Note that the nonzero elements of form a cyclic multiplicative group. Let be the generator of this group. Then 0, , , … are the squares in this field. Obviously, if an element is a square itself, then it can be expressed as sum of two squares. Now we shall prove that a non-square can also be expressed as sum of two squares.

Suppose is a non-square. The existence of the expression of sum of two squares of is equivalent to the solvability of the equation in the field, which is in turn equivalent to (note that ). Let and . We can rewrite the equation as . is a non-square because it is a product of a square and a non-square.

Suppose on the contrary that admits no solutions. That means whenever is a square, is also a square. Now, given a square , we can get squares, namely , , , …( since the field is of characteristic ). It is easy to verify that given two elements and , either the two sets of elements and are the same or they are disjoint. Since the squares are closed under the addition of identity, the number of squares in the field should be divisible by . But by the previous argument, this number is , which is not divisible by , a contradiction.$[/dollar]