1. ## 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: http://www.site.uottawa.ca/~chouinar/Handout_CSI4138_ECC_2002.pdf 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: 3.4 QUIZ 2 ~ Solutions Solution Number 5.

Help is much appreciated.

Thank you

2. Originally Posted by Hemi08
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).
Best not to use decimals in calculations like this. Write 5/20 as 1/4. You are working in the field of numbers mod 23, and in that field you want to find the inverse of 4 (most people would actually write it as $4^{-1}\!\!\!\pmod{23}$).

The inverse of a number is the number that you multiply it by to get 1. In a finite number field there are no fractions, just integers (reduced by the modulus). In the field of numbers mod 23, we ignore multiples of 23. So the inverse of 4 is going to be an integer x (between 1 and 22) such that 4x is equal to 1 (plus a multiple of 23). Since 4×6 = 24 = 1 + 23, it follows that 6 is the inverse of 4 in that number field.