Carmichael Number has 3 Odd Prime Factors

Theorem
Let $n$ be a Carmichael number.

Then $n$ has at least $3$ distinct odd prime factors.

Proof
By Korselt's Theorem, $n$ is odd.

Therefore $n$ has at least $1$ odd prime factor.

By Korselt's Theorem, for each prime factor of $n$:
 * $p^2 \nmid n$
 * $\paren {p - 1} \divides \paren {n - 1}$

Suppose $n = p^k$ for some odd prime $p$.

By Korselt's Theorem, $k = 1$.

However by definition of a Carmichael Number, $n$ cannot be prime.

Therefore $n$ has at least $2$ distinct odd prime factors.

Suppose $n = p^a q^b$ for distinct odd primes $p$ and $q$.

By Korselt's Theorem, the following holds:
 * $a = b = 1$
 * $n = p q$
 * $\paren {p - 1} \divides \paren {n - 1}$
 * $\paren {q - 1} \divides \paren {n - 1}$

Hence:

Swapping $p$ and $q$ yields $\paren {q - 1} \divides \paren {p - 1}$.

Hence $p - 1 = q - 1$ and $p = q$, which is a contradiction.

Therefore $n$ has at least $3$ distinct odd prime factors.