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