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We have to find an elliptic curve $E:= y^2=x^3+Ax+B $ (A, B are integers) which has points $P, Q$ with rational coordinates and satisfy $P=[n]Q, n>1$. So,

Input: Rational coordinates = $P$.

Output: Integers $A, B, n$ (when $n>1$), and rational coordinates $Q,$ $$OR,$$

$$Return \; 0.$$

What is the best algorithm or what kind approach should be taken to get the output? is there anything in the literature related to the above problem?

Note:

  1. Previously I thought that it is not needed to assume something about the rational point. For example, if rational point $(x,y)=(p/q,r/2)$ where $q$ and $r$ odd, then we consider such $A$ and $B$ that makes $x^3 + Ax + B - y^2 = (p^3+Apq^2+Bq^3)/q^3 - r^2/4$ equal $0$, in this case, $(p^3+Apq^2+Bq^3)/q^3 - r^2/4=0 \implies Apq^2+Bq^3 = p^3-r^2\times q^3/4 \cdots$, but it is not true, in that case, please put necessary and sufficient condition on $P$, if $P$ does not satisfy that we return $0$.

  2. It might not be possible to have such Integers $A, B, n>1, Q$ for a given $P$ satisfying above condition, in that case we return $0.$

  3. I am not considering Torsion Group or points of finite order only, $P, Q$ could be points of infinite order.

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  • $\begingroup$ I'm struggling to understand your question. Fix $P\in E(\mathbb Q)$. Are you asking what percentage of points $Q\in E(\mathbb Q)$ can be written as $[n]P$ for some $n$? If so, what do you main by percentage if $E(\mathbb Q)$ is infinite? Also, surely the answer depends a lot on the choice of $E$ and $P$ – for example if $E(\mathbb Q)\cong \mathbb Z$ and $P$ is chosen to be a generator, the answer will surely be different to if $E(\mathbb Q) \cong \mathbb Z^5$.... $\endgroup$
    – Mathmo123
    May 7 at 19:04
  • $\begingroup$ @Mathmo123 let me put it in an input-output problem, if $P$ is given, can you always provide an $E$ where $Q$ is a multiple of $P$ on $E$ where $Q$ has rational x-y components? I was searching an algorithm but could not find, it made me think it is not possible always. $\endgroup$
    – Consider Non-Trivial Cases
    May 7 at 19:13
  • $\begingroup$ I'm still confused. Does $P$ have rational coordinates? Is $Q$ fixed? $\endgroup$
    – Mathmo123
    May 7 at 23:24
  • $\begingroup$ @Mathmo123 $P$ has rational coordinates (as written in the post), it is given as input, $Q$ is the output, thus it is not known or fixed. Also if the question makes sense to you when we consider points on finite order, do consider that case, discarding the infinite order case, thank you. $\endgroup$
    – Consider Non-Trivial Cases
    May 8 at 17:10
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    $\begingroup$ Your input is a point $P$ with two rational coordinates. Write it with a minimal $d$ as $P(a/d^2, \ b/d^3)$. Then we can find $A,B\in \Bbb Z$ with $y^2=x^3+Ax+B$ if( and only i)f we can do so for:$$\frac{b^2-a^3}{d^6} =A\cdot\frac a{d^2}+B\ .$$This depends on the L.H.S. - if it has - after simplifying it - denominator dividing $d^2$, and so that the resulted diophantine equation of degree one in $A,B$ has solutions - then we can. Else not. If we can solve and find $A,B$, i see no point in allowing an $n$ also to be part of an output, why not simply take $n=1$ and $Q=P$ ?! $\endgroup$
    – dan_fulea
    May 10 at 15:02

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