About Fermat's Little Theorem?

[Tigerrabbit]

If p is a prime and a is any integer not divisible by p, then ap − 1 − 1 is divisible by p.

This is equivalent to the following two statements:
\(\displaystyle a^{p-1} \equiv 1 \: \: \text{(mod p)}\) and \(\displaystyle a^p \equiv a \: \: \text{(mod p)}\)

May I ask one question? Is \(\displaystyle a^{(p-1)/2}=-1 \) right? How can I prove that? Could you help me?
Is there one theorem can prove that [imath]a^{(p-1)/2}=-1[/imath]? Somebody show me this and some examples are also given to satisfy the result. Like that:
\(\displaystyle a=2, p=5 \rightarrow a^{(5-1)/2}=-1(mod\ 5)\)
How can I prove that?

 

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[imath]\;[/imath]

 

[topsquark]

May I ask one question? Is \(\displaystyle a^{(p-1)/2}=-1 \) right? How can I prove that? Could you help me?
Is there one theorem can prove that [imath]a^{(p-1)/2}=-1[/imath]? Somebody show me this and some examples are also given to satisfy the result. Like that:
\(\displaystyle a=2, p=5 \rightarrow a^{(5-1)/2}=-1(mod\ 5)\)
How can I prove that?

To start with, is [imath]2^{(5-1)/2} \equiv -1 \text{ (mod 5)}[/imath]?

As to a proof, can you square both sides? Under what conditions can you do so?

-Dan

 

[blamocur]

May I ask one question? Is \(\displaystyle a^{(p-1)/2}=-1 \) right? How can I prove that? Could you help me?
Is there one theorem can prove that [imath]a^{(p-1)/2}=-1[/imath]? Somebody show me this and some examples are also given to satisfy the result. Like that:
\(\displaystyle a=2, p=5 \rightarrow a^{(5-1)/2}=-1(mod\ 5)\)
How can I prove that?

[imath]2^{(7-1)/2} \neq -1 (\mod 7)[/imath]

 

[Tigerrabbit]

[imath]2^{(7-1)/2} \neq -1 (\mod 7)[/imath]

Thank you give me the answer.

 

[Tigerrabbit]

To start with, is [imath]2^{(5-1)/2} \equiv -1 \text{ (mod 5)}[/imath]?

As to a proof, can you square both sides? Under what conditions can you do so?

-Dan

Someone gives the answer. What I say just an coincide....