Student ID: 102566527
Name: Mohammad omair alam
Character
M
o
h
a
m
m
a
d
o
m
a
i
r
a
l
a
m
Total
ASCII Value in
Dec
77
111
104
97
109
109
97
100
111
109
97
105
114
97
108
97
109
1751
“ASCII values Reference: http://www.asciitable.com/”
There are 23 questions in Chapter 9. Public-Key and RSA in the book (Cryptography and
Network Security Principles and Practices, Fourth edition by William Stallings)
Therefore,
Total mod (number of questions)
1751 mod 23 = 3
The question number I have to solve is the result + 1, i.e. Question number 3
Q) In an RSA system, the public key of a given user is e = 31, n = 3599. What is the
private key of this user?
Solution:
The private key can be written as {d, n}
To calculate d, we have to 1st calculate Ф(n)
Where Ф(n) is the number of prime factors of n
1
e = 31, n = 3599
It means that e is relatively prime to Ф(n), i.e. gcd (Ф(n), e) = 1
To find Ф(n), we have to first check whether n is prime or not, and if it is not, what are
the prime factors of n.
Step1: To find prime factors of n
By using trial and error we find that n is not prime and the prime factors of n are
n = 3599 = 61 * 59
Thus p = 61 and q = 59 are both prime numbers.
Step 2: To calculate Ф(n)
Here we find the Euler’s totient function written as Ф(n), where Ф(n) is the number of
positive integers less than n and relatively prime to n.
It is clear that for a prime number p,
Ф(p) = p-1
Now since we have two prime numbers p and q, with p ≠ q, Then for n= pq,
Ф(n) = Ф(pq) = Ф(p) * Ф(q) = (p-1) * (q-1)
Therefore
Ф(3599) = Ф(61) * Ф(59) = 60 * 58
= 3480
Step 3: Calculate d
We know that if gcd (Ф(n), e) = 1, then e has a multiplicative inverse modulo Ф(n)
Therefore
d ≡ e-1modФ(n)
i.e. d is multiplicative inverse of e mod Ф(n)
So now we use the extended Euclid’s algorithm to find the multiplicative inverse of e.
EXTENDED EUCLID (Ф(n), e)
1. (A1, A2, A3) ← (1, 0, m); (B1, B2, B3) ← (0, 1, b)
2
2. if B3 = 0
return A3 = gcd(Ф(n), e); no inverse
3. if B3 = 1
return B3 = gcd(Ф(n), e); B2 = e-1 mod m
4. Q = [A3/B3]
5. (T1, T2, T3) ← (A1 - QB1, A2 - QB2, A3 - QB3)
6. (A1, A2, A3) ← (B1, B2, B3)
7. (B1, B2, B3) ← (T1, T2, T3)
8. goto 2
Throughout the computation, the following relationships hold:
Ф(n)T1 + eT2 = T3
Ф(n)A1 + eA2 = A3
Ф(n)B1 + eB2 = B3
Now we construct a multiplication table
Q
A1
A2
A3
B1
B2
B3
-
1
0
3480
0
1
31
112
0
1
31
1
-112
8
3
1
-112
8
-3
337
7
1
-3
337
7
4
-449
1
Now we need to check whether -449 is multiplicative inverse of 31 modulo 3480
Therefore
-449 * 31 = -13919
-13919 mod 3480 = -3479 mod 3480 = 1 mod 3480
Hence it is proved that d = -449 is multiplicative inverse of e = 31 mod Ф(n)
So the private key of this user will be {-449, 3599}
3