help with abstract algebra please?

Question

If f is a primitive root mod 37 which of the numbers f^2, f^3,...,f^8 are primitive roots mod 37?

Answer 1

Hint:
(f^2)^18=f^36=1 mod 37
(f^3)^12=f^36=1 mod 37
...
So f^{2,3,4,6,8} are all not primitive roots.
f^5 and f^7 are both primitive since gcd(5,36) and gcd(7,36) are both 1.

Answer 2

The answer is f^k, where k = 5 or 7.
In general, f^k will be a primitive root mod 37
if and only if k is relatively prime to 36.

Answer 3

f^3