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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.

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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.

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  • $\begingroup$ Thank you very much, it's clear! $\endgroup$ – gigi Jan 9 at 17:59
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    $\begingroup$ 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?) $\endgroup$ – David Loeffler Jan 9 at 18:12
  • $\begingroup$ @DavidLoeffler yeah sure, I'm glad that it is useful for someone :) $\endgroup$ – gigi Jan 9 at 18:17
  • $\begingroup$ And, of course, you are very welcome to use this example. $\endgroup$ – Ari Shnidman Jan 9 at 18:27
  • $\begingroup$ Another good point! I made the change. $\endgroup$ – Ari Shnidman Jan 9 at 18:56

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