Wilson's Theorem/Necessary Condition/Proof 2
Theorem
Let $p$ be a prime number.
Then:
- $\paren {p - 1}! \equiv -1 \pmod p$
Proof
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 $\dfrac {2 \pi} p$.
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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:
- $p \divides \paren {\paren {p - 1}! + 1}$
$\blacksquare$
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 $1793$.
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-3}$ Wilson's Theorem: Theorem $\text {3-5}$