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A Carmichael number is a composite number $N>1$ for whichEvery unit modulo $N$ satisfies the same modular congruence supplied by Fermat's little theorem for a prime number. The name refers to Robert Daniel Carmichael. The coprime qualification is essential: a nonunit can expose a prime factor through a greatest common divisor calculation. A Carmichael number is therefore a Fermat pseudoprime to every coprime base, rather than to every nonzero residue class.^[ref]

For example,The factors are distinct prime numbers. Their predecessor numbers $2,10,16$ all divide $560$, so Korselt criterion applies. More explicitly, for any integer $a$ coprime to $561$, Fermat's little theorem givesThe Chinese remainder theorem combines these into $a^{560}\equiv 1(\mathrm{mod}561)$. This proves all coprime bases pass from three short divisibility checks, rather than from a long table of sampled bases.

Equivalently, a composite $N$ is a Carmichael number exactly when $a^N\equiv a(\mathrm{mod}N)$ for every integer $a$, including nonunits. The equivalence follows from Korselt criterion: modulo each prime factor, either $a$ is zero or its nonzero power returns to one, and squarefreeness combines those congruences. This does not say a nonunit satisfies $a^{N-1}\equiv 1$. For instance $a=3$ gives $3^{560}\equiv 375(\mathrm{mod}561)$, while $3^{561}\equiv 3(\mathrm{mod}561)$; the greatest common divisor already supplies the factor three.^[ref]

The Carmichael function supplies another compact description: a composite $N$ is Carmichael exactly when $\lambda (N)\mid N-1$. For $561$, $\lambda (561)=\mathrm{lcm}(2,10,16)=80$ and $560=7\cdot 80$. In contrast, Euler's totient function gives $\phi (561)=320$, which does not divide $560$. The least common return exponent, rather than the number of units, controls this test.

The obstruction is useful to know when interpreting a Fermat primality test: even checking every coprime base cannot certify primality. The passing-base fraction for the Fermat test makes this failure quantitative. Stronger primality tests require additional structure, such as the squaring conditions in a Miller-Rabin primality test.

The historical record predates the name. Václav Šimerka's 1885 paper lists $561,1105,1729,2465,2821,6601,8911$ on printed page 224, together with factorizations. This is an early published list of examples, independently of their later name.^[ref]

For a prime number $p$ and an integer $x$ coprime to $p$, multiplication by $x$ permutes the nonzero residue classes modulo $p$. Indeed, $ax\equiv bx(\mathrm{mod}p)$ implies $a\equiv b(\mathrm{mod}p)$, and none of the products is zero. Taking the product over the $p-1$ classes givesThe factorial is coprime to $p$ and can be cancelled, proving Fermat little theorem: $\boxed{x^{p-1}\equiv 1(\mathrm{mod}p)}$.

Now $n=mp$ and $x$ is coprime to $n$, so it is coprime to $p$. Since $(n-1)-(m-1)=m(p-1)$, the proved theorem givesThus the two specified integer congruences modulo $p$ are equivalent. In this particular step the additional hypothesis $(m,p)=1$ is not needed; it will hold for the squarefree factorization used next.

If $n$ is a squarefree integer, its distinct prime factors are pairwise coprime. Consequently $n$ divides $x^{n-1}-1$ exactly when every prime factor does. Applying the preceding equivalence with $m=n/p$ provesFinally retain this squarefree hypothesis. If $p-1$ divides $n-1$, write $n-1=k_p(p-1)$. Fermat little theorem gives $x^{n-1}=(x^{p-1}{)}^{k_p}\equiv 1(\mathrm{mod}p)$ for every prime factor. Their pairwise coprimality then gives $\boxed{x^{n-1}\equiv 1(\mathrm{mod}n)}$. For composite $n$ this is the sufficient direction of the Korselt criterion for a Carmichael number; primality is not excluded by the requested implication.

The prime factorization is $561=3\cdot 11\cdot 17$, so it is composite and squarefree. Moreover $3-1=2$, $11-1=10$ and $17-1=16$ all divide $560=561-1$. The Korselt criterion therefore proves that 561 is a Carmichael number.

By the Korselt criterion, the prime factors are distinct. There cannot be just one prime factor, since a Carmichael number is composite. If there were exactly two, write $n=pq$ with $p<q$. Then $q-1\mid pq-1=p(q-1)+(p-1)$, so $q-1\mid p-1$, impossible because $0<p-1<q-1$. Hence at least three distinct prime factors are necessary.

An odd composite integer $n>1$ coprime to $b$ is a Fermat pseudoprime to base $b$ if $b^{n-1}\equiv 1(\mathrm{mod}n)$. A Carmichael number is a composite integer satisfying this congruence for every integer $b$ coprime to $n$. The Korselt criterion says that a composite $n$ is a Carmichael number exactly when it is squarefree and $p-1$ divides $n-1$ for every prime divisor $p$ of $n$.

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.