Difference between revisions of "Wilson's Theorem"
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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>. | + | 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 11:40, 18 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
![$1, 2, 3, \dots, p-1$](http://latex.artofproblemsolving.com/0/d/f/0dfe24d3485a7f19fa17e2195c625959c00dc811.png)
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,
![$2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}$](http://latex.artofproblemsolving.com/b/6/a/b6a67ca261ea87615017b0a10b367e711c9b801b.png)
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?>