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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 $$\begin{bmatrix}m_{11}&m_{12}&m_{13} \\ m_{21}&m_{22}&m_{23} \\ m_{31}&m_{32}&m_{33}\end{bmatrix}\cdot \begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$

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    $\begingroup$ Rational solutions to your equation ($a^3+b^3+c^3+d^3=0$ in projective form, after switching some numbers) are parametrized by a projective triple (a pair, plus a line at infinity) of rationals: math.harvard.edu/~elkies/4cubes.html $\endgroup$ May 15 '17 at 6:06
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    $\begingroup$ There's the permutation matrices (with added $-1$s) but somehow I don't think that's what you want... $\endgroup$ May 15 '17 at 8:13
  • $\begingroup$ @T.Amdeberhan: Actually, there is something analogous for cubics. Recall that if $a^2+b^2 = 1$, then $$\left(\frac{-a + 2 b + 2}{-2 a + 2 b + 3}\right)^2 + \left(\frac{-2 a + b + 2}{-2 a + 2 b + 3}\right)^2 = 1$$ But if $x^3+y^3 = 1$, then $$\left(\frac{x^3y + y}{x y^3 + x}\right)^3+\left(\frac{x^3 - y^3}{x y^3 + x}\right)^3=1$$ which is quite useful when dealing with cubic eta or theta functions. Kindly see this post. $\endgroup$ Jun 13 '17 at 13:09
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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.

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    $\begingroup$ Yes I know; this is exactly what my answer addresses. It tells you that there are exactly $6$ such matrices given by permuting the $(a,b,c)$. $\endgroup$ May 15 '17 at 12:49
  • $\begingroup$ There seems to be some serious misunderstandings somewhere. I thought my answer was clear enough already, but I will add some more details. $\endgroup$ May 15 '17 at 18:06
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    $\begingroup$ @T.Amdeberhan The point of this answer is that there is no such matrix available in this case; the structure of the solution set is different in the cubic case than it is in the quadratic case. $\endgroup$ May 15 '17 at 22:56
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    $\begingroup$ @T.Amdeberhan Firstly: That $M$ has infinite order was not stipulated in the question. Secondly: If you read my answer, then you will see that no such matrix $M$ of infinite order exists! The group of integer matrices which preserves the equation is finite of order $6$, isomorphic to the symmetric group $S_3$. $\endgroup$ May 15 '17 at 22:57
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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 $v_{max}=(-\frac 1 8 (1+ \sqrt{85}),-\frac 1 8 (1- \sqrt{85}), 1)$, $v_1=(-16,-16,43)$, $v_{min}=(-\frac 1 8 (1- \sqrt{85},-\frac 1 8 (1+ \sqrt{85}),1)$.

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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.

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    $\begingroup$ "Let's see if we can prove that it works" : out of curiosity, do you have a proof of that fact in this case? You've offered a couple of examples here but no mention of either where your $M$ came from (other than 'tedious experimentation) or why you believe it works for all iterations from your starting position (other than the same experimentation)... $\endgroup$ May 16 '17 at 19:16
  • $\begingroup$ At first glance, incidentally, this is related to the fact that the eigenvector of $M$ corresponding to the dominant eigenvalue $\frac12(83+9\sqrt{85})$, namely the vector $\langle \frac2{21}(\sqrt{85}-1), \frac1{42}(43-\sqrt{85}), 1\rangle$, satisfies $x^3+y^3=z^3$; my first guess would be that $M$ is related to generating simultaneous convergents to the quadratics here and that these convergents satisfy the same equation approximately, that this is effectively a cubic Pell equation, but it would be better if this were explicit rather than needing to read between the lines... $\endgroup$ May 16 '17 at 19:25
  • $\begingroup$ Thanks for your intimate look into the example. I have a proof (once the matrix is found), but I did not include it so as not to counteract the "accepted answer" which is a direct response to my question. Although I had other thoughts when I did ask the question in line with my constructed example. So, in short, I can provide proof to my claim if there is enough interest by people here. Otherwise, I can write it up and send it to you personally. $\endgroup$ May 16 '17 at 19:58
  • $\begingroup$ The matrix M can be found in: jstor.org/stable/27642956 Another Look at an Amazing Identity of Ramanujan Jung Hun Han and Michael D. Hirschhorn Mathematics Magazine Vol. 79, No. 4 (Oct., 2006), pp. 302-304 $\endgroup$
    – Favst
    May 17 '17 at 17:26

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