It is well-known that in a finite field every element is a sum of two squares (Problem 1). It is however not true that every element of a finite field is a sum of two cubes. For example, in we cannot write or as a sum of two cubes because and so the only elements of that are a sum of two cubes are

But if, in a finite field, for some non-zero element of the field, then we can show that every element of the field is a sum of two cubes (Problem 2).

**Problem 1**. Show that every element of a finite field is a sum of two squares.

**Solution**. Let be a finite field. So we want to show that if then for some We can actually be more specific if we consider two cases. Let

Case 1: for some integer Then, since for all we get So in this case, every element of the field is a square.

Case 2: for some integer Since is finite, the multiplicative group is cyclic.

So Let and consider the sets

Clearly and Thus and hence

Therefore i.e. there exist such that and the result follows.

**Remark 1**. Regarding the second case in the solution of Problem 1, notice that, in fact, we have

and so The reason is that if for some integers then and hence must be divisible by implying that is even.

**Problem 2**. Let be a finite field and suppose that there exists such that Show that every element of is a sum of two cubes.

**Solution**. So we want to show that if then for some Let and let’s consider three cases.

Case 1: for some integer Then for all

Case 2: for some integer Then for all and clearly

So, in both cases 1 and 2, for every there exists such that

Case 3: for some integer Since is finite, the multiplicative group is cyclic. So Let and consider the sets

Clearly and So and

So at least two of the sets have non-empty intersection. If or then for some and we are done.

Now suppose that So there exist such that and so Since, as given in the problem, for some we have and Hence

**Remark 2**. Regarding the third case in the solution of Problem 2, notice that, in fact, we have

and so The reason is that if for some integers then and hence must be divisible by implying that is divisible by