Difference between revisions of "Wilson's Theorem"

Revision as of 16:53, 19 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?>

Example #2

If p is a prime $>2$ , define $p=2q+1$ . Prove that $(q!)^2 + (-1)^q$ is divisible by p.
$q!^2=(2q)!(-1)^q=(p-1)!(-1)^q$ then the claim follows due to Wilson's Theorem. (see here)

@@ Line 7: / Line 7: @@
 Hence, <math>\gcd( (p - 1)!, p) > 1</math>, so  <math>(p-1)! \neq -1 \pmod{p}</math>.
-However, if <math>{p}</math> is prime, then each of the above integers are relatively [[prime]] to <math>{p}</math>. So, for each of these integers a, there is another <math>b</math> such that <math>ab \equiv 1 \pmod{p}</math>. It is important to note that this <math>b</math> is unique [[modulo]] <math>{p}</math>, and that since <math>{p}</math> is prime, <math>a = b</math> if and only if <math>{a}</math> is <math>1</math> or <math>p-1</math>. Now, if we omit 1 and <math>p-1</math>, then the others can be grouped into pairs whose product is congruent to one,
+However, if <math>{p}</math> is prime, then each of the above integers are [[relatively prime]] to <math>{p}</math>. So, for each of these integers a, there is another <math>b</math> such that <math>ab \equiv 1 \pmod{p}</math>. It is important to note that this <math>b</math> is unique [[modulo]] <math>{p}</math>, and that since <math>{p}</math> is prime, <math>a = b</math> if and only if <math>{a}</math> is <math>1</math> or <math>p-1</math>. Now, if we omit 1 and <math>p-1</math>, then the others can be grouped into pairs whose product is congruent to one,
 <center><math>2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}</math></center>  <br>

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