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



carel
Posts:
161
Registered:
12/12/04


Wilson's theorem
Posted:
Dec 6, 2005 3:04 AM


Dear Group
Have a look at my proof of Wilson's theorem.
Wilson's Theorem
Wilson's theorem states that p if prime divides (p1)! + 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 ab
Note that if a == b mod c then a == (ab)/c + b mod (c+1) (Note: dont know if this is known)
or in general a == (ab)d/c + b mod (cd)
For instance 6 == 2 mod 4 , so 6 ==  (62)/4 + 2 mod 5 , so 6 == 1 mod 5
Now (p1)! == p1 mod (p1)
So (p1)! == [ (p1)! (p1) ]/(p1] + p1 mod p
So (p1)! == (p2)! +1 + p 1 mod p , so (p1)! == (p2)! mod p
so (p1)! + (p2)! == 0 mod p as in 5 divides 3!+4! and 7 divides 6! + 5! and so on
We have (p1)! == x mod p where 0< x <p We have (p2)! == y mod p where 0<y<p
So (p1)(p2)! = (p1)! == y(p1) == y mod p
So x = y
So (p1)! == x mod p and (p2)! == x mod p as in 4! == 4 mod 5 and 3! == 1 == 15 == 4 mod 5
But (p1)! = (p1)(p2)! == x mod p , so x / (p1) must have a remainder of x
So x == x mod (p1) , so 2x == 0 mod (p1) , so p1 divides x as p does not divide 2 , but x<p so x = p1
So (p1)! == p1 == 1 mod p , so (p1)! + 1 == 0 mod p
Therefore p divides (p1)! + 1



