rational parameterization of quartic

Question

With the curve $x^4 - 6x^2 - y^2 + 1 = 0$ in the range of $x$ inside of $(-1,1)$, I can only identify two rational points $(0,1)$ and $(0,-1)$. Is it possible to determine if there are others?

Answer 1

Let $A=1$; $B=0$; $C=-6$; $D=0$; $q=1$; $Q=2$;

then let $x = (Q(X+C)-D^2/Q)/Y$

$y = -q + x(xX-D)/Q$

$x = 2/YX - 12/Y$

$y = 2X^3/Y^2 - 24X^2/Y^2 + 72X/Y^2 - 1$

then using Ian Conell's quartic to cubic reduction formuli[ref] we obtain an elliptic curve

(1) $y^2 = x^3 - 6x^2 - 4x + 24$

Running mwrank shows that

Regulator = 1

The rank and full Mordell-Weil basis have been determined unconditionally. [[0],[]]

(0.020706 seconds)

Answer: There are no rational points on the curve other than those you already found. There are 2 affine points at infinity [1:1:0] and [1:-1:0] for the quartic, but you probably are not interested in them.

ref: https://ask.sagemath.org/question/50319/i-have-the-hyperelliptic-curve-y2x4814731024x2-1-how-do-i-convert-this-curve-to-an-elliptic-curve-in-a-short-weierstrass-form/

Answer 2

Yes, it is possible.

You curve is an elliptic curve over $\mathbb Q$, written in Jacobi quartic form.

The standard form for an elliptic curve is Weierstrass cubic form.

It is fairly straightforward to convert from one to the other. If you do this with your curve, you will get a Weierstrass cubic whose coefficients are small enough that it will be listed on standard tables, and then you can look it up and find out how many rational points it has.

I can't do this for you at the moment; perhaps someone else will post an answer doing it. It is fairly easy to do in any case, and I would guess that SAGE can make these sorts of conversions.