- (A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2

(B) {[(p-1)!+1]^z - 1}^2 = 0modp

If you expand B, you get [(p-1)!+1]^(p-1) - 2[(p-1)!+1]^z + 1

as 2z=p-1.

Jon Perry

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-----Original Message-----

From: Marcel Martin [mailto:znz@...]

Sent: 04 November 2001 23:25

Cc: primenumbers@yahoogroups.com

Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

>Therefore (A-B):

How do you obtain this using (A) and (B)?

>2.[(p-1)!+1]^z - 1 = -1modp

Marcel Martin

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Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ - I think I have fixed all remaining bugs:

http://www.users.globalnet.co.uk/~perry/maths/wilsonfermat/wilsonfermat.htm

(or:

http://www.users.globalnet.co.uk/~perry/maths/othermaths.htm

[Wilson's from Fermat's]

Jon Perry

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BrainBench MVP for HTML and JavaScript

http://www.brainbench.com

-----Original Message-----

From: Marcel Martin [mailto:znz@...]

Sent: 05 November 2001 21:05

Cc: primenumbers@yahoogroups.com

Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

>(A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2

That's the first thing I did. But that doesn't answer my question:

>(B) {[(p-1)!+1]^z - 1}^2 = 0modp

>If you expand B, you get [(p-1)!+1]^(p-1) - 2[(p-1)!+1]^z + 1

>as 2z=p-1.

How do you obtain this using (A) and (B)?>2.[(p-1)!+1]^z - 1 = -1modp

Marcel Martin

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Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ - I've changed the proof.

"(n-1)!+1=0modn iff n is prime",

is known. If n is prime, then (p-1)!+1=0modp, if not, (n-1)!+1 does not

divide n.

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry

BrainBench MVP for HTML and JavaScript

http://www.brainbench.com

-----Original Message-----

From: Marcel Martin [mailto:znz@...]

Sent: 10 November 2001 11:51

Cc: primenumbers@yahoogroups.com

Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

>(B) [(p-1)!+1]^(2z) - 2[(p-1)!+1]^z + 1 = jp

Once more, how do you get it? (I assume the variable x is not the one

>Now consider (A)-(B). As 2z = p-1, we get:

>2[(p-1)!+1]^z + 1 = xp + 1

you previously defined to be equal to [(p-1)!+1]^z).

[(p-1)!+1]^(2z) - 2[(p-1)!+1]^z + 1 = jp

[(p-1)!+1]^(2z) - 2[(p-1)!+1]^z + 1 = 0 mod p

1 - 2[(p-1)!+1]^z + 1 = 0 mod p (using (A))

- 2[(p-1)!+1]^z + 1 = -1 mod p

2[(p-1)!+1]^z - 1 = 1 mod p

2[(p-1)!+1]^z - 1 = xp + 1

I am very curious to see how you got

2[(p-1)!+1]^z + 1 = xp + 1

Moreover, in order to state "(n-1)!+1=0modn iff n is prime", you

must also prove that the congruence implies the primality. Up to now,

you only try to prove 'the primality implies the congruence'.

Marcel Martin

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