5

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.

gigi
  • 1,333
  • 1
    It would be better to say that you're looking for rational points on the elliptic curves, since your equation is homogeneous in $x,y,z$, so you're asking about points $[x,y,z]\in\mathbb P^2(\mathbb Q)$ satisfying the equation. Of course, you can always clear denominators to make $x,y,z\in\mathbb Z$. But usually when one talks about integer points, it's for an affine equation: Example: Find all integer solutions to $y^2=x^3+17$. – Joe Silverman Jan 09 '21 at 19:07
  • 4
    Related : https://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation and http://publikacio.uni-eszterhazy.hu/2858/1/AMI_43_from29to41.pdf – Chris Wuthrich Jan 09 '21 at 19:16
  • See https://observablehq.com/@robinhouston/a-remarkable-diophantine-equation and also https://math.stackexchange.com/questions/402537/find-integer-in-the-form-fracabc-fracbca-fraccab and also https://math.stackexchange.com/questions/2779545/ask-for-the-rational-roots-of-fracabc-fracbac-fraccab-4 and also https://math.stackexchange.com/questions/3016720/technique-behind-solving-4-fracabc-fracbac-fraccab and also https://shainer.github.io/math/2017/09/12/finding-solutions-to-weird-equations.html and many others. – Gerry Myerson Jan 10 '21 at 12:21

1 Answers1

12

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.

Ari Shnidman
  • 2,481
  • 22
  • 23
  • Thank you very much, it's clear! – gigi Jan 09 '21 at 17:59
  • 3
    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 09 '21 at 18:12
  • @DavidLoeffler yeah sure, I'm glad that it is useful for someone :) – gigi Jan 09 '21 at 18:17
  • And, of course, you are very welcome to use this example. – Ari Shnidman Jan 09 '21 at 18:27
  • Another good point! I made the change. – Ari Shnidman Jan 09 '21 at 18:56
  • @DavidLoeffler Would your students be entertained by the fruit version? https://qph.fs.quoracdn.net/main-qimg-5b0690e302a38cf2a8068158199e7a21.webp – Will Sawin Jan 10 '21 at 01:40