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(Note: The following post is an offshoot of this earlier MSE question: Can a multiperfect number be a perfect square?.)

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

A number $m$ satisfying $$\sigma(m)=2m$$ is said to be perfect.

More generally, we call any number $n$ satisfying $$\sigma(n)=kn$$ for $k \in \mathbb{N}$ to be multiperfect (or $k$-perfect).

It is known that multiperfect numbers cannot be squares.

Furthermore, it is also known that perfect numbers cannot be perfect powers.

I found a reference to the last statement in Walter Nissen's Concise, remarkable facts about perfect numbers:

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Perfect Naturals
are not
Perfect Powers ( e.g. , perfect squares , perfect cubes , etc. )

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Here is my question:

Can a multiperfect number be a perfect power?

Update (August 9, 2020 - 12:04 PM Manila time) I have posted a closely related question in MO here.

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    $\begingroup$ Just to mention it : $1$ is not considered to be a multi-perfect number. Currently, I am checking the cubes upto some limit, no idea yet for a proof. $\endgroup$
    – Peter
    Aug 7, 2020 at 22:07
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    $\begingroup$ No cube upto $10^{21}$ is multi-perfect. No $5$ th power upto $10^{30}$ is multi-perfect. $\endgroup$
    – Peter
    Aug 7, 2020 at 22:13
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    $\begingroup$ I checked all the multi-perfect numbers in the b-file of A007691, and none of them were perfect powers. This is all the way up to $n \approx 1.8 \cdot 10^{303}$ $\endgroup$
    – Varun Vejalla
    Aug 7, 2020 at 22:35
  • $\begingroup$ @VarunVejalla: Please post your last comment as an answer, so that I can upvote it. Thank you. $\endgroup$
    – Jose Arnaldo Bebita Dris
    Aug 8, 2020 at 0:24

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I checked all the multiperfect numbers from this site, which include all the multiperfect numbers in the b-file of A007691, as well as additional ones. The largest number that was in the list was $\approx 10^{34850339}$. None of those values were perfect powers.

What I did find for all multiperfect numbers is that there was at least one prime factor with an exponent of exactly $1$ in the prime factorization. If this could be proved for all multiperfect numbers, then the conjecture that there is no number that is both multiperfect and a perfect power could be proven.

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  • $\begingroup$ Just curious, @VarunVejalla, what software did you use to check this? $\endgroup$
    – Jose Arnaldo Bebita Dris
    Aug 8, 2020 at 23:41
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    $\begingroup$ I just used plain old Python. The numbers from Flammenkamp's site were easy to check since they were prime factorized. $\endgroup$
    – Varun Vejalla
    Aug 8, 2020 at 23:42
  • $\begingroup$ Okay thanks a lot! =) $\endgroup$
    – Jose Arnaldo Bebita Dris
    Aug 8, 2020 at 23:45

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