4. Proof
Consider a set of positive integers less than ‘p’ :
• {1,2,3,…..,(p-1)} and multiply each element by ‘a’ and
‘modulo p’ , to get the set
• X = {a mod p, 2a mod p,…, (p-1)a mod p}
No elements of X is zero and equal, since p doesn’t
divide a.
Multiplying the numbers in both sets (p and X) and
taking the result mod p yields
5. Proof (Contd.,)
• Multiplying the numbers in both sets (p and X) and taking
the result mod p yields
• a * 2a *…* (p-1)a [1 * 2 * 3 *…* (p-1)] (mod p)
• a p 1
( p 1 ) ! ( p 1) ! ( mod p)
• Thus on equating (p-1)! term from both the sides, since it
is relatively prime to p, result becomes,
An alternative form of Fermat’s Theorem is given as
1 ( m o d p)
a p
a ( m o d p )
a p 1
6. EULER TOTIENT FUNCTION : φ (n)
(n) : How many numbers there are between 1 and n-
1 that are relatively prime to n.
(4) = 2 (1, 3 are relatively prime to 4).
(5) = 4 (1, 2, 3, 4 are relatively prime to 5).
(6) = 2 (1, 5 are relatively prime to 6).
(7) = 6 (1, 2, 3, 4, 5, 6 are relatively prime to 7).
7. EULER TOTIENT FUNCTION : φ (n)
From (5) and (7), (n) will be n-1
whenever n is a prime number.
This implies that (n) will be easy to calculate when n
has exactly two different prime factors:
(P * Q) = (P-1)*(Q-1)
if P and Q are prime.
9. Above equation is true if n is prime because then,
and Fermat’s theorem holds.
Consider the set of such integers, labeled as,
Here each element xi of R is unique positive integer less than n
with GCD( xi ,n ) = 1.
n ( n 1)
10. Proof
• Multiply each element by a, modulo n :
The set S is permutation of R :
• Because a and xi is relatively prime to n, so multiplication
is also be relatively prime to n. Thus the elements of S are
integers that are less than n and that are relatively prime
to n.
• There are no duplicates in S.
18. Quiz
• Who am I
• I can scramble the text
• One key is not enough for me
• Its more secure if you exploit me
Asymmetric Key Encryption / Public key
encryption
19. • Who am I
• I was designed by 3 people
• Fermats theorem is my basis
• I can perform encryption
• I need more than one key for my operation
RSA Algorithm
Quiz
20. • Who am I
• I was designed by 2 people
• One key is not enough for me to do my operation
• Discrete Logarithm is my basis
• I am just used to share the personal data
• I can’t hold the identity
Diffie Hellman Key Exchange Algorithm
Quiz