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Thread: Primitive roots

  1. #1
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    Primitive roots

    If g is a primitive root mod 13 then I can show that g^11 is also a primitive root.

    My question is how can I deduce that the product of all the primitive roots mod 13 are congruent to 1 (mod 13) from this result?

    Is there a corresponding result for the product of all the primitive roots mod 169?

    Thanks in advance.
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  2. #2
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    find the condition for k such that $\displaystyle g^k$ is also a primitive root mod 13.
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  3. #3
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    I'm aware of the condition for k to such that g^k is a primitive root mod 13. In fact, I have shown that g^11 is a primitive root mod 13.

    But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13.

    I was then curious if a similar result holds for the product of all primitive roots mod 169.
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  4. #4
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    Quote Originally Posted by Cairo View Post
    I'm aware of the condition for k to such that g^k is a primitive root mod 13. In fact, I have shown that g^11 is a primitive root mod 13.

    But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13.

    I was then curious if a similar result holds for the product of all primitive roots mod 169.
    Given that $\displaystyle g $ is a primitive root of $\displaystyle 13 $, all the primitive roots are given by $\displaystyle g^k$, where $\displaystyle (k,12)=1$; so the primitive roots of $\displaystyle 13 $ are $\displaystyle g^1, g^5, g^7,$ and $\displaystyle g^{11}$.

    Then the product of all the primitive roots of $\displaystyle 13 $ is congruent to $\displaystyle g^{1+5+7+11}=g^{24}$ modulo $\displaystyle 13 $. By Fermat's Theorem, $\displaystyle g^{24}=(g^{12})^2\equiv1\pmod {13}$.

    The general result is the following: Let $\displaystyle m $ be a positive integer which has a primitive root. Then the product of the primitive roots of $\displaystyle m $ is congruent to $\displaystyle 1 $ modulo $\displaystyle m $; the exceptions are $\displaystyle m=3, 4,$ and $\displaystyle 6 $.

    By the way, the integers $\displaystyle m $ that have primitive roots are $\displaystyle 1, 2, 4, p^k,$ and $\displaystyle 2p^k$, where $\displaystyle p $ is an odd prime.
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  5. #5
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    Thank you both for your help with this.

    I managed to prove the first result in the end. I never thought to use Fermat's Theroem until It struck me at the last minute.

    I would like some clarification on your second point melese if that is possible?

    If g is a primitive root mod 169, then so is g+13. So how are the rest of the primitive roots obtained and how does their product become congruent to 1 mod 169?
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