Four squares theorem for finite fields

Question

We know that there exists four squares theorem for the integers, i.e. any integer can be written as a sum of four squares of integers. I am wondering if there are similar results for integers in finite fields $\mathbb F_{p^n}$. Anything that is even tangentially related to the four squares theorem would be greatly appreciated. Also, are there any proof of the four squares theorem that doesn't use complex analysis?

Answer 1

I am not going to give a complete answer. But it will be more than tangential you asked for.

Instead of sum of four squares we can try to find elements which are sums of two squares. I'll indicate two ways (not going to complete them).

As zero is always a square, we focus on the nonzero elements which is called multiplicative group $F^*$. Being abelian (actually cyclic) the function $f(x) = x^2$ is a homomorphism on that group to itself.

Case 1: Fields with $2^n$ elements: Here $F^*$ is of odd order, a number coprime to 2. So there are no elements of order 2. So the homomorphism $f$ above is one-to-one and hence an automorphism (one-one functions from a finite set to itself are also onto).

Surjectivity of $f$ shows that every nonzero element is a square. and hence every element is a sum of atmost 1 square! (in this case)

Case 2: Fields of odd order. In this case $f$ has a kernel, namely elements satisfying $x^2=1$ or roots of the equation $x^2-1=0$. In a field quadratic equations can have at most 2 roots,; here $\pm1$ are the two roots. This shows $f$ is 2-to-1 homomorphism. Hence by first isomorphism theorem the image (which consists of squares) has half the elements of $F^*$. The image being a subgroup of $F^*$ of index 2, we see that product of two nonsquares is a square.[quotient group being order 2 means square of any element is identity].

You can use Brahmagupta's identity which in words can be expressed as "if two elements are sum of two squares then their product is also a sum of two squares." So this can give you a larger set of elements which are sums of two squares.

You can avoid all the above and try the following which is direct.

For a given $a\in F$. how many solutions does the equation $x^2+y^2 = a$ in the field F?