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RSA Cryptography Taravat Moshtagh Department of Mathematics and Statistics York University January 19th 2006 Context Introduction to RSA cryptography Basic terminology Mathematics behind the RSA How the algorithm work How secure the algorithm is Conclusion References Introduction to cryptography Today Internet is inseparable part of our life and millions of people will be using the Internet. Reading the news, chatting with friends, purchasing a new product, researching for a paper … the number of uses of the Internet is endless. One of the attractions of the Internet is that one can do almost anything from the comfort of his/her own home and with a relative sense of anonymity. Unfortunately, the data going across the Internet may not be as secure as we would like to think. It is not especially difficult for a person with the right technical skills to intercept the data going from one computer to another. Usually this is not a problem; people don’t really care if someone knows that they went to google.com and started researching Number Theory. However, if the intercepted data contains a credit card number, password, social security number, or some other private information – it becomes a whole different story. Online banking and a host of other services rely heavily upon the security of credit card numbers, PINs, and other private information as it goes across the network. But if it is easy to intercept these numbers, how do these services work? The answer: Cryptography. What is meant by cryptography? Cryptography comes from Greek word kryptós means "hidden", and gráphein means“to write“. So cryptography is ‘secret‘or ‘hidden‘writing. Cryptology is as old as writing itself, and has been used for thousands of years to safeguard military and diplomatic communications. Cryptography is the art of keeping information secret from all but those who are authorized to see Cryptography has become a widely-used tool in communications and computer security. Basic terminology Plaintext: Data that can be read and understood without any special measures .It is also the message. Encryption: Encoding the contents of the message in such a way that hides its contents from outsiders. Ciphertext: The encrypted message. Decryption: The process of retrieving the plaintext from the ciphertext. 2 Public Key: The user releases a copy of this key to the public to allow anyone to use it for encrypting messages to be sent to the user and for decrypting messages received from the user. Private Key: The user keeps the private key secret and uses it to encrypt outgoing messages and decrypt incoming messages Note that: A key is a value that works with a cryptographic algorithm to produce a specific ciphertext. Keys are basically really, really, really big numbers. Key size is measured in bits; the number representing a 1024-bit key is darn huge. In public key cryptography, the bigger the key, the more secure the ciphertext. Encryption and Decryption For our scenarios we suppose that Alice and Bob are two users. Now we would like to know how Bob can send a private message to Alice in a cryptosystem. Suppose Bob encrypts the message by replacing every “A” in his messages with a “D”, every “B” with an “E”, and so on through the alphabet. Only someone who knew the "shift by 3" rule could decrypt his messages. Figure 1 illustrates this process. Encryption and Decryption Key Encryption Decryption Bob Alice Plaintext Ciphertext Plaintext Figure1. Encryption and decryption For a sender and recipient to communicate securely using above encryption and decryption method, they must agree upon a key and keep it secret between themselves. If they are in different 3 physical locations, they must trust a courier, the Bat Phone, or some other secure communication medium to prevent the disclosure of the secret key during transmission. Anyone who overhears or intercepts the key in transit can later read, modify, and forge all information encrypted or authenticated with that key. Broadly speaking, the problem is key distribution: how do you get the key to the recipient without someone intercepting it? Public key cryptography The problems of key distribution are solved by public key cryptography, the concept of which was introduced by Whitfield Diffie and Martin Hellman in 1975. Public key cryptography is an asymmetric scheme that uses a pair of keys for encryption: a public key, which encrypts data, and a corresponding private, or secret key for decryption. You publish your public key to the world while keeping your private key secret. Anyone with a copy of your public key can then encrypt information that only you can read. Even people you have never met. Figure 2 illustrates this process. Public-key Cryptography Encryption key Decryption key Plaintext Ciphertext Original plaintext Bob Encryption Decryption Alice Figure2. Public key cryptography In the original description, the Diffie-Hellman exchange by itself does not provide authentication of the parties. 4 RSA public key cryptography In 1977 Ronald L. Rivest, Adi Shamir, and Leonard Adleman, came up with a new public key encryption scheme with implementation of digital signature .It is called RSA which stands for the first letter in each of its inventors' last names. RSA is an elegant algorithm based on the product of two large prime numbers that exactly fit the requirement for a practical public key cryptography implementation. According to this scheme when Bob wants to send a secure message to Alice; Bob uses Alice’s public key (for example her email address) to encrypt the message. Alice then uses her private key to decrypt it. . Figure 3 illustrates this process. RSA public key for sending and receiving message Alice public key Alice private key Plaintext Ciphertext Plaintext Encryption Decryption Bob Alice Figure 3.RSA Public key encryption and decryption How does RSA work? Essentially, the public key is the product of two randomly selected large prime numbers ‘p’ and ‘q’, and the secret key is the two primes