Re: Elliptic Curve Y**2 = x**3 + a * x**2 + b

[Justin Card <Wyntermute@worldnet.att.net>]

Tom Rollins wrote:
> 
> Hello all,
> 
> I have a math question concerning implementation of elliptic
> curve systems. In coding some elliptic curve source, I
> need to pick a random point on the following elliptic
> curve in field F_p where p is a prime number.
> 
>        Y**2 = x**3 + a * x**2 + b
>        where 4a**3 + 27b**2 is not equal to 0 mod p
> 
> In selecting a random point, I pick a random value for
> x in the range 0 < x < p, compute the right hand side
> of the equation and find myself needing to take the
> square root for the two solutions.

I can't remember the elliptic curve system well, but if the parameters
of the curve are not standard for everyone (which I am afraid they are)
one method is to pick the point first, then solve for the a & b.

If this is not the case, finding the square root may be nice or tricky.

if p=3 mod 4, then the sqrt is
X^(P+1) mod P, where X is the number you are trying to find the sqrt
of.  It can be extended to X=5(mod 8) and a few others, but I'm not sure
how.  There is also a form for X=1 mod 4,but I can't find reference to
it. Hope this helps

-- 
  Wyntermute   
     -----BEGIN GEEK CODE BLOCK-----
     Version: 3.1
     GE d@ s++:+ a? C++++ UL++ P+ L++ E W+++ N+++ o? K--? w !O M-- V?
PS+++
     PE++ Y+ PGP++ t+++ !5 X+++ R++* tv++ b+++ DI++ D++ G++ e h r- !y 
     ------END GEEK CODE BLOCK------