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point tripling over characteristics three field not getting proper result

asked 2019-08-01 03:43:49 -0600

santoshi gravatar image

updated 2019-08-01 06:40:59 -0600

FrédéricC gravatar image

import time p=(3^151)

F1.<x>=GF(3)[]
F.<a>=GF(3^151,'a',modulus=x^151+2*x^2+1);F.modulus();F

A=1;A
B=(0x1fc4865afe00a9216b0b5fd32c6300c4bed0707ae4072a03e55299f157b);B
E = EllipticCurve(F, (0,A,0,0,B));E
P = E.random_point();#P
Q = E.random_point();#Q
print "\n point1 =",P# P.xy()
print "\n point2 =",Q #Q.xy()
x1 = P[0]
y1 = P[1]
x2 = Q[0]
y2 = Q[1]

#x31=((y1^6)*(A^2*(x1^3+B)^2)^(-1))
#x32=((A*x1^3)*((x1+B)^(-1)))
#X3=(x31-x32);X3
#y31=((y1^9)*((A^3*(x1^3+B)^3))^(-1));y31 
#y32=((y1^3)*((x1^3+B)^(-1)));y32
#Y3=(y31-y32)

X3=F(((x1^3+B)^3-(B*x1^3))*((x1+B)^2)^(-1));X3

Y3=F(((y1^9)-((y1^3)*(x1^3+B)^2)*((x1+B)^3)^(-1)));Y3

R=E(X3,Y3);R
#print "\n H = P + Q has the components:\n

after executing the code the error is point are not on the curve. I have written the equation for x3 and y3 by different ways.

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Please edit your post to format the code: select the code and press the 101010 button. Also, what is the source of these formulae?

rburing gravatar imagerburing ( 2019-08-01 04:55:27 -0600 )edit

Paper: Efficient Arithmetic on Elliptic Curves over Fields of Characteristics three

santoshi gravatar imagesantoshi ( 2019-08-01 05:39:15 -0600 )edit

You can vote on answers using the arrows on the left of them, and accept an answer by clicking the checkmark below the arrows; please do this (in general).

rburing gravatar imagerburing ( 2019-08-04 02:08:16 -0600 )edit

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answered 2019-08-01 09:23:31 -0600

rburing gravatar image

First of all the paper is wrong: the denominator should have $x^3$ instead of $x$. Secondly the parentheses are not placed correctly in your second expression. The following seems to be right:

X3=((x1^3+B)^3-B*x1^3)*(x1^3+B)^(-2)
Y3=(y1^9-y1^3*(x1^3+B)^2)*(x1^3+B)^(-3)
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Asked: 2019-08-01 03:43:49 -0600

Seen: 31 times

Last updated: Aug 01