# 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.

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Has anyone come up with any solutions such that $A^3$, $B^3$, and $C^3$ aren't zero? – Ben Millwood Jun 2 '12 at 22:26
@benmachine I put one in with the matrices all non-nilpotent. – rschwieb Jun 3 '12 at 0:58

## 5 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) $\begin{pmatrix}n&0&0\\ 0&p&0\\0&0&0 \end{pmatrix}^3+\begin{pmatrix}0&0&0\\ 0&0&0\\0&0&m\end{pmatrix}^3=\begin{pmatrix}n&0&0\\ 0&p&0\\0&0&m\end{pmatrix}^3$

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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 $$R^3 + S^3 = T^3.$$

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+1 for a solution that even has all three matrices non-singular! – Steven Stadnicki Jun 3 '12 at 4:39
I think your post might be clearer to read if you put the matrices in each solution side-by-side. (it might also not be, I'm not sure) – Ben Millwood Jun 3 '12 at 8:45
You really had fun with this one :P – rschwieb Jun 3 '12 at 11:26
@rschwieb, yes. The original idea was to build in commutativity. If $R$ is a companion matrix, 3 by 3, then all matrices that commute with $R$ are of the form $xI + y R + z R^2.$ Then take a cyclic cubic field $\mathbb Q[x] / ((f(x))$ in which $a^3 + b^3 = c^3$ has solutions, and let $R$ be the companion matrix for $f(x).$ When I noticed I could take $f(x) = x^3 - 2$ actual matrix solutions began to appear. – Will Jagy Jun 3 '12 at 16:53

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 $X^3 = O$.

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Hint: some nilpotent matrices should do the trick.

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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.

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This factorization typically only holds if $B$ and $C$ commute. In general $(C - B)(C^2 + CB + B^2) = C^3 - BC^2 + C^2 B - BCB + CB^2 - B^3$. – Qiaochu Yuan Jun 2 '12 at 21:09
Good point! Luckily, I don't think this matters, as our choices of $B$ and $C$ at the end do commute. – Potato Jun 2 '12 at 21:09
Your final choice of $A$ also isn't non-negative. (And $A, B, C$ are diagonal.) – Qiaochu Yuan Jun 2 '12 at 21:11
Let me think about it some more... – Potato Jun 2 '12 at 21:12
@potato Fortunately Fermat's Last Theorem is not a barrier if any of the bases are 0 :) – rschwieb Jun 3 '12 at 0:43