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August 16th, 2009, 06:40 AM   #1
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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
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August 20th, 2009, 08:54 PM   #2
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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.
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