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> **I. Elliptic curves**
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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** 
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**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?*** 



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> **III. Example**
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For simplicity, assume the special case of $b = -a$ which makes $m\_1=1$ as the *[congruent number problem](http://mathworld.wolfram.com/CongruentNumber.html)*. 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 $m\_k > 1$? 



On elliptic curves, $\sqrt{x^2-101y^2} ,\sqrt{x^2+101y^2}$, and their ilk
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).


