Proof of a^{\frac{p-1}{2}}=-1 mod p

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If p is a prime and a an integer coprime with p, why is

a^{\frac{p-1}{2}}\equiv -1 mod p ?
 
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It's not true. Consider p=5 and a=4.
 
Yes, i was trying some examples and I think it only works with primitive roots... But why?
 
Are you aware of Fermat's little theorem?? Try to use that...
 
Hah, that's basically the definition of a primitive root. :smile:

Any co-prime "a" has a "period" indicating how soon it gets to "1".
Fermat's little theorem guarantees that (p-1) will bring it back to "1".
The actual period will be a divider of (p-1).
Only for primitive roots, the period is exactly (p-1).
 

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