finite field with characteristic not equal to 2

Question

How many elements are squares for a finite field with characteristic $p \ne 2$?

Not sure how to begin this problem.

Answer 1

Let $K$ be that finite field of prime characteristic $p > 2,$ and let $K^*$ be its multiplicative group. Then $K$ has $p^n$ elements for some positive integer $n,$ and $K^*$ is a cyclic group with $p^n - 1$ elements. Consider the set $H := \{t^2 \mid t \in K^*\}.$ It consists precisely of the non-zero square numbers in $K.$ Show that $H$ is a subgroup of $K^*$ of index $2.$

Answer 2

Let $\Bbb F$ be a finite field with characteristic $> 2$. Then, $\Bbb F^\times = \Bbb F \setminus \{0\}$ is an abelian group. Thus, the map $$\varphi : \Bbb F^\times \to \Bbb F^\times$$ defined by $\varphi(x) = x^2$ is a homomorphism.

Now, note that $\varphi(x) = 1$ has two distinct solutions $1$ and $-1$. (Why are they distinct?)
Since $\Bbb F$ is a field, $x^2 = 1$ cannot have more solutions.

Thus, $\ker \varphi$ has cardinality exactly $2$ and thus, $\varphi(\Bbb F^\times)$ has cardinality $\frac{1}{2}\lvert \Bbb F^\times \rvert.$
Note that $\varphi(\Bbb F^\times)$ is precisely the set of non-zero squares. Since $0$ is indeed a square, the total number of squares are $$\frac{1}{2}\lvert \Bbb F^\times \rvert + 1 = \frac{1}{2}(\lvert \Bbb F \rvert - 1) + 1 = \boxed{\frac{1}{2}(\lvert \Bbb F \rvert + 1)}.$$