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Suppose $a$ and $b$ are elements of a finite field of order $2^n$ with $n$ odd and $a^2+ab+b^2=0$. Is it necessary that both $a$ and $b$ must be zero ?

I understand that the field has characteristic $2$ but don't know how to use the fact that $n$ is odd, please help.

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    @KCd, you should post your comment as an answer: it is$a$very nice and accurate hint for the OP.2012-12-23

2 Answers 2

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If $b$ were not 0 then $a/b$ would be a root of $x^2 + x + 1$, which is irreducible over ${\mathbf F}_2$. Look at the size of the field ${\mathbf F}_2(a/b)$ and the size of the field you are working in that has order $2^n$.

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If $a=b$, then $a^2+ab+b^2=3a^2=a^2=0$ and $a=0=b$, we are done.

Suppose that $a\neq b$.

Observe that $0=(a-b)(a^2+ab+b^2)=a^3-b^3$. Thus, $a^3=b^3$.

We claim that $a=0$ and $b=0$.

If $a\neq 0$, then $(a^{-1}b)^3=1$ and the multiplicative order of $a^{-1}b$ in the multiplicative group $F-\{0\}$ is $1$ because $3\not\mid |F-\{0\}|=2^n-1$. Hence, $a^{-1}b=1$ and $a=b$, a contradiction. Therefore, $a=0$. By a similar argument, we have $b=0$.

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    And $3\nmid 2^n-1$ because $n$ is odd. Right.2015-05-24