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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 $m_k > 1$?

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    $\begingroup$ "... 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
    Commented Mar 19, 2016 at 23:23
  • $\begingroup$ @JoeSilverman: Wording changed as suggested. Thanks. $\endgroup$
    – Tito Piezas III
    Commented Mar 19, 2016 at 23:30
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    $\begingroup$ "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
    Commented Mar 20, 2016 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
    Commented Mar 20, 2016 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
    Commented Mar 20, 2016 at 9:47

1 Answer 1

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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).

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  • $\begingroup$ 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
    Commented Mar 20, 2016 at 16:06
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    $\begingroup$ @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
    Commented Mar 20, 2016 at 16:29
  • $\begingroup$ This related question might be of interest. $\endgroup$
    – Tito Piezas III
    Commented Oct 15, 2016 at 2:43

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