Cryptology Assignment - Cryptography Techniques and Applications

This document provides a comprehensive solution to a cryptology assignment. The assignment covers various aspects of cryptography, including encryption and decryption techniques using cipher text and keys. It explores prime number verification, specifically testing whether a given number is prime. Furthermore, the assignment delves into elliptic curve cryptography, defining curves and calculating points on them. The solution also involves the application of cryptographic principles to verify digital signatures and message authenticity. Overall, the assignment demonstrates a practical understanding of cryptographic concepts and their applications in securing information and communications. The solution includes detailed steps and calculations to arrive at the final answers for each question.

Running head: Cryptology in practices
Cryptology in practices
Name of the Student
Name of the University
Author Note

Cryptology in practices 1
Question 1:
A 0
B 1
C 2
D 3
E 4
F 5
G 6
H 7
I 8
J 9
K 10
L 11
M 12
N 13
O 14
P 15
Q 16
R 17
S 18
T 19
U 20
V 21
W 22
X 23
Y 24
Z 25
a)

2Cryptology in practices
Cipher text: VYEHVCWIQTBGUFKTMWOVQUNB
Key; k = (2, 17, 24, 15, 19, 14)
Plain text: THIS COURSE IS SOMETIMES FUN
b)
Chosen key: image
c)
Plain text: He is a good person
Cipher text: Pqiyeoaojtmdsur
Question 2:
a)
n= 659041
n-1 = 25×3×5×1373
m = 3×5×1373 = 20595
b = 2^20595 mod 659041 = -1
So, the number 659041 is not a prime.
659041 is divisible by 743 and 887. So it is not a prime number.
b)
n = 657412
a = 2, B = 50
b0 = a

3Cryptology in practices
b1 = b0 (mod 657412) = 2 and gcd(b1 – 1,n) = 1
b2 = 4 mod 657412 and gcd (b2 – 1,n) = 1
……………
B4 = 16777216 mod 657412 = 341916
So, n has a factor of p = 341916
q = n/p = 1.92
c)
Plain text:
I will pass this test
Key:
MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCK8dCaRnbI1NWvYdieNDj5kd
AvF/vT6l4ysN/Q+4k1xTXOkz8QnZQDnIPanjRJQP30WrFNaCJfuq/
HBlDDKlcKlspCU8N9tnNzzavaCVU5RDMiNTTVZGh8Lw0Gs+8gwH9UAFwjkFbOZj7Q
V0YtCPSqWa+vDewviDWCefHlqtZD+wIDAQAB
Cipher text:
ASXkYPHLLOzUEZQWZlOH0uSLJUoUmp30Yt8lVVm6m7stYqKTSFqgtWPz+VBjt69KJ
WiPxINNcpbDHqa9tmAPgfe8HH+9r8+zhT95qpcs1tgRa3Ezmf6SUYAE3ExxaZeSiEaWCj8
kKn0ZdOCtyxQvLcfxJLiMmRLq626pK5+HpEc=
Question 3:
a)
Elliptic Curve defined by y^2 = x^3 + 8*x + 3 over GF(23)

4Cryptology in practices
x=0,1,2,…,22
For example with x=0, we get y2=3(mod23)..
Thus, (±7)2≡3(mod23).
Thus (0,7) and(0,16) are two points on the curve.
A1 = (0 : 1)
A2 = (0 : 7)
A3 = (0 : 16)
A4 = (1 : 9)
A5 = (1 : 14)
b)
i. (0+1)+2*(1+9)+(1+14) = 36
ii. 3*(0+7)+(0+16) = 37
Question 4:
Given E: y 2 = x3 +2x+2 mod 19
and point P=(5,1)
Goal: Compute 2P = P+P = (5,1)+(5,1)= (x3 ,y3 )
s = (2·1) −1 (3·52 +2) = 2−1 ·9 ≡ 9·9 ≡ 13 mod 19
x3 = 132 − 5 − 5 = 159 ≡ 7 mod 19
y3 = s(x1−x3 ) − y1 = 13(5 − 6) − 1= −14 ≡ 5 mod 19

5Cryptology in practices
Finally 2P = (1,5) + (5,1) = (7,5)
So, (1,5) lies on the curve. ------ (a)
Verify that (7,5) is a point on the curve:
52 =25, 73 + 14 +2=49*7 +16 ≡ 2•7+14 = 28 ≡ 9 mod 19
k= (7-1)/(7-5)=3 (1) −1 ≡ 3 mod 19
So, k(1,5) = (9,3) ------ (b)
x3 = 10-12=-2 ≡ 17 mod 19 (1,5) ----- (c)
Question 5:
a)
r = 10
s = 55
b)
kE = 59
x = 83
p = 113
g = 67
(x1,r1,s1) = (38,12,77)
(x2,r2,s2) = (72,92,76) from Bob.
Hence, it is verified that Alice verify whether these messages originated actually from Bob.

6Cryptology in practices