Trace function equation Let $a,b\in\mathbb{F}_{2^{m}}$ (a field of characteristic $2$, m odd), with $a,b\neq 0$.
I need to prove that
$$\sum_{i=1}^{(m-1)/2}\operatorname{tr}(a^{2^{i}}b+b^{2^{i}}a)=0\qquad \text{ iff }\qquad a=b,$$
where $\operatorname{tr}:\mathbb{F}_{2^{m}}\longrightarrow\mathbb{F}_2$ is the trace function.
 A: Since the form takes values in the prime field it is nigh impossible for it to vanish only when $a=b$. Indeed, for a given $a$ it vanishes for at least half of the possible values of $b$ (if not all) by bilinearity alone. Hence I assume that the question is really to prove that the
bilinear form vanishes whenever $a=b$.
I give two proofs. The latter is more trivial, but the former is suggested by
my answer to another question about this form. There I derive the formula
$$
B(a,b)=tr(ab)+tr(a)tr(b).
$$
Using the fact $tr(a)=tr(a^2)$ (the Frobenius conjugates have the same trace) we can deduce the present claim easily.
If $a=b$ we have
$$
B(a,a)=tr(a^2)+tr(a)^2=tr(a)+tr(a)^2.
$$
As $tr(a)$ is either $0$ or $1$ this is equal to zero.

Oh dear, I'm sluggish. Here is a straightforward proof:
If $a=b$, then $a^{2^i}b+ab^{2^i}=a^{1+2^i}+a^{1+2^i}=0$ for all $a$ and all $i$.
The claim follows immediately from this.
A: Some ideas (too long for a comment):
1) $\,tr(x+y)=tr(x)+tr(y)\,$ ;
2) By Hilbert's Theorem 90 (additive form) (page 4) , we get that 
$$tr(x)=0\Longleftrightarrow x=\alpha-\alpha^2\;\;,\;\;\alpha\in\Bbb F_{2^{m-1}}$$
3) Every element in $\,\Bbb F_{2^{m-1}}\,$ is a square
