The elliptic curve,

$$y^2 = 23328x^3-890273x^2+14755570x-7^7 \tag{1}$$

has the small solution $x = 58$. I know how to find other rational points, but the number of digits in the denominator gets large fast.

Question: Does (1) have other rational points of small height, maybe where the numerator or denominator has only 5 digits or less?

P.S. I routinely come across situations like this. Is there an online Alpertron equivalent for elliptic curves, where given $ax^4+bx^3+cx^2+dx+e = y^2$, you just input {$a,b,c,d,e$} into the applet, and it outputs, if any, rational "x" with small height below a bound? (The Alpertron is a very useful site.)

  • $\begingroup$ Do you know about accepting answers? $\endgroup$ – Antonio Vargas Apr 27 '12 at 15:28
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    $\begingroup$ Just curious, where is this elliptic curve coming from? $\endgroup$ – Álvaro Lozano-Robledo Apr 28 '12 at 14:34
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    $\begingroup$ Thanks so much, Alvaro, for your answer below. The curve comes from the assumption that the octic $z^8-z^7+(x/2)z^2+(x/2) = 0$ (1) is solvable if x is a rational point of the given elliptic curve (which is its square-free discriminant). The point $x = 58$ is solvable, but the new point is not, though if there is another one, I believe it must fall in the curve. Incidentally (1) at x = 58 is solvable either by a 7th degree equation, or by the 29th root of unity. $\endgroup$ – Tito Piezas III Apr 28 '12 at 16:44
  • $\begingroup$ If I may clarify, the assumption is it is a necessary, but not sufficient, condition that if that octic is solvable, then x is a rational point on the curve. $\endgroup$ – Tito Piezas III Apr 28 '12 at 17:44
  • $\begingroup$ Nice! Thanks for the background. $\endgroup$ – Álvaro Lozano-Robledo Apr 28 '12 at 18:24

The answer to your question is no, there are no other points with numerator or denominator of $5$ digits or less.

If you have access to a Linux system, you may want to try Michael Stoll's "ratpoints", which can be found here. The documentation for ratpoints, and a description of the algorithm is here. This program tries to find all rational points within a given height bound on a hyperelliptic curve in the most efficient way possible.

Otherwise, ratpoints can be accessed through Sage. In the Sage command line, type

from sage.libs.ratpoints import ratpoints


ratpoints([$a_0$,$a_1$,$a_2$,...,$a_n$], H, verbose=False, max=0)

finds all the rational points on $y^2=a_0+a_1x+\cdots+a_nx^n$, where H is the bound for the denominator and the absolute value of the numerator of the x-coordinate. When I type,


the answer is

[(1, 0, 0), (58, 49109, 1), (58, -49109, 1), (5170922, 1182208159673289, 344763), (5170922, -1182208159673289, 344763)]

meaning that after $\pm P =(58,\pm 49109)$ the next point with lowest height is

$$\pm Q =\left(\frac{5170922}{344763} , \pm \frac{387482189339}{38958219}\right),$$

and there are no other points (other than $\pm P$ or $\pm Q$) such that the $x$ coordinate has a numerator (in absolute value) or denominator less than $10^7$.

  • $\begingroup$ Thanks again, Alvaro. (For those interested where this curve came from, kindly see the comments in the original post.) $\endgroup$ – Tito Piezas III Apr 28 '12 at 17:08

Another approach to find all solutions to this elliptic curve would be to determine the full Mordell-Weil group of rational points, using a $2$-descent.

First, one would have to find a minimal model in Weierstrass form for your curve. This is easy, here is one:

$$E : y^2 = x^3 + x^2 + 80022598784x + 49433175273149108$$

But the conductor of this curve is big:

$$N_E =3020151629712 = 2^4\cdot 3 \cdot 7 \cdot 3947 \cdot 2277311,$$

and the $2$-descent is a very lengthy process. I had my computer running for a while, trying to calculate the rank (or just some bounds for the rank), but it didn't get anywhere. Maybe more patient users can let their computer run and see if they can find generators for the group of rational points of this elliptic curve.


There is John Cremona's mwrank. A more friendly implementation comes with SAGE. Expect a steep learning curve.

  • $\begingroup$ Thanks, Julian, for addressing my postscript. I've vaguely heard of Sage. (I'm more familiar with Mathematica.) But I believe my primary question has not yet been answered conclusively. $\endgroup$ – Tito Piezas III Apr 27 '12 at 17:54

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