Groups of Rational Points on Gaussian Circles Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R, (x,y) \in \mathbb{R}^2$$, where $R$ is the norm of a gaussian integer ($R=a^2+b^2, (a,b) \in \mathbb{Z}^2$). IF $R$ contains $n$ prime factors, it is not hard to show that $C_R$ contains $2^{n+2}$ integer points organized in $2^n$ classes of 4 elements. For example, with $R=65=5\times 13$ we have the only 4 integer points classes: $(\pm4,\pm7), (\pm8, \pm1), (\pm7,\pm4)$ and $(\pm1, \pm8)$.
We define the "non-abelian" addition law $\times$ on $C_R$ as follows:
$$(x_0, y_0) \times (x_1, y_1) = \frac{1}{R}\left(x_0\left(x_1^2-y_1^2\right)-2y_0x_1y_1, 2x_0x_1y_1+y_0 \left(x_1^2-y_1^2\right)\right),$$
the set of all rational points on $C_R$ forms a semigroup $G_R$, and if $P$ is an integer point on $C_R$ the set generated by repeated addition is dense on $C_R$.
Suppose now that $R$ factorization is unknown. Is it possible to find any non-trivial rational point on $C_R$? 
I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on $C_R$, we have an infinity to choose from ...
 A: Well, there are some easy points on $C_R$. The point $((1+R)/2, (1-R)/(2i))$, for example. 
More generally, the equation $x^2+y^2=R$ is equivalent to $(x+iy)(x-iy) = R$. So finding a rational point on $C_R$ is exactly equivalent to finding $u$ and $v$ in $\mathbb{Q}(i)$ with $uv=R$. You can do that by choosing $u$ at random, and computing $v=R/u$, $x=(u+v)/2$ and $y=(u-v)/(2i)$. I took $u=1$ above.
But, as you say, if we had any way to find nontrivial solutions with $u$ and $v$ Gaussian integers, then we would have a way to factor Gaussian integers, and that is believed to be difficult.
A: Perhaps the situation becomes clearer by looking at a similar problem. Replacing the circle by a hyperbola, we can look at rational points on $H: XY = R$. This conic has the obvious point $N = (1,R)$, and among the many rational points on $H$ there are a few integral points,
each corresponding to a factorization of $R$. You can define a group law on $H$ with neutral element, but this does not help at all at finding "nontrivial" rational points (that is, integral points). Neither does it seem to help to replace the rationals by an algebraic number field. In fact the geometric picture does not seem to add to our understanding in this case (on this elementary level, at least).
A: Answering mainly about "any nontrivial rational point".
Don't think factoring becomes easy even if an oracle gave you one integral point.
For some numbers such oracle come for free, e.g. for $R=u^2+1$ the point is $(u,1)$. 
Fermat numbers are of such form and they are hard to factor.
Since your problem is a genus 0 curve over $\mathbb{Q}(i)$ you can parametrize all rational points:
$$  x,y=(- i R+ i t^2)/(2t),(R+t^2)/(2 t)$$
