Korselt's Theorem

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Theorem

Let $n \ge 2$ be an integer.

Then $n$ is a Carmichael number if and only if:

$(1): \quad n$ 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.


Proof

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:

$(1): n$ 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 Integers with Primitive Roots there exists a primitive root $a_i$ modulo $p_i^{k_i}$ for each $i$.

In particular, we have that $a_i \perp p_i$.

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:

\(\displaystyle a^n\) \(\equiv\) \(\displaystyle a \pmod n\) $\quad$ from $(4)$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{n - 1}\) \(\equiv\) \(\displaystyle 1 \pmod n\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{n - 1}\) \(\equiv\) \(\displaystyle 1 \pmod {p_i^{k_i} }\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle {a_i}^{n - 1}\) \(\equiv\) \(\displaystyle 1 \pmod {p_i^{k_i} }\) $\quad$ $\quad$


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

Thus:

\(\displaystyle \map {\operatorname {ord}_{p_i^{k_i} } } {a_i}\) \(=\) \(\displaystyle \map \phi {p_i^{k_i} }\) $\quad$ where $\phi$ is the Euler $\phi$ function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle p_i^{k_i - 1} \paren {p_i - 1}\) $\quad$ $\quad$
\(\displaystyle \) \(\divides\) \(\displaystyle n - 1\) $\quad$ $\quad$


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:

$(1): n$ is odd

and for all primes $p$ such that $p \divides n$:

$(2): \quad p^2 \nmid n$
$(3): \quad \paren {p - 1} \divides \paren {n - 1}$

$\Box$


Necessary Condition

Suppose $n$ is such that:

$(1): n$ 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.

$\blacksquare$

Also known as

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


Source of Name

This entry was named for Alwin Reinhold Korselt.


Historical Note

Alwin Reinhold Korselt first demonstrated this result in $1899$, but did not actually identify any integers that fulfilled the criterion.

That did not happen until Robert Daniel Carmichael found that $561$ was such a number in $1910$.

It turns out that Václav Šimerka had already found the first $7$ Carmichael numbers in $1885$, but his work was not noticed at the time.


Sources