Sum of squares in finite fields

Let K be a finite field. Let us prove that every element of K is the sum of two squares. Let us distinguish two cases.

K has characteristic 2. Let us consider the map

Φ:K—>K, x |—>x² .

Let us observe that Φ is an injective homomorphism. In fact it is a homomorphism having the field characteristic 2 and it is injective because the kernel is an ideal of K that is not the whole field (because Φ(1)=1), therefore, since K is a field, the kernel is trivial. Now since K is finite Φ is also surjective. It follows that if x ∈ K then x=a²+0² for some a in K.

K has characteristic an odd prime

Lemma 1:

If F is a finite field then F* = F \ {0} is a cyclic group.

Lemma 2:

Let G be a fake group and A,B be subsets of G such that |A|+|B| > |G| then G=AB.

The proof of lemma 1 can be found in any group theory book. For lemma 2 the proof is very basic and I refer you to the link https://groupprops.subwiki.org/wiki/Product_of_subsets_whose_total_size_exceeds_size_of_group_equals_whole_group.

Now suppose that |K| = n. Since the characteristic of K is odd n is odd. Then the cyclic group K* = K \ {0} has order n-1 which is even. Let a ∈ K* : K*=. Then A={0,1,a²,…,aⁿ⁻³} is a set of elements of K sum of two squares (the fact that 0 is a sum of squares is obvious, for the remaining ones it follows because they are to the power of an even power) and we have |A|=(n+1)/2. Now let’s prove that K=A+A. Let’s think of K as an additive group with the sum. Then A is a subset of K: |A|+|A|=n+1>n=|K|. From lemma 2 it follows that K=A+A from which the thesis. □