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Primitive element modulo p

  1. Dec 1, 2009 #1
    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.
     
  2. jcsd
  3. Dec 1, 2009 #2
    That sounds like a homework problem?? Do you think?
     
  4. Dec 2, 2009 #3
    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.
     
    Last edited: Dec 2, 2009
  5. Dec 3, 2009 #4
    thanks alot i really appreciate your help
     
  6. Dec 3, 2009 #5
    hope2009: thanks alot 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 [tex]a^q \equiv -1 \bmod p [/tex] Since by Fermat's little theorem for a prime, [tex]a^q \equiv a \bmod q [/tex]
     
    Last edited: Dec 3, 2009
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