Roots of $X^{l-1}+1$ in a quadratic extension $F_q$, $q=l^2$ Consider a finite field $F_q$ where $q=l^2$ ($l$ can be of the form $p^m$). Does $F_q$ has a root of $X^{l-1}+1$?
As $X^l+X = X(X^{l-1}+1)$ we can show that $X^{l-1}+1$ splits if it has a root. This is simply because if $\alpha$ is a root of $X^{l-1}+1$ and $a \in F_l \subseteq F_q$ then $(a \alpha)^l =-a \alpha$.
So how can we show that $X^{l-1}+1$ has a root in $F_q$?
 A: This is true.
The multiplicative group of $\Bbb{F}_q$ is cyclic of order 
$$
q-1=\ell^2-1=(\ell-1)(\ell+1).
$$
If $\ell$ is odd (or, equivalently, if $p$ is an odd prime) then $\ell+1$ is even. Therefore $2(\ell-1)\mid q-1$. This means that in $\Bbb{F}_q^*$ there is an element of order $2(\ell-1)$. Let $a$ be such an element. Then we know that $a^{\ell-1}\neq1$ but $a^{2(\ell-1)}=1$. Consequently
$$
0=a^{2(\ell-1)}-1=(a^{\ell-1}-1)(a^{\ell-1}+1),
$$
and we know that the first factor is non-zero. Therefore the second factor is zero, which is what we want to prove.
OTOH if $\ell$ is even (so $p=2$), then for all the non-zero elements $a$ of the subfield $\Bbb{F}_\ell$ we have $a^{\ell-1}=1$. Therefore
$$
a^{\ell-1}+1=1+1=0,
$$
so we again have solutions.

It is easy to see that in both cases your equation actually has the maximum number, $\ell-1$, solutions in the field $\Bbb{F}_q$. IOW all the solutions are in that field. In the even characteristic case we already showed this. In the odd characteristic case the solutions are exactly the non-squares within a multiplicative cyclic subgroup of order $2(\ell-1)$.  
