TITLE: Carmichael proof of at least 3 prime factors
QUESTION [6 upvotes]: I'm having some trouble while trying to prove the well known fact that a Carmichael number has at least 3 prime factors.
Basically, how I see it, I have 2 options:

building a number b that will satisfy $b^{n-1}\not\equiv 1(n)$ thus creating a fermat witness
trying to contradict the condition of: $p|n \iff\ p-1|n-1$

Maybe I'm missing something very simple but I can't seem to get this done.
Addition: we are talking about the case where $n=pq$, the case $n=p$ is trivial.

REPLY [11 votes]: Assume that $n=pq$, with $p<q$ two distinct primes, is a Carmichael number. Then we have
$$
q\equiv1\pmod{q-1}\implies n=pq\equiv p\mod{q-1}\implies n-1\equiv p-1\pmod{q-1}.
$$
Here $0<p-1<q-1$, so $n-1$ is not divisible by $q-1$. A contradiction.
The same argument can be extended to prove that any prime factor of a Carmichael number $n$ is less than $\sqrt n$. Namely, as above we get for all primes $p\mid n$
$$
n-1\equiv p\frac{n}{p}-1\equiv \frac{n}{p}-1\pmod{p-1}.
$$
Therefore $n/p-1$ must be a (positive) multiple of $p-1$. Thus
$$
\frac{n}{p}-1\ge p-1\implies\frac{n}{p}\ge p\implies n\ge p^2.
$$