Field with 4 Elements has only Order 2 Elements

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

Then the characteristic of $F$ is $2$.

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

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

Proof
By Order of Element Divides Order of Finite Group, any element $a \in F$ has order $1$, $2$ or $4$ in the additive group $\left({F, +}\right)$.


 * : $\S 2.4$: The rational numbers and some finite fields: Exercise $4$