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Topic: Wilson's theorem
Replies: 6   Last Post: Dec 7, 2005 9:27 PM

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carel

Posts: 161
Registered: 12/12/04
Wilson's theorem
Posted: Dec 6, 2005 3:04 AM
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Dear Group

Have a look at my proof of Wilson's theorem.

Wilson's Theorem

Wilson's theorem states that p if prime divides (p-1)! + 1

For p = 2 , 3 , 5 , 7 it follows quite readily. So we are going to
prove for p>5

Note a == b mod c means c divides a-b

Note that if a == b mod c then a == -(a-b)/c + b mod (c+1) (Note: dont
know if this is known)

or in general a == (a-b)d/c + b mod (c-d)

For instance 6 == 2 mod 4 , so 6 == - (6-2)/4 + 2 mod 5 , so 6 == 1 mod
5

Now (p-1)! == p-1 mod (p-1)

So (p-1)! == -[ (p-1)! -(p-1) ]/(p-1] + p-1 mod p

So (p-1)! == -(p-2)! +1 + p -1 mod p , so (p-1)! == -(p-2)! mod p

so (p-1)! + (p-2)! == 0 mod p as in 5 divides 3!+4! and 7 divides
6! + 5! and so on

We have (p-1)! == x mod p where 0< |x| <p
We have (p-2)! == y mod p where 0<|y|<p

So (p-1)(p-2)! = (p-1)! == y(p-1) == -y mod p

So x = -y

So (p-1)! == x mod p and (p-2)! == -x mod p as in 4! == 4 mod 5 and 3!
== 1 == 1-5 == -4 mod 5

But (p-1)! = (p-1)(p-2)! == x mod p , so x / (p-1) must have a
remainder of -x

So x == -x mod (p-1) , so 2x == 0 mod (p-1) , so p-1 divides x as p
does not divide 2 , but x<p so x = p-1

So (p-1)! == p-1 == -1 mod p , so (p-1)! + 1 == 0 mod p

Therefore p divides (p-1)! + 1




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