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Math Help - Primitive Root Question

  1. #1
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    Primitive Root Question

    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.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that x=r^{\tfrac{p-1}{2}} \equiv -1 (\bmod. p) since x^2 \equiv 1 (\bmod. p) and so -since p is prime- x\equiv 1 (\bmod. p) or x \equiv -1 (\bmod. p), but the former is not possible since r is a primitive root.

    Thus -r \equiv r^{\frac{p+1}{2}}(\bmod. p) so your problem is equivalent to showing that \displaystyle\frac{p+1}{2} is coprime to p-1.

    But 2\cdot \left(\displaystyle\frac{p+1}{2}\right) - (p-1) = 2 and so \text{gcd}\left(\tfrac{p+1}{2}, p-1\right) is either 1 or 2.

    Now note that 4 doesn't divide p+1, since p-1 is divisible by 4, and so (p+1)/2 must be odd.
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  3. #3
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    Re: Primitive Root Question

    Quote Originally Posted by PaulRS View Post
    Note that x=r^{\tfrac{p-1}{2}} \equiv -1 (\bmod. p) since x^2 \equiv 1 (\bmod. p) and so -since p is prime- x\equiv 1 (\bmod. p) or x \equiv -1 (\bmod. p), but the former is not possible since r is a primitive root.

    Thus -r \equiv r^{\frac{p+1}{2}}(\bmod. p)
    why -r \equiv r^{\frac{p+1}{2}}(\bmod. p) is true?
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