# Action of the symmetric group $S_3$ on an elliptic curve $E$ defined over $\mathbb{Z}$

I came up with the following question on a facebook group: find the positive integer solutions of the equation $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=4$$ Now clearly this is very difficult, indeed it is equivalent to find the integer solutions of an elliptic curve $$E$$ defined over $$\mathbb{Z}$$, in particular $$E$$ is the smooth plane cubic curve in $$\mathbb{P}^2$$ given by the equation $$E=(x^3 - 3 x^2 y - 3 x^2 z - 3 x y^2 - 5 x y z - 3 x z^2 + y^3 - 3 y^2 z - 3 y z^2 + z^3=0)$$

Anyway I started playing with this: since $$E$$ is symmetric with respect to the permutations of the variables $$x,y,z$$ we have an action $$S_3 \times E \rightarrow E \\ (\sigma,[x,y,z]) \rightarrow [\sigma(x),\sigma(y),\sigma(z)]$$ where I've identified the set $$\{x,y,z\}$$ with the set $$\{1,2,3\}$$. We can form the quotient of $$E$$ with respect to this action yielding a map $$f: E \rightarrow \mathbb{P}^1$$ of degree $$6$$. By Hurwitz's formula we have that the degree of the ramification divisor $$R$$ is $$12$$ and since the ramification points consists in those points with $$2$$ coordinates equal we have $$3$$ ramification points of order $$4$$. My questions now are the following:

1. From this description are we able to say that $$E$$ has infinite integer solutions, maybe noting that this type of elliptic curve is particular?
2. It seems to me that $$S_3 \subset Aut(E)$$ but searching on the internet I found that the automorphism group of elliptic curve can only be of type $$\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}$$ and $$\mathbb{Z}/6\mathbb{Z}$$. What I'm missing?

I'm sorry if I said something wrong, I'm not an expert on this field. Thanks in advance for the help.

Technically speaking, an elliptic curve is a genus 1 curve with a choice of rational point. The automorphism group of an elliptic curve is the subgroup of automorphisms of the genus 1 curve that fix that rational point. So there is no contradiction with the facts you looked up, it just means that no point is fixed by all of $$S_3$$, and indeed (1:1:1) is not on the curve.

You can check in sage or Magma that this genus 1 curve does have infinitely many integral points (which are the same as the rational points, since the curve is projective). I chose $$P = (1:-1:0)$$ to be the identity. Then the elliptic curve has rank 1 and the group of rational points is generated by $$Q = (4:-1:11)$$ and the 6 "obvious" points with one coordinate equal to 0, which are the six torsion points.

This answers your two questions, but doesn't answer the original question from the facebook group because $$Q$$ doesn't have positive coordinates. I found that the point $$9Q$$, or explicitly, $$(154476802108746166441951315019919837485664325669565431700026634898253202035277999: 36875131794129999827197811565225474825492979968971970996283137471637224634055579: 4373612677928697257861252602371390152816537558161613618621437993378423467772036)$$ does have positive coordinates. There are several other such points, enough to make me think that there are infinitely many, but I don't know how to prove it.

To do these computations yourself in sage, use the EllipticCurve_from_cubic function.

• Thank you very much, it's clear! – gigi Jan 9 at 17:59
• For a smooth curve over $\mathbf{Q}$, either $C(\mathbf{Q})$ is finite or its closure in $C(\mathbf{R})$ is a subset of the components of $C(\mathbf{R})$. So, since $C(\mathbf{Q})$ is infinite and contains one point with all coordinates positive, it must contain infinitely many. (This example is wonderful -- do you mind if I steal it for the elliptic curves lecture course I'm starting next week?) – David Loeffler Jan 9 at 18:12
• @DavidLoeffler yeah sure, I'm glad that it is useful for someone :) – gigi Jan 9 at 18:17
• And, of course, you are very welcome to use this example. – Ari Shnidman Jan 9 at 18:27
• Another good point! I made the change. – Ari Shnidman Jan 9 at 18:56