Transforming a Diophantine equation to an elliptic curve

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 $x/y + y/z + z/x = 3$.

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