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Find all possible rational solutions pairs $(z,a)$ of the equation

$a^6 + a^4 (-18368 + 9184 z - 2912 z^2) + a^2 (61702144 - 61702144 z + 36814848 z^2 - 10694656 z^3 + 1748992 z^4) - z^2 (44281626624 - 44281626624 z + 13332971520 z^2 - 1131282432 z^3 + 28901376 z^4) = 0$.

I have the following solutions for $a = (0, \pm 24,\pm48,\pm72,\pm 96,\pm 144,\pm 192)$ and $z = (-2,0,1,4)$ which I think is the complete set. A second question is to find all possible rational $a$ or $z$ such that this equation is reducible in either parameter. I have the set $z =(-44,-2,0,1,44/23,67/19,4,134/29)$ which I also think is complete. In this case for the $z$ values I get four pairs of cubics and one quartic all in parameter $a$.

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    $\begingroup$ Your equation defines a curve of genus $7$. Setting $b = a^{2}$ gives a curve (in terms of $b$ and $z$) of genus $4$, which has an involution. The quotient by this involution is a rank $3$ elliptic curve $E$. This does not give a good method of proving your set of rational solutions is complete, but it does give an efficient way to search for points. I looked for preimages on your original curve of points on $E$ of the form $aP_{1} + bP_{2} + cP_{3}$, where $P_{1}$, $P_{2}$ and $P_{3}$ are generators of the MW-group and $|a|, |b|, |c| \leq 7$ and only found the 19 points you know about. $\endgroup$ Commented May 27, 2015 at 1:28
  • $\begingroup$ Can you show the genus 3 hyperellipic cure that you deriived for this problem. $\endgroup$ Commented May 28, 2015 at 4:39
  • $\begingroup$ There's no genus 3 hyperelliptic curve that I found. Do you want the elliptic curve $E$ and the generators $P_{1}$, $P_{2}$ and $P_{3}$? $\endgroup$ Commented May 28, 2015 at 15:28
  • $\begingroup$ Yes, that is what I need, the elliptic curve. I misunderstood what you said and thought there was any hyper elliptic curve. $\endgroup$ Commented May 28, 2015 at 17:26
  • $\begingroup$ The elliptic curve is $E : y^{2} = x^{3} - 159600x + 314825000$. Generators are $(-2600/9,-197450/27)$, $(-1015/4,59675/8)$, and $(700/9,-119350/27)$. I'm guessing you probably also need the map from your curve to the elliptic curve. $\endgroup$ Commented May 29, 2015 at 12:00

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It seems hopeless to try to provably find all the rational points on the genus $7$ curve $C$ mentioned in the question using either Chabauty's method, or etale descent. I will simply explain how to find a map from the genus $7$ curve to the rank three elliptic curve mentioned in the comment.

First, I noticed that all of the exponents on $a$ are even, and so setting $b = a^{2}$, we get an equation $b^{3} + b^2 (-18368 + 9184z - 2912z^{2}) + \cdots$. This equation defined a curve of $X$ genus $4$. I computed the automorphism group of this curve modulo several primes, and noticed that the order was $2$ (for $p > 17$ or so). I then had Magma compute the automorphism group $G$ over $\mathbb{Q}$ and found it had order $2$. If we realize $X \subseteq \mathbb{P}^{3}$ in its canonical model, the automorphism can be given by a simple linear change of variables. This allows one to easily compute the curve quotient, which results in a plane cubic. Magma's built-in routines can then get one to the minimal Weierstrass model of $X/G$.

Searching for nice equations for the maps at each step results in the map from the original curve (in terms of $a$ and $z$), to the homogenous form of the elliptic curve ($y^{2} z = x^{3} - 159600xz^{2} + 31482500z^{3}$) given by the equations

$$ x = a^{2} (470957175a^2 - 403726629600z^2 + 1240537082400z - 2481074164800) $$

$$ y = 6385126825a^4 - 6377874641600a^2z^2 + 33875545316800a^2z - 67751090633600a^2 + 216129127244800z^4 - 8459911552153600z^3 + 99706100436096000z^2 - 331145109327155200z + 331145109327155200 $$

$$ z = a^{2} (1220184a^2 - 875686464z^2). $$

If I hadn't tried several tricks to manually simplify the equations, they would have been several pages long.

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  • $\begingroup$ Fantastic, that is what I needed. This does help in my work on a family of polynomials. There are several more equations like this one, so with your solution I should be able to reduce them to an elliptic or hyper elliptic curve. Thanks again. $\endgroup$ Commented Jun 3, 2015 at 1:47

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