An Equality Involving Character Sums in Characteristic 2 For context: 
Let $E = \mathbb{F}_{2^n}$, and let $\alpha \in E$. Define $\chi(a) := (-1)^{Tr(a)}$, where $Tr$ is the absolute trace from $E$ to $\mathbb{F}_2$. For the purposes of this question, we are using the convention $Tr(\frac{1}{0}) = 0$. 

The Problem:
I was reading a proof that contained a large-ish (~ 15 lines) calculation, and mid-way through was the following equality:
$$\sum\limits_{c \in E^* \text{, } Tr(c) = 0} \chi\left(\frac{\alpha}{c}\right) - \sum\limits_{c \in E} \chi\left(\frac{\alpha}{c}\right) = \frac{1}{2}\sum\limits_{\beta \in E \backslash \mathbb{F}_2} \chi\left(\frac{\alpha}{\beta^2+\beta}\right)$$
I know that the additive Hilbert 90 tells us that $Tr(c) = 0 \Leftrightarrow c = \beta^2+\beta$ for some $\beta \in E$, but beyond that I don't see what's going on here. 
I feel like I'm missing something blatantly obvious, as every other step in the calculation makes perfect sense to me.
 A: As you observe, the association $b^2+b$ to elements of trace zero is precisely 2 to 1 (if $b$ is a solution to $c=b^2+b$, the only other solution is $b+1$, and is always distinct from $b$).  Furthermore, the pair $0$ and $1$ is the single pair representing $0$ as such a sum.  Therefore,
$$\sum_{b\neq 0,1}\chi(\frac{a}{b^2+b}) = 2\sum_{c\in E^*, Tr(c)=0}\chi(\frac{a}{c})  $$ which is the first summand in your left hand side.
The second summand must therefore be zero.  Here is why: basically, it is a character sum of a non-trivial character over the entire group, so it must be zero.  More precisely:
As $c$ runs through non-zero elements of $E$, $\frac{a}{c}$ runs through non-zero elements of $c$.  For the zero element, by the convention you are stating, $$\chi(0) =1= \chi(\frac{a}{0}).$$  Therefore,
$$\sum_{c\in E}\chi(\frac{a}{c}) =\sum_{c\in E}\chi(c)=S$$ which by the usual argument (choose a $d$ with $\chi(d)=-1$  and multiply $S$ by $\chi(d)$ which just permutes $E$ additively to get $-S=S$) gives $S=0$.
