Every element of a finite field is expressible as a sum of two squares

History
This result is attributed to Henry Mann.

Statement
Suppose $$k$$ is a finite field. Then, every element $$x \in k$$ can be expressed in the form $$x = a^2 + b^2$$, where $$a,b \in k$$.

Facts used

 * 1) uses::Multiplicative group of a finite field is cyclic: We actually only need the weaker statement that, for a field of odd characteristic, exactly half the elements of the multiplicative group are squares.
 * 2) uses::Product of subsets whose total size exceeds size of group equals whole group: If $$A,B$$ are subsets of a finite group $$G$$, where $$|A| + |B| > |G|$$, then $$G = AB$$.

Case of characteristic two
In this case, the square map is surjective and every element is a square, because the multiplicative group is of odd order.

Case of odd characteristic

 * 1) Reasoning in the multiplicative group: Suppose $$k$$ has $$q$$ elements. Then its multiplicative group $$k^\times$$ has $$q - 1$$ elements. By fact (1), the multiplicative group is cyclic of order $$q - 1$$, which is even. Thus, exactly half the elements (corresponding to even powers of the generator) are squares. Since $$0$$ is also a square, we obtain $$(q + 1)/2$$ elements of $$k$$ that are squares.
 * 2) Reasoning in the additive group: We now apply fact (2) with $$G$$ as the additive group of $$k$$, which has size $$q$$, and both $$A$$ and $$B$$ as equal to the set of (multiplicative) squares, which has size $$(q + 1)/2$$.