How do I eliminate the repeating cases? Let $K$ be the field with exactly $7$ elements. Let $\mathscr M$ be the set of all $2×2$ matrices with entries in $K$. How many elements of $\mathscr M$ are similar to the following matrix? 
$ \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}$
My attempt: Answer is given is $56$. We need to find the cardinality of $\{\begin{pmatrix}
a & b \\
c & d
\end{pmatrix} a,b,c,d\in K:a+d=1 \wedge ad=bc\}=\{\begin{pmatrix}
a & b \\
c & 1-a
\end{pmatrix} a,b,c,d\in K: ad=bc\}.$ I got $ad=bc \implies a(1-a)=bc \implies a-a^2=bc\implies $ $a$ as a function of $bc$. So, there are $7^3$ possibilities.How do I eliminate the repeating cases?
 A: Thanks to the comment by @ancientmathematician , any matrix similar to the given matrix would have $0$ and $1$ as eigenvalues with corresponding eigenvectors similar upto a nonzero multiple of the column vectors (as they are cancelled in the inverse i.e. since $P^{-1}AP=M$). Thus, as the first and second columns of any matrix in $\mathbb{F}_p$ can be chosen in $(p^2-1)(p^2-p), p=7$ ways, therefore the desired number of eigenvectors are $\frac{(p^2-1)(p^2-p)}{(p-1)(p-1)}$ where $p=7$. Hence, the answer is $56$
A: You can continue as below:
$$a-a^2=bc \Longrightarrow a^2-a+bc=0 \Longrightarrow a=\frac{1\pm\sqrt{1-4bc}}{2}\overset{1/2 = 4}{===}4\pm4\sqrt{1-4bc}$$
So $(1-4bc)$ must be square:
$$(1-4bc) \in \{ 0^2,(\pm)1^2,(\pm2)^2,(\pm3)^2 \} = \{ 0,1,4,2 \} \Longrightarrow 4bc \in \{ 1,0,-3,-1 \} \overset{1/4=2}{=\Longrightarrow}$$
$$\overset{1/4=2}{=\Longrightarrow} bc \in \{ 2,0,-6,-2 \}$$
Now we can count:
$$\begin{array}{l}
bc \overset{\text{Has 6 solutions}}{=======} {\color{white}-}2 & \Longrightarrow & a\overset{\text{Has 1 solution}}{=======}{\color{white}-}4  & \Longrightarrow & \text{Has ${\color{white}6}$6 Solutions} \\
bc \overset{\text{Has 13 solutions}}{=======} {\color{white}-}0 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}{\color{white}-}0,1  & \Longrightarrow & \text{Has 26 Solutions} \\
bc \overset{\text{Has 6 solutions}}{=======} -6 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}{\color{white}-}3,5  & \Longrightarrow & \text{Has 12 Solutions} \\
bc \overset{\text{Has 6 solutions}}{=======} -2 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}-1,2 & \Longrightarrow & \text{Has 12 Solutions} \\
\end{array}$$
So totally we have:
$$6+26+12+12=56 \ \text{ Solutions }$$
Also you may don't calculate $a$ and just notice the relation between "Number of solution & Sign of $\Delta$" is hold in any field.
