Lecture9

Lecture9 - Lecture 9 Public Key Cryptography Public Key...

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1 Lecture 9 Public Key Cryptography Public Key Algorithms CNT 5412 Network Security
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2 Euler’s Totient function φ (n) - again φ (n) is the number of positive integers less than n and relatively prime to n. Let p be a prime. Then φ (p) = p – 1 Let n = pq for p and q prime. Then φ (n) = φ (pq) = φ (p) φ (q) Proof outline: look at numbers relatively prime to pq. Must take all the numbers less than pq but eliminate multiples of p and multiples of q. (pq – 1) – (q – 1) – (p – 1) = (p – 1) (q – 1) Example: let n = 15, Then φ (15) = φ (5) φ (3) = 8 Z 15 * = {1,2,4,7,8,11,13,14}
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3 Euler’s Theorem - extension Euler’s theorem : for every a and n that are co-prime, a φ ( n ) ≡ 1 mod n Corollary : for every a and n that are co-prime, k ≥ 0, a k φ ( n )+1 a mod n Euler’s extension : For n a product of primes p and q , then for a in Z n and k ≥ 0 , a k φ ( n )+1 a mod n (note however that this does not imply that a φ (n) = 1.) Proof of extension outline: Consider three cases, gcd( a,n ) =1, a multiple of p , and a multiple of q . If a is a multiple of p , a = cp . Then since gcd( a,q ) = 1, a φ ( q ) φ ( p ) ≡ 1 mod q , and by multiplying both sides by a = cp we can get the result. Similarly for a multiple of q .
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x y 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 4 8 6 2 4 8 6 2 4 8 6 3 1 3 9 7 1 3 9 7 1 3 9 7 1 4 1 4 6 4 6 4 6 4 6 4 6 4 6 5 1 5 5 5 5 5 5 5 5 5 5 5 5 6 1 6 6 6 6 6 6 6 6 6 6 6 6 7 1 7 9 3 1 7 9 3 1 7 9 3 1 8 1 8 4 2 6 8 4 2 6 8 4 2 6 9 1 9 1 9 1 9 1 9 1 9 1 9 1 Table for exponentiation mod 10
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5 Looking at the mod 10 table Note that φ (10) = 4. Also, 10 is the product of distinct primes so Euler’s theorem applies (for any x ) – that is x 1 + k φ ( n ) = x mod n where n = 10, for all x – thus φ
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Lecture9 - Lecture 9 Public Key Cryptography Public Key...

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