Bruno
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![[SOLVED] Wilson's theorem revisited Empty](https://2img.net/i/empty.gif) | | Subject: [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
Admin
Posts : 184
Join date : 2009-09-15
Age : 36
Location : the infinite, frictionless plane of uniform density
| | Subject: 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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Bruno
Admin
Posts : 184
Join date : 2009-09-15
Age : 36
Location : the infinite, frictionless plane of uniform density
| | Subject: Re: [SOLVED] Wilson's theorem revisited Sat Jan 09, 2010 5:37 am | |
| Here is the solution (sorry, don't feel like typing LaTeX  ): 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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| | Subject: Re: [SOLVED] Wilson's theorem revisited | |
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