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Math Help - Primitive roots

  1. #1
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    Exclamation Primitive roots

    Hi, another question again, pls help!!!

    Let r be a primitive root of the prime p with p congruent to 1 (mod 4). Show that -r is also a primitive root.

    Thanks for your help!!!
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  2. #2
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    Quote Originally Posted by suedenation View Post
    Hi, another question again, pls help!!!

    Let r be a primitive root of the prime p with p congruent to 1 (mod 4). Show that -r is also a primitive root.

    Thanks for your help!!!
    Let, r be a primitive root.
    What is the order of -r?
    Assume, 0<k<p and arrive at contradiction.
    We have,
    (-r)^k\equiv 1(\mbox{ mod }p)

    If, k is even then we can drop the negative sign,
    r^k\equiv 1(\mbox{ mod }p)
    Contradiction.

    If, k is odd then some difficultly arises.
    So we cannot drop the negative sign,
    r^k\equiv -1(\mbox{ mod }p)
    Multiply through by r,
    r^{k+1}\equiv -r(\mbox{ mod }p)
    Since, k odd we have k+1 even.
    Thus, we can write,
    (r^{\frac{k+1}{2}})^2\equiv -r (\mbox{ mod }p)
    We see that -r is a quadradic residue of p.
    Hence the Legendre symbol,
    (-r/p)=1
    Since the Legendre symbol is multiplicative,
    (-1/p)(r/p)=1
    Since, p\equiv 1(\mbox{ mod }4)
    We know that, (-1/p)=1
    Thus,
    (r/p)=1
    Then by Euler's criterion we have,
    r^{\frac{p-1}{2}}\equiv 1(\mbox{ mod }p)
    But, r is a primitive root
    And, \frac{p-1}{2}<p.
    Thus, we have another contradiction.
    Thus k can never be odd.

    That means the only possible choice is k=p.
    Last edited by ThePerfectHacker; November 21st 2006 at 06:19 PM.
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