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Starting with,

$$x^3+y^3+z^3 = (z+1)^3$$

we do the substitution $z = 3 n^2x + (3 n^2 + 1)(y - 1)$ like Adam Bailey in this post to get the ellipse,

$$x^2 - x y + y^2 - (27 n^4 - 1) x - (27n^4 + 18 n^2 + 2 ) y + (27n^4 + 9 n^2 + 1 ) = 0$$

after removing a trivial factor. For general integer $n$, this family of ellipses seems to have nice properties. The case $n=1$ was in the linked post. For $n=2$,

$$x^2 - x y + y^2 - 431 x - 506 y + 469 = 0$$

which yields the ellipse,

enter image description here

graphed by the Desmos calculator. Using the $6$ blue lattice points (and flipping $x,y$ as needed), we get the nice $6$-cycle of equalities,

\begin{align} \color{blue}{(-66)}^3\; +\; \color{blue}{97}^3 + 456^3 &= 457^3\\ \color{blue}{97}^3\; +\; \color{blue}{594}^3 + 8376^3 &= 8377^3\\ \color{blue}{594}^3 + \color{blue}{1003}^3 + 20154^3 &= 20155^3\\ \color{blue}{1003}^3 + \color{blue}{840}^3 + 23106^3 &= 23107^3\\ \color{blue}{840}^3 + \color{blue}{343}^3 + 14526^3 &= 14527^3\\ \color{blue}{343}^3 + (-66)^3 + 3654^3 &= 3655^3\end{align}

thus the sextuple $a_i = (-66, 97, 594, 1003, 840, 343)$ and similarly for the given $24$ lattice points. But the Alpertron calculator says this ellipse has a total of $48$, with the others being,

enter image description here

So we in fact have $48/6 = 8$ sextuples,

\begin{align}a_i & = (-66, 97, 594, 1003, 840, 343)\\ b_i & = (-72, 115, 618, 1009, 822, 319)\\ c_i & = (90, -47, 294, 847, 984, 643)\\ d_i & = (72, -41, 318, 865, 978, 619)\\ e_i & = (-78, 139, 648, 1015, 798, 289)\\ f_i & = (264, 823, 990, 673, 114, -53)\\ g_i & = (-42, 49, 522, 979, 888, 415)\\ h_i & = (24, -17, 390, 913, 954, 547)\end{align}

each one yielding a sextuple system $x^3+y^3+z^3 =(z+1)^3$ as demonstrated above for $a_i$. But notice the curious relationship,

$$\alpha = a_1 + a_3 + a_5 = b_1 + b_3 + b_5 = \dots = h_1 + h_3 + h_5 = 1368$$

$$\beta = a_2 + a_4 + a_6 = b_2 + b_4 + b_6 = \dots = h_2 + h_4 + h_6 = 1443$$

Question: Using the theory of conic sections, what is the geometric interpretation of \begin{align}\alpha &= 3(27n^4+6n^2)\\ \beta &= 3(27n^4+12n^2+1)\end{align} and why is it invariant using different lattice points?

P.S. For this family, I've tested it with other $n$ and the same relationship seems to hold.

Addendum: As Jan-Magnus pointed out in his answer $\big(\dfrac{\alpha}3,\dfrac{\beta}3\big)$ should be the center of the ellipse. So,

\begin{align} \alpha &= a_1 + a_3 + a_5 = -66 + 594 + 840 = 1368 = 3\times456\\ \beta &= a_2 + a_4 + a_6 = \;97 + 1003 + 343 = 1443 = 3\times481\end{align}

Yup, looks like $(456,481)$ is indeed the center.

enter image description here

Tito Piezas III
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1 Answers1

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It seems they average to the $x$ and $y$ coordinates of the center of the ellipse which is $$(x,y)=(27n^4+6n^2,27n^4+12n^2+1)=(\frac{\alpha}3,\frac{\beta}3),$$ as you can find by solving $$\frac{\partial}{\partial x}(x^2 - x y + y^2 - (27 n^4 - 1) x - (27n^4 + 18 n^2 + 2 ) y + (27n^4 + 9 n^2 + 1 )) = 0\\\frac{\partial}{\partial y}(x^2 - x y + y^2 - (27 n^4 - 1) x - (27n^4 + 18 n^2 + 2 ) y + (27n^4 + 9 n^2 + 1 )) = 0.$$

Edit: If translated to the center (by Jan-Magnus Økland):

Let $x\to x+\dfrac{\alpha}3,\; y\to y+\dfrac{\beta}3$, then,

$$x^2-xy+y^2=9n^2(3n^2+1)(27n^4+9n^2+1)$$

Edit: (More important given the solution for the other Q.)

