Is the prime factorisation of Carmichael Numbers always Square Free?

Question

Is the prime factorisation of carmichael Numbers always square free?

I saw the next theorem in my textbook which would say yes.

Answer 1

Consider a prime factor $p$ of $n$ such that $p^2 \mid n$. Let $t$ be a primitive root modulo $p^2$, thus, $p \mid o_{p^2}(t)$. And hence, $p^2 \nmid t^{n}-t \iff p^2 \nmid t^{n-1}-1 $. Latter one being true by our choice of $t$.