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 ...



Groups of Rational Points on Gaussian Circles
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:


$$ x,y=(- i R+ i t^2)/(2t),(R+t^2)/(2 t)$$


