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11
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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$$C_R$.
Suppose now that $R$ is a gaussian norm, but its factorization is unknown. Is it possible to find any non-trivial rational point on $C_R$?
I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on $C_R$, we have an infinity to choose from ...
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10
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R, (x,y) \in \mathbb{R}^2$$, where $R$ is the norm of a gaussian integer ($R=a^2+b^2, (a,b) \in \mathbb{Z}^2$). IF $R$ contains $n$ prime factors, it is not hard to show that $C_R$ contains $2^{n+1}$ 2^{n+2}$ integer points organized in $2^{n-1}$ 2^n$ classes of 4 elements. For example, with $R=65=5\times 13$ we have the only 2 4 integer points classes: $(\pm4,\pm7)$ (\pm4,\pm7), (\pm8, \pm1), (\pm7,\pm4)$ and $(\pm8, (\pm1, \pm1)$.pm8)$.
If we define the "non-abelian" group law $\times$ on $C_R$ as follows:
$$(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$$
Suppose now that $R$ is a gaussian norm, but its factorization is unknown. Is it possible to find any non-trivial rational point on $C_R$?
I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on $C_R$, we have an infinity to choose from ...
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9
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R, (x,y) \in \mathbb{R}^2$$, where $R$ is the norm of a gaussian integer ($R=a^2+b^2, (a,b) \in \mathbb{Z}^2$). IF $R$ contains $n$ prime factors, it is not hard to show that $C_R$ contains $2^{n+1}$ integer points organized in $2^{n-1}$ classes of 4 elements. For example, with $R=65=5\times 13$ we have the only 2 integer points classes: $(\pm4,\pm7)$ and $(\pm8, \pm1)$.
If we define the "non-abelian" group law $\times$ on $C_R$ as follows:
$$(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$$
Suppose now that $R$ is a gaussian norm, but its factorization is unknown. Is it possible to find any non-trivial rational point on $C_R$?
I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on $C_R$, we have an infinity to choose from ...
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8
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Group Groups of Rational points Points on Gaussian circlesCircles
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7
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R, (x,y) \in \mathbb{R}^2$$, where $R$ is the norm of a gaussian integer . ($R=a^2+b^2, (a,b) \in \mathbb{Z}^2$). IF $R$ contains $n$ prime factors, it is not hard to show that $C_R$ contains $2^{n+1}$ integer points organized in $2^{n-1}$ classes of 4 elements. For example, with $R=65=5\times 13$ we have the only 2 integer points classes: $(\pm4,\pm7)$ and $(\pm8, \pm1)$.
If we define the "non-abelian" group law $\times$ on $C_R$ as follows:
$$(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$$
Suppose now that $R$ is gaussian, but its factorization is unknown. Is it possible to find any non-trivial rational point on $C_R$?
I suspect the answer is no, otherwise we could easily factor large Gaussian integers. On the other hand, since rational points are dense on $C_R$, we have an infinity to choose from ...
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6
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R$$R, (x,y) \in \mathbb{R}^2$$, where $R$ is a gaussian integer. 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 gaussian, 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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5
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R$$, where $R$ is a gaussian integer. 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 gaussian, 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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4
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R$$, where $R$ is a gaussian integer. 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 gaussian, but its factorization is unknown. Is it possible to find any 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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3
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R$$, where $R$ is a gaussian integer. 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 "natural" non-abelian" group law $\times$ on $C_R$ as follows:
$$(x_0, y_0) \times (x_1, y_1) = (x_0x_1-y_0y_1, x_0y_1+x_1y_0),$$\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$$
Suppose now that $R$ is gaussian, but its factorization is unknown. Is it possible to find any rational point on $C_R$?
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2
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Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R$$, where $R$ is a gaussian integer. It IF $R$ contains $n$ prime factors, it is not hard to show that $C_R$ contains $2^{n+1}$ integer points organized in $2^{n-1}$ classes of 4 elements. For example, with $R=65=5\times 13$ we have the only 2 integer points classes: $(\pm4,\pm7)$ and $(\pm8, \pm1)$.
If we define the "natural" group law $\times$ on $C_R$ as follows:
$$(x_0, y_0) \times (x_1, y_1) = (x_0x_1-y_0y_1, x_0y_1+x_1y_0)$$
The x_0y_1+x_1y_0),$$
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 gaussian, but its factorization is unknown. Is it possible to find any rational point on $C_R$?
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1
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Group of Rational points on Gaussian circles
Let a gaussian circle $C_R$ be any circle defined by the equation:
$$x^2+y^2 = R$$, where $R$ is a gaussian integer. 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 "natural" group law $\times$ on $C_R$ as follows:
$$(x_0, y_0) \times (x_1, y_1) = (x_0x_1-y_0y_1, x_0y_1+x_1y_0)$$
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 gaussian, but its factorization is unknown. Is it possible to find any rational point on $C_R$?
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