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Number theory: primitive roots

  1. Dec 16, 2009 #1
    Find a primitive root modulo 101. What integers mod 101 are 5th powers? 7th powers?

    -I tested 2.
    -2 and 5 are the prime factors dividing phi(101)=100 so i calculated 2^50 is not congruent to 1 mod 101 and 2^20 is not congruent to 1 mod 101.
    -Therefore 2 is a primitive root modulo 101

    I guess this means to find find all m such that there exists x such that
    x^5 = m (mod 101)

    If this is the case, then how do I find these solutions?
    I found that the congruence x^5 = 1 (mod 101) has gcd(5,100)=5 solutions. I also know that 2^100 = 1 (mod 101) so that ((2^20)^5) = 1 (mod 101).
    I don't know where to go from there.
    Last edited: Dec 16, 2009
  2. jcsd
  3. Dec 16, 2009 #2
    To ask what integers modulo 101 are 5th or 7th powers does not mean to find all [tex]x[/tex] such that [tex]x^5 \equiv 1 \pmod{101}[/tex] (or 7), but to find all [tex]m[/tex] such that there exists [tex]x[/tex] such that [tex]x^5 \equiv m \pmod{101}[/tex] (or 7).
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