 at 1:06
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    $\begingroup$ I do agree the $(s-1)$-fold definition is suspect. However, having just skimmed over the mentioned theorem, they restrict $k$ under the condition $\nu(k)\geq1$, where $\nu(m)$ is the highest power of 2 dividing $m$. Does this not imply $k$ is even? $\endgroup$ – fruitegg Sep 25 '16 at 1:09
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    $\begingroup$ Yes it does. However, I would just like to point out the paragraph on page 3, immediately before Theorem 2.2, which says: "As an application of Theorem 2.1, we will establish the explicit structure theorem of $k$-perfect numbers for any integer $k \geq 2$." $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 25 '16 at 1:18
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    $\begingroup$ @fruitegg, I would suggest that you contact Chen and Luo directly, as the published version of their paper might contain some difference(s) from the preprint. =) $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 25 '16 at 1:40

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