Let $q$ be the power of a prime $p$ and suppose $\lambda\in \mathbb{F}_{q}$ and $\lambda=-\lambda^{p}$. Do there exists elements $b,c\in \mathbb{F}_{q}$ such that the following hold?
$$2bc=1-\lambda^{2}$$ $$b^{p}=-b, \quad c^{p}=-c$$
I know that this implies $bc\in \mathbb{F}_{p}$, but I am not sure how to proceed next. Computations suggest this is true.
Thanks in advance for any help.
Assuming $\lambda\neq0$ and $p>2$.
If $\lambda^p=-\lambda$, then $\lambda^{p^2}=(-\lambda)^p=-\lambda^p=\lambda$. Therefore $\lambda\in\Bbb{F}_{p^2}$. In particular, we can deduce that $q$ must be an even power of $p$. Furthermore, the minimal polynomial of $\lambda$ over the prime field is $$m(T)=(T-\lambda)(T-\lambda^p)=(T-\lambda)(T+\lambda)=T^2-\lambda^2\in\Bbb{F}_p[T].$$ Therefore $\lambda^2\in\Bbb{F}_p$. Obviously $a=\lambda^2$ is a non-square in $\Bbb{F}_p$.
The elements $z\in\Bbb{F}_{p^2}$ with the property $z^p=-z$ form a 1-dimensional vectorsubspace over the prime field (the Frobenius has two eigenvalues, $\pm1$, in the two-dimensional extension field $\Bbb{F}_{p^2}$). Therefore we have no choice but to select $b$ and $c$ as multiples of $\lambda$. In other words, we are looking for $x,y\in\Bbb{F}_p$ such that $b=x\lambda$, $c=y\lambda$, and (your first condition) $$ 2xy \lambda^2=1-\lambda^2\Longleftrightarrow 2xy a=1-a. $$ As $a\in\Bbb{F}_p\setminus\{0,1\}$ this system has plenty of solutions. To each non-zero choice of $x$ we get a unique solutions $y$.