Oracle finding all integral points on genus 0 curves is a factoring oracle (e.g. $xy=n$ and $x^2-y^2=n$


I asked [Can the number of solutions $xy(x−y−1)=n$ for x,y,n∈Z be unbounded as n varies?](https://mathoverflow.net/questions/50479/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-z-be-unbounded-as-n-var) and occurred to me that an oracle giving all integral points may find nontrivial factor of $n$. Drama is this will not work for all $n$.


Would an oracle for finding all integral points on genus 1 curves (in whatever model) be:


1. (loosely defined) Weak factoring oracle which finds at least one nontrivial factor
2. Strong factoring oracle which finds all prime factors?


The factoring oracle must work for all integers if it exists.


(EDIT): Intuitively if I had genus 0 oracle for integral points I could factor general integers. If the oracle were for genus 1 I don't see a way for general integers but I would be lucky with integers of the form $xy(x-y-1)$ (just an example)



Would an oracle for integral points on elliptic curves be a factoring oracle?
An integral point (actually, a rational point in the affine plane will do) on an elliptic curve $y^2 = x(x^2 + ax + b)$ comes (by the standard technique of simple 2-descent) from a rational point on some quartic 
$$ N^2 = b\_1M^4 + aM^2e^2 + b\_2e^4, $$
where $b\_1b\_2 = b$. Thus if you want to factor an integer $N$, ask the oracle for (rational) points on the curve $y^2 = x(x^2 + ax + N)$ for sufficiently many values of $a$ until you find a point that gives you a nontrivial factorization $N = b\_1b\_2$. If you choose $a$ in such a way that the parity conjecture predicts an odd rank, you will know in advance that such a point exists. 


A similar technique works for the Pell equation and shows that solving the Pell equation 
$T^2 - dU^2 = 1$ is at least as hard as factoring $d$.


