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We are given that $a, b \in F_{2^n}$ where $n$ is an odd +ve integer. Suppose $a^2 + ab + b^2 = 0$ then we have either $a = 2^n-b^2$ or $a+b = 2^n - b^2$. Which implies that $\sqrt{2^n -a} = +-b $ or $b = 2^n - a$. Thus $b = 0$ and $a= 2^n= 0$

Is it correct? Is there a simple proof?

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    $\begingroup$ I think you confuse $\mathbb{Z}_{2^n}$ with $\mathbb{F}_{2^n}$. There is no such element as $2^n$ in the finite field $\mathbb{F}_{2^n}$. $\endgroup$ – Leif Sabellek May 11 '14 at 22:49
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    $\begingroup$ I don’t understand what follows the words “then we have”. $\endgroup$ – Lubin May 11 '14 at 22:56
  • $\begingroup$ Note that $\mathbb{F}_{2^n}$ has characteristic $2$. $\endgroup$ – Omnomnomnom May 11 '14 at 23:03
  • $\begingroup$ The major problem is that you have is that you don't know what $F_{2^n}$ looks like. So this is non-sensical. Once you learn what they are, then here's a hint. If $a=b$, then the given equation implies $a=0=b$. Otherwise multiplying it by $a-b$ gives $a^3=b^3$. If $b\neq0$ this implies $(a/b)^3=1$. But in the group $\Bbb{F}_{2^n}^*$ there are no elements of order three, when $n$ is odd. $\endgroup$ – Jyrki Lahtonen May 12 '14 at 5:21

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