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 August 16th, 2009, 05:40 AM #1 Newbie   Joined: Aug 2009 Posts: 1 Thanks: 0 Modulus Help I am currently in university, busy with an encryption project, more specifically Elliptic Curves over prime integers. In order to double a point on the elliptic curve, one has to use the formula: lambda = [3*(x1)^2 + a] / 2*(y1) mod p, where x1 and y1 is a point on the curve, and 'a' is part of the equation(y^2 = x^3 +a*x+b mod p), and p is the prime modulus. Below is an example I found on the Internet for computing lambda: Location:[color=#0000BF] http://www.site.uottawa.ca/~chouinar/Ha ... C_2002.pdf[/color] on page 4 Point: (3,10) a = 1 p = 23 lambda = [3*(3)^2 + 1] / [2*10] mod 23 = 5 / 20 mod 23 = 0.25 mod 23 = 6 mod 23 The problem I am having is: How do they convert the decimal result(0.25) to an integer result(6). Another example of this can be found at:[color=#0000BF] http://www.certicom.com/index.php/34-quiz-2--solutions[/color] Solution Number 5. Your help is much appreciated. Thank you  August 20th, 2009, 07:54 PM #2 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Modulus Help 0.25 * 4 = 1 = 1 (mod 23) 6 * 4 = 24 = 1 (mod 23) So 0.25 = 6 (mod 23) Since p = 23 is prime, you are doing arithmetic in F_23, a finite field. So there is a unique multiplicative inverse of 4, which is 6. I personally would not write it 0.25, but either 4^-1 or 1/4 for clarity. Tags modulus Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bilano99 Calculus 2 March 15th, 2013 12:48 PM omega Complex Analysis 2 December 17th, 2011 02:51 PM jsmith613 Algebra 3 September 14th, 2011 01:25 PM julian21 Number Theory 1 March 23rd, 2010 08:27 PM musicmental Complex Analysis 1 October 9th, 2007 07:08 AM

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