Linear dependency in certain fields EDIT: I made a vital mistake switching $4$ by $3$
Let $q$ be a prime number and $n\in\mathbb{N}$.
Suppose that $q^n\equiv 1\pmod{3}$.
Is it possible that there are $0\neq a,b\in \mathbb{F}_{q^n}$ and $0\neq\lambda,\mu\in\mathbb{F}_q$
$$\lambda a+\mu b=0,\lambda a^4+\mu b^4=0,a\neq b$$
 A: Solve the equation:
$$\lambda a +\mu b=\lambda a^3+\mu b^3\implies \lambda a(a-1)(a+1)+\mu b(b-1)(b+1)=0$$
and thus you can easily check that $\;a=0, b=1\;$ solve the question, for example...and there are several more solutions, and it isn't that important that $\;q^n=1\pmod3\;$ 
If it must be that $\;ab\neq0\;$ , then choose $\;a=1\,,\,\,b=-1\;$ , which are different unless $\;q=2\;$ .
A: With all the variables non-zero the former equation gives
$$
\lambda/\mu=-a/b
$$
and the latter equation gives
$$
\lambda/\mu=-a^4/b^4.
$$
So for solutions to exist we must have $-a/b=-a^4/b^4$, or $a^3/b^3=1$ or, equivalently
$$a=\omega^j b,$$
where $\omega\in\Bbb{F}_{q^n}$ is a third root of unity (exists, because $3\mid q^n-1$), and $j=0,1$ or $2$.
The case $a=b$ was excluded, so we are left with $a=\omega b$, and $a=\omega^2 b$. 
But, we also have $\lambda/\mu=-\omega^j$, $j=1,2$.
Because $\lambda/\mu\in\Bbb{F}_q^*$ we therefore need the third roots of unity to reside in $\Bbb{F}_q$. For solutions to exist it is thus necessary that $q\equiv1\pmod 3$.
On the other hand, $\omega\in\Bbb{F}_q$ is clearly a sufficient condition for the existence of solutions. We can simply set $b=\omega a$, $\lambda=-\omega\mu$, and both equations will hold.

Solutions $(a,b,\mu,\lambda)$ such that all the variables are non-zero and $a\neq b$ exist if and only if $q\equiv1\pmod3$.

