# RSA Prepared by SITI ZAINAH ADNAN If you

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RSA Prepared by: SITI ZAINAH ADNAN If you do have any feedback or comment, please feel free to email me at [email protected] com Your cooperation is very much appreciated !

RSA Is a famous implementation of public key protocol. n Deviced by Ronald Rivest, Adi Shamir and Len Adleman of MIT in 1977. n Algorithm involves multiplication of large (100 digits) prime numbers to produce keys. n Difficult to break the product of two large primes into its two prime factors, hence very secure. n

Number Theory A number of concepts from number theory are essential in the design of public key cryptography n. Divisors n. Prime Number n. Relatively Prime Number n. Modular Arithmetic n

Divisors We say that b =/ 0 divides a if a = mb for some m, where a, b, and m are integers n That is, b divides a if there is no remainder on division n The notation b|a is commonly used to mean b divides a n Also, if b|a, we say that d is a divisor of a n eg. The positive divisor of 24 are 1, 2, 3, 4, 6, 8, 12, and 24 n

Prime Number An integer p > 1 is a prime number if its only divisors are 1 and itself n eg. 2 3 5 7 1 13 17 19 23 29 31 37 41 43 47 53… n Any integer a > 1 can be factored in a unique way as n a = p 1 m p 2 m…. pnm where p 1 > p 2 > …. . pn are prime numbers and where each m > 0 eg. n 91 = 7 x 13 11011 = 7 x 112 x 13

Prime Number Therefore, if P is the set of all prime numbers, then any positive integer can be written uniquely as the product of all possible prime numbers n eg. 216 = 12 x 18 = (2 x 6) x (2 x 9) = (2 x 3) x (2 x 3) = 2 3 x 32 n

Relative Prime Numbers We will use the notation gcd(a, b) to mean the greatest common divisor of a and b n The positive integer is said to be the greatest common divisor of a and b if nc is a divisor of a and of b nany divisor of a and b is a divisor of c n An equivalent definition is the following: n n gcd(a, b) = max[ k, such that k|a and k|b]

Relative Prime Numbers We always take the positive value as the greatest common divisor n It is easy to determine the greatest common divisor of two positive integers if we express each integer as the product of primes n eg. 300 = 22 x 31 x 52 18 = 21 x 32 n Therefore, gcd(18, 300) = 21 x 31 x 50 = 6

Relative Prime Numbers The integers a and b are relatively prime if they have no prime factors in common, that is, if their only common factors is 1. n This is equivalent to saying that a and b are relatively prime if gcd(a, b) = 1 n eg. 8 and 15 are relatively prime because the divisors of 8 are 1, 2, 4, and 8, and the divisors of 15 are 1, 3, 5, and 15. So 1 is the only number on both lists n

Modular Arithmetic Given any positive integer n and any integer a, we get a quotient q and a remainder r that obey the following relationship: n a = qn + r 0 < r < n; q = [a/n] where [x] is the largest integer less than or equal to x eg. a = 11; n = 7; 11 = 1 x 7 + 4; n Therefore we can say that r=4 11 = 1 x 7 + 4

RSA - key generation Select p, q n. Calculate n = p x q n. Calculate m = (p-1)(q-1) n. Select integer e n. Calculate d n. Public key = { e, n } n. Private key = { d, n } n Keep p, q, m, and d secret n. Make e and n public n p and q both large prime gcd(e, m) = 1, 1 < e < m de mod m = 1

RSA Implementation RSA encrypt RSA decrypt C = Pe mod n P = Cd mod n

RSA - example 1 Steps of RSA implementation: 1. Select two prime numbers, p = 7 and q = 17 2. Calculate n = pq = 7 x 17 = 119 3. Calculate m = (7 - 1)(17 - 1) = 6 x 16 = 96 4. Select e such that e is relatively prime to m = 96 and less than m, 1 < e < 96 gcd (e, 96) = 1 96 = 1 x 6 x 16 =1 x 2 x 3 x 8 = 1 x 2 x 3 x 23 = 1 x 24 x 3 Assume that we take the smallest prime number in between of 3 to 96, so e = 5 You also can take other value as long as it fulfils the conditions – it must be relatively prime with 96 and it is less than 96

