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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 $(\pm1, \pm8)$.

If we

We define the "non-abelian" group 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 group semigroup $G_R$, and if $P$ is an integer point on $C_R$ the group set generated by $P$ repeated addition is a subgroup.

$dense on$<{P}> \subset G_R$$C_R. Suppose now that R is a gaussian norm, but its factorization is unknown. Is it possible to find any non-trivial rational point on C_R? I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on C_R, we have an infinity to choose from ... 10 added 23 characters in body; [made Community Wiki] Let a gaussian circle C_R be any circle defined by the equation:$$x^2+y^2 = R, (x,y) \in \mathbb{R}^2$$, where R is the norm of a gaussian integer (R=a^2+b^2, (a,b) \in \mathbb{Z}^2). IF R contains n prime factors, it is not hard to show that C_R contains 2^{n+1} 2^{n+2} integer points organized in 2^{n-1} 2^n classes of 4 elements. For example, with R=65=5\times 13 we have the only 2 4 integer points classes: (\pm4,\pm7) (\pm4,\pm7), (\pm8, \pm1), (\pm7,\pm4) and (\pm8, (\pm1, \pm1).pm8). If we define the "non-abelian" group law \times on C_R as follows:$$(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),$$the set of all rational points on C_R forms a group G_R, and if P is an integer point on C_R the group generated by P is a subgroup.$$<{P}> \subset G_R$$Suppose now that R is a gaussian norm, but its factorization is unknown. Is it possible to find any non-trivial rational point on C_R? I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on C_R, we have an infinity to choose from ... 9 added 7 characters in body Let a gaussian circle C_R be any circle defined by the equation:$$x^2+y^2 = R, (x,y) \in \mathbb{R}^2$$, where R is the norm of a gaussian integer (R=a^2+b^2, (a,b) \in \mathbb{Z}^2). IF R contains n prime factors, it is not hard to show that C_R contains 2^{n+1} integer points organized in 2^{n-1} classes of 4 elements. For example, with R=65=5\times 13 we have the only 2 integer points classes: (\pm4,\pm7) and (\pm8, \pm1). If we define the "non-abelian" group law \times on C_R as follows:$$(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),$$the set of all rational points on C_R forms a group G_R, and if P is an integer point on C_R the group generated by P is a subgroup.$$<{P}> \subset G_R

Suppose now that $R$ is a gaussian norm, but its 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 ...

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