Let $F$ be a finite field of order $q$ where $q=2^{n}$ and fix $l\in F\setminus{0}$ with $Tr(l)=0$. I want to determine the number of $a$ such that $$Tr(la)=Tr(la^{-1})=1,$$
where $Tr$ denotes the trace of a field element.
Could anyone give any hints on how to start?
Many thanks.
Let $\chi:\Bbb{F}_q\to\{\pm1\}$ be the canonical additive character $$ \chi(x)=(-1)^{tr(x)}. $$ Fix an element $\ell\in\Bbb{F}_q^*$. Consider the function $f(x)=(1-\chi(\ell x))(1-\chi(\ell x^{-1}))$. It follows that for $a\in\Bbb{F}_q^*$ we have $f(a)=4$, if $tr(\ell a)=1=tr(\ell a^{-1})$ and $f(a)=0$ otherwise.
Let $N(\ell)$ be the number of elements $a$ with the prescribed properties. The above consideration shows that $$ \begin{aligned} 4N(\ell)&=\sum_{x\in\Bbb{F}_q^*}(1-\chi(\ell x))(1-\chi(\ell x^{-1}))\\ &=\sum_{x\in\Bbb{F}_q^*}\left(1-\chi(\ell x)-\chi(\ell x^{-1})+\chi(\ell x+\ell x^{-1})\right). \end{aligned} $$ Here, trivially $$ \sum_{x\in\Bbb{F}_q^*}1=q-1 $$ and by the elementary properties of character sums $$ \sum_{x\in\Bbb{F}_q^*}\chi(\ell x)=-1=\sum_{x\in\Bbb{F}_q^*}\chi(\ell x^{-1}). $$ The last sum on the other hand is not trivial. But it is a so called Kloosterman sum. They are known to be bounded from above as follows $$ \left\vert\sum_{x\in\Bbb{F}_q^*}\chi(\ell x+\ell x^{-1})\right\vert \le 2\sqrt q. $$ The bound is of the same type as the Weil bounds on multiplicative character sums and the Carlitz-Uchiyama bound on additive character sums. It is mentioned (and IIRC also proven) in Lidl & Niederreiter as well as many other texts. Because we are in characteristic two we can also get this bound from the Hasse-Weil bound on the number of points on an elliptic curve over a finite field. Namely, it is easy to show that the number of $\Bbb{F}_q$-rational points on the curve $$ y^2+xy=\ell x^3+\ell x $$ is equal to $q+1+\sum_{x\in\Bbb{F}_q^*}\chi(\ell x+\ell x^{-1})$.
Anyway, we can deduce that $$ |4N(\ell)-(q+1)|\le 2\sqrt q, $$ and consequently $$ N(\ell)\ge\frac{q+1-2\sqrt q}4. $$ So if $q\ge16$ we deduce that $N(\ell)\ge 9/4$. As $N(\ell)$ is obviously an integer, you get the desired $N(\ell)\ge3$. I suggest that you use brute force to see what happens, when $q=8$.