>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

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

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