I try to prove that ${2^{n-1}}$ elements of the field $\mathbf{F}_{2^{n}}$ have a Trace with value 1, while the other ${2^{n-1}}$ elements have a Trace with value 0.
I started to show that Trace(1) = 1, and I tried to use the additivity of the Trace but I wasn't successful. Any advice ?
On
Because $tr(x)$ is a polynomial of degree $2^{n-1}$ it can take the value zero at most $2^{n-1}$ times. Because it is linear it takes the value zero at least $2^{n-1}$ times (rank-nullity tells that the kernel has dimension $\ge n-1$). Because $tr(x)$ is either zero or $1$, it has to take both values equally often.
This is just linear algebra: the trace map is a linear functional $\;\Bbb F_{2^n}\to\Bbb F_2\;$, and since the extension $\;\Bbb F_{2^n}/\Bbb F_2\;$ is separable it is not the zero functional (or just show there's some element with trace different from zero), from where it follows that it is onto (this much is true for any linear functional over any vector space) and thus $\;\dim\ker\,Tr.=n-1\;$, and this means that
$$|\ker Tr.|=|\Bbb F_{2^{n-1}}|=2^{n-1}$$
and we're done.