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It is known that a positive composite integer $n$ is a Carmichael number if and only if $n$ is square-free, and for all prime divisors $p$ of $n$, it is true that $(p-1)\mid (n-1)$.

Question: Is it true that if $n$ is a Carmichael number then $n-1$ cannot be square free? I am looking for a proof or disproof of this.

Bill Dubuque
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Nilotpal Sinha
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    Just in case anyone else is wondering about the same thing I was: $n=8911=7\cdot19\cdot67$ is a Carmichael number, but $n-1$ is not a multiple of four. It is divisible by nine though. – Jyrki Lahtonen Aug 27 '16 at 14:19
  • @JyrkiLahtonen - So in this case $n-1$ is not sqaure free which is exactly what the claim is. Whats new? – Nilotpal Sinha Aug 27 '16 at 14:47
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    Nothing. It is just that my first idea was to prove that we always have $n\equiv1\pmod4$. A bit of searching put an end to that approach. I wanted to save the next viewer the trouble. – Jyrki Lahtonen Aug 27 '16 at 14:50

2 Answers2

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No, it is false!

Counterexample:

$ C = 139952671 $ is a Carmichael number and $C-1$ is square-free.

$C = 131 \dot\ 571 \dot\ 1871$ with $130,570$ and $1870$ divides $C-1$.

$C-1 = 2 \dot\ 3 \dot\ 5 \dot\ 11 \dot\ 13 \dot\ 17 \dot\ 19 \dot\ 101$ is square-free.

See also: A257643.

Martin Hopf
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There are infinitely many Carmicael Numbers which are multiples of a prime $p$ (see https://en.wikipedia.org/wiki/Carmichael_number and https://en.wikipedia.org/wiki/Dirichlet%27s_theorem) By Dirichlet's theorem, there are infintely many primes $p$ congruent to $d$ $\pmod n$ if $d$ and $n$ are relatively prime. Since $1$ is relatively prime to all numbers $n$, it follows that there are infinitely many primes $p$ congruent to $1$ $\pmod n$, therefore $n$ does not have to be squarefree. Since $p-1$ is not squarefree and $c$ is a Carmichael number that is a multiple of $p$, by Korselt's Criterion, $p-1$ divides $c-1$ and $p-1$ is not squarefree, and neither is $c-1$.

J. Linne
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