1. Codes, Ciphers, and Cryptography-RSA Encryption
Michael A. Karls
Ball State University

2. 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!

3. 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

4. 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.

5. 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.

6. 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).

7. 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.

8. 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.

9. 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.

10. 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 00space; 01A; 02B, … , 26Z (or use ASCII).
¨Plaintext
Plain # HE
0805 LL
1212 0_
1500

11. 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

12. 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_.