# Existence of rational points on some genus 3 curves

Do there exist a pair of rational numbers $$(x,y)$$ such that $$(a) \quad x^4+x^3+y^4+y-1=0$$ $$(b) \quad x^4+x^3+y^4+y^2-1=0$$ Magma function IsLocallySoluble returns that both equations are everywhere locally soluble, but a search for small rational solutions returned no results. Both equations define genus 3 curves. I assume that the first step should be computing the rank of the Jacobian, because there are known methods for small ranks. However, I am not aware of any code (e.g. Magma) that can compute the rank of general genus 3 curves, no talking about rational points.

In a similar recent question points on non-hyperelliptic curves of genus 3 the specific equation(s) are not given, and a comment indicates that writing the equation of the curve would help a lot, because different tricks are available for different curves. So, are there tricks available for these two curves?

• For the first curve, one could try to use the techniques of this paper to find the (rank of) the Mordell-Weil group. But this would likely be a somewhat iinvolved endeavor. Commented Apr 14, 2023 at 10:39
• It looks like the bad primes are 149, 283 and 163991, and the conductor is $149 \cdot 283^2 \cdot 163991$ (simple nodes for the first and last, a cusp for the second). Getting an upper bound for the rank under BSD might be borderline possible. Anyone wants to take up from here? Commented Apr 14, 2023 at 10:53

There are no rational solutions to curve (b). This curve has the automorphism $$(x,y) \mapsto (x,-y)$$ and the quotient is the genus one curve $$-yz + w^{2} = 0 \quad x^{2} + y^{2} + xz - z^{2} + yw = 0.$$ The first equation $$-yz + w^{2} = 0$$ can be thought of as a genus $$0$$ curve in the variables $$y$$, $$z$$ and $$w$$ and its parametrization is $$y = s^{2}$$, $$z = t^{2}$$ and $$w = st$$. This allows us to rewrite the genus one curve as $$x^{2} = -4s^{4} - 4s^{3} t + 5t^{4}.$$ This curve is one of the two-covers of the elliptic curve $$E : y^{2} + y = x^{3} + 5x + 1$$. (This elliptic curve has rank zero and, assuming BSD, a Shafaraveich-Tate group of order $$4$$.) Indeed, performing a 2-descent on $$E$$ finds the curve $$y^{2} = -4x^{4} - 4x^{3} + 5$$ as a two-cover. This two-cover has non-trivial Cassels-Tate pairing with $$y^{2} = 2x^{4} + 6x^{3} + 6x^{2} + 10x - 4$$, which shows that $$y^{2} = -4x^{4} - 4x^{3} + 5$$ is locally solvable everywhere, but has no rational points. The same must be true of the original curve.
EDIT: At the request of the OP, I am providing some information and references about using the Cassels-Tate pairing to show that a genus $$1$$ curve is everywhere locally solvable, but has no rational points. This pairing was originally defined by Cassels (in the paper "Arithmetic of Curves of genus 1: III. The Tate-Šafarevič and Selmer groups" published in the Proceedings of the London Mathematical Society, Issue 1, 1962, pages 259-296). In the instance in question, there is a pairing on the $$2$$-Selmer group $${\rm Sel}_{2} \times {\rm Sel}_{2} \to \mathbb{Z}/2\mathbb{Z}$$ whose kernel consists of elements in $${\rm Sel}_{2}$$ that are the image of an element in $${\rm Sel}_{4}$$ the $$4$$-Selmer group. (This follows from Theorem 1.2 of Cassels's paper.) So if $$C_{1}$$ and $$C_{2}$$ are two $$2$$-covers of an elliptic curve and the pairing of $$C_{1}$$ and $$C_{2}$$ is non-trivial, it follows that neither $$C_{1}$$ nor $$C_{2}$$ are the image of an element in the $$4$$-Selmer group. This implies that neither $$C_{1}$$ nor $$C_{2}$$ can have a rational point.
The Cassels-Tate pairing on the $$2$$-Selmer group has been implemented in Magma by Steve Donnelly, but it appears that a paper documenting this implementation hasn't been published. Cassels's article linked above is behind a paywall. For recent treatments of the Cassels and Cassels-Tate pairings, see the articles here (which is open access) and here (which is an arXiv preprint).
• The L-factor of the first curve at $p=5$ is irreducible, so its Jacobian is simple. In fact, since no (nontrivial) quotient of two of the roots of the L-factor is a root of unity, it is even absolutely simple. This means that no reduction to a lower-genus curve or lower-dimensional abelian variety is possible. Commented Apr 14, 2023 at 10:36
• The L-factor at $p = 11$ is also irreducible, and the number fields defined by the two polynomials are linearly disjoint. This implies that the endomorphism ring of the Jacobian is just the ring of integers, so there is also no extra structure there. Commented Apr 14, 2023 at 10:42