Korselt criterion ($N\text{ squarefree},p-1\mid N-1(p\mid N)$)

A composite integer $N>1$ is a Carmichael number if and only if it is a squarefree integer andThe criterion is named for Alwin Korselt. The two requirements are different: the first excludes repeated prime factors, while the second makes the exponent $N-1$ a common multiple of the prime-field return exponents.^[ref]

For $1105=5\cdot 13\cdot 17$, the numbers $4,12,16$ divide $1104$ with quotients $276,92,69$. Thus 1105 is another Carmichael number. By contrast $341=11\cdot 31$ is squarefree, but $30\nmid 340$; a single failed divisor condition excludes it. This illustrates the appeal of the criterion: a statement about every unit becomes a small calculation using a known prime factorization.

The composite condition must be retained. A prime number is squarefree and satisfies the divisibility condition, but is excluded from the definition of a Carmichael number. The resulting certificate also depends on the supplied factors really being prime numbers; merely finding a product decomposition is insufficient.

For sufficiency, let $N$ be composite and squarefree, with $p-1\mid N-1$ for every prime factor $p$. For any integer $a$ coprime to $N$, Fermat's little theorem gives $a^{N-1}\equiv 1(\mathrm{mod}p)$ for every such $p$. Their product is $N$, so the Chinese remainder theorem gives the required congruence modulo $N$.

For necessity, first rule out a repeated factor. If $N=p^{k}m$ with $k\ge 2$ and $\gcd (p,m)=1$, the Chinese remainder theorem supplies $a\equiv 1+p(\mathrm{mod}{p}^k)$ and $a\equiv 1(\mathrm{mod}m)$. This $a$ is coprime to $N$. But the binomial theorem givescontradicting the Carmichael number condition. Hence $N$ is squarefree.

Now choose a primitive root $g$ modulo a prime factor $p$. Lift it to a unit modulo $N$ by prescribing $a\equiv g(\mathrm{mod}p)$ and $a\equiv 1(\mathrm{mod}N/p)$. The Chinese remainder theorem applies because $N$ is squarefree. The Carmichael congruence implies $g^{N-1}\equiv 1(\mathrm{mod}p)$. Since $g$ has multiplicative order $p-1$, that order divides $N-1$. This proves the other condition.^[ref]