<s>[INST] If $(A-B)^2=O_2$ then $\det(A^2 - B^2)=(\det(A) - \det(B))^2$ 
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.
Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.

OBS. $O_2$ is the zero matrix

Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $tr(A)=tr(B)=t$ and, from here, $A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2$
This is all I could get.
UPDATE
The matrices are from $M_2(\mathbb{R})$. Sorry for misleading you.
 [/INST] Since $D=A-B$ is nilpotent, it is similar to the matrix $\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. So without loss of generality assume $D = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. If $B = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$, then $A = B+D = \left(\begin{matrix} a & b+1 \\ c & d \end{matrix}\right)$. Now
$$ A^2-B^2 = (B+D)^2 - B^2 = (B^2 + BD + DB + D^2) - B^2 = BD+DB$$
since $D^2 = 0$. Notice that
$$ BD+DB = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right) + \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)\left(\begin{matrix} a & b \\ c & d \end{matrix}\right) = \left(\begin{matrix} c & a + d \\ 0 & c\end{matrix}\right) $$
and hence
$$ \det(A^2-B^2) = \det(BD+DB) = c^2. $$
On the other hand,
$$ \det(A) = \det\left(\begin{matrix} a & b+1 \\ c & d \end{matrix}\right) = ad - (b+1)c = \det(B) - c$$
and hence $(\det(A) - \det(B))^2 = c^2$ as well.
Thus $\det(A^2-B^2) = c^2 = (\det(A)-\det(B))^2$, as desired.
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