Number theory: primitive roots

  • #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:

Answers and Replies

  • #2
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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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