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Rachana Y. Patil

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Modula Arithmetic Rachana Y. Patil * Cryptographic Theory Primality: Two nos are relatively prime if they have no factors common in them other than 1. – PowerPoint PPT presentation

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Title: Rachana Y. Patil


1
Modula Arithmetic
  • Rachana Y. Patil

2
Cryptographic Theory
  • Primality
  • Two nos are relatively prime if they have no
    factors common in them other than 1.
  • i.e gcd(a,n) 1
  • gcd (7, 78) 1

3
Euclids Alorithm
  • What is gcd of 21 and 45???
  • gcd(a,b) gcd(b, a mod b)

4
Modular Arithmetic
  • Says that 23 and 11 are equivalent ??????
  • 23 mod 11 12
  • Or 23 11 mod 12 ..

5
Cont
  • a b mod n if a b kn for some integer k.
  • If a gt 0 and 0 lt b lt n then b is the remainder of
    the division a/n.

6
Properties of Modulo operator
  • a b mod n if n/a-b
  • a b mod n -gt b a mod n
  • a b mod n and b c mod n implies
  • a c mod n

7
Modular Arithmetic
  • (a mod n) (b mod n) ( a b ) mod n
  • (a mod n) x (b mod n) (a x b) mod n
  • (a b) (a c) mod n then b c mod n

8
Eulers theorem
  • Eulers Toient function Ø(n)
  • Ø(n) is the set of ve integers less than n and
    relatively prime to n
  • n 6 What is Ø(n) ????
  • n 7 Ø(n) ????

9
Eulers theorem.cont
  • For any prime no Ø(n) n-1.
  • Suppose p and q are two prime nos.
  • For npq we have
  • Ø(n) Ø(pq) Ø(p) x Ø(q) (p-1) x (q-1)
  • n21 p3 and q 7
  • Ø(21) Ø(3) x Ø(7) 2 x 6 12.

10
Fermats Theorem
  • If p is prime and a is a ve integer not
    divisible by p then
  • a p-1 1 (mod p)
  • Let a 3 and p 5
  • a 5-1 a 4 34 81 1 (mod 5)
  • proved..

11
The Theorem
  • For every a and n which are relatively prime
  • a Ø(n) 1 (mod n)
  • a 3 n 10
  • Ø(n) Ø(10) 1,3,7,9 4
  • a Ø(n) 3 4 81 1 (mod 10)
  • hence proved

12
Modular Exponentiation
  • xy mod n xy mod ø(n) mod n
  • if y 1 mod ø(n) then xy mod n x mod n

13
Modular exponentiation
  • One way function used in cryptography
  • ax mod n
  • Can u find x where ax b mod n???
  • That is the discrete logarithm problem
  • If 3x 15 mod (17) find x.

14
Discrete Logarithm problem
  • Solution.easy enough
  • Solve 3x mod 15 17
  • x 6
  • For large nos solving this is difficult!!!
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