Trace function equation

Question

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.

Answer 1

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

Answer 2

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.

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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.