Difference between revisions of "Wilson's Theorem"

Revision as of 23:40, 17 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 ${p}<math> by 4 leaves the remainder 1. Show that there is an integer <math>{n}$ such that $n^2+1$ is divisible by ${p}$ .

<Solutions?>

@@ Line 9: / Line 9: @@
 Finally, multiply this equality by <math>p-1</math> to complete the proof.
-==Example Problem utilizing Wilson's==
+==Example==
-<Maybe steal something from AoPS 2 with the Admin's permission?>
+Let <math>{p}</math> be a prime number such that dividing <math>{p}<math> by 4 leaves the remainder 1. Show that there is an integer <math>{n}</math> such that <math>n^2+1</math> is divisible by <math>{p}</math>.
+<Solutions?>
 == See also ==

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Difference between revisions of "Wilson's Theorem"

Revision as of 23:40, 17 June 2006

Contents

Statement

Proof

Example

See also