let p^n be a prime power (n>=1) and let F be a field with p^n elements
Let p^n be a prime power (n greater or equal to 1) and let F be a field with p^n elements.
(a) If p=2 show that all of the elements of F* are squares (i.e. every element is of the form a^2 for some a in F)
(b) If p is odd how many elements of F* are squares? Why?
Re: let p^n be a prime power (n>=1) and let F be a field with p^n elements
I assume you mean F* is the set of non-zero elements of F. Don't you know the multiplicative group of F is cyclic? So this is really a question about finite cyclic groups.
1. If the order is 2n-1, then if x is a generator of F*, so also is x2.
2. If p is odd, consider the homomorphism which sends every element to its square. The image is then the set of squares; thus you should be able to figure out the order of this image from the fundamental theorem of homomorphisms.