A Problem on Finite Fields Here is the theorem I am trying to prove: 
Let F be a finite field of odd order. Prove that exactly half of nonzero elements of F are squares in F.
Here are my ideas:
(1) Consider the squaring homomorphism $\phi: F \to F$ defined by $x \to x^2$. First I determine the kernel of the homomorphism (which I believe to be {-1, 1}, but please let me know if this is correct)
(2) Maybe use that to find the order of the $Im(\phi)?$. But I do not know how to do this. Any help will be appreciated.
 A: Your ideas are both good. Here are some hints/minor adjustments. 
For (1): you really want to view $\phi$ as a map from $F^{\times}$ to $F^{\times}$, so that $\phi$ is a group homomorphism. The kernel of this morphism is indeed $1, -1$ - why? (Uncover the spoiler for one way to see this.)

 As Arturo points out in the comments, it's probably best to argue that any element of the kernel is a root of $X^{2}-1$ over $F$. Since $1$ and $-1$ are clearly roots, and there are at most $2$ roots, this completely determines the kernel. Alternatively, if one wants to use a harder result than necessary: $F^{\times}$ is cyclic of even order, and so there is a unique cyclic subgroup of order $n$ for each $n$ dividing $|F^{\times}|$. The kernel of $\phi$ consists of the subgroup of $F^{\times}$ of elements of order dividing $2$.

For (2): use the first isomorphism theorem for groups. Again, uncover the spoiler for further details. 

 $\mathrm{Im}(\phi) \cong F^{\times}/\ker(\phi)$, so $|\mathrm{Im}(\phi)| = |F^{\times}|/|\ker(\phi)| = |F^{\times}|/2$, where the last equality follows from (1).  

