Chapter8Notes - Chapter 8 Notes Public Key cryptography The...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 8 Notes Public Key cryptography: The basic idea is to do away with the necessity of a secure key exchange, which is necessary for all private key encryption schemes. The idea is as follows: 1) Bob creates two keys, a public key, E and a private key D. 2) Bob posts the public key in a location that anyone can access. The important thing here is that the knowledge of E does not compromise the value of D in any way shape or form. 3) Now, anyone who wants to send a message to Bob encrypts it using the public key E. 4) Bob can now read the message using his private key D. However, since the value of E gives no information as to the value to D, all others can not read the message. The idea seems easy enough, but the difficulty is in finding some mathematical function to use in this scheme such that the knowledge of E does not compromise the secrecy of D. Clearly in all the other schemes we have seen, knowledge of the encrypting key all but completely gives away the decrypting key. One thing to note however is that if you use a system outlined above with nothing extra, although Bob can decipher a message sent to him, he can not be sure of who the sender is, because the whole public has the ability to encipher a message, so someone could easily indicate in their message that they were someone else and Bob would not have any way of knowing. But, the person sending the message can be confident that no one read can read the plaintext except for Bob, the only person with the private key. Number Theory Background for RSA. First, we need to go over some mathematics before we look at RSA: Euler Phi Function First, let’s define the Euler φ (phi) function: φ (n) = the number of integers in the set {1, 2, . .., n-1} that are relatively prime to n. φ (p) = p –1 , for all prime numbers φ (pq) = (p-1)(q-1), where p and q are distinct primes. Here is a derivation of that result: We want to count all values in the set {1, 2, 3, . .., pq –1} that are relatively prime to pq. Instead, we could count all value in the set NOT relatively prime to pq. We can list these values:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
p, 2p, 3p, . .. , (q-1)p q, 2q, 3q, . .. (p-1)q Note that each of these values are distinct. To notice this, see that no number of the first row is divisible by q and no number on the second row is divisible by p. This ensures that there are no repeats on both rows. since p and q are relatively prime, in order for q to be a factor of a number on the first row, it would have to divide evenly into either 1, 2, 3, . .. q- 1. But clearly, it does not. The same argument will show that none of the values on the second row are divisible by p. Finally, we can count the number of values on this list. It’s (q-1) + (p-1) = p + q – 2. Now, in order to find
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 11

Chapter8Notes - Chapter 8 Notes Public Key cryptography The...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online