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MathGroup Archive 2004

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Re: Elliptic Curves and Cryptography Questions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47635] Re: Elliptic Curves and Cryptography Questions
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Mon, 19 Apr 2004 04:33:05 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <c5tepb$hv0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <c5tepb$hv0$1 at smc.vnet.net>,
 "flip" <flip_alpha at safebunch.com> wrote:

> I am working on a presentation with Elliptic Curves/Cryptography and had a
> few questions I was hoping someone could answer.  I use Mathematica version 
> 4.0 (wish I could afford 5.0).
> 
> The elliptic curve I am using is E: y^2 == x^3 + x + 4 mod 23 (defined over
> F{23}, that is p = 23).
> 
> Questions:
> 
> 1.  There are 29 solution set points ((x, y) pairs plus the point at
> infinity) to this elliptic curve.
> 
> {{0,2},
> {0,21},{1,11},{1,12},{4,7},{4,16},{7,3},{7,20},{8,8},{8,15},{9,11},{9,12},{1
> 0,5},{10,18},{11,9},{11,14},{13,11},{13,12},{14,5},{14,18},{15,6},{15,17},{1
> 7,9},{17,14},{18,9},{18,14},{22,5},{22,19}}
> 
> Manually, one can check the quadratic residues and then determine if y^2 is
> in that set (a cumbersome approach).
> 
> Is there a command in Mathematica to find this solution set?

The last point in your solution set is not correct. Here is the correct 
answer:

  Reduce[y^2 == x^3 + x + 4, {x, y}, Modulus -> 23]

  pts = {x, y} /. {ToRules[%]}

{{0, 2}, {0, 21}, {1, 11}, {1, 12}, {4, 7}, {4, 16}, {7, 3}, {7, 20}, 
{8, 8}, {8, 15}, {9, 11}, {9, 12}, {10, 5},  {10, 18}, {11, 9}, {11, 
14}, {13, 11}, {13, 12}, {14, 5}, {14, 18}, {15, 6}, {15, 17}, {17, 9}, 
{17, 14}, {18, 9},  {18, 14}, {22, 5}, {22, 18}}

Checking the solution is straighfoward:

  f = Function[{x, y}, Mod[y^2 - (x^3 + x + 4), 23] == 0]; 

  f @@@ pts // Union

> ...  text omitted ...
>
> 4.  When I do,
> 
> << Graphics`ImplicitPlot`
> 
> ImplicitPlot[y^2 == x^3+x+4, {x, -2,2}]
> 
> I am wondering if there is a way to show the elliptic curve E and to
> superimpose the points above and their values so the points are visible

Actually, because you want to show a family of curves, ContourPlot is 
probably better:

  ContourPlot[y^2 - ( x^3 + x + 4), {x, 0, 12}, {y, 0, 23},     
   Contours -> 23 Range[-60, 20], ContourShading -> None,     
   Epilog -> {Hue[1], AbsolutePointSize[5], Point /@ pts}];

> and the plot is useable in a powerpoint slide?

Others will tell you how to do this, and it has appeared in this group 
previously: do a google search on powerpoint at

http://groups.google.com/groups?hl=en&lr=&group=comp.soft-sys.math.mathem
atica

Personally, I prefer to use the Mathematica SlideShow mechanism, 
especially if you have a number of Mathematica graphics and equations. 
You can obtain the SlideShow tools for free for Mathematica 4.2 from the 
Wolfram website.

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


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