# Are all known $k$-multiperfect numbers (for $k > 2$) not squarefree?

I asked the following question in MSE four ($4$) days ago, but so far nobody has posted an answer.

The gist of the question is as follows:

Are all known $k$-multiperfect numbers (for $k > 2$) not squarefree?

From my own computational verifications using WolframAlpha, it appears that a closely related question is:

Are all known (even) $k$-multiperfect numbers (for $k > 2$) divisible by $4$?

Thank you!

• Wow! A very quick downvote - I wonder who you are ... =) – Jose Arnaldo Bebita-Dris Oct 5 '13 at 23:14
• I removed the attempted italics from the title (titles don't support that kind of formatting), and the crosspost notification from there; thank you (sincerely) for indicating this in the question. Also, the downvote seems to have been taken back. – Arturo Magidin Oct 5 '13 at 23:16
• Thank you for editing the title @ArturoMagidin, appreciate it! =) – Jose Arnaldo Bebita-Dris Oct 5 '13 at 23:16
• By the way: You may want to add a note in the math.SE post as well indicating that you have crossposted it here (perhaps as an addendum at the top or at the end)... – Arturo Magidin Oct 5 '13 at 23:17
• Aren't all the known multiperfects tabulated somewhere, so anyone can work out whether they are squarefree? – Gerry Myerson Oct 5 '13 at 23:31

## 1 Answer

It's easy to prove that if $n$ is squarefree with largest prime factor $p \geq 5$, then $n$ is not multiply perfect. Indeed, in that case $p$ is larger than any of the primes dividing $\sigma(n)$, so that $n \nmid \sigma(n)$. So the only squarefree multiply perfect number is $n=6$ (or also $n=1$, if you allow $k=1$).

• Don, I'm afraid but I don't follow your argument. Why would $p$ be larger than all the primes dividing $\sigma(n)$, if $p$ is the largest prime factor of a squarefree integer $n$? – Jose Arnaldo Bebita-Dris Oct 6 '13 at 1:49
• Alternatively, can you cite the reference where you obtained this argument? – Jose Arnaldo Bebita-Dris Oct 6 '13 at 1:51
• If $n$ is squarefree, then $\sigma(n) = \prod_{q \mid n}(q+1)$, where $q$ runs over the primes dividing $n$. Now unless $q=2$, all of the primes dividing $q+1$ are smaller than $q$. And when $q=2$, the largest prime factor of $q+1$ is $3$. So if the largest prime factor of $n$ is $p\geq 5$, then no term in the product $\prod_{q \mid n}(q+1)$ can have a prime factor as large as $p$. – so-called friend Don Oct 6 '13 at 2:02
• Thank you for your detailed explanation, Don! I think I get it now. – Jose Arnaldo Bebita-Dris Oct 6 '13 at 2:11