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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.

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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    $\begingroup$ That's the general quadratic twist $Dy^2 = x^3 - x$ of the "congruent-number curve" $y^2 = x^3 - x$, subject to the condition that it have a point of infinite order (the point cannot be torsion for any rational $c \neq 0, \pm 1$). The behavior of the rank is a well-known open question. The current record is $7$; Nick Rogers found the first example some years ago. $\endgroup$
    – Noam D. Elkies
    Commented Jun 18, 2012 at 2:34

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First you use $c$ as a parameter, i.e., consider your equation as elliptic curve over $\mathbb{Q}(c)$. You can also consider this equation as an equation of an elliptic surface $S$. Now one easily proves that $S$ is a K3 surface, and that is the quotient of $E\times E$ by a group of order 2, with $E$ the elliptic curve $y^2=x^3-x$. From this it follows that the Picard number of $S$ is 20. Using the Shioda-Tate formula it follows that the $\overline{\mathbb{Q}}(c)$ rank of $y^2c(c^2-1)=x(x^2-1)$ is two. Since this elliptic surface has 4 $I_0^*$ fibers, the torsion group has to be a subgroup of $(\mathbb{Z}/2\mathbb{Z})^2$, therefore $(x,y)=(c,1)$ is a point of infinite order. (Of course you could prove this by showing that $2(c,1)$, $4(c,1)$, $6(c,1)$ and $8(c,1)$ are nonzero.)

Note that $(x,y)=(-c,\sqrt{-1})$ is also a point of infinite order, and that this point is not in the subgroup generated by $(c,1)$. This implies $E(\mathbb{Q}(c))$ is generated by the four two-torsion points and $(c,1)$. So the point $(c,1)$ is of infinite order and this is the only point you get for free.

For special values of $c$ the rank can be higher. If you were able to control in the way you ask in your question then this would be very great, because it means more or less that you are able to control the rank of elliptic curves under quadratic twisting.

However, there is a weaker results, using 2-descent you can get a bound on the rank depending on the number of primes dividing $2c(c^2-1)$, see e.g. Section X.5 of Silverman's book on the Arithmetic of Elliptic Curves.

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    $\begingroup$ $c$ is rational but not necessarily integral, so it's not just primes dividing $c(c^2-1$). Instead write $c = m/n$ and $|mn(m^2-n^2)| = r^2 d$ with $d \in \bf Z$ squarefree, and ask for the number of prime factors in $2d$. If I remember right the actual bound is something like the number of prime factors plus the number of prime factors congruent to $1 \bmod 4$. $\endgroup$
    – Noam D. Elkies
    Commented Jun 18, 2012 at 5:17
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Not an answer, just an image of Weierstrass curves for several values of $c \in (1,3]$:
       Weierstrass curves
(I was curious to see how the curve varies with $c$.)

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  • $\begingroup$ @Joseph: Which software did you use to draw these curves? $\endgroup$
    – Srilakshmi
    Commented Jun 18, 2012 at 5:28
  • $\begingroup$ @Srilakshmi: Mathematica + Adobe Illustrator. $\endgroup$
    – Joseph O'Rourke
    Commented Jun 18, 2012 at 12:16

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