Trace function in $GF(2^k)$

Question

Trace function is defined as follows $$ Tr(a) = a + a^2 + \dots + a^{2^{k-1}}$$

I'm asked to show if $Tr(a) = 0$ then $a = z^2 + z$ has two roots: $\theta$ and $\theta + 1$.

Don't know where to begin. How shall I approach to it ?

Answer 1

Hints.

• There is always some root in some extension field $GF(2^{k+s})$ with $s\geq 0$, call it $\theta$. If $\theta$ is a root, then so is $\theta+1$ because of $(*)$.
• Remember that $\theta\in GF(2^k)$ if and only if $\theta^{2^k}=\theta$. You should actually be able to verify this! Write $\theta^2=\theta+a$ (which is equivalent to $a=\theta^2+\theta$) and square, reduce, square...

$(*)$ Do not forget $(\alpha+\beta)^2=\alpha^2+\beta^2$ in char $2$.

Comment. This is a somewhat different approach to the Linear Algebra one in the comments by Jyrki Lahtonen. If you like, we are showing that $x^2+x+a$ divides $x^{2^k}+x$ when ${\text{Tr}}(a)=0$.