let p be an odd prime number, and suppose that p-1 is not divisible by 3. prove that, for every integer a, there is an integer x, such that
x^3 = a (mod p)
My approach is to use fermat's little theorem's proof, and get a^p=a(mod p)
from there we can compare a^p=a(mod p) to x^3 = a (mod p), so for every enteger a, there is an integer x, such that a=x.
What do you guys think?