Difference between revisions of "Wilson's Theorem"

(Proof)
(Statement)
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== Statement ==
 
== Statement ==
If and only if <math>{p}</math> is a prime, then <math>(p-1)! + 1</math> is a multiple of <math>{p}</math>.  In other words <math>(p-1)! \equiv -1 \pmod{p}</math>.
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If and only if <math>{p}</math> is a [[prime]], then <math>(p-1)! + 1</math> is a multiple of <math>{p}</math>.  In other words <math>(p-1)! \equiv -1 \pmod{p}</math>.
  
 
== Proof ==
 
== Proof ==

Revision as of 10:40, 18 June 2006

Statement

If and only if ${p}$ is a prime, then $(p-1)! + 1$ is a multiple of ${p}$. In other words $(p-1)! \equiv -1 \pmod{p}$.

Proof

Wilson's theorem is easily verifiable for 2 and 3, so let's consider $p>3$. If ${p}$ is composite, then its positive factors are among

$1, 2, 3, \dots, p-1$.


Hence, $\gcd( (p - 1)!, p) > 1$, so $(p-1)! \neq -1 \pmod{p}$.

However if ${p}$ is prime, then each of the above integers are relatively prime to ${p}$. So for each of these integers a there is another $b$ such that $ab \equiv 1 \pmod{p}$. It is important to note that this $b$ is unique modulo ${p}$, and that since ${p}$ is prime, $a = b$ if and only if ${a}$ is $1$ or $p-1$. Now if we omit 1 and $p-1$, then the others can be grouped into pairs whose product is congruent to one,

$2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}$


Finally, multiply this equality by $p-1$ to complete the proof.

Example

Let ${p}$ be a prime number such that dividing ${p}$ by 4 leaves the remainder 1. Show that there is an integer ${n}$ such that $n^2 + 1$ is divisible by ${p}$.

<Solutions?>

See also

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