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I am sorry if this is a basic question but I am pretty new to elliptic curves. More precisely I am trying to understand elliptic divisibility sequence. While I was searching online I came across to:
''Given an elliptic curve $E$ in short Weierstrass form, $E:y^2=x^3+ax+b$ with $a,b \in \mathbb{Z}$, let $P \in E(\mathbb{Q})$ be non-torsion. The shape of the equation forces the expression of the point $P$ to be in the form $P=(\frac{A}{B^2},\frac{C}{B^3})$ where $A,B,C \in \mathbb{Z}$ such that gcd$(AC,B)=1$.
However I don't see why every point of $E$ is of this form. In fact when I tried to place the coordinates of $P$ into $E$ it didn't seem to satisfy the equation and when I found a point of $E$ it wasn't of this form.
Can anyone help me see what I am missing in here?
Thank you in advance

Kate Jns
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  • Please show us your work and give an example of E and P so that we can tell what you are missing. – Somos Apr 17 '21 at 18:16
  • the curve that I am working on is on a general form,e.i. $E:y^2=x^3+ax+b$ and is over $\mathbb{Q}$. So I took an arbitrary point, which would be of the form P(k/l,m/n) with k,l,m,n on the integers (without loss of generality I took l,n to be positive). Then I have $m^2/n^2=k^3/l^3+ak/l+b$ and I don't how to get that $ P(k/l,m/n)$ is actually of the form $P=(A/B^2,C/B^3)$. I am working on something general, hence I don't have an example of $E,P$ since I want to see why in any point of an elliptic curve with short W. eq over $\mathbb{Q}$ is the form $P=(A/B^2,C/B^3)$ – Kate Jns Apr 17 '21 at 18:32
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    Continuing with your equation, you see that $\frac{k^3 + akl^2 + bl^3}{l^3}$ is a square. Since, by assumption, $(l, k) = 1$, the fraction is in reduced form. But the fraction is equal to $(m/n)^2$, so $l^3$ and hence $l$ must be a square - say $l = B^2$. You end up with $A = k$ and $C = \sqrt{k^3 + akl^2 + bl^3}$ which is an integer by the assumption that the point you started with was a point on the curve. – Mathmo123 Apr 17 '21 at 20:30

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