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Let $a$ be a primitive element of $\mathbb{F}_{16}$ that satisfies the equation $a^4=1+a$.

The logarithm of $1+a+a^2$ in $\mathbb{F}_{16}$ with base $a$ is the integer $i$ such that $0≤i<15$ and $1+a+a^2=a^i$.

Give the logarithm of $1+a+a^2$ with base $a$.

From the first line we know that the Zech logarithm of $1$ is $4$, i.e. $Zech(1)=4$.

Also $a+a^2=a^{1+Zech(2-1)}=a^{5}$.

Now I am unsure of how to proceed. Any tips?

Community
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falkon
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  • Many ways to continue. One way is to observe that as $a^5$ is a cubic root of unity, it belongs to the subfield of four elements consisting of fifth powers. The element $1+a+a^2=1+a^5$ must then also belong to this subfield, so it has to be $a^{10}$. In other words we must have $Zech(5)=10$. – Jyrki Lahtonen Jun 02 '16 at 20:42
  • The "easiest" way is to take a peek. – Jyrki Lahtonen Jun 02 '16 at 20:43

1 Answers1

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$a^4 = a +1$ so $1+a + a^2 = a^4+a^2$ but since we are in characteristic 2 we have that $a^4+a^2=(a^2+a)^2=(a*(a+1))^2=(a*a^4)^2=a^{10}$

Sean Ballentine
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