2. Działania na krzywej eliptycznej
Zad.1
Y2+a1XY+a3Y-(X3+a2x2+a4X+a6)=0
y = Dx + �
l:
E(x,Dx + �) = 0 = -(x - x1)(x - x2)(x - x3)
(Dx + �)2 + a1x(Dx + �) - a2x2 (x1 + x2 + x3) - x2
x2(D2 + a1D - a2)
x3 = D2 + a1D - a2 - x1 - x2
y = Dx3 + � = Dx3 + y1 - Dx1 = D(x3 - x1) + y1
- y3 = D(x3 - x1) + y1 + a1x3 + a3
y3 = -D(x3 - x1) - y1 - a1x3 - a3
y2 - y1
D = dla _ P `" Q
x2 - x1
3x2 - a1y + 2a2x + a4
D = dla _ P = Q
2y + a1 + a3
zatem 1) E:y2=x3-x mod 5
p+Q& .=(3,3)
Zad.2
Oblicz 2*(P+Q)
3x2 - a1y + 2a2x + a4 3x2 -1 27 -1 26 1
D = = = = = = 1
2y + a1 + a3 2y 6 6 1
x3 = D2 + a1D - a2 - x1 - x1 = 12 - 2 *3 = 1- 6 = -5 = 0
y3 = D(x1 - x3) - y1 - a1x3 - a3 = 1(3 - 0) - 3 = 0
2(P+Q)=(0,0)
Ł,(0,0), (1.0), (2.1), (3,3), (2,4)
Inny sposób P=(1,0) Q=(2,1) P+Q=(3,3) 2*(P+Q)=(0,0)
Znajdz
generator grupy:
(0,0)
nie dają się mno\
yć 2*(0,0)= Ł
(1,0)
(2,1)+(2,4)= Ł
P=(2,1) 2*P
12 -1 11 1
D = = = (mod 5) = 1 = 2-1(mod 5) = 3
2 2 2
x3=9+4=0
y3=3*(2-0)-1=0
P=(2,4) 2*P
12 -1 11
D = = = 1*3-1 = 2
9 3
x3=4-4=0
y3=2*(2-0)-4=0