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Given the rational Diophantine equation,

$$t^3 - t^2 - \tfrac{1}{3}(n^2 + n)t - \tfrac{1}{27}n^3=w^3\tag1$$

Two points are, $$t_0 = 0\tag2$$ $$t_2 = \frac{-(1 + 2 n) (1 + 11 n + 42 n^2 + 14 n^3 + 13 n^4)}{9 (7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4)}\tag3$$

Question: How do we find an intermediate point $\color{blue}{t_1}$ between $t_0$ and $t_2$; that is, one with numerator and denominator less than a quartic?

It's the missing piece in this family,

$$\sqrt[3]{t_0+x_1}+\sqrt[3]{t_0+x_2}+\sqrt[3]{t_0+x_3}= \sqrt[3]{z_0}\tag4$$

$$\sqrt[3]{\color{blue}{t_1}+x_1}+\sqrt[3]{\color{blue}{t_1}+x_2}+\sqrt[3]{\color{blue}{t_1}+x_3}= \sqrt[3]{z_1}\tag5$$

$$\sqrt[3]{t_2+x_1}+\sqrt[3]{t_2+x_2}+\sqrt[3]{t_2+x_3}= \sqrt[3]{z_2}\tag6$$

where,

$$z_0=-(2n+1)+3\sqrt[3]{\tfrac{n(n^2+n+1)}{3}}\tag7$$

$$z_2=\frac{-3(2 + n)^3 (1 + n + n^2)}{7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4}\tag8$$

and the $x_i$ are the three roots of,

$$x^3 + x^2 - \tfrac{1}{3}(n^2 + n)x + \tfrac{1}{27}n^3=0\tag9$$

P.S. Once $\color{blue}{t_1}$ is found, and assuming it also has additional properties, it is easy to find $z_1$ as a rational root of a nonic. (The relations found by davidoff303 was just the special case $n=-3$.) There are infinitely many rational points $t_i$ but I am interested in those with the smallest height.

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  • $\begingroup$ However, $t_1$ must not be either of the 3 points, $$t_1 \neq \frac{1-n}{9},\;\frac{n+2n^2}{3(-1+n)},\;\frac{-n^2}{1+2n}$$ $\endgroup$ – Tito Piezas III Jul 4 '16 at 16:04
  • $\begingroup$ why do you think there is a missing piece ? $\endgroup$ – mercio Jul 5 '16 at 10:03
  • $\begingroup$ @mercio. From davidoff303's solutions for $n=-3$ as $t_1= 74/43$ and $t_2 = 5105/11349$, I had assumed that his $t_1$ would belong to a family as well. I guess that point is just a peculiarity for the case $n=-3$. $\endgroup$ – Tito Piezas III Jul 7 '16 at 19:26
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The cubic \begin{equation*} u^3-u^2-(n^2+n)u/3-n^3/27-w^3=0 \end{equation*} can be shown to be equivalent to the elliptic curve \begin{equation} y^2=x^3+1296n^2(n^2+n+1)^2 \end{equation} by using Nagell's algorithm and a computer algebra package. It does not have to be state-of-the-art software since I used an ancient MS-DOS version of Derive. The curve has torsion points of order $3$ when $x=0$.

The transformations are straightforward but lengthy, so are not given.

Tests suggest that the rank is greater than $0$ unless $n=1$. They suggest the rank is often greater than $1$.

Using simple experimental results it is possible to show that \begin{equation} (-4(n-1)(2n+1), \, 4(n+2)(7n^2+n+1)) \end{equation} gives a simple non-torsion point on the curve, unless $n=1$.

Applying the transformations to this point and adding the torsion points just gives the values of $t$ not wanted, the quartic one given in the original question and two very similar points \begin{equation*} t=\frac{n^2(19n^4+17n^3+24n^2+14n+7)}{(1-n)(31n^4+41n^3+15n^2-7n+1)} \end{equation*} and \begin{equation*} t=\frac{n(n-1)(31n^4+41n^3+15n^2-7n+1)}{3(2n+1)(13n^4+14n^3+42n^2+11n+1)} \end{equation*}

Doubling the point above gives a much more complex form. There might be massive cancellations reducing the corresponding value of $t$ but I did not try it.

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  • $\begingroup$ is there a value of $n$ where those points generate the entire group of rational points for that curve ? $\endgroup$ – mercio Jul 6 '16 at 17:43
  • $\begingroup$ Thanks. I found those points also, but had hoped there would be simpler ones to explain davidoff303's $t_1 = 74/43$ for the case $n=-3$. $\endgroup$ – Tito Piezas III Jul 7 '16 at 19:30

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