[((n-1)/2)]^2 = -(-1)^[(n-1)/2] (mod n)

xax · Apr 5, 2008

p is prime, not 2. Thanks in advance

mathman · Apr 5, 2008

xax said:

p is prime, not 2. Thanks in advance

There is no p in your expression!

matticus · Apr 5, 2008

he had to have meant (mod p)

xax · Apr 6, 2008

p = n

robert Ihnot · Apr 6, 2008

This is not a difficult problem, but you must understand Wilson's Theorem. Wilson, by the way, was never a mathematician, and as a student went on to become a lawyer.

He is credited with noticing the theorem, but was unable to prove it. So much for the immortal glory of a name theorem!

Pere Callahan · Apr 6, 2008

xax said:

p = n

So what is your question and what have you tried to solve it?

xax · Apr 7, 2008

I thought it was clear what the question was(since the others understood): prove that this is true for every n prime, n!= 2. Using the Wilson theorem, I didn't go far:
((n-1)/2)!*((n+1)/2)*((n+3)/2)*...*(n-1) = -1 (mod n).

xax · Apr 7, 2008

Nevermind guys, I figured it out. Thanks for your input.

[((n-1)/2)]^2 = -(-1)^[(n-1)/2] (mod n)

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[((n-1)/2)]^2 = -(-1)^[(n-1)/2] (mod n)

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