Field with 4 Elements has only Order 2 Elements

Theorem
Let $\left({F, +, \times}\right)$ be a field which has exactly $4$ elements.

Then:
 * $\forall a \in F: a + a = 0_F$

where $0_F$ is the zero of $F$.

Proof
By definition of field, both the algebraic structures $\left({F, +}\right)$ and $\left({F^*, \times}\right)$ are (abelian) groups, where $F^* := F \setminus \left\{{0}\right\}$.

By definition:
 * $\left({F, +}\right)$ is of order $4$
 * $\left({F^*, \times}\right)$ is of order $3$.

From Classification of Groups of Order up to 15, there are only two possibilities:
 * $(1):\quad \left({F, +}\right) \cong \Z_4$ and $\left({F^*, \times}\right) \cong \Z_3$
 * $(2):\quad \left({F, +}\right) \cong \Z_2 \times \Z_2$ and $\left({F^*, \times}\right) \cong \Z_3$.

In case $(2)$:
 * $\forall a \in F: a + a = 0_F$

but in case $(1)$:
 * $\exists a, b \in F: a + a = b \ne 0_F$ where $b + b = 0_F$

as there exists an element of $\left({F, +}\right)$ of order $4$.

So suppose $(1)$ describes a field.

Then:

But then:

As $\left({F, +}\right) \cong \Z_4$, both $a + a = b$ and $\left({-a}\right) + \left({-a}\right) = b$.

The only way for $\left({b \times b}\right) + \left({b \times b}\right) = 0_F$ is for $b \times b = b$ or $b \times b = 0_F$.

The second case is eliminated as $0_F \notin \left({F^*, \times}\right)$.

So it must be the case that $b \times b = b$.

So as $\left({F^*, \times}\right) \cong \Z_3$ it must follow that $a \times b = a, a \times \left({-a}\right) = b$ and then $a \times a = -a$.

It follows that the Cayley tables of $\left({F, +}\right)$ and $\left({F^*, \times}\right)$ must therefore be as follows:


 * $\begin{array}{c|cccc}

+ & 0_F & a & b & -a \\ \hline 0_F & 0_F & a & b & -a \\ a & a & b & -a & 0_F \\ b & b & -a & 0_F & a \\ -a & -a & 0_F & a & b \\ \end{array} \qquad \begin{array}{c|cccc} \times & 0_F & b & a & -a \\ \hline 0_F & 0_F & 0_F & 0_F & 0_F \\ b & 0_F & b & a & -a \\ a & 0_F & a & -a & b \\ -a & 0_F & -a & b & a \\ \end{array}$

But:

So:
 * $\left({a + b}\right) \times a \ne \left({a \times a}\right) + \left({b \times a}\right)$

demonstrating that $\times$ is not distributive over $+$.

Thus $F$ as has been defined:
 * $\left({F, +}\right) \cong \Z_4$ and $\left({F^*, \times}\right) \cong \Z_3$

is not a field.

It follows that our supposition that $\left({F, +}\right) \cong \Z_4$ was false.

Thus, if $F$ is a field, then $\left({F, +}\right) \cong \Z_2 \times \Z_2$.

That is:
 * $\forall a \in F: a + a = 0_F$

as needed to be proved.

Note
It has still not been demonstrated that such an $F$:
 * $\left({F, +}\right) \cong \Z_2 \times \Z_2$ and $\left({F^*, \times}\right) \cong \Z_3$

is actually a field.

What has been demonstrated here is that if $F$ is a field, then it would have to be of that form.