# Transforming a Diophantine equation to an elliptic curve

I heard that the following problem lead to determine the rational points of an elliptic curve:

For which integers $n$ there are integers $x,y,z$ such that $x/y+y/z+z/x=n$. Could anyone show me why this question leads to the theory of elliptic curves?

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If you multiply your equation by xyz you get a homogeneous cubic equation. Standard textbooks on elliptic curves will then tell you how to transform that into the equation of an elliptic curve and what to do with it. You may start with Silverman-Tate. – Felipe Voloch Jun 4 '10 at 21:30

Nearly 10 years ago, I gave a talk at Wesleyan, and a gentleman named Roy Lisker asked me the same question: Fix an integral solution $(x, \ y, \ z)$ and make the substitution

$$u = 3 \ \frac {n^2 z - 12 \ x}z \qquad v = 108 \ \frac {2 \ x \ y - n \ x \ z + z^2}{z^2}$$

Then $(u, \ v)$ is a rational point on the elliptic curve $E_n: \ v^2 = u^3 + A \ u + B$ where $A = 27 \ n \ (24 - n^3)$ and $B = 54 \ (216 - 36 \ n^3 + n^6)$. (It actually turns out that $E_n$ is an elliptic curve whenever $n$ is different from 3, but I’ll discuss this case separately.)

Let me say a little about the structure of this curve for the experts:

This curve has the “obvious” rational point $T=(3 n^2, 108)$ which has order 3, considering the group structure of $E_n$.  It actually turns out that these three multiples correspond to the cases $x = 0$ and $z = 0$, so if such an integral solution $(x, \ y, \ z)$ exists then the rational solution $(u, \ v)$ must correspond to a point on $E_n$ not of order 3. (Of course, I don’t care about the cyclic permutation $x \to y \to z \to x$.)

In the following table I’m computing the Mordell-Weil group of the rational points on the elliptic curve i.e. the group structure of the set of rational solutions $(u, \ v)$:

$$\begin{matrix} n & E_n(\mathbb Q) \\ \\ 1 & Z_3 \\ 2 & Z_3 \\ 3 & \text{Not an elliptic curve} \\ 4 & Z_3 \\ 5 & Z_6 \\ 6 & Z_3 \oplus \mathbb Z \\ 7 & Z_3 \\ 8 & Z_3 \\ 9 & Z_3 \oplus \mathbb Z \\ \end{matrix}$$

Hence when $n =$ 1, 2, 4, 7 or 8 we find no integral solutions $(x, \ y, \ z)$.  When $n = 5$, there are only six rational points on $E_n$, namely the multiples of $(u,v) = (3, 756)$ which all yield just one positive integral point $(x,y,z) = (2,4,1)$.

Something fascinating happens when $n = 6$... The rank is positive (the rank is actually 1) so there are infinitely many rational points $(u, \ v)$.  But we must be careful: not all rational points $(u, \ v)$ yield positive integral points $(x, \ y, \ z)$.  Clearly, we can scale $z$ large enough to always choose $x$ and $y$ to be integral, but we might not have $x$ and $y$ to both be positive.  You’ll note that $x > 0$ if only if $u < 3 \ n^2$, so we only want rational points in a certain region of the graph.  Since the rank is 1, this part of the graph is dense with rational points! Let me give some explicit numbers.  The torsion part of $E_n( \mathbb Q)$ is generated by $T = (75, 108)$ and the free part is generated by $(u,v) = (-108, 2052)$.  By considering various multiples of this point we get a lot of positive integral -- yet unwieldy! --  points $(x,y,z)$ such that $x/y + y/z + z/x = 6$:

\begin{aligned} (x,y,z) & = (12, 9, 2), \\ & = (17415354475, 90655886250, 19286662788) \\ & = (260786531732120217365431085802, 1768882504220886840084123089612, 1111094560658606608142550260961) \\ & = (64559574486549980317349907710368345747664977687333438285188, 70633079277185536037357392627802552360212921466330995726803, 313818303038935967800629401307879557072745299086647462868546) \end{aligned}

I’ll just mention in passing that when $n = 9$ the elliptic curve $E_n$ also has rank 1.  The generator $(u,v) = (54, 4266)$ corresponds to the positive integral point $(x,y,z) = (63, 98, 12)$ on $x/y + y/z = z/x = 9$.

What about $n = 3$?  The curve $E_n$ becomes $v^2 = (u – 18) (u + 9)^2$.  This gives two possibilities: either $u = -9$ or $u \geq 18$.  The first corresponds to $x = z$ while the second corresponds to $(z/x) \geq 4$.  By cyclically permuting $x$, $y$, and $z$ we find similarly that either $x = y = z$ or $x/y + y/z + z/x \geq 6$.  The latter case cannot happen by assumption so $x = y = z$ is the only possibility i.e. $(x,y,z) = (1,1,1)$ is the only solution to $x/y + y/z + z/x = 3$.

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The gentleman Roy Lisker happens to be the translator of (parts of) Grothendieck's "Recolte et Semailles" into the english language: fermentmagazine.org/rands/recoltes1.html – Hans Stricker Jul 7 '14 at 16:30

Allan MacLeod's website Elliptic Curves in Recreational Number Theory considers this problem among many other interesting ones.

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