0
$\begingroup$

I. Elliptic curves

Given integers $a,b,m_k$. Let,

$$x^2+a = m_1u_1^2\\x^2+b = m_1u_2^2\tag1$$

If there is a rational point $x_i$, then the pair (after a transformation) is birationally equivalent to an elliptic curve, call it $E_1$, and frequently has infinitely many rational points. Assume a second $m_2$,

$$x^2+a = m_2v_1^2\\x^2+b = m_2v_2^2\tag2$$

with a different rational point $x_j$, and yielding an elliptic curve $E_2$.

II. Question

Q1: What is the relationship between $m_1,m_2$ given that ALL the rational points on $(1),(2)$ are on the single elliptic curve $E_3$,

$$(x^2+a)(x^2+b) = y^2\tag3$$

Or knowing the generators of $E_3$, can we predict what and how many square-free integer $m_k>1$ are permissible?

III. Example

For simplicity, assume the special case of $b = -a$ which makes $m_1=1$ as the congruent number problem. Let $a=101$. For,

$$(x^2+a)(x^2-a) = y^2\tag4$$

two solutions are $x_1 = \frac{2015242462949760001961}{118171431852779451900}$, and $x_2 =\frac{2125141}{63050}$. As applied to,

$$x^2+101 = m_k u_1^2\\x^2-101 = m_k u_2^2\tag5$$

the point $x_1$ also solves $m_1 = 1$, while $x_2$ also solves $m_2 = 101$. (What other $m_k$ is permissible?)

Q2: In general, if $(5)$ is rationally solvable for one integer $m_k$, does it imply finitely (or infinitely) many other square-free integer $m_k > 1$?

$\endgroup$
  • 2
    $\begingroup$ "... is birationally equivalent to an elliptic curve, call it $E_1$, and generally has infinitely many rational points." Really? Current perceived wisdom seems to be that unless your family has a section, then the probability that a random element in your family has infinitely many rational points (i.e., has positive rank) is 50%. So maybe "frequently" would be a better word than "generally". $\endgroup$ – Joe Silverman Mar 19 '16 at 23:23
  • $\begingroup$ @JoeSilverman: Wording changed as suggested. Thanks. $\endgroup$ – Tito Piezas III Mar 19 '16 at 23:30
  • 1
    $\begingroup$ "the probability . . . is 50%" assuming the signs are $+1$ and $-1$ with equal frequency. That's usually the case (though hard to prove, indeed often beyond current machinery), but there are families with constant sign. $\endgroup$ – Noam D. Elkies Mar 20 '16 at 1:01
  • $\begingroup$ @NoamD.Elkies Indeed there are such families. But since the OP didn't construct his family with an eye toward skewing the FE signs, it's likely that it's a 50-50 family (although, as you note, proving it might be hard or currently impossible). $\endgroup$ – Joe Silverman Mar 20 '16 at 1:43
  • $\begingroup$ The elliptic curve corresponding to (3) is $Y^2 = (X+a+b)(X^2-4ab)$. And possible $m_k$'s correspond to possible values of $X+a+b \bmod \mathbb{Q^*}^2$. So, the number of possible values for $m_k$ is closely related to the rank of $E(\mathbb{Q})$. In particular, there are only finitely many possible $m_k$'s. $\endgroup$ – duje Mar 20 '16 at 9:47
4
$\begingroup$

The possible $m_k$, when assumed to be squarefree, must be divisors of the resultant of $x^2 + a$ and $x^2 + b$, which is $(a-b)^2$; so $m_k \mid a-b$. (Note that bot factors must be in the same square class; if a prime $p$ divides the squarefree representative of this class, then the binary forms $x^2 + a z^2$ and $x^2 + b z^2$ have a common root mod $p$.)

This is the usual argument used in 2-descent on elliptic curves or hyperelliptic Jaobians that one uses to show that the Selmer group is contained in a finite subgroup of the multiplicative group of the relevant etale algebra modulo squares (the algebra here is just ${\mathbb Q} \times {\mathbb Q}$; we are looking at a 2-isogeny).

Put differently, you get a homomorpism $E_3(\mathbb Q) \to {\mathbb Q}^\times/{\mathbb Q}^{\times 2}$ (taking one of the points at infinity as the origin) whose image gives you the $m_k$ for which there are rational points on the covering curves. To figure out exactly which $m_k$ occur may be difficult, since this is essentially equivalent to determining the Mordell-Weil group of $E_3$ (for which no method is known so far that could be shown to work in all cases).

$\endgroup$
  • $\begingroup$ Thanks. I knew the squarefree $m_k$ had to have a relationship. So they divide $a-b$. For the special case $a=-b=p$ and prime $p$, then possible (not actual) $m_k$ would be just $m_k = 1,2,p$ which, in general, answers my Q1 and Q2. $\endgroup$ – Tito Piezas III Mar 20 '16 at 16:06
  • 1
    $\begingroup$ @TitoPiezasIII ... and, a priori, $-1, -2, -p, 2p, -2p$. But of course, $x^2 + p$ is positive, so we can reduce to $1, 2, p, 2p$. By considering $x^4 - p^2 = y^2$ modulo powers of $p$ and powers of $2$, we can rule out $p$ and $2p$ if $p \not\equiv 1 \bmod 4$, and we can rule out $2$ and $2p$ if $p \equiv \pm 3 \bmod 8$ (and there are possibly some more cases). This is basically computing the 2-isogeny Selmer group. $\endgroup$ – Michael Stoll Mar 20 '16 at 16:29
  • $\begingroup$ This related question might be of interest. $\endgroup$ – Tito Piezas III Oct 15 '16 at 2:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.