Pairs $(a,b)\in F^2$ such that $a^6+b^6=1$ where $F$ is a finite field of 25 elements Let $F$ be a field of 25 elements, and consider the group $G$ of all $2\times 2$ matrices $A$ with entries in $F$ satisfying $A_5A=I$, where $I$ is the identity matrix and 
    $A_5=\begin{pmatrix} a^5 & c^5 \\ b^5 & d^5 \end{pmatrix}$ if $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. 
I am asked to find the number of pairs $(a,b)\in F^2$ such that $a^6+b^6=1$, and conclude that $|G|=720$.
Note that $(a+b)^6=(a+b)^5(a+b)=(a^5+b^5)(a+b)=a^6+b^6+a^5b+ab^5$, so it suffices to show find the number of pairs $(a,b)$ such that $a^5b+ab^5=-1$. How do I have to proceed next?

One needs to recognise that $a^6$ is the norm map from $F=\Bbb F_{25}$ to the field
$\Bbb F_5$ of order $5$, so $a^6\in\{0,1,2,3,4\}$. $a^6=0$ only has the solution $0$
but for $c\in\{1,2,3,4\}$ $a^6=c$ has six solutions in $F$. We need to count
the solutions of $(a^6,b^6)=(c,d)$ where $(c,d)$ is one of the pairs $(0,1)$, $(1,0)$,
$(2,4)$, $(3,3)$ and $(4,2)$. There are six solutions for each of the first two pairs
and thirty-six for the other three. So overall $a^6+b^6=1$ has $120$ solutions
in $F$.
Once you have done that you observe that $(c,d)$ must be "orthogonal" to $(a^5,b^5)$ i.e., $ca^5+db^5=0$. The solutions to this form a one-dimensional
subspace of $F^2$ and one needs to show exactly six of these $(c,d)$
also satisfy $c^6+d^6=1$.