themselves. The algorithm encrypts data using the product, and decrypts it with the two primes, and vice versa. A mathematical description of the encryption and decryption expressions is shown below: Encryption: C=Me (mod n) Decryption: M=Cd (mod n) 5 where: M = the plain-text message expressed as an integer number. C = the encrypted message expressed as an integer number. n = the product of two randomly selected, large primes p and q. d = a large, random integer relatively prime to (p-1)*(q-1). e = the multiplicative inverse of d, that is: ( e * d ) =1 ( mod ( p - 1 ) * ( q - 1 ) ) The public key is the pair of numbers ( n, e ). The private key is the pair of numbers ( n, d ). It is computationally infeasible to deduce the private key from the public key. Anyone who has a public key can encrypt information but cannot decrypt it. Only the person who has the corresponding private key can decrypt the information. Why does it work? To encrypt a message using the integers n and e for a public key, first encode the message as an integer M relatively prime to n. Let C denote the encrypted version of the message, where C is e defined as C = M (m o d n ) .This integer C can be made available to anyone but it can only be decrypted back into the original message M by someone knowing the corresponding private key. To decrypt the message, recall Euler's formula which says that for any integers m and a (m ) with g c d ( a ,m ) = 1 that a 1 (m o d m ) .Since M was chosen relatively prime to n it follows then (n) that M 1 (m o d n ) . Using this information leads to a method for decrypting C .First note that since e .d 1 ( m o d ( n ) ) for some integer k we get e .d 1 k ( n ) therefore, decrypting of encrypted message lead us to compute 6 1 k ( n ) (n) (M ) M M M .( M M d e d ed k C ) (m o d p ) M M ed ed and thus p M .Arguing similarly for q yields q M .Together these last two equations 1 k ( n ) M M ed imply that for all M, M (m o d n ) .Since p and q are distinct primes, it follows that M ed also their product, means pq is a divisor of M . Note this computation recovers the original message M and makes use of information in the private key set. RSA Public Key algorithm Example Choose prime numbers p and q. Choose 11 and 13 Find their product n = pq. Calculate n = Calculate Phi(n) = (p-1)(q-1). Calculate Phi(n) = Select an integer ”e “, in which the Let e = 7. gcd (e, Phi(n)) = 1. Calculate d such that e*d = 1 mod (p-1)(q-1). Calculate d The public key is (e, n). The public key is The private key is (d, n). The private key is Plaintext can be any number M, where Let the numerical representation of M be M = 5, M < n, and neither p nor q divides M for example. The ciphertext is C=Me (mod n) The ciphertext is The plaintext is Cd =Med (mod n) The plaintext is 7 Is RSA secure? The security of the RSA cryptosystem depends on the difficulty of factoring n. It is currently difficult to obtain the private key‘d’ from the public key (n, e). However if one could factor n into p and q, then one could obtain the private key‘d’. To make it clear, note that e . d 1 (m o d ( n )) . Now e and n are in the public key set, so‘d’ can be computed by computing the multiplicative inverse of ‘e’ modulo ( n ) . Since ( n ) ( p q ) ( p ) . ( q ) ( p 1) ( q 1) it is not a problem for someone who knows (n ) to find’d’. So the problem of unearthing the private key ‘d’ boils down to computing ( n ) , which in turn is equivalent to factoring n. Therefore breaking RSA system by computing (n ) is not easier than breaking RSA system by factoring n. (This is why n must be composite; (n ) is easy to compute if n is prime.) If a method is discovered for factoring arbitrary integers quickly, then any RSA private key could be discovered and the system would become insecure. Factoring n: The fastest known factoring algorithm developed by Pollard is the General Number Field Sieve, which has running time for factoring a large number of size n, of order 64 1 2 exp ( lo g n ) ( lo g lo g n ) 3 3 9 The method relies upon the observation that if integers x and y are such that x ≠ y (mod n) and x 2 = y 2 (m o d n ) then gcd(x − y, n) and gcd(x+y, n) are non-trivial factors of n. The following table gives the number of operations needed to factor n with GNFS method, and the time required if each operation uses one microsecond, for various lengths of the number n (in decimal digits) 8 Digits Number of operations Time 100 9.6× 108 16 minutes 200 3.3 × 1012 38 days 300 1.3 × 1015 41 years 400 1.7 × 1017 5313 years 500 1.1 × 1019 3.5 × 105 years 1024 1.3 × 1026 4.2 × 1012 years 2048 1.5 × 1035 4.9 × 1021 years Computing (n ) without Factoring “n”: Assume that n pq, p q . Since ( q p ) (q p ) 4 pq 4n (q p ) 4n (q p ) q p 4n (q p ) 2 2 2 2 2 , then so ; guess q p and then find q p ,so (n) n ( p q ) 1 . Example: Suppose n 221 (4 n 884) q p (q p ) 4n (q p ) 2 2 q p 4n (q p ) 2 1 885 29.7489· · · 2 888 29.7993· · · 3 893 29.8831· · · 4 900 30 So, q p 4 and q p 30 then (n) 221 30 1 192 and p 13 , q 17 , n 13 17 9 Review One can break RSA by Factoring n Computing Phi(n) Compute d given e and n o Still need to know n or Phi(n) Computing e-th roots modulo n (C= Me (mod n); then M= C1/e (mod n)) o It is computationally intractable Conclusion Prime numbers play an essential role in the art of Public Key Cryptography. Public Key Cryptosystem is secure and strong. Factorization is a fast-moving field. If no new methods are developed, then 2048-bit RSA keys will always be safe from factorization, but one can't predict the future. References 1. W. Diffie and M. E. Hellman, New directions in cryptography, IEEE Transactions on Information Theory IT-22 (1976), 644-654. 2. L. M. Adelman, R. L. Rivest and A. Shamir, A method for obtaining digital signatures and public-key cryptosystems, Communications of the ACM, 21 (1978), 120-126. 3. L. M. Adelman, R. L. Rivest and A. Shamir, How public key cryptography works, Retrieved from the web sit: http://www.livinginternet.com/i/is_crypt_pkc_inv.htm#diffie 10

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