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 [SOLVED] Wilson's theorem revisited

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Bruno
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PostSubject: [SOLVED] Wilson's theorem revisited   Fri Nov 27, 2009 7:58 pm

Show that if p is an odd prime then {[(p-1)/2]!}^2 = (-1)^{(p+1)/2} (mod p).


Last edited by Bruno on Mon Jan 18, 2010 10:15 am; edited 1 time in total
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Bruno
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PostSubject: Re: [SOLVED] Wilson's theorem revisited   Sun Dec 20, 2009 4:02 pm

Hint : factor the polynomial x^{(p-1)/2}-1 over the field with p elements and set x=0.
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PostSubject: Re: [SOLVED] Wilson's theorem revisited   Sat Jan 09, 2010 5:37 am

Here is the solution (sorry, don't feel like typing LaTeX Embarassed ):

The set of squares in Z/pZ is S={ 1^2, 2^2, ..., [(p-1)/2]^2) }. So the roots of the polynomial x^{(p-1)/2}-1 are precisely the elements of S. Since the constant term -1 must be (-1)^{(p-1)/2} times the product of the roots, we get (-1)^{(p+1)/2} = {[(p-1)/2]!}^2.
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