# Wilson's Theorem/Necessary Condition

## Theorem

Let $p$ be a prime number.

Then:

$\paren {p - 1}! \equiv -1 \pmod p$

## Proof 1

If $p = 2$ the result is obvious.

Therefore we assume that $p$ is an odd prime.

Let $p$ be prime.

Consider $n \in \Z, 1 \le n < p$.

As $p$ is prime, $n \perp p$.

From Law of Inverses (Modulo Arithmetic), we have:

$\exists n' \in \Z, 1 \le n' < p: n n' \equiv 1 \pmod p$

By Solution of Linear Congruence, for each $n$ there is exactly one such $n'$, and $\paren {n'}' = n$.

So, provided $n \ne n'$, we can pair any given $n$ from $1$ to $p$ with another $n'$ from $1$ to $p$.

We are then left with the numbers such that $n = n'$.

Then we have $n^2 \equiv 1 \pmod p$.

Consider $n^2 - 1 = \paren {n + 1} \paren {n - 1}$ from Difference of Two Squares.

So either $n + 1 \divides p$ or $n - 1 \divides p$.

Observe that these cases do not occur simultaneously, as their difference is $2$, and $p$ is an odd prime.

$p - 1 \equiv -1 \pmod p$

Hence $n = 1$ or $n = p - 1$.

So, we have that $\paren {p - 1}!$ consists of numbers multiplied together as follows:

in pairs whose product is congruent to $1 \pmod p$
the numbers $1$ and $p - 1$.

The product of all these numbers is therefore congruent to $1 \times 1 \times \cdots \times 1 \times p - 1 \pmod p$ by modulo multiplication.

$\paren {p - 1}! \equiv -1 \pmod p$

$\blacksquare$

## Proof 2

If $p = 2$ the result is obvious.

Therefore we assume that $p$ is an odd prime.

Consider $p$ points on the circumference of a circle $C$ dividing it into $p$ equal arcs.

By joining these points in a cycle, we can create polygons, some of them stellated.

From Number of Different n-gons that can be Inscribed in Circle, the number of different such polygons is $\dfrac {\paren {p - 1}!} 2$.

When you rotate these polygons through an angle of $\dfrac {2 \pi} p$, exactly $\dfrac {p - 1} 2$ are unaltered.

These are the regular $p$-gons and regular stellated $p$-gons.

That there are $\dfrac {p - 1} 2$ of them follows from Number of Regular Stellated Odd n-gons.

The remaining $\dfrac {\paren {p - 1}!} 2 - \dfrac {p - 1} 2$ polygons can be partitioned into sets of $p$ elements: those which can be obtained from each other by rotation through multiples of $\dfrac {2 \pi} p$.

The total number of such sets is then:

$\dfrac {\paren {p - 1}! - \paren {p - 1} } {2 p}$

Thus we have that:

$2 p \divides \paren {\paren {p - 1}! - \paren {p - 1} }$

That is:

$p \divides \paren {\paren {p - 1}! + 1}$

$\blacksquare$

## Proof 3

Let $p$ be prime.

Consider $\struct {\Z_p, +, \times}$, the ring of integers modulo $m$.

From Ring of Integers Modulo Prime is Field, $\struct {\Z_p, +, \times}$ is a field.

Hence, apart from $\eqclass 0 p$, all elements of $\struct {\Z_p, +, \times}$ are units

As $\struct {\Z_p, +, \times}$ is a field, it is also by definition an integral domain, we can apply:

From Product of Units of Integral Domain with Finite Number of Units, the product of all elements of $\struct {\Z_p, +, \times}$ is $-1$.

But the product of all elements of $\struct {\Z_p, +, \times}$ is $\paren {p - 1}!$

The result follows.

$\blacksquare$

## Also known as

Some sources refer to Wilson's Theorem as the Wilson-Lagrange theorem, after Joseph Louis Lagrange, who proved it.

## Source of Name

This entry was named for John Wilson.

## Historical Note

The proof of Wilson's Theorem was attributed to John Wilson by Edward Waring in his $1770$ edition of Meditationes Algebraicae.

It was first stated by Ibn al-Haytham ("Alhazen").

It appears also to have been known to Gottfried Leibniz in $1682$ or $1683$ (accounts differ).

It was in fact finally proved by Lagrange in $1793$.