RSA - example 1 5. Determine d such that de = 1 mod 96 and d < 96. de mod m = 1 d x 5 = ( ? X 96) + 1 1 x 96 + 1 = 97 97/5 = 19. 4 2 x 96 + 1 = 192 + 1 = 193/5 = 38. 6 3 x 96 + 1 = 288 + 1 = 289/5 = 57. 8 4 x 96 + 1 = 388 + 1 = 385/5 = 77 Find the value which when the results mod 5 with no remainder The correct value is d = 77, because 77 x 5 = 385 = 4 x 96 + 1 You also can take other value as long as it is a positive integer

RSA - example 1 Therefore, n = 119, m = 96, e = 5, d = 77 The resulting keys are: Public key = { e, n } and Private key = { d, n} Public key = {5, 119} and Private key = {77, 119} Assume that P is the plaintext and C is the ciphertext The encryption is C = P 5 mod 119 The decryption is P = C 77 mod 119

RSA - example 1 Public key = { e, n } and Private key = { d, n} Public key = {5, 119} and Private key = {77, 119} Given P = 4 The encryption is C = P 5 mod 119 C = 45 mod 119 = 1024 mod 119 = 72 The decryption is P = C 77 mod 119 P = 7227 mod 119 =4

RSA - example 2 Steps of RSA implementation: 1. Select two prime numbers, p = 7 and q = 11 2. Calculate n = pq = 7 x 11 = 77 3. Calculate m = (7 - 1)(11 - 1) = 6 x 10 = 60 4. Select e such that e is relatively prime to m = 60 and less than m, 1 < e < 60 gcd (e, 60) = 1 60 = 1 x 10 x 6 =1 x 2 x 5 x 2 x 3 = 1 x 22 x 3 x 5 Assume that we take any prime number in between of 7 to 60, so e = 37 You also can take other value as long as it fulfils the conditions – it must be relatively prime with 60 and it is less than 60

RSA - example 2 5. Determine d such that de = 1 mod 60 and d < 60 de mod m = 1 d x 37 = ( ? X 60) + 1 1 x 60 + 1 = 61 2 x 60 + 1 = 121 3 x 60 + 1 = 181 4 x 60 + 1 = 241 5 x 60 + 1 = 301 6 x 60 + 1 = 361 7 x 60 + 1 = 421 8 x 60 + 1 = 481 61/37 = 1. 6 121/37 = 3. 2 181/37 = 4. 8 241/37 = 6. 5 301/37 = 8. 1 361/37 = 9. 7 421/37 = 11. 3 481/37 = 13

RSA - example 2 Find the value which when the results mod 5 with no remainder The correct value is d = 13, because 13 x 37 = 481 = 8 x 60 + 1 You also can take other value as long as it is a positive integer

RSA - example 2 Therefore, n = 77, m = 60, e = 37, d = 13 The resulting keys are: Public key = { e, n } and Private key = { d, n} Public key = {37, 77} and Private key = {13, 77} Assume that P is the plaintext and C is the ciphertext The encryption is C = P 37 mod 77 The decryption is P = C 13 mod 77

RSA - example 2 Public key = { e, n } and Private key = { d, n} Public key = {37, 77} and Private key = {13, 77} Given P = 5 The encryption is C = P 37 mod 77 C = 537 mod 77 = 1024 mod 77 = 47 The decryption is P = C 77 mod 119 P = 4713 mod 77 =5

RSA - example 3 Steps of RSA implementation: 1. Select two prime numbers, p = 3 and q = 11 2. Calculate n = pq = 3 x 11 = 33 3. Calculate m = (3 - 1)(11 - 1) = 2 x 10 = 20 4. Select e such that e is relatively prime to m = 20 and less than m, 1 < e < 20 gcd (e, 20) = 1 20 = 1 x 2 x 10 =1 x 2 x 2 x 5 = 1 x 22 x 5 Assume that we take the smallest prime number in between of 2, 5 to 20, so e = 3 You also can take other value as long as it fulfils the conditions – it must be relatively prime with 20 and it is less than 20

RSA - example 3 5. Determine d such that de = 1 mod 20 and d < 20 de mod m = 1 d x 3 = ( ? X 20) + 1 1 x 20 + 1 = 21 2 x 20 + 1 = 41 21/3 = 7 41/3 = 13. 6 Find the value which when the results mod 3 with no remainder The correct value is d = 7, because 7 x 3 = 21 = 1 x 20 + 1 You also can take other value as long as it is a positive integer

