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

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p is prime, not 2. Thanks in advance
 
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xax said:
p is prime, not 2. Thanks in advance
There is no p in your expression!
 
he had to have meant (mod p)
 
p = n
 
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!
 
xax said:
p = n


:smile:

So what is your question and what have you tried to solve it?
 
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).
 
Nevermind guys, I figured it out. Thanks for your input.
 

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