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John R Ramsden
John R Ramsden
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Can anything be said in general about the rank etc over $\mathbb{Q}$ of the family of Weierstrass equations (in slightly non-standard form) $x (x^2 - 1) = c (c^2 - 1) y^2$ for various given rational values of $c$ ?

  I have a good reason for asking, so this isn't idle curiosity. In fact the values of $c$ which interest me are themselves solutions of $(c^2 - 1)^2 - 3 = d^2$, the first few of these being $c = 3/4, 45/52, 127/96, ...$

So naturallyNaturally, it would be simplest if the Weierstrass equation has the same behaviour for each of these values of $c$, or least with a manageable (finite) amount of variation. Obviously there is always a solution $x, |y| = c, 1$; but that might be a trivial solution of a rank 0 case.

Can anything be said in general about the rank etc over $\mathbb{Q}$ of the family of Weierstrass equations (in slightly non-standard form) $x (x^2 - 1) = c (c^2 - 1) y^2$ for various given rational values of $c$ ?

  I have a good reason for asking, so this isn't idle curiosity. In fact the values of $c$ which interest me are themselves solutions of $(c^2 - 1)^2 - 3 = d^2$, the first few of these being $c = 3/4, 45/52, 127/96, ...$

So naturally, it would be simplest if the Weierstrass equation has the same behaviour for each of these values of $c$, or least with a manageable (finite) amount of variation. Obviously there is always a solution $x, |y| = c, 1$; but that might be a trivial solution of a rank 0 case.

Can anything be said in general about the rank etc over $\mathbb{Q}$ of the family of Weierstrass equations (in slightly non-standard form) $x (x^2 - 1) = c (c^2 - 1) y^2$ for various given rational values of $c$ ? I have a good reason for asking, so this isn't idle curiosity.

Naturally, it would be simplest if the Weierstrass equation has the same behaviour for each of these values of $c$, or least with a manageable (finite) amount of variation. Obviously there is always a solution $x, |y| = c, 1$; but that might be a trivial solution of a rank 0 case.

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John R Ramsden
John R Ramsden
  • 1.5k
  • 13
  • 20

Rank of $x (x^2 - 1) = c (c^2 - 1) y^2 $ over $\mathbb{Q}$ for given rational values of $c$

Can anything be said in general about the rank etc over $\mathbb{Q}$ of the family of Weierstrass equations (in slightly non-standard form) $x (x^2 - 1) = c (c^2 - 1) y^2$ for various given rational values of $c$ ?

I have a good reason for asking, so this isn't idle curiosity. In fact the values of $c$ which interest me are themselves solutions of $(c^2 - 1)^2 - 3 = d^2$, the first few of these being $c = 3/4, 45/52, 127/96, ...$

So naturally, it would be simplest if the Weierstrass equation has the same behaviour for each of these values of $c$, or least with a manageable (finite) amount of variation. Obviously there is always a solution $x, |y| = c, 1$; but that might be a trivial solution of a rank 0 case.