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Thread: 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, $\displaystyle r$ be a primitive root.
    What is the order of $\displaystyle -r$?
    Assume, $\displaystyle 0<k<p$ and arrive at contradiction.
    We have,
    $\displaystyle (-r)^k\equiv 1(\mbox{ mod }p)$

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

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

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