Difference between revisions of "Wilson's Theorem"
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Hence, <math>\gcd( (p - 1)!, p) > 1</math>, so <math>(p-1)! \neq -1 \pmod{p}</math>. | 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> | <center><math>2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}</math></center> <br> | ||
Revision as of 13:12, 19 June 2006
Contents
Statement
If and only if 
is a prime, then 
is a multiple of . In other words 
.
Proof
Wilson's theorem is easily verifiable for 2 and 3, so let's consider 
. If is composite, then its positive factors are among
.
Hence, 
, so 
.
However, if is prime, then each of the above integers are relatively prime to . So, for each of these integers a, there is another 
such that 
. It is important to note that this is unique modulo , and that since is prime, 
if and only if 
is 
or 
. Now, if we omit 1 and , then the others can be grouped into pairs whose product is congruent to one,
Finally, multiply this equality by to complete the proof.
Example
Let be a prime number such that dividing by 4 leaves the remainder 1. Show that there is an integer 
such that 
is divisible by .
<Solutions?>
