Obtaining elliptic curve solution in integers from solution in quadratic field
$begingroup$
Are there any methods or known tricks to obtain elliptic curve solutions in the integers from a solution in a quadratic field?
Starting with a Mordell curve:
$$y^2 = x^3 + k$$
Consider an integer $p$ such that $k-p = -c^3$. This let's us write:
$$(y+sqrt{p})(y-sqrt{p}) = (x-c)(x^2 + cx + c^2)$$
Since a number has roughly 1/log(n) chance of being a prime, in practice it isn't hard to search possible values of $c^3$ til we get $p$ a prime. And because the class number of $mathbb{Q}(sqrt{p})$ is often 1 for prime p, we can even search until we find one that gives us a unique factorization domain in case that helps.
So now we have this nice factorization in a unique factorization domain, giving us a point on the curve in $mathbb{Q}(sqrt{p})$. And since it is on an elliptic curve, we can find new solutions by adding together old solutions.
Is there a way to leverage any of this to obtain solutions in $mathbb{Q}$ and hopefully even $mathbb{Z}$?
number-theory elliptic-curves mordell-curves
asked Jan 10 at 16:14
PineDoorsPineDoors
335
$endgroup$
add a comment |
$begingroup$
Are there any methods or known tricks to obtain elliptic curve solutions in the integers from a solution in a quadratic field?
Starting with a Mordell curve:
$$y^2 = x^3 + k$$
Consider an integer $p$ such that $k-p = -c^3$. This let's us write:
$$(y+sqrt{p})(y-sqrt{p}) = (x-c)(x^2 + cx + c^2)$$
Since a number has roughly 1/log(n) chance of being a prime, in practice it isn't hard to search possible values of $c^3$ til we get $p$ a prime. And because the class number of $mathbb{Q}(sqrt{p})$ is often 1 for prime p, we can even search until we find one that gives us a unique factorization domain in case that helps.
So now we have this nice factorization in a unique factorization domain, giving us a point on the curve in $mathbb{Q}(sqrt{p})$. And since it is on an elliptic curve, we can find new solutions by adding together old solutions.
Is there a way to leverage any of this to obtain solutions in $mathbb{Q}$ and hopefully even $mathbb{Z}$?
number-theory elliptic-curves mordell-curves
asked Jan 10 at 16:14
PineDoorsPineDoors
335
$endgroup$
add a comment |
$begingroup$
Are there any methods or known tricks to obtain elliptic curve solutions in the integers from a solution in a quadratic field?
Starting with a Mordell curve:
$$y^2 = x^3 + k$$
Consider an integer $p$ such that $k-p = -c^3$. This let's us write:
$$(y+sqrt{p})(y-sqrt{p}) = (x-c)(x^2 + cx + c^2)$$
Since a number has roughly 1/log(n) chance of being a prime, in practice it isn't hard to search possible values of $c^3$ til we get $p$ a prime. And because the class number of $mathbb{Q}(sqrt{p})$ is often 1 for prime p, we can even search until we find one that gives us a unique factorization domain in case that helps.
So now we have this nice factorization in a unique factorization domain, giving us a point on the curve in $mathbb{Q}(sqrt{p})$. And since it is on an elliptic curve, we can find new solutions by adding together old solutions.
Is there a way to leverage any of this to obtain solutions in $mathbb{Q}$ and hopefully even $mathbb{Z}$?
number-theory elliptic-curves mordell-curves
asked Jan 10 at 16:14
PineDoorsPineDoors
335
$endgroup$
Are there any methods or known tricks to obtain elliptic curve solutions in the integers from a solution in a quadratic field?
Starting with a Mordell curve:
$$y^2 = x^3 + k$$
Consider an integer $p$ such that $k-p = -c^3$. This let's us write:
$$(y+sqrt{p})(y-sqrt{p}) = (x-c)(x^2 + cx + c^2)$$
Since a number has roughly 1/log(n) chance of being a prime, in practice it isn't hard to search possible values of $c^3$ til we get $p$ a prime. And because the class number of $mathbb{Q}(sqrt{p})$ is often 1 for prime p, we can even search until we find one that gives us a unique factorization domain in case that helps.
So now we have this nice factorization in a unique factorization domain, giving us a point on the curve in $mathbb{Q}(sqrt{p})$. And since it is on an elliptic curve, we can find new solutions by adding together old solutions.
Is there a way to leverage any of this to obtain solutions in $mathbb{Q}$ and hopefully even $mathbb{Z}$?
number-theory elliptic-curves mordell-curves
number-theory elliptic-curves mordell-curves
asked Jan 10 at 16:14
PineDoorsPineDoors
335
asked Jan 10 at 16:14
PineDoorsPineDoors
335
asked Jan 10 at 16:14
PineDoorsPineDoors
335
asked Jan 10 at 16:14
PineDoorsPineDoors
335
asked Jan 10 at 16:14
PineDoorsPineDoors
335
335
add a comment |
add a comment |
1 Answer
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Maybe there are other ways to proceed, but at least the naive approach of adding the initial point hoping to find a rational point will not work.
The initial point found will be linearly independent of any rational point. So changing to the quadratic field to get an initial point just increases the rank, or adds to the torsion group.
This can be seen by writing out the addition formulas. (I started with the explicit addition formulas from here: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html )
I am unsure of the terminology for the parts of a value in a real quadratic number field, so for terminology convenience and without loss of generality let's consider an imaginary quadratic field.
