Is α a Primitive Element Modulo p Given α^q ≡ -1 mod p?

[hope2009]

Suppose that p and q are odd primes and p=2q+1. Suppose that
α∈ Z_p^*,α≢±1 mod p.

Prove that α is primitive element modulo p if and only if α^q≡-1 mod q.

 

[robert Ihnot]

That sounds like a homework problem?? Do you think?

 

[robert Ihnot]

I don't get any response on this. But since q is a "Sofie Germain Prime," then q=(p-1)/2.

By Fermat's Little Theorem z^(p-1) ==1 Mod p. IF a number is a primitive root--which I take is what he means here- no extensions are mentioned; then all its powers Modulo p must differ, giving us the p-1 residue system. ;

To do this, a^(p-1)/2 ==-1 Mod p. Since if it was 1, then a would not generate all the elements.

 

[hope2009]

I don't get any response on this. But since q is a "Sofie Germain Prime," then q=(p-1)/2.

By Fermat's Little Theorem z^(p-1) ==1 Mod p. IF a number is a primitive root--which I take is what he means here- no extensions are mentioned; then all its powers Modulo p must differ, giving us the p-1 residue system. ;

To do this, a^(p-1)/2 ==-1 Mod p. Since if it was 1, then a would not generate all the elements.

thanks a lot i really appreciate your help

 

[robert Ihnot]

hope2009: thanks a lot i really appreciate your help

Happy to hear you are satisifed. However, there is an error I see now:

Prove that α is primitive element modulo p if and only if α^q≡-1 mod q.

You mean $$a^q \equiv -1 \bmod p$$ Since by Fermat's little theorem for a prime, $$a^q \equiv a \bmod q$$