Carmichael number is not semiprime

Statement

Suppose $n$ is a composite natural number that is a Carmichael number, i.e., it is a Fermat pseudoprime to every base relatively prime to it. Equivalently, the universal exponent $\lambda (n)$ divides $n-1$.

Then, $n$ is not a semiprime.

Facts used

1. Carmichael number is square-free

Proof

We prove the contrapositive: a semiprime cannot be a Carmichael number.

By Fact (1), it suffices to restrict attention to a Carmichael number of the form $pq$ where $p,q$ are distinct primes.

We have:

$\lambda (n)=\mathrm{l}\mathrm{c}\mathrm{m}\{p-1,q-1\}$

In particular, for $\lambda (n)$ to divide $n-1$, we must have:

$p-1\mid n-1\text{ and }q-1\mid n-1$

The first condition tells us that:

$n\equiv 1(\mathrm{mod}p-1)$

Since we already have $p\equiv 1(\mathrm{mod}p-1)$, and since $n=pq$, this gives $q\equiv 1(\mathrm{mod}p-1)$. In particular, this gives $q\ge p$.

Similarly, the second condition tells us that $p\equiv 1(\mathrm{mod}q-1)$. In particular, this gives $p\ge q$.

Combining, we get $p=q$, contradicting our requirement that $n$ be a product of distinct primes.