How to prove that in a field with order $4$, $x^2 = x +1$ for $x$ different from $0$ and $1$?

Question

For a field $\mathbb{F}$ with order 4, prove that elements in $\mathbb{F}$ (except for $0$ and $1$) satisfies the equation:

$$ x^2 = x +1 $$

My thought is as followings:

Set $G = \{0,1,a,b\}$. We can prove $Char \mathbb{F} = 2$ and consider $ab$ since the field has no zero divisors then $ab \neq 0$ and if $ab = a$ then $b = 1$ but $b \neq 1$, which implies $ab = 1$. Then consider $a^2$, since $a \neq b$ then $a^2 \neq 1$, by the same reason $a^2 \neq a,0$ then $a^2 = b$, then use the same way to consider $a+1$ we can get the answer.

But does there exist a easier way?

Answer 1

This is obviously not true for $x=0$ or $x=1$. But then the question looks a bit weird. Anyway,

Suppose that $x$ is neither $0$ nor $1$. We have $(x+1)(x^2 - x - 1) =(x+1)( x^2 - x + 1 ) = x^3 -1$. Since $F-\{0\}$ is cyclic of order $3$, $x^3 - 1 = 0$. Hence $x = -1 = 1$ or $x^2 = x+1$. So $x^2 = x+1$.

Answer 2

The multiplicative group of $F$ has order $3$ and so $x^3=1$ for all $x\ne 0$. Now $$ 0 = x^3-1 = (x-1)(x^2+x+1) $$ Therefore, if $x\ne1$, then $x^2+x+1=0$ and $x^2=-x-1=x+1$, because $-1=1$.