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.
On
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 $c-1$.
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.