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Let $a$ be an element in $GF(2^9)$ which satisfies $a^{9}+a^{8}=-1$. If $H=\langle{a\rangle}$, then what is the order of $H$ in $GF(2^9)$. I have no idea how to start. $a^{9}+a^{8}=-1$, tried taking powers of this expression, to see if it gives anything, but I am lost. A brief solution would be of great help.

Jyrki Lahtonen
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Ramana
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1 Answers1

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This field is of characteristic $2$, so $(x+y)^2=x^2+y^2$ for all $x,y\in GF(2^9)$, and $-1=1$. Thus:

$$a^9=a^8+1=(a+1)^8$$

and

$$a^8(a+1)=1$$

In other words, $a^9=b^8$ and $a^8=b^{-1}$ where $b=a+1$. Thus, $a^9\times (a^8)^8=a^{73}=1$. As $73$ is prime, the order of $a$ must be $73$.

  • Good job! Since I'm cursed with too good a recollection, I remembered this right away from this oldie where the point was to find "non-trivial" roots of unity $\alpha$ such that $\alpha+1$ has the same order. – Jyrki Lahtonen Nov 13 '20 at 14:21