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 :(
 A: You already showed 
$$
AB + BA = 0\\
A^2B+B^2A = 0
$$
Continuing from there,
\begin{align*}
&A^2B+B^2A = 0\\[4pt]
\implies\;&A(AB)+B(BA) = 0\\[4pt]
\implies\;&A(AB)+B(-AB) = 0\\[4pt]
\implies\;&(A-B)(AB) = 0\\[4pt]
\implies\;&AB = 0&&\text{[since $A-B$ is invertible]}\\[4pt]
\end{align*}
Note that $A-B\;$invertible implies $B-A\;$invertible. Thus, the hypothesis is symmetric in the variables $A,B.\;$Hence, by switching the variables, we get $BA=0$.

It follows that
$$(A+B)^m = A^m + B^m$$
for all $m \ge 2,\;$since in the expansion of 
$$(A+B)^m = (A+B)(A+B)\cdots(A+B)$$
all the "middle terms" vanish.

Trivially, $(A+B)^m = A^m + B^m$ holds for $m=1,\;$hence
$$(A+B)^m = A^m + B^m$$
for all positive integers $m$.
A: 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 $k$?
