Difference between revisions of "Wilson's Theorem"

Revision as of 16:23, 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 16: / Line 16: @@
 <Solutions?>
+==Example #2==
+If ''p'' is a prime <math>>2</math>, define <math>p=2q+1</math>.
+Prove that <math>(q!)^2 + (-1)^q</math> is divisible by ''p''.
+<math>q!^2=(2q)!(-1)^q=(p-1)!(-1)^q</math> then the claim follows due to Wilson's Theorem. ''(see [http://www.mathlinks.ro/Forum/viewtopic.php?&t=21733 here])''
 == See also ==

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