For $a,b \in \mathbb {F}_{{2^n}}$, $n$ is odd; $a^2+b^2+ab=0$ implies that $a=b=0$.

Question

Let $a,b$ be two elements of the finite field $\mathbb {F}_{{2^n}}$, where $n$ is odd. If $a^2+b^2+ab=0$ then I have to show that $a=b=0$.

Enough to show at least one of $a,b\in \mathbb F_2$. If $a\in \mathbb F_2$ then $b$ satisfy a polynomial of degree at most $2$. Since $n$ is odd, $b$ must also be in $\mathbb F_2$. If none of $a,b$ is in $\mathbb F_2$ I can't draw a contradiction. Please help me.

Answer 1

If $b\ne 0$ then let $x= a/b$ and now we have $$x^2+x+1=0\implies x^3=1$$

Since $x^{2^n-1}=1$ and $\gcd (3,2^n-1)=1$ we have $x=1$ so $a=b$. But then $3a^2=0$ and thus $a=0$

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$\gcd (3,2^n-1)=1$ since: $$2^n-1 \equiv_3 (-1)^n-1 \equiv_3 1\implies 3\nmid 2^n-1$$