Our initial point has $y$ purely imaginary, and $x$ real valued. The group structure on the Mordell curve preserves this, so this will never lead to a rational point.
Following the doubling algorithm, simplified for a Mordell curve, is:
$$
begin{eqnarray}
lambda &=& frac{3x_1^2}{2y_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - 2 x_1 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
And the addition formula:
$$
begin{eqnarray}
lambda &=& frac{y_2 - y_1}{x_2-x_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - x_1 - x_2 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
answered Jan 12 at 8:22
PineDoorsPineDoors
335
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Maybe there are other ways to proceed, but at least the naive approach of adding the initial point hoping to find a rational point will not work.
The initial point found will be linearly independent of any rational point. So changing to the quadratic field to get an initial point just increases the rank, or adds to the torsion group.
This can be seen by writing out the addition formulas. (I started with the explicit addition formulas from here: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html )
I am unsure of the terminology for the parts of a value in a real quadratic number field, so for terminology convenience and without loss of generality let's consider an imaginary quadratic field.
Our initial point has $y$ purely imaginary, and $x$ real valued. The group structure on the Mordell curve preserves this, so this will never lead to a rational point.
Following the doubling algorithm, simplified for a Mordell curve, is:
$$
begin{eqnarray}
lambda &=& frac{3x_1^2}{2y_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - 2 x_1 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
And the addition formula:
$$
begin{eqnarray}
lambda &=& frac{y_2 - y_1}{x_2-x_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - x_1 - x_2 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
answered Jan 12 at 8:22
PineDoorsPineDoors
335
$endgroup$
add a comment |
$begingroup$
Maybe there are other ways to proceed, but at least the naive approach of adding the initial point hoping to find a rational point will not work.
The initial point found will be linearly independent of any rational point. So changing to the quadratic field to get an initial point just increases the rank, or adds to the torsion group.
This can be seen by writing out the addition formulas. (I started with the explicit addition formulas from here: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html )
I am unsure of the terminology for the parts of a value in a real quadratic number field, so for terminology convenience and without loss of generality let's consider an imaginary quadratic field.
Our initial point has $y$ purely imaginary, and $x$ real valued. The group structure on the Mordell curve preserves this, so this will never lead to a rational point.
Following the doubling algorithm, simplified for a Mordell curve, is:
$$
begin{eqnarray}
lambda &=& frac{3x_1^2}{2y_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - 2 x_1 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
And the addition formula:
$$
begin{eqnarray}
lambda &=& frac{y_2 - y_1}{x_2-x_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - x_1 - x_2 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
answered Jan 12 at 8:22
PineDoorsPineDoors
335
$endgroup$
add a comment |
$begingroup$
Maybe there are other ways to proceed, but at least the naive approach of adding the initial point hoping to find a rational point will not work.
The initial point found will be linearly independent of any rational point. So changing to the quadratic field to get an initial point just increases the rank, or adds to the torsion group.
This can be seen by writing out the addition formulas. (I started with the explicit addition formulas from here: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html )
I am unsure of the terminology for the parts of a value in a real quadratic number field, so for terminology convenience and without loss of generality let's consider an imaginary quadratic field.
Our initial point has $y$ purely imaginary, and $x$ real valued. The group structure on the Mordell curve preserves this, so this will never lead to a rational point.
Following the doubling algorithm, simplified for a Mordell curve, is:
$$
begin{eqnarray}
lambda &=& frac{3x_1^2}{2y_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - 2 x_1 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
And the addition formula:
$$
begin{eqnarray}
lambda &=& frac{y_2 - y_1}{x_2-x_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - x_1 - x_2 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
answered Jan 12 at 8:22
PineDoorsPineDoors
335
$endgroup$
Maybe there are other ways to proceed, but at least the naive approach of adding the initial point hoping to find a rational point will not work.
The initial point found will be linearly independent of any rational point. So changing to the quadratic field to get an initial point just increases the rank, or adds to the torsion group.
This can be seen by writing out the addition formulas. (I started with the explicit addition formulas from here: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html )
I am unsure of the terminology for the parts of a value in a real quadratic number field, so for terminology convenience and without loss of generality let's consider an imaginary quadratic field.
Our initial point has $y$ purely imaginary, and $x$ real valued. The group structure on the Mordell curve preserves this, so this will never lead to a rational point.
Following the doubling algorithm, simplified for a Mordell curve, is:
$$
begin{eqnarray}
lambda &=& frac{3x_1^2}{2y_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - 2 x_1 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
And the addition formula:
$$
begin{eqnarray}
lambda &=& frac{y_2 - y_1}{x_2-x_1} & text{(purely imaginary)}\
x_3 &=& lambda^2 - x_1 - x_2 & text{(real)} \
y_3 &=& - lambda x_3 + lambda x_1 - y_1 quad & text{(purely imaginary)} \
end{eqnarray}
$$
answered Jan 12 at 8:22
PineDoorsPineDoors
335
edited Jan 12 at 8:29
answered Jan 12 at 8:22
PineDoorsPineDoors
335
answered Jan 12 at 8:22
PineDoorsPineDoors
335
answered Jan 12 at 8:22
PineDoorsPineDoors
335
335
add a comment |
add a comment |
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