## Proof of Wilson's Theorem using Fermat's

Expand Messages
• If (p-1)!+1 is co-prime to p, then: [(p-1)!+1]^[p-1]=1modp So, j(p-1)!+1=kp+1 j(p-1)!=kp (p-1)!=xp which is a contradiction, therefore (p-1)!+1 always divides
Message 1 of 11 , Nov 1, 2001
If (p-1)!+1 is co-prime to p, then:

[(p-1)!+1]^[p-1]=1modp

So, j(p-1)!+1=kp+1

j(p-1)!=kp

(p-1)!=xp

which is a contradiction, therefore (p-1)!+1 always divides p.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• Fermat s Little Theorem, the well-known a^(p-1)=1modp, holds iff a and p are co-prime. So if a and p are not co-prime then the result is NOT 1modp, but xmodp,
Message 2 of 11 , Nov 2, 2001
Fermat's Little Theorem, the well-known a^(p-1)=1modp, holds iff a and p are
co-prime.

So if a and p are not co-prime then the result is NOT 1modp, but xmodp,
where x<>1.

[(p-1)!+1]^[p-1] when expanded gives a whole load of powers of (p-1)!, and
1, so j in:

j(p-1)!+1=kp+1

represents a collection of the (p-1)!s. e.g. if p was 3, then we have:

[(p-1)!+1]^2 = (p-1)!^2 + 4(p-1)! + 1

so in this case j=(p-1!)+4.

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: 01 November 2001 21:54
Cc: Prime Numbers
Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

The proof requires
(1) The congruence implies the primality
(2) The primality implies the congruence

It is not clear what you want prove, (1) or (2)?
I suppose this is (2), otherwise
>[(p-1)!+1]^[p-1] = 1 mod p
would have no reason to be true.

But where does this come from?
>So, j(p-1)!+1=kp+1

Marcel Martin

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• Yes, but p is a prime. Therefore iff (p-1)!+1 is co-prime to p will the RHS be 1modp. ... Ok. I skipped a bit. Jon Perry perry@globalnet.co.uk
Message 3 of 11 , Nov 3, 2001
Yes, but p is a prime. Therefore iff (p-1)!+1 is co-prime to p will the RHS
be 1modp.

>j(p-1)! = 0 mod p doesn't imply (p-1)! = 0 mod p

Ok. I skipped a bit.

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: 02 November 2001 20:00
Cc: Prime Numbers
Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

>Fermat's Little Theorem, the well-known a^(p-1)=1modp, holds iff a and p
are
>co-prime.

No, not "if and only if". For instance, 3^(8-1) mod 8 is not equal
to 1 whereas GCD(3,8)=1.

>[(p-1)!+1]^[p-1] when expanded gives a whole load of powers of (p-1)!, and
>1, so j in:
>j(p-1)!+1=kp+1

Ok. We can go further.

j(p-1)! = 0 mod p doesn't imply (p-1)! = 0 mod p

Using your notation, j(p-1)!=kp =/=> (p-1)!=xp

For instance, 21 mod 3 = 0 doesn't imply 7 mod 3 = 0.
If you want use (p-1)!=xp as a contradiction, you have to prove
first that p doesn't divide j.

Marcel Martin

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• ... This is going too fast.( and BTW it s lose, not loose. lose means to waste, loose means slack, as in My shirt is loose, but my jeans are tight. ) *** If
Message 4 of 11 , Nov 4, 2001
>Yes but do not loose time to try to prove j <> 0 mod p. Here,
>we can say p divides j precisely because, p being prime, it
>cannot divide (p-1)!

This is going too fast.( and BTW it's lose, not loose. lose means to waste,
loose means slack, as in 'My shirt is loose, but my jeans are tight.')

***
If (p-1)!+1 is co-prime to p, then: [1]

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

So, j(p-1)!+1=kp+1 [3]

j(p-1)!=kp [4]

(p-1)!=xp [5]
***

[1][2] is true, with p a prime. OK, this doesn't do n composite, but then
this isn't very difficult.

[3] is true, although we do not know anything about j.

[4] is just [3] reduced by +1.

[5] contains the assumptive step. Here I say (p-1)! obviously doesn't
contain p as a factor. But I have ignored the fact that in getting from [4]
to [5] involves assuming that j<>0modp.

But all hope is not lost. If j was to contain a factor of p, then the proof
would be saved, but to what purpose? There is an obvious contradiction
somewhere. So, we may 'safely' assume that j does not contain p as a factor.
But this is ESTD.

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 00:09
Cc: Prime Numbers
Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

>>j(p-1)! = 0 mod p doesn't imply (p-1)! = 0 mod p

>Ok. I skipped a bit.

Yes but do not loose time to try to prove j <> 0 mod p. Here,
we can say p divides j precisely because, p being prime, it
cannot divide (p-1)!

Marcel Martin

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• ... OK, I propose a conjugate theorem to accompany your proof. Phil s Theorem of Binary Bogosity: forall prime p 3, p|2 ... If (2+1) is coprime to p (true
Message 5 of 11 , Nov 4, 2001
On Sun, 04 November 2001, "Jon Perry" wrote:
> >Yes but do not loose time to try to prove j <> 0 mod p. Here,
> >we can say p divides j precisely because, p being prime, it
> >cannot divide (p-1)!

OK, I propose a conjugate theorem to accompany your proof.

