Field with 4 Elements has only Order 2 Elements

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

Then:
 * $\forall a \in \GF: a + a = 0_\GF$

where $0_\GF$ is the zero of $\GF$.

Also see

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


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


 * is actually a field.