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An odd composite $N$ coprime to $b$ is a Fermat pseudoprime to base $b$ when $b^{N-1}\equiv 1(\mathrm{mod}N)$. It is a Carmichael number when this congruence holds for every $b$ coprime to $N$.

Suppose a Carmichael number has $p^a\Vert N$ with $a\ge 2$. The group $(\mathbb{Z}/p^a\mathbb{Z}{)}^{\ast }$ is cyclic, so choose a residue of orderBy the Chinese remainder theorem, extend it to an integer $b$ that is congruent to $1$ modulo every other prime-power factor of $N$. Then $\gcd (b,N)=1$, and the Carmichael congruence forces the order of $b$ modulo $p^a$ to divide $N-1$. In particular $p\mid N-1$, contradicting $p\mid N$. Therefore every Carmichael number is square-free.

Now assume $N$ is square-free. If $p-1\mid N-1$ for every $p\mid N$, then for every $b$ coprime to $N$, Fermat little theorem givesfor each $p\mid N$. The Chinese remainder theorem then gives the congruence modulo $N$, so $N$ is Carmichael.

Conversely, if $N$ is Carmichael, fix $p\mid N$. Choose a primitive root $g$ modulo $p$ and use the Chinese remainder theorem to choose $b\equiv g(\mathrm{mod}p)$ and $b\equiv 1(\mathrm{mod}N/p)$. Then $b^{N-1}\equiv 1(\mathrm{mod}p)$, so the order $p-1$ of $g$ divides $N-1$. This proves the Korselt criterion in the required form:

A Carmichael number cannot have one prime factor because it is composite. If $N=pq$ with distinct primes $p<q$, the condition $q-1\mid pq-1$ implies $q-1\mid p-1$, which is impossible. Hence every Carmichael number has at least three prime factors.

The converted source omits $r$ from the product; the PDF asks about $N=pqr$. Fix the odd prime $r$, and suppose $pqr$ is Carmichael. Its three prime factors are distinct by the Korselt criterion. Relabel $p,q$ so that $p<q$.

The divisibility conditions includeWritefor a positive integer $b$. Since $p<q$,so $1\le b<r$. Moreover,Multiplying the second divisibility by $b$ and reducing modulo $p-1$ givesThe positive integer on the right is at most $(r-1)(2r-1)$. ConsequentlyThere are only finitely many possible primes $p$ and integers $b\in \{1,\dots ,r-1\}$, and each pair determines at most one $q=(pr+b-1)/b$. Therefore

A Carmichael number is a composite $N$ such that $a^{N-1}\equiv 1(\mathrm{mod}N)$ for every $a$ coprime to $N$. For $p\mid N$, choose by CRT a number whose residue mod $p$ is a primitive root; its order $p-1$ divides $N-1$. If $p^2\mid N$, choose instead a primitive root modulo $p^2$; then $p(p-1)\mid N-1$, impossible because $p\mid N$. Thus $N$ is square-free. Write $N=pm$. Since $p-1\mid N-1$, also $p-1\mid m-1$; compositeness gives $m>1$, hence $m\ge p$ and $p\le \sqrt{N}$. This is Korselt criterion.