If 2 matrices are such that $(A+B)^k=A^k+B^k$ for $k=2,3$, show that $(A+B)^m=A^m+B^m $ for all $m \in \mathbb{N}$ Let $A,B \in M_n(C) $. The matrix  $A-B$ is invertible and $(A+B)^k=A^k+B^k $, $k \in {2,3} $. Prove that $(A+B)^m=A^m+B^m $ for every $m \in N $.
PS. I obtained $AB+BA=0$ and $A^2B+B^2A=0$, but I need your help, please :(

Hint For $k = 4$, we have
$$(A + B)^4 = (A + B)^3 (A + B) = (A^3 + B^3)(A + B) = A^4 + \color{red}{A^3 B + B^3 A} + B^4 .$$
On the other hand, using both of the identities you deduced gives
$$B^3 A = B (B^2 A) = B (-A^2 B) = BABA = -B^2 A^2 = A^2 B A = -A^3 B ,$$ so the red terms vanish.
Can you generalize this manipulation to arbitrary $k$?
