
The RSA Algorithm  http://www.claymath.org/posters/primes/

RSA Algorithm Demo 

The task of cryptography (Greek kryptos hidden, graphein to write) is to turn messages into gibberish so that they cannot be read by persons other than the intended recipient.
If I send my credit card number and PIN over the internet to an online bookstore, the bookstore should be able to read it, but no one else should.
A cryptographer starts with plain text such as m="ATTACKATDAWN" and produces cipher text such as c="OTMMGKLHDTIR." This is encryption.
Decryption is the reverse: it produces plain text from cipher text.
In the old days, to send a secret message, you first had to send a key secretly to the recipient. This you did by meeting in person, or through a trusted courier. In the example above, the key was "OATMEAL" and the encryption method is due to Vigenère. But the internet is not a trusted medium, so how do you get started?
You can't use the internet to send the key, then your financial information. A bad person could capture your internet traffic with the bookstore. He would use the key to read your credit card number and PIN, and would then charge expensive travel to exotic places to your account.
This dilemma was solved in 1978 by Rivest, Shamir, and Adelman. Their method, now known as RSA, depends on some marvelous properties of prime numbers. One of these is that it is rather easy to generate large prime numbers, but much harder to factor large numbers into primes. Another is Fermat's little theorem.
 Note: "a mod N" is the remainder of a/N.  RSA starts first with translating the message letters into blocks of numbers and back again. We can do this using a random table like the one below, where A corresponds to 11, B to 12, etc.
Conversion Table  0 1 2 3 4 5 6 7 8 9  1SP A B C D E F G H I 2J K L M N O P Q R S 3T U V W X Y Z a b c 4d e f g h i j k l m 5n o p q r s t u v w 6x y z . , : ; ' " ` 7! @ # $ % ^ & *  + 8( ) [ ] { } ? / < > 90 1 2 3 4 5 6 7 8 9 
Thus the word Attack becomes 115656373947 = message = m.
From now on encrypting and decrypting becomes a matter of computing with large integers.
Let p and q be large prime numbers Let N = p*q.
For encryption use: c = f(m) = m^e mod N,
where e is a positive integer which has no factor in common with (p1)(q1).
For decryption use: m = g(c) = c^d mod N,
where d is a positive integer such that (e*d  1) is divisible by (p1)(q1).
What makes RSA so hard to break? Well, think first about what Alice, the person who designs the code, does. First she generates the large primes p and q, then she chooses e. Finally she solves an equation to find d:
x*e + (p1)(q1)*c = 1, where all the letters are integers.
Alice makes e and N public. This is all anyone needs to send secret messages to her.
Now think about what Bob the spy, who wants to intercept and decrypt messages sent to Alice, needs to know. First he must factor N to find p and q so as to set up the equation:
x*e + (p1)(q1)y = 1.
Then he solves the equation to find x and decrypt Alice's messages using g(c) = c^x mod N.
The problem is that the factoring problem takes huge amounts of computer time, for large enough p and q so much time that it would take millions of years, given current mathematical theory and current computer technology, to crack the code.
The fact that g decrypts messages encrypted by f, is a consequence of Fermat's little theorem:
If a is not divisible by p, where p is prime, then a^(p1)  1 is divisible by p.
Fermat (16051661) was a French mathematician whose work is the foundation of modern number theory.

Download the demo from http://www.heliwave.com/RSADemo.zip
and download its source code from http://www.heliwave.com/RSADemo.Source.zip


EXAMPLE from http://en.wikipedia.org/wiki/RSA  Here is an example of RSA encryption and decryption. The parameters used here are artificially small, but one can also use OpenSSL to generate and examine a real keypair.
1. Choose two distinct prime numbers, such as
p = 61 and q = 53.
2. Compute n = p * q giving
n = 61 * 53 = 3233.
3. Compute the totient of the product as r = (p  1)(q  1) giving
r(3233) = (61  1)(53  1) = 3120.
4. Choose any number 1 < e < 3120 that is coprime to 3120.
Choosing a prime number for e leaves us only to check that e is not a divisor of 3120.
Let e = 17.
5. Compute x, the modular multiplicative inverse of e(mod r(n)) yielding
x = 2753.
The public key is (n = 3233, e = 17). For a padded plaintext message m, the encryption function is m^17(mod 3233).
The private key is (n = 3233, x = 2753). For an encrypted ciphertext c, the decryption function is c^2753(mod 3233).
In order to encrypt letter a with ASCII = 65, let m = 65, then we calculate: c = 65^17(mod 3233) = 2790.
To decrypt c = 2790, we calculate: m = 2790^2753(mod 3233) = 65 which in ASCII is letter a.
Both of these calculations can be computed efficiently using the squareandmultiply algorithm for modular exponentiation. In real life, the primes p and q would be much much larger. In our example it would be relatively trivial to factor n (3233) found in the public key back to the secret primes p and q. And given e from the public key, we could then compute the private key x. 
Ali Adams God > 
