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.
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.
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) $\begin{pmatrix}n&0&0\\ 0&p&0\\0&0&0 \end{pmatrix}^3+\begin{pmatrix}0&0&0\\ 0&0&0\\0&0&m\end{pmatrix}^3=\begin{pmatrix}n&0&0\\ 0&p&0\\0&0&m\end{pmatrix}^3$
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. $
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 $X^3 = O$.
Hint: some nilpotent matrices should do the trick.
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.