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## RE: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

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• (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.
Message 1 of 11 , Nov 5, 2001
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(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
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: 04 November 2001 23:25
Cc: primenumbers@yahoogroups.com
Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

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

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

Marcel Martin

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• I think I have fixed all remaining bugs: http://www.users.globalnet.co.uk/~perry/maths/wilsonfermat/wilsonfermat.htm (or:
Message 2 of 11 , Nov 10, 2001
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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
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: 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
>(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.

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

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

Marcel Martin

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• 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
Message 3 of 11 , Nov 10, 2001
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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
>Now consider (A)-(B). As 2z = p-1, we get:
>2[(p-1)!+1]^z + 1 = xp + 1

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