Written about the center and multiplied by two, the equation is $$2(x-3n^2(9n^2+2))^2-2(x-3n^2(9n^2+2))(y-(3n^2+1)(9n^2+1))+2(y-(3n^2+1)(9n^2+1))^2=\\18n^2(3n^2+1)(27n^4+9n^2+1)$$ and the Smith normal form is $$\left[ \begin {array}{ccc} 1&0&0\\ 0&3&0 \\ 0&0&-3M \end {array} \right] =\\ \left[ \begin {array}{ccc} 1&1&0\\ 1&2&0 \\ M & 2M &1\end {array} \right] \cdot \left[ \begin {array}{ccc} 2&-1&0\\ -1&2&0 \\ 0&0&-3M \end {array} \right] \cdot \left[ \begin {array}{ccc} 1&-1&0\\ 0&1&0 \\ 0&1&1\end {array} \right] $$

where $M = 6n^2(3n^2+1)(27n^4+9n^2+1).$

enter image description here

Tito Piezas III
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Jan-Magnus Økland
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  • Ah, I understand how you did it. We get the partial derivatives using Wolfram so \begin{align} -27n^4 + 2x - y + 1 &= 0\ -27n^4 - 18n^2 - x + 2y - 2 &= 0 \end{align} thus \begin{align} x &= 27n^4+6n^2 = \frac{\alpha}3\ y &= 27n^4+12n^2+1 = \frac{\beta}3 \end{align} But what led you to the idea of using partial derivatives? – Tito Piezas III Jan 15 '24 at 14:34
  • @TitoPiezasIII It’s the standard way to find the center of an ellipse or a hyperbola. – Jan-Magnus Økland Jan 15 '24 at 14:51
  • I see. Let me refine my question. What was your clue that $(\alpha, \beta)$ would involve the center of the ellipse? – Tito Piezas III Jan 15 '24 at 14:58
  • Actually it finds for a hyperbola where the asymptotes cross or for an ellipse where the only real point is (where the imaginary lines cross) because the equation is a constant added from the equation for those cases anyway – Jan-Magnus Økland Jan 15 '24 at 14:59
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    @TitoPiezasIII Honestly, it was the first thing I tried, because it’s the first thing you do in searching for the canonical form. – Jan-Magnus Økland Jan 15 '24 at 15:01
  • I added a third image showing $(\alpha/3,,\beta/3)$. It seems to be at the center indeed. P.S. For this family of ellipses, one can calculate the center from 3 carefully chosen lattice points. I wonder if it is possible to do the same for this ellipse with 11 lattice points. – Tito Piezas III Jan 15 '24 at 16:19
  • I made a new question about how I found $(\alpha, \beta)$ without knowing about the partial derivatives method. Might be of interest. – Tito Piezas III Jan 16 '24 at 08:36
  • What was the transformation you did on $(x,y)$ to move the center of the family to the origin $(0,0)$? – Tito Piezas III Jan 16 '24 at 11:51
  • @TitoPiezasIII On the equation you have to use the inverse translation $x\to (x+x(S)), y \to (y+y(S)).$ – Jan-Magnus Økland Jan 16 '24 at 11:57
  • I understand it now. Your transformation was $x\to x+\dfrac{\alpha}3,,y\to y+\dfrac{\beta}3$. But to maintain the Diophantine $x^3+y^3+z^3=(z+1)^3$ relation, one has to reverse the transformation. – Tito Piezas III Jan 16 '24 at 12:19
  • @TitoPiezasIII See the edit – Jan-Magnus Økland Jan 16 '24 at 12:25
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    I tweaked your edit to make it more aesthetic, and added your pic. I hope it is ok. – Tito Piezas III Jan 16 '24 at 12:46
  • You made a transformation such that the center of the ellipse coincides with the origin $(0,0)$. Is there any general transformation such that the vertices of triangle ADE and triangle BCF will coincide? – Tito Piezas III Jan 17 '24 at 00:03
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    A little remark about the triangular correspondences in the last figure : they can be "rendered" by the $2 \times 2$ matrix $A:=\pmatrix{0&-1\1&-1}$ with order $3$ ($A^3=I_2$), explaining the 3-cycles, for example $\pmatrix{366&162&-528&366&162&-528\ -162&528&-366&-162&528&-366}$ where each column is the image of the previous one by $A$ : $\pmatrix{366\-162} \stackrel{A}{\rightarrow} \pmatrix{162\528} \stackrel{A}{\rightarrow}...$ – Jean Marie Mar 11 '24 at 11:00
  • [Ctd] which doesn't come as a surprise because if we denote by $Q=\pmatrix{1&−1/2\−1/2&1}$ the matrix of the quadratic form $x^2−xy+y^2$, we have $A^TQA=Q$ : this matrix $A$ preserves the quadratic form (in other terms, $A$ belongs to the orthogonal group associated with $Q$). – Jean Marie Mar 11 '24 at 11:24
  • Any comment about my remarks ? – Jean Marie Mar 12 '24 at 15:46