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I have the following hyperelliptic curve of genus $2$: $$ y^2 = 561 x^6 - 41904 x^5 + 627264 x^4 + 11860992 x^3 - 197074944 x^2 + 124416^2 $$ I need to find all the rational points on this curve. There is one obvious point at $(0, 124416)$. I have some experience in finding rational points on elliptic curves. I also have the reference Handbook of Elliptic and Hyperelliptic Curve Cryptography (Discrete Mathematics and Its Applications). Most of this work uses genus $2$ curves reduced to quintic form. So a related question is how to transform this sextic to the quintic form.

Thank you

Lorenz Menke

A question as to where this problem originates. The nuclear electric quadrupole Hamiltonian characteristic equation for spin $9/2$ is $$ E_p^5 - 99 (3 + z)E_p^3 - 1188(1 - z)E_p^2 + 1188(3 + z)^2 E_p + 11664(3 + z)(1 - z) $$ where the energy states are $E = 2 E_p$ and $z = \eta^2$ where $\eta$ is the asymmetry parameter. Now solve this equation for $z$. The above sextic is the square root term. Thus when the sextic is a perfect square for a rational energy state the resultant asymmetry parameter squared is also a rational. This set of rational points of this sextic are the values that result in the quadrupole quintic Galois factoring into a linear times a quartic. The only other values of $z$ that reduce this quintic are $z = 0, 1, -3,$ and $9$. The goal is to find all the points $z$ such that the quintic factors over the rationals.

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    $\begingroup$ You can't transform it into a quintic over the rationals as the sextic has no rational roots. To find rational points, you can try Ratpoints: mathe2.uni-bayreuth.de/stoll/programs $\endgroup$ – Felipe Voloch Dec 2 '14 at 3:45
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    $\begingroup$ Why this particular curve? In general, if $(x,y)$ is a solution then so is $(x,-y)$; so having found $(0,124416)$ there's also $(0,-124416)$. Stoll's ratpoints (suggested by Felipe Voloch) quickly finds 8 more rational point pairs, with $x = -1/6$, $1/12$, $-1/18$, $-1/24$, $1/36$, $1/48$, $-1/72$, and $11/216$. Taking $x=1/24X$ and clearing square factors simplifies the equation to $Y^2 = 561X^6-1746X^5+1089X^4+858X^3-594X^2+81$, when the small points are at $X = -3, 1, 0, 2, 1/2, 3/2, -3/4, 1/4, 9/11$, and ratpoints says there are no others with $X =m/n$ with $|m|,|n| < 10^4$. $\endgroup$ – Noam D. Elkies Dec 2 '14 at 4:03
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    $\begingroup$ The sextic has maximal Galois group, so it might still be out of reach to prove that the list of rational points is complete; but the discriminant is very smooth (no prime factors larger than $7$) so there might be hope. The smooth discriminant, as well as the moderately large harvest of points, again leads me to wonder how this curve arose. $\endgroup$ – Noam D. Elkies Dec 2 '14 at 4:05
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By now there is a fairly rich literature on computing the set of rational points on curves of higher genus, see for example my survey paper on "Rational points on curves".

What one can do for your concrete curve is this. First, one can simplify the curve equation (even more than Noam did); your curve is isomorphic to $$y^2 = x^6 - 12 x^5 - 6 x^4 + 82 x^3 + 1461 x^2 - 630 x + 225.$$ Then most methods aiming at finding the set of rational points use the group of rational points on the Jacobian variety of the curve (which by the Mordell-Weil theorem is a finitely generated abelian group). In your case, this group turns out to be free abelian of rank 2. Which is bad news, since the method of choice ("Chabauty's method") only works when this rank is strictly less than the genus. Since (as Noam already mentioned) the defining polynomial has large (even maximal) Galois group, the next method one could try ("Elliptic curve Chabauty") will be infeasible. So despite all the progress that was made in the last couple of decades or so, a curve like yours (with rank $\ge$ genus and no special geometric or arithmetic properties [the smoothness of the discriminant does not help here]) is still likely to resist our efforts.

So what can be done regarding your question? Using explicit generators of the Mordell-Weil group (which takes some computation to provably find), one can fairly efficiently show that there are no rational points beyond the 18 that Noam found up to any given (even quite large) height bound, for example by using the "Mordell-Weil sieve". And if you are interested in integral points, then one can combine the Mordell-Weil sieve with bounds coming from Baker's method (linear forms in logarithms) to prove that you know all integral points.

If desired, I can add details on any or all of the methods mentioned above.

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  • $\begingroup$ This does help. I am interested in how you reduced the equation to the above form. $\endgroup$ – Lorenz H Menke Dec 3 '14 at 0:02
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    $\begingroup$ There are the notions of "minimizing" and "reducing" equations, for example for hyperelliptic curves. Minimization tries to eliminate redundant powers of primes from the discriminant, while keeping the coefficients integral. Qing Liu has worked this out in detail for hyperelliptic curves. Reduction tries to make the coefficients small unsing a unimodular transformation (which keeps the coefficients integral and the discriminant the same). This is considered in a paper by Cremona and myself. --> $\endgroup$ – Michael Stoll Dec 3 '14 at 8:37
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    $\begingroup$ Both are implemented in Magma. Assuming f is the polynomial in your equation, you do "SimplifiedModel(ReducedMinimalWeierstrassModel(HyperellipticCurve(f)))" to get the equation in my answer. The minimization part reduces the discriminant of the curve ($= 2^8 \operatorname{disc}(f)$) from $2^{113} 3^{57} 5^{10} 7^5$ down to $2^{23} 3^{17} 5^{10} 7^5$. If you don't have access to Magma, you can try the above out with the online Magma calculator at magma.maths.usyd.edu.au/calc. $\endgroup$ – Michael Stoll Dec 3 '14 at 8:41

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