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Let's say I wanted to find all the rational solutions to $$-2(x^2 + 4x + 3)(x^2 - 2x - 7) = y^2$$

Clearly this is a genus 1 curve with a rational point, so it is therefore an elliptic curve. I tried attempting this, and I found a page in a book of Mordell.

Mordell outlines a way to convert equations of the form $y^2 = f(x)$ where $deg(f) = 4$ but many of the initial assumption he makes, such as the leading coefficient and the constant term are perfect square, I don't understand.

Can someone help me with converting this to a elliptic curve and with the overall problem of finding rational points on this curve?

Thanks in advance.

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Ravikanth Athipatla
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  • That's essentially what I want... to write the equation of all rational points that would be a solution. I am more concerned about finding the elliptic cubic this is birationally equivalent to. – Ravikanth Athipatla Sep 27 '24 at 22:20
  • Why do you say Clearly this is a genus $1$ curve? The graph of the affine curve appears with none singular points so the genus could not be equal to $1$ – Ataulfo Sep 27 '24 at 23:40
  • A plane curve, whatever its degree be, is birationally equivalent to an elliptic cubic if its genus is equal to one. I am not sure your quartic has genus $1$. If its genus is greater than $1$ then the curve has only a finite number of rational points (Mordell's theorem) Unfortunately "finite" can be "very big". – Ataulfo Sep 28 '24 at 01:19
  • There are many posts on transforming $y^2 = $ a quartic into standard form: 1, 2, 3 (another question of yours), 4, 5, 6. Do any of these help? – Viktor Vaughn Sep 28 '24 at 04:09
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    Using Magma, I found that your curve is isomorphic to the elliptic curve $E: y^2 = x^3 - 8*x$. This has only $2$ rational points, namely $\infty$ and $(0,0)$, which correspond to the obvious rational points $(-1,0), (-3,0)$ of your original equation. – Viktor Vaughn Sep 28 '24 at 04:10
  • @Piquito the projective closure has a singular point at infinity. Also, the result you quote is not Mordell's theorem, it is the Mordell conjecture = Faltings' theorem – Mummy the turkey Sep 28 '24 at 21:02
  • @ViktorVaughn how did Magma help you find the isomorphism? That elliptic curve is exactly what I wanted. – Ravikanth Athipatla Sep 28 '24 at 21:13
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    I just used the command EllipticCurve as described here. Here are all the commands I used: R<x,y> := PolynomialRing(Rationals(),2); f := y^2 - (-2*(x^2 + 4*x + 3)*(x^2 - 2*x - 7)); C := Curve(Spec(R),f); Cbar := ProjectiveClosure(C); E := EllipticCurve(Cbar);. – Viktor Vaughn Sep 29 '24 at 02:16

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