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# Elliptic Curve

Section A

Question No: 1

Elliptic curve defined as

y2=x3+3x+7

Standard elliptic curve equation E:

y2=x3+ Ax+B

Comparing both equation

a)

A= 3

B = 7

m = y1-yx1-x

σ = 3 x2+ A2Y

P = ( 5 , 2)

σ = 3 52+ 32 x 2

σ = 78/4 = 78 x 4-1   mod (7)

σ = 78 x 3.5

σ =273

y2=x3+3x+7

 X Y^2 = a ap-1/2 0 0 -1 1 3 -1 2 4 -1 3 0 1 4 6 1 5 0 – 6 3 –

b)

P+Q

P (5, 2)

Q (8,9)

m = 9-28-5 = 7/3

2.3 mod (7)

y2=x3+3x+7

x3 = y2 -3x – 7

X3= 1

Y= 0

And

(P+Q)+R

R ( 10 , 5) m = 5-110-0

m = 410

m= 0.4

x3 = y2 -3x – 7

X3= 9

Y= 9

c)

Q+ R

Q (8, 9)

R (10, 5)

m = 5-910-8 = -2/2

= -1

x3 = y2 -3x – 7

X3= 8

Y3= 2

(Q+ R)+ P

(Q + R) = (8, 2)

P (5, 2)

m = 2-25-8 = 0/3

x3 = 9

y3  = 9

d)

Find 2P

n= 2

P = (5, 2)

2 (5, 2)

2P =2(5, 2)

2P = (10, 5)

Find 3P

3P = 3(10, 5)

= (30, 15)

Question No: 2

This method includes successive division first of the smaller of given all two numbers into larger and given resulting remainder divided into the divisor until remainder is equal to zero. Here the remainder of previous division is greatest common divisor.

a)

gcd (354 , 93)

Dividing 354 by 93

= 35493

93 x 2 = 279

While 75 is remainder.

Divide 93 by 75

75 x 1 = 75

93/75 = 1

While 18 is remainder

Divide 75 by 18

= 75/18

= 4

While 3 is remainder.

As gcd (354 , 93) = 354 x + 93y

3 = 354 x + 93y ——1)

According to remainder equation 354x/ 93y = 3

354x = 279y ——2)

x = 279/354 y

Putting value of x in given equation

3= 279 y + 93 y

y = 0.0086

According to equation no 1

3 = 354 x + 93 x 0.0086

X= 0.006355

( x, y) = (0.006355, 0.0086)

After successive division process 3 is remainder that is gcd of (354 ,93) by Euclidean algorithm.

b)

x = 2 mod 11

x = 9 mod 17

x = 14 mod 29

All the moduli are relatively prime

m = 11 x 17 x 29

= 5423

(mm1)-1 = (542311)-1

= (493)-1

= 493 mod 11

= 9

(mm1)-1 = (542317)-1

= (317)-1

= (317)-1   mod 17

= 11

(mm1)-1 = (542329)-1

= (187)-1

= (187)-1   mod 29

= 13

x = 9, 11, 13

c)

Multiplicative order of 8 modulo 37

1 = 1  (mod 37)

10 = 10   (mod 37)

102 = 26 (mod 37)

103 = 1 (mod 37)

104 = 10 (mod 37)

105 = 26 (mod 37)

106 = 1 (mod 37)

107 = 10 (mod 37)

Multiplicative order of 8 modules 37 is given as

d)

Smallest positive integer n for which n63 = 2 mod 89

89 is prime number while 63 is also a prime number

n63 = 2 mod 89

e)

Finding Jacobi symbol

(191229)

= (191229)

= (38191)    since 38 = 19 mod (9)

= (19191)

= (19191)  (-1)
= (19191)  (-1)

= (119)  (-1)

= (-1) 