Matrix multiplication proof: $(A+B)^k$ for matrices such that $A^2=B^2=0$ and $AB=BA$ I have the following details about the matrices $A,B$:

$$A_{n\times n},B_{n\times n}$$ $$A^2 = B^2 = 0$$ $$AB=BA$$

I need to find $x \in \mathbb{N}$ so $(A+B)^x=0$
What implications can I make from the given details?


*

*$A,B$ doesn't have to equal $0$ so $A^2 = B^2 = 0$

*Can I imply from the 2nd and 3rd given details that $A=B$?

*Any other leads? 


$$\mbox{ let } x > 2\mbox{ and }  x\in \mathbb{N}$$
$$\textbf{(A+B)}^2 = \textbf{A}^2+\textbf{B}^2+2\textbf{AB} = \textbf{2AB}$$
$$\textbf{(A+B)}^x = \textbf{(A+B)}^2\textbf{(A+B)}^{x-2}$$
$$\Rightarrow \textbf{(A+B)}^x =2\textbf{AB}\textbf{(A+B)}^{x-2}$$
$$\Rightarrow \textbf{(A+B)}^x =2\textbf{AB}\textbf{(A+B)}\textbf{(A+B)}^{x-3}$$
$$\Rightarrow \textbf{(A+B)}^x =2(\textbf{ABA} + \textbf{ABB})\textbf{(A+B)}^{x-3}$$
$$\Rightarrow \textbf{(A+B)}^x =2(\textbf{AAB} + \textbf{ABB})\textbf{(A+B)}^{x-3}\quad(\because \textbf{BA = AB})$$
$$\Rightarrow \textbf{(A+B)}^x = \textbf{0} (\because\hspace{2pt}\textbf{A}^2 = \textbf{B}^2 = 0)$$
$$\text{If x = 2, I can not show }\textbf{(A+B)}^2 = 0$$
$$\text{ can anyone to prove that or give a counter example}$$
