有限域的元素都可以写成两个平方的和

We denote a prime $p$, $q = p^n$ and $F$ is a field with $q$ element.

Lemma 1

$x = y^2 \text{ for some } y \in F \Leftrightarrow x^{\frac{q-1}{2}} = 1$

Lemma 2

For any generator $g$ of multiplication group of $F^\times$, there exist a element $x \in F$ such that $(g - x^2) = y^2$ for some $y \in F$.

Consider
$$
\begin{array}{l}
\sum_{x \in F^\times} \left(g - x^2\right)^{\frac{q-1}{2}}
\end{array}
$$
$$
= \sum_{k=1}^{q-1} \left(g - (g^{k})^2\right)^{\frac{q-1}{2}}
$$
$$
=g^{\frac{q-1}{2}} \sum_{k=1}^{q-1} \left(\sum_{i=0}^{\frac{q-1}{2}}\binom{\frac{q-1}{2}}{i} (-g)^{(2k-1)i} \right)
$$
$$
= - \sum_{i=1}^{\frac{q-1}{2}-1}\binom{\frac{q-1}{2}}{i}\sum_{k=1}^{q-1} (-g)^{(2k-1)i} -\binom{\frac{q-1}{2}}{0}\sum_{k=1}^{q-1} (-g)^{0}-\binom{\frac{q-1}{2}}{\frac{q-1}{2}}\sum_{k=1}^{q-1} (-g)^{(2k-1)\frac{q-1}{2}}
$$
$$
= -\sum_{i=1}^{\frac{q-1}{2}-1}\binom{\frac{q-1}{2}}{i}\frac{1-(-g)^{(2q-1)i}}{1-(-g)^{2i}} (-g)^i - (q-1) - (q-1)
$$
$$
= -\sum_{i=1}^{\frac{q-1}{2}-1}\binom{\frac{q-1}{2}}{i} (-g)^i +2
$$
$$
= -(1-g)^{\frac{q-1}{2}} + 1 + (-g)^{\frac{q-1}{2}} + 2
$$
$$
= 3 - (1-g)^{\frac{q-1}{2}} - (-1)^{\frac{q-1}{2}} \in {1,3,5}
$$

If $p \not = 2$, then $\sum_{x \in F^\times} \left(g - x^2\right)^{\frac{q-1}{2}} \not = q-1$ (calculate in $F$). Thus, there at least one element $x$ of $F$ such that $(g-x^2) = y^2$ for some $y \in F$.
If $p = 2$, then every element of $F$ are square.
we finish our proof.

Thm

Every element of $F$ is a sum of two square.

Consider $g$ as a generator of $F^\times$,
By Lemma 2, there exist $a,b \in F$ such that $g = a^2 + b^2$.
for any $x = g^r \in F$, if $r$ is even ,then $x = (g^{\frac{r}{2}})^2 + 0^2$.
If $r$ is odd, $x = (g^{\frac{r-1}{2}})^2 g= ((g^{\frac{r}{2}})^2 + 0^2)(a^2+b^2) = (g^{\frac{r}{2}}a)^2+(g^{\frac{r}{2}}b)^2$.

Hence, every element of $F$ can write as a sum of two square.