"Nearly" Fermat triples: case cubic

Question

Suppose $a^2+b^2-c^2=0$ are formed by a (integral) Pythagorean triple. Then, there are $3\times3$ integer matrices to generate infinitely many more triples. For example, take $$\begin{bmatrix}-1&2&2 \\ -2&1&2 \\ -2&2&3\end{bmatrix}\cdot \begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$

One may wish to do the same with $a^3+b^3-c^3=0$, but Fermat's Last Theorem forbids it!

Alas! one settles for less $a^3+b^3-c^3=\pm1$. Here, we're in good company: $9^3+10^3-12^3=1$, coming from Ramanujan's taxicab number $1729=9^3+10^3=12^3+1^3$. There are plenty more.

Question. Does there exist a concrete $3\times3$ integer matrix $M=[m_{ij}]$ such that whenever $a^3+b^3-c^3\in\{-1,1\}$ (integer tuple) then $u^3+v^3-w^3\in\{-1,1\}$ provided $$\begin{bmatrix}m_{11}&m_{12}&m_{13} \\ m_{21}&m_{22}&m_{23} \\ m_{31}&m_{32}&m_{33}\end{bmatrix}\cdot \begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$

Answer 1

Any such matrix $M$ would give rise to an automorphism of the cubic surfaces $$a^3 + b^3 = c^3 \pm d^3 \quad \subset \mathbb{P}^3.$$ These are both just different ways of writing the Fermat cubic surface $$x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0 \quad \subset \mathbb{P}^3.$$ It is well-known that the automorphism group of the Fermat cubic surface over $\mathbb{C}$ is generated by the "obvious automorphisms", namely by permuting coordinates and multiplying coordinates by a third root of unity (See Table 9.6 of "Dolgachev - Classical algebraic geometry" for this claim).

Hence, as you are interested with matrices with integer entries, we see that the only such $M$ are the 6 permutation matrices which permute $a,b,c$, with also possibly multiplying some variable by $-1$ to fix the signs.

For example \begin{bmatrix}0&1&0 \\ 1&0&0 \\ 0&0&1\end{bmatrix} is an example of such a matrix. The other matrices can be written down analogously.

Answer 2

This writeup addresses the orginal question, as stated, just marginally. However, I hope to answer in more detail Amdeberhan's "other thoughts when he did ask the question, as he wrote.

Actually, for every rational integer solution $(a,b,c)$ to $a^3+b^3-c^3=1$ there is a matrix $M$ of infinite order, depending on $(a,b,c)$, such that for all integer $n$ we have that $(a',b',c')=M^n\cdot (a,b,c)^T$ is another triple of integers satisfying the equation.

To see this, we can observe that there are parametrizations $x_i=f_i(p,q)$ of solutions to $x_1^3+x_2^3+x_3^3+x_4^3=0$, such that:

• $f_1$, $f_2$, $f_3$, $f_4$ are quadratic polynomials in $p,q$;
• $f_4(p,q)=p^2-Dq^2$ for some squarefree $D$;
• $f_1(1,0)=-a$, $f_2(1,0)=-b$, $f_3(1,0)=c$.

Then we solve the quadratic Pell equation $f_4(p,q)=1$. In other words, we find a primitive solution $(p_1,q_1)$ and we generate the sequence $(p_n,q_n)$ by $p_n+\sqrt D q_n = (p_1+\sqrt D q_1)^n$.

We plug in the parametrization and we find $(a_n,b_n,c_n)$. Now, the key observation is that the transition from $n$ to $n+1$ is a linear transformation of the triple, so we get our matrix that does the job.

Below, I give a numerical example. I use the parametrization (5) given in this MSE post, with $a=-10, b=12, c=-9, d=1$, after the linear change of coordinates $p=x+22y$, $q=2y$. Unfortunately a solution to the Pell equation with $D=85$ is quite big: $p_1=285769$, $q_1=30996$. But it does give a matrix of integers $$M=\begin{pmatrix}-331671644135 & -512714878704& -314190334080\\ 266762362608 & 412374813481 & 252702205056\\ 259637126112 & 401360260896 & 245952516097\end{pmatrix} $$ that predictably works: $M\cdot (-10,12,-9)= (-8149096378, 6554290188, 6379224759)$, and $(-8149096378)^3+6554290188^3=(-6379224759)^3+1$.

To help comparing with Stadnicki's analysis, I report that $M$ has characteristic polynomial $(x - 1) \cdot (x^2 - 326655685442 x + 1)$ and eigenvectors $v_{max}=(-\frac 1 8 (1+ \sqrt{85}),-\frac 1 8 (1- \sqrt{85}), 1)$, $v_1=(-16,-16,43)$, $v_{min}=(-\frac 1 8 (1- \sqrt{85},-\frac 1 8 (1+ \sqrt{85}),1)$.

Answer 3

Daniel Loughran: I apologize for the confusion. You were right. I was only focusing on a single family of solutions to $x^3+y^3-z^3=\pm1$, for which of course there is tedious experimental finding that does a parametrization. Not for all solutions.

Let's see if we can prove that it works: start with $[9,10,12]^T$ with $a^3+b^3-c^3=\pm1$ then I claim that $[u,v,w]^T=M^n\cdot[9,10,12]^T$ also satisfies $u^3+v^3-w^3=\pm1$.

Take the matrix to be $$M=\begin{bmatrix}63&104&-68\\64&104&-67 \\80&131&-85\end{bmatrix}.$$

For example, if $[a,b,c]^T=[9,10,12]^T$ we have $9^3+10^3-12^3=1$ and $$[u,v,w]^T=M\cdot[9,10,12]^T=[791,812,1010]^T$$ satisfies $791^3+812^3-1010^3=-1$. If we continue, $$M\cdot[791,812,1010]^T=[65601,67402,83802]^T$$ satisfies $65601^3+67402^3-83802^3=1$. This process generates and infinite family of solutions.