Field with 4 Elements has only Order 2 Elements

Theorem
Let $\struct {F, +, \times}$ be a field which has exactly $4$ elements.

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

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

Also see

 * Galois Field of Order $4$, where it is demonstrated that such an $F$:


 * $\struct {F, +} \cong \Z_2 \times \Z_2$ and $\struct {F^*, \times} \cong \Z_3$


 * is actually a field.