RSA Cryptography KC LG

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Cryptography Terms

Cryptography – The practice and study of encryptionand decryption – encoding data so that it can onlybe encoded by specific individuals.

Cryptosystem – system for encrypting anddecrypting data. Security of cryptosystem dependson secrecy of the keys rather than the secrecy of thealgorithm. It is important to have a large range of possible keys, so that it is not possible to do a ―brute

force‖ approach in cracking the algorithm. Plaintext – Unencrypted data.

Ciphertext – Encrypted data. ―unreadable jibberish‖1



How Cryptography Works

Uses cryptographic algorithms called―ciphers‖ and ―deciphers.‖ Mathematicalfunctions that work with cryptographic

keys to encrypt and decrypt plaintext. The larger the ―key,‖ the more secure the

ciphertext.

The same plaintext encrypts to differentciphertexts using different keys.



Public Key Cryptography

Traditionally, cryptography was done with just a secret key,which would have to be known to everyone, and this wasinsecure.

The challenge would be that parties would have to agree on asecret key without anyone else finding out.

The secret key method is generally faster, but less secure.

The public key cryptosystem was introduced in 1976 byWhitfield Diffie and Martin Hellman.

Uses public key for encryption, as well as a private key for decryption.

Each user gets two keys: one public and one private. The public

key is published; the private key is secret. This eliminates theneed to share the private key.

Private key is always linked mathematically to the public key. If the system, is hacked, then the secret key can be found.

The solution is to make it so that the hacker has to factor a verylarge number to get the private key, so that it would become

―computationally infeasible‖ to perform the derivation.



Advantages and Disadvantages

of Public Key Cryptography Advantages:

Increased security and convenience.

Provides digital signatures that cannot be

repudiated.

Disadvantages:

Slower than secret key method, but can be

used in conjunction with the secret key tomake it more efficient.

Can be vulnerable to impersonation if hacked.



RSA Public-Key Cryptography

Named after the inventors, Ron Rivest, Adi

Shamir, and Len Adleman in 1977.

One of the interesting things about RSA is thatyou can tell anyone about how the encryption

works; however, this knowledge is not sufficient

to be able to decrypt the ciphertext. Only the

chosen few who have extra information can

decrypt the message.



Computing the RSA Public Key

RSA public key: Modulus (also called n): The product of two

large, odd, prime numbers p and q, of about

the same size. Therefore, n = pq.

Exponent (e): 3 < e < n

and e and ( p – 1)(q – 1) have not

common factors except 1 (ie. e is relatively

prime to (p –

1)(q –

1) ).

RSA Private key d :

Number d such that (ed – 1) is divisible

by (p –

1)(q –

1).



Meet Alice & Bob

Bob wants to send Alice the test answers

without Professor Eve knowing. Bob decides to

use RSA cryptography to send Alice his

message. Bob encrypts his message byperforming the following formula:

c = me mod n

where c = ciphertext, m = message, e =

exponent, n = modulus (ie. pq)



Alice receives the ciphertext c , and

needs to decrypt it using the followingformula:

m = c d mod n

where m = the test answers Bob sent to Alice

c = the ciphertext of the message, d = Alice’s

private key, n = pq (the modulus).



Bob Alice

Test Answers (M) Test Answers (M)

Me % n Alice’s

Public Key

(e, n)

Ciphertext (c) Ciphertext (c)

Cd % n Alice’s

Private Key

(d)

transmission

n = p * q



Mathematical Example2

Let’s say that. . .

p = 61 — first prime number (to be kept secret or deleted securely)

q = 53 — second prime number (to be kept secret or deleted securely)

n = pq = 3233 — modulus (to be made public)

e = 17 — public exponent (to be made public)

d = 2753 — private exponent (to be kept secret)

The encryption function is:

encrypt(m) = me mod n = m17 mod 3233

where m is the plaintext.

The decryption function is:

decrypt(c ) = c d mod n = c 2753 mod 3233

where c is the ciphertext.

To encrypt the plaintext value 123, we calculate

encrypt(123) = 12317 mod 3233 = 855

To decrypt the ciphertext value 855, we calculate

decrypt(855) = 8552753 mod 3233 = 123



Digital Signature

Hash (M)

Alice’s

Public Key

(e, n)

Ciphertext (c) Ciphertext (c)

decrypt

Alice’s

Private Key

(d)

n = p * q

Digital Signature

+

Test Answers (M)

Bob’s

Private

Key (d)

M + Digital

Signature

Decrypt with Bob’s

public keyHash (M) ==



Padding

RSA is usually combined with a padding scheme inorder to make sure the message m never has aninsecure ciphertext.

RSA implementations typically embed some form of structured, randomized padding into the message m

before encrypting it. This will ensure that it does notfall into the range of insecure plaintexts.

The padded encryption can have a large number of possible ciphertexts, thus decreasing the risk of anattacker building a dictionary by encrypting likelyplaintexts under the public key and storing theresulting ciphertexts. When matching ciphertexts areobserved on a communication channel, the attacker can use this dictionary in order to learn the content of the message.2



References

1 http://www.codeproject.com/dotnet/RSACryptoPad.asp

2 http://en.wikipedia.org/wiki/Rsa

Cormen, Leiserson, Rivest, Stein. Introduction to

Algorithms. McGraw-Hill.

http://www.codeproject.com/dotnet/RSACryptoPad.asp

http://en.wikipedia.org/wiki/Rsa

http://en.wikipedia.org/wiki/Rsa

http://www.codeproject.com/dotnet/RSACryptoPad.asp

RSA Cryptography KC LG

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