Matrices with $A^3+B^3=C^3$ Problem: Find infinitely many triples of nonzero $3\times 3$ matrices $(A,B,C)$ over the nonnegative integers with 
$$A^3+B^3=C^3.$$
My proposed solution is in the answers.
 A: Hint: If $X = \left(\begin{matrix} 0 & 0 & n \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right)$ for some $n \in \mathbb{N}_+$ then $X \neq O$ and $X^3 = O$.
A: Hint: some nilpotent matrices should do the trick.
A: Due to Fermat's last theorem, trying to be cheap and using diagonal matrices won't work. We need to be more subtle. Assume we can make $C$ and $B$ commute.
Then we have the factorization
$$A^3=(C-B)(C^2+CB+B^2)$$
and decide to see if we can set $(C-B)=A$. Then we need
$A^2=C^2+CB+B^2\Leftrightarrow C^2-2CB+B^2=C^2+CB+B^2 \Leftrightarrow CB=0$.
So we just need to generate infinitely many $C$ and $B$ with $CB=0$, from which we can generate the required $A$. But this is easy. For example, for all $n>0$, $C=$
\begin{bmatrix}
       0 & 0 & n+1          \\[0.3em]
       0 & 0           & 0 \\[0.3em]
       0           & 0 & 0
     \end{bmatrix}
and $B=$
\begin{bmatrix}
       0 & 0 & n          \\[0.3em]
       0 & 0           & 0 \\[0.3em]
       0           & 0 & 0
     \end{bmatrix}
work, because $C-B$ is nonnegative and they commute.
A: Nobody suggested this scheme? Maybe it is hidden in the details of someone's answer, but this is the first thing I would suggest:
(EDIT: used to be 2x2, but my example obviously generalizes to any size matrices)
$\begin{pmatrix}n&0&0\\ 0&p&0\\0&0&0 \end{pmatrix}^3+\begin{pmatrix}0&0&0\\ 0&0&0\\0&0&m\end{pmatrix}^3=\begin{pmatrix}n&0&0\\ 0&p&0\\0&0&m\end{pmatrix}^3$
A: OR OR OR, given 
$$ x, y > 0,   $$
let
$$ R \; = \;  
 \left(  \begin{array}{ccc}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  x & 0  & 0  
\end{array} 
  \right)  , \; \;
 S \; = \;  
 \left(  \begin{array}{ccc}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  y & 0  & 0  
\end{array} 
  \right)  , \; \;
 T \; = \;  
 \left(  \begin{array}{ccc}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  x + y & 0  & 0  
\end{array} 
  \right)  ,
  $$
then
$$ R^3 = x I, \; \; S^3 = y I, \; \; T^3 = (x+y) I    $$
and
$$ R^3 + S^3 = T^3.  $$
OR
$$ S \; = \;  
 \left(  \begin{array}{rrr}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  2n^2 & 0  & 0  
\end{array} 
  \right)  , \; \;
 T \; = \;  
 \left(  \begin{array}{rrr}
  0 & 0 & 1 \\
  2 n  & 0 & 0 \\
  0 & 2 n  & 0  
\end{array} 
  \right) .
  $$
Then 
$$  S^3 = 2 n^2 I, \; \;  T^3 = 4 n^2 I,   $$
and
$$ S^3 + S^3 = T^3.  $$
OR OR, given a Pythagorean triple
$$ a^2 + b^2 = c^2,   $$
let
$$ R \; = \;  
 \left(  \begin{array}{rrr}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  a^2 & 0  & 0  
\end{array} 
  \right)  , \; \;
 S \; = \;  
 \left(  \begin{array}{rrr}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  b^2 & 0  & 0  
\end{array} 
  \right)  , \; \;
 T \; = \;  
 \left(  \begin{array}{rrr}
  0 & 0 & 1 \\
  c  & 0 & 0 \\
  0 & c  & 0  
\end{array} 
  \right) ,
  $$
then
$$ R^3 = a^2 I, \; \; S^3 = b^2 I, \; \; T^3 = c^2 I    $$
and
$$ R^3 + S^3 = T^3.  $$
