Number theory: primitive roots

[timjones007]

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.

 

[ystael]

To ask what integers modulo 101 are 5th or 7th powers does not mean to find all $$x$$ such that $$x^5 \equiv 1 \pmod{101}$$ (or 7), but to find all $$m$$ such that there exists $$x$$ such that $$x^5 \equiv m \pmod{101}$$ (or 7).