Find a primitive root modulo 101. What integers mod 101 are 5th powers? 7th powers?(adsbygoogle = window.adsbygoogle || []).push({});

-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.

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# Homework Help: Number theory: primitive roots

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