Korselt's Theorem

Theorem
Let $n \ge 2$ be an integer.

Then $n$ is a Carmichael number :
 * $(1): \quad n$ is odd

and the following conditions hold for every prime factor $p$ of $n$:
 * $(2): \quad p^2 \nmid n$
 * $(3): \quad \paren {p - 1} \divides \paren {n - 1}$

where:
 * $\divides$ denotes divisibility
 * $\nmid$ denotes non-divisibility.

Sufficient Condition
Let $n$ be a Carmichael number:


 * $(4): \quad \forall a \in \Z: a \perp n: a^n \equiv a \pmod n$

where $\perp$ denotes coprimality.

Suppose $n$ is even.

Set $a = -1$ in $(4)$.

Then $\paren {-1}^n = 1$ and so:
 * $1 \equiv -1 \pmod n$

resulting in $n = 2$.

But as $2$ is not a Carmichael number, it follows that for $n$ to be Carmichael it must be odd.

Thus:
 * $(1): \quad n$ is odd

holds.

From the Fundamental Theorem of Arithmetic, let $n$ be expressed as:
 * $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

where:
 * the $p_i$'s are distinct odd primes
 * for all $i$, $k_i \ge 1$.

By Conditions for Integer to have Primitive Root there exists a primitive root $a_i$ modulo $p_i^{k_i}$ for each $i$.

In particular:
 * $a_i \perp p_i$

By the Chinese Remainder Theorem:
 * $\exists a: \forall i: a \equiv a_i \pmod {p_i^{k_i} }, a \perp n$

Consider a particular $i$.

We have:

We have that $a_i$ is a primitive root modulo $p_i^{k_i}$.

Thus:

But because $p_i \divides n$:
 * $p_i \perp n - 1$

Thus:
 * $k_i = 1$

giving that $n$ is square-free, and:
 * $p_i - 1 \divides n - 1$

As $i$ is arbitrary, both conditions $(1)$ and $(2)$ of the statement of the theorem are thus fulfilled:

That is:
 * $(1): \quad n$ is odd

and for all primes $p$ such that $p \divides n$:
 * $(2): \quad p^2 \nmid n$
 * $(3): \quad \paren {p - 1} \divides \paren {n - 1}$

Necessary Condition
Suppose $n$ is such that:
 * $(1): \quad n$ is odd

and for all primes $p$ such that $p \divides n$:
 * $(2): \quad p^2 \nmid n$
 * $(3): \quad \paren {p - 1} \divides \paren {n - 1}$

Thus:
 * $n = p_1 p_2 \cdots p_r$

is the product of $r$ distinct odd primes such that:
 * $\paren {p_i - 1} \divides \paren {n - 1}$

for all $i$.

Let $a \in \Z$.

Let $a \perp p_i$.

Then by Fermat's Little Theorem:
 * $a^{p_i - 1} \equiv 1 \pmod {p_i}$

Thus:
 * $a^{n - 1} \equiv 1 \pmod {p_i}$

and so:
 * $a^n \equiv a \pmod {p_i}$

Let $\gcd \set {a, p_i} > 1$.

Then as $p_i$ is prime:
 * $p_i \divides a$

and so:
 * $a^n \equiv a \equiv 0 \pmod {p_i}$

Thus for all $i$ and for all $a \in \Z$:
 * $a^n \equiv a \pmod {p_i}$

From the Chinese Remainder Theorem:
 * $a^n \equiv a \pmod n$

for all $a$.

Thus, by definition, $n$ is a Carmichael number.

Also known as
Korselt's theorem is also seen referred to as Korselt's criterion.