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Do there exists rational numbers $x$ and $y$ such that $$ y^3 = x^4 + x + 2 ? $$

Context: There are a lot of publications about computing rational points on elliptic and hyperelliptic curves, and these problems has been solved in a number of special cases. The "next simplest" case are Picard curves, which can be described by equations in the form $y^3=P(x)$, where $P(x)$ is a polynomial of degree 4. An important parameter measuring the complexity of this problem is the rank of the Jacobian of the curve. The rank can be computed using RankBounds function in Magma, and this particular curve has rank 0. There is a magma function (Chabauty0) for computing rational points on rank 0 hyperelliptic curves, but Picard curves are not hyperelliptic. On the other hand, Hashimoto and Morrison https://arxiv.org/abs/2002.03291 recently showed how to compute rational points on Picard curves of rank 1. Based on this, it looks like the case of Picard curves of rank 0 should be tractable, but I cannot find any reference, and also cannot solve this myself, hence the question. The particular equation above is chosen because it is the simplest example of such curve for which there is a rational point everywhere locally but there are no obvious rational points.

Bogdan Grechuk
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    After doing a quick computation, I don't think there are rational solutions. To do this, I determined what the torsion of the Jacobian of the curve could be using local computations of zeta functions and found that it is trivial. Since the Jacobian has rank 0, this implies that there should not be any rational points. I need to run now, but I can upload the code later (once I double check it) if you would like. – Jackson Morrow Nov 03 '21 at 18:50
  • Thank you! I hope the code is general and works for any equation of this form of rank 0? – Bogdan Grechuk Nov 03 '21 at 19:58

1 Answers1

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The below Magma code determines the size of $J_C(\mathbb{F}_p)$ for various primes $p$, and finally compute the GCD of their orders, which gives you a bound on the size of $J_C(\mathbb{Q})$. For a discussion of this, see Section 4.1 of this paper. To determine $J_C(\mathbb{F}_p)$, you can just evaluate the numerator of the zeta function of the reduction of $C$ mod $p$ (see e.g., Hindry--Silverman Diophantine Geometry Exercise C.4).

The below code does this for primes up to 30 and tells you that the GCD of $\#J_C(\mathbb{F}_p)$ for these primes is 1, and hence $J_C(\mathbb{Q})$ is just the trivial group. The code is general and can work for any equation of this form of rank 0. If you do have some non-trivial torsion in $J_C(\mathbb{Q})$, then you will need to work harder to determine the rational points.

I will also note that you can run this on the online magma calculator.

P2<x,y,z> := ProjectiveSpace(Rationals(),2);
C := Curve(P2,z*y^3 - (x^4 + x*z^3 + 2*z^4));
bad := {2,3};
bool := false; p := 2;
torsOrders := {@@};
while  p le 30 do
  p :=  NextPrime(p);        
  p := p in bad select NextPrime(p) else p;
  Cp := Curve(Reduction(C,p));
torsOrders := torsOrders join  {@ Evaluate(Numerator(ZetaFunction(Cp)),1) @};
end while;
GCD(torsOrders); //1
Jackson Morrow
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  • On the other hand, it is also possible the gcd works out to be greater than 1, but there is no nontrivial torsion (see this old question of mine about elliptic curves). I suspect there may be more sure-fire ways to compute torsion like there are in genus 1 case, but your method is still likely to work in practice most of the times. – Wojowu Nov 03 '21 at 20:35
  • @Wojowu This is a good point! What one should really do is take the "GCD of the groups" $J_C(\mathbb{F}_p)$. This is explained in the paper I linked to, and this usually provides better bounds than just looking at the GCD of $#J_C(\mathbb{F}_p)$. – Jackson Morrow Nov 03 '21 at 21:08
  • Thank you! I have tried on several examples and the code works, although for some equations I needed to increase the bound from 30 to 50. Just one question - how to determine the set of bad primes for other equations? (I run the code as it is, with bad := {2,3}; for all equations). – Bogdan Grechuk Nov 03 '21 at 22:16
  • @JacksonMorrow The example in the answer I link shows that "GCD of groups" doesn't necessarily always work either (but, I agree, it is still a better invariant). – Wojowu Nov 03 '21 at 22:19
  • @BogdanGrechuk I just guessed that these were the bad primes. I don't have a good way to get this in general. Sorry about that! – Jackson Morrow Nov 04 '21 at 00:11
  • Hm, but this seems crucial! I tested your code on the equation $y^3=x^4+x$ that has obvious solution (0,0) and correctly get GCD=9, not 1. But I tested it as it is, with bad={2,3}. Now assume that I "guessed" that bad={2,5} (or any other set not containing 3). Then your program returns 1, incorrectly claiming that no rational point exists! So, it is critical to "guess" bad primes correctly. As a minimum, can "bad" should contain bad primes (and possibly some extra primes). In this case if the code returns 1 this is still a certificate of no rational solutions. – Bogdan Grechuk Nov 04 '21 at 12:01
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    Ok, the bad primes are prime divisors of the conductor, but I do not see how to compute conductors of such curves in magma. However, there is a function DiscriminantOfTernaryQuartic that returns the discriminant. And the discriminant of a curve is necessarily divisible by every prime that divides the conductor. So, we can compute the discriminant and let bad be the set of its prime factors. For the curve in the question this gives bad={3,43,47} and the code returns 1. For $y^3=x^4+x$ this gives bad={3} and the code returns 9, so all seems to work! – Bogdan Grechuk Nov 04 '21 at 12:08