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

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This result is attributed to Henry Mann.


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. 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. 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.