Phil's Theorem of Binary Bogosity:
\forall prime p>3, p|2

> ***
> If (p-1)!+1 is co-prime to p, then: [1]

If (2+1) is coprime to p
(true from p>3)

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

(2+1)^(p-1) == 1 (mod p)

> So, j(p-1)!+1=kp+1 [3]

So, j(2)+1 = kp+1

> j(p-1)!=kp [4]

j(2) = kp

> (p-1)!=xp [5]

2 = xp

> ***
>
> [1][2] is true, with p a prime. OK, this doesn't do n

2!+1 = 3 = 1*3
4!+1 = 25 = 5*5
6!+1 = 721 = 103*7

When we can't agree on what 'true' means, then we're going to come up with very different conclusions.

> composite, but then
> this isn't very difficult.
>
> [3] is true, although we do not know anything about j.
>
> [4] is just [3] reduced by +1.
>
> [5] contains the assumptive step. Here I say (p-1)! obviously doesn't
> contain p as a factor. But I have ignored the fact that in getting from [4]
> to [5] involves assuming that j<>0modp.

Never ASSUME, as it makes an ASS of U and ME
(Benny Hill, I believe.)

> But all hope is not lost. If j was to contain a factor of p, then the proof
> would be saved, but to what purpose? There is an obvious contradiction
> somewhere. So, we may 'safely' assume that j does not contain p as a factor.

I don't like that kind of safety...

> But this is ESTD.

Endemic Sexually Transmitted Disease?

More number-theoretic precautions are required, methinks.

Phil

Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com
• ... I didn t mean to - it was an accident. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry BrainBench MVP for HTML and JavaScript
Message 6 of 11 , Nov 4, 2001
>If you wanted to kill all the mathematicians on this list, you
>succeeded :-)

I didn't mean to - it was an accident.

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 18:37
Cc: Prime Numbers
Subject: Re: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

>>Yes but do not lose time to try to prove j <> 0 mod p. Here,
>>we can say p divides j precisely because, p being prime, it
>>cannot divide (p-1)!

>This is going too fast.

Yes, you're right. I forgot that gcd((p-1)!+1,p) <> 1. If you could
prove that p doesn't divide j, your "half" proof could be correct.

>There is an obvious contradiction somewhere. So, we may 'safely'
>assume that j does not contain p as a factor.

In short, the fact we know this is provable otherwise would allow us
to accept lame proofs?
If you wanted to kill all the mathematicians on this list, you
succeeded :-)

Marcel Martin

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• ... Funny. Easier Stated Than Demonstrated. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry BrainBench MVP for HTML and JavaScript
Message 7 of 11 , Nov 4, 2001
> But this is ESTD.

>Endemic Sexually Transmitted Disease?

Funny. Easier Stated Than Demonstrated.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Phil Carmody [mailto:fatphil@...]
Sent: 04 November 2001 12:44
Subject: RE: [PrimeNumbers] Proof of Wilson's Theorem using Fermat's

On Sun, 04 November 2001, "Jon Perry" wrote:
> >Yes but do not loose time to try to prove j <> 0 mod p. Here,
> >we can say p divides j precisely because, p being prime, it
> >cannot divide (p-1)!

OK, I propose a conjugate theorem to accompany your proof.

Phil's Theorem of Binary Bogosity:
\forall prime p>3, p|2

> ***
> If (p-1)!+1 is co-prime to p, then: [1]

If (2+1) is coprime to p
(true from p>3)

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

(2+1)^(p-1) == 1 (mod p)

> So, j(p-1)!+1=kp+1 [3]

So, j(2)+1 = kp+1

> j(p-1)!=kp [4]

j(2) = kp

> (p-1)!=xp [5]

2 = xp

> ***
>
> [1][2] is true, with p a prime. OK, this doesn't do n

2!+1 = 3 = 1*3
4!+1 = 25 = 5*5
6!+1 = 721 = 103*7

When we can't agree on what 'true' means, then we're going to come up with
very different conclusions.

> composite, but then
> this isn't very difficult.
>
> [3] is true, although we do not know anything about j.
>
> [4] is just [3] reduced by +1.
>
> [5] contains the assumptive step. Here I say (p-1)! obviously doesn't
> contain p as a factor. But I have ignored the fact that in getting from
[4]
> to [5] involves assuming that j<>0modp.

Never ASSUME, as it makes an ASS of U and ME
(Benny Hill, I believe.)

> But all hope is not lost. If j was to contain a factor of p, then the
proof
> would be saved, but to what purpose? There is an obvious contradiction
> somewhere. So, we may 'safely' assume that j does not contain p as a
factor.

I don't like that kind of safety...

> But this is ESTD.

Endemic Sexually Transmitted Disease?

More number-theoretic precautions are required, methinks.

Phil

Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• ... (A) [(p-1)!+1]^(p-1) = 1modp = kp+1 = x^2 [(p-1)!+1]^(p-1) - 1 = x^2-1 = (x-1)(x+1) Therefore either x-1 or x+1 contains p as a factor. As p is odd,
Message 8 of 11 , Nov 4, 2001
Mark II:

---

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

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

Therefore either x-1 or x+1 contains p as a factor. As p is odd, let 2z=p-1:

{[(p-1)!+1]^z - 1}{[(p-1)!+1]^z + 1} = (x-1)(x+1)

Therefore x=[(p-1)!+1]^z.

Note that:

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

if (x-1) contains the factor of p.

Therefore (A-B):

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

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

as [(p-1)!+1] doesn't contain a factor of p.

And :

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

if (x+1) contains the factor of p.

Therefore (A-C):

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

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

as [(p-1)!+1] doesn't contain a factor of p.

Q.e.d

---

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• (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 9 of 11 , Nov 5, 2001
(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
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

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

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:
Message 10 of 11 , Nov 10, 2001
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
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

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

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
Message 11 of 11 , Nov 10, 2001
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
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

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
Your message has been successfully submitted and would be delivered to recipients shortly.