# Each Element in a Finite Field is the Sum of Two Squares

## Problem 511

Let $F$ be a finite field.

Prove that each element in the field $F$ is the sum of two squares in $F$.

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## Proof.

Let $x$ be an element in $F$. We want to show that there exists $a, b\in F$ such that

\[x=a^2+b^2.\]

Since $F$ is a finite field, the characteristic $p$ of the field $F$ is a prime number.

If $p=2$, then the map $\phi:F\to F$ defined by $\phi(a)=a^2$ is a field homomorphism, hence it is an endomorphism since $F$ is finite.( The map $\phi$ is called the Frobenius endomorphism).

Thus, for any element $x\in F$, there exists $a\in F$ such that $\phi(a)=x$.

Hence $x$ can be written as the sum of two squares $x=a^2+0^2$.

Now consider the case $p > 2$.

We consider the map $\phi:F^{\times}\to F^{\times}$ defined by $\phi(a)=a^2$. The image of $\phi$ is the subset of $F$ that can be written as $a^2$ for some $a\in F$.

If $\phi(a)=\phi(b)$, then we have

\[0=a^2-b^2=(a-b)(a+b).\]
Hence we have $a=b$ or $a=-b$.

Since $b \neq 0$ and $p > 2$, we know that $b\neq -b$.

Thus the map $\phi$ is a two-to-one map.

Thus, there are $\frac{|F^{\times}|}{2}=\frac{|F|-1}{2}$ square elements in $F^{\times}$.

Since $0$ is also a square in $F$, there are

\[\frac{|F|-1}{2}+1=\frac{|F|+1}{2}\]
square elements in the field $F$.

Put

\[A:=\{a^2 \mid a\in F\}.\]
We just observed that $|A|=\frac{|F|+1}{2}$.

Fix an element $x\in F$ and consider the subset

\[B:=\{x-b^2 \mid b\in F\}.\]
Clearly $|B|=|A|=\frac{|F|+1}{2}$.

Observe that both $A$ and $B$ are subsets in $F$ and

\[|A|+|B|=|F|+1 > |F|,\]
and hence $A$ and $B$ cannot be disjoint.

Therefore, there exists $a, b \in F$ such that $a^2=x-b^2$, or equivalently,

\[x=a^2+b^2.\]

Hence each element $x\in F$ is the sum of two squares.

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