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Can an odd perfect number $n$ be a Carmichael number?

We know that all Carmichael numbers are odd and square-free.

But is there a Carmichael number that is also a perfect number?

We all know that if odd perfect numbers exists, they must be form of either $1\pmod{12}$,$81\pmod{324}$, or $117\pmod{468}$.

So if a perfect odd Carmichael number exists, then it must be $1\pmod{12}$, as $81, 117, 324$, and $468$ are all divisible by $9$ and therefore not squarefree.

And a positive integer $n$ is Carmichael number if and only $a^{n-1}\equiv 1\pmod{n}$.

Cameron L. Williams
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Thirdy Yabata
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    See e.g. https://arxiv.org/abs/hep-th/0401052, but be aware that this proof may not be entirely correct. AFAIK this is the oldest known open problem in mathematics. – R. J. Mathar Dec 07 '23 at 20:39
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    Maybe , I miss something , but does Euler's necessary criterion not already establish that an odd perfect number cannot be squarefree ? – Peter Dec 07 '23 at 22:03
  • Should there be no proof , the large lower bound ($10^{1500}$) would make it anyway extremely unlikely that an odd perfect number (if existent at all) can be a Poulet- number let alone a Carmichael number. – Peter Dec 07 '23 at 22:08
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    See this question for a proof that odd perfect numbers could not be square free. – lulu Dec 08 '23 at 00:04

1 Answers1

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Euler proved that an odd perfect number $N$ must have the form $$N = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In particular, since $n > 1$, then $N$ cannot be squarefree.

But all Carmichael numbers are squarefree. (This means that, if $C$ is a Carmichael number, then $C$ must be squarefree.)

Therefore, an odd perfect number cannot be a Carmichael number.

We conclude that all perfect numbers cannot be Carmichael.

Jose Arnaldo Bebita Dris
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