Trace 0 and Norm 1 elements in finite fields Let $\mathbb{F}_{q^\ell}/\mathbb{F}_{q}$ be the extension of finite filed $\mathbb{F}_{q}$, where $\ell$ be a odd prime and $(\neq q)$. Take $\zeta\in\mathbb{F}_{q^\ell}$. Does there exist different $i,j$ where 
$1\le i,j<\ell$ such that 
$\hspace{5cm} Tr_{\mathbb{F}_{q^\ell}/\mathbb{F}_{q}}(\zeta^{1-q^i})=0$
and
$\hspace{5cm}Tr_{\mathbb{F}_{q^\ell}/\mathbb{F}_{q}}(\zeta^{1-q^j})=0$
 A: No, not for every $q,\ell,\zeta$. In fact in general there does not even exist one solution.
Trivial example: $\zeta \in \mathbb F_q^\times$ implies $\zeta^{1- q^i} = 1$ and hence $Tr( \zeta^{1-q^i})=\ell \neq 0$. for all $i$.
Existence of nontrivial examples: If we multiply $Tr( \zeta^{1- q^i})$ by the norm of $\zeta$, it becomes a polynomial function of degree $\ell$ in $\zeta, \zeta^q, \dots, \zeta^{q^{\ell-1}}$. Indeed, we can write the trace as $\sum_{k=0}^{\ell-1} \zeta^{ q^k} \zeta^{ - q^{k +j}}$ where the sum $k+j$ is taken modulo $\ell$. The norm is $\prod_{k=0}^{\ell-1} q^k$ and by multiplying them we may clear denominators.
Each of $\zeta, \zeta^q, \dots, \zeta^{q^{\ell-1}}$ is a linear function of the $\ell$ coordinates of $\zeta$ in some $\mathbb F_q$ basis of $\mathbb F_{q^\ell}$. So the trace times the norm is a polynomial function of degree $\ell$ in those coordinates. Because it is not identically zero (as it is novanishing on elements of $\mathbb F_q)$ it has at most $\ell q^{\ell-1}$ solutions. Hence the total number of $\zeta$ for which there is some $i$ where this equation holds is at most $(\ell-1) \ell q^{\ell-1}$. For $q> \ell (\ell-1)$, this is much less than $q^\ell$ and so there are many points with no solution.
