[Cryptography] Fun crypto puzzles, one easy, two impossible

[Bill Cox]

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
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