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**Codes, Ciphers, and Cryptography-RSA Encryption**

Michael A. Karls Ball State University

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**The RSA Encryption Scheme**

We now look at the public key cryptography scheme developed by Rivest, Shamir, and Adleman (RSA) in 1977. In order to understand this scheme, we need some definitions!

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Definition of Divisor Let a and b be integers, with b 0. We say that b divides a or b is a divisor of a if a = b x c for some integer c. Notation: b|a Example 1: 3|24 since 24 = 3 x 8. Divisors of 12 are: -12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 12

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**Definition of Greatest Common Divisor (GCD)**

Let a and b be integers, not both zero. The greatest common divisor (GCD) of a and b is the largest integer that divides both a and b. Notation: (a,b) Example 2: Divisors of 6: -6, -3, -2, -1, 1, 2, 3, 6 Divisors of 8: -8, -4, -2, -1, 1, 2, 4, 8 Thus, (6,8) = 2 Since the divisors of 7 are -7, -1, 1, 7, (7,8) = 1.

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**Definition of Relatively Prime**

Two integers whose GCD is 1 are said to be relatively prime. Example 3: Since (7,8) = 1, 7 and 8 are relatively prime.

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**Definition of Prime Number**

A positive integer p is said to be prime if p>1 and the only positive divisors of p are 1 and p. Example 4: 2, 3, and 7 are prime. 6, 8, 10, 100 are not prime (composite).

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**RSA Scheme (with Alice and Bob!)**

Step 1: Alice chooses two huge prime numbers p and q. Note: Alice keeps p and q secret! Example 5: p = 47 and q = 59.

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**RSA Scheme (with Alice and Bob!) (cont.)**

Step 2: Alice computes N = p x q. Then she computes k = (p-1)(q-1). Finally, she chooses an integer e such that 1<e<N and (e,k) =1. Example 5 (cont.): N = 47 x 59 = 2773. k = 46 x 58 = 2668. e = 17. Choice of e is o.k., since 1<17<2773 and (17,2668)= 1.

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**RSA Scheme (with Alice and Bob!) (cont.)**

Step 3: Alice computes d = e-1 mod k. Alice publishes her public key: N, e. Alice keeps secret her private key: p, q, d, k. Example 5 (cont.): d = 17-1 mod 2668 = 157. Alice’s public key: N = 2773; e = 17. Alice’s private key: p = 47; q = 59; d = 157; k = 2668.

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**RSA Scheme (with Alice and Bob!) (cont.)**

Step 4: Suppose Bob wants to send a message to Alice. To do so, he looks up Alice’s public key, converts the message into numbers M<N. Example 5 (cont.): Plaintext is HELLO HELLO HE LL O_ Assign 00space; 01A; 02B, … , 26Z (or use ASCII). ¨Plaintext Plain # HE 0805 LL 1212 0_ 1500

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**RSA Scheme (with Alice and Bob!) (cont.)**

Step 4 (cont.): Next Bob computes: C = Me mod N (1) for each plaintext number M to get ciphertext number C. Example 5 (cont.): mod 2773 = 542. mod 2773 = 2345. mod 2773 = 2417. Encrypted message is

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**RSA Scheme (with Alice and Bob!) (cont.)**

Step 5: Bob s Alice the encrypted message. To decrypt, Alice uses her private key and computes: M = Cd mod N (2) Example 5 (cont.): mod 2773 = 805. mod 2773 = 1212. mod 2773 = 1500. Decrypted message is HE LL 0_.

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