[Cryptography] Fun crypto puzzles, one easy, two impossible
Bill Cox
waywardgeek at gmail.com
Sat Apr 23 17:54:26 EDT 2016
On Sat, Apr 23, 2016 at 3:30 AM, Bill Cox <waywardgeek at gmail.com> wrote:
> Puzzle 3:
>
> Find functions X(x, y) and Y(x, y) such that:
>
> X(x, y)^2 - Y(x, y)^2 = X((x^2 - y^2)/(2 - x^2 - y^2), 2xy/(x^2 + y^2))
> 2X(x, y)Y(x, y) = Y(x^2 - y^2)/(2 - x^2 - y^2), 2xy/(x^2 + y^2))
>
>
Note that if you restrain this to the circle, it works out:
x^2 + y^2 = 1, X(x, y) = x, Y(x, y) = y
This is a sub-case of an Edwards curve, with d = 0. For d != 0, an
alternate form of the equations is:
X(x, y)^2 - Y(x, y)^2 = X((x^2 - y^2)/(1 - dx^2y^2), 2xy/(1 + dx^2y^2))
2X(x, y)Y(x, y) = Y((x^2 - y^2)/(1 - dx^2y^2), 2xy/(1 + dx^2y^2))
Bill
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://www.metzdowd.com/pipermail/cryptography/attachments/20160423/1f6e3ba9/attachment.html>
More information about the cryptography
mailing list