RSA - example 3 Therefore, n = 33, m = 20, e = 3, d = 7 The resulting keys are: Public key = { e, n } and Private key = { d, n} Public key = {3, 33} and Private key = {7, 33} Assume that P is the plaintext and C is the ciphertext The encryption is C = P 3 mod 33 The decryption is P = C 7 mod 33

RSA - example 3 Public key = { e, n } and Private key = { d, n} Public key = {3, 33} and Private key = {7, 33} Given P = 6 The encryption is C = P 3 mod 33 C = 63 mod 33 = 216 mod 33 = 18 The decryption is P = C 7 mod 33 P = 187 mod 33 =6

RSA - example 4 Steps of RSA implementation: 1. Select two prime numbers, p = 5 and q = 11 2. Calculate n = pq = 5 x 11 = 55 3. Calculate m = (5 - 1)(11 - 1) = 4 x 10 = 40 4. Select e such that e is relatively prime to m = 40 and less than m, 1 < e < 40 gcd (e, 40) = 1 40 = 1 x 4 x 10 =1 x 2 x 2 x 2 x 5 = 1 x 23 x 5 Assume that we take the smallest prime number in between of 2, 5 to 40, so e = 3 You also can take other value as long as it fulfils the conditions – it must be relatively prime with 40 and it is less than 40

RSA - example 4 5. Determine d such that de = 1 mod 40 and d < 40 de mod m = 1 d x 3 = ( ? X 40) + 1 1 x 40 + 1 = 41 2 x 40 + 1 = 81 41/3 = 13. 6 121/3 = 27 Find the value which when the results mod 3 with no remainder The correct value is d = 27, because 27 x 3 = 81 = 2 x 40 + 1 You also can take other value as long as it is a positive integer

RSA - example 4 Therefore, n = 55, m = 40, e = 3, d = 27 The resulting keys are: Public key = { e, n } and Private key = { d, n} Public key = {3, 55} and Private key = {27, 55} Assume that P is the plaintext and C is the ciphertext The encryption is C = P 3 mod 55 The decryption is P = C 27 mod 55

RSA - example 4 Public key = { e, n } and Private key = { d, n} Public key = {3, 55} and Private key = {27, 55} Given P = 4 The encryption is C = P 3 mod 55 C = 43 mod 55 = 64 mod 55 =9 The decryption is P = C 27 mod 55 P = 927 mod 55 =4

RSA - example 5 Steps of RSA implementation: 1. Select two prime numbers, p = 11 and q = 17 2. Calculate n = pq = 11 x 17 = 187 3. Calculate m = (11 - 1)(17 - 1) = 10 x 16 = 160 4. Select e such that e is relatively prime to m = 160 and less than m, 1 < e < 160 gcd (e, 160) = 1 160 = 1 x 10 x 16 = 1 x 2 x 10 x 2 x 8 =1 x 2 x 2 x 5 x 2 x 2 x 2 = 1 x 26 x 5 Assume that we take the smallest prime number in between of 2, 5 to 160, so e = 3 You also can take other value as long as it fulfils the conditions – it must be relatively prime with 160 and it is less than 160

RSA - example 5 5. Determine d such that de = 1 mod 160 and d < 160 de mod m = 1 d x 3 = ( ? X 160) + 1 1 x 160 + 1 = 161 2 x 160 + 1 = 321 161/3 = 53. 6 321/3 = 107 Find the value which when the results mod 3 with no remainder The correct value is d = 107, because 107 x 3 = 321 = 2 x 160 + 1 You also can take other value as long as it is a positive integer

RSA - example 5 Therefore, n = 187, m = 160, e = 3, d = 107 The resulting keys are: Public key = { e, n } and Private key = { d, n} Public key = {3, 187} and Private key = {107, 187} Assume that P is the plaintext and C is the ciphertext The encryption is C = P 3 mod 187 The decryption is P = C 107 mod 187

RSA - example 5 Public key = {3, 187} and Private key = {107, 187} Given P = 7 The encryption is C = P 3 mod 187 C = 73 mod 187 = 343 mod 187 = 156 The decryption is P = C 107 mod 187 P = 156107 mod 187 =7

RSA of trap-door function A trap-door function is a function that is easy to compute ‘forwards’ but hard to compute ‘backwards’ n Specifically, it should be fairly cheap to compute the output given the input, but computationally infeasible to recover the input given the output n It is also called as a one-way-function is one that maps a domain into a range such that every function value has a unique inverse, with the condition that the calculation of the function is easy whereas the calculation of the inverse is infeasible n

How secure is your RSA ? With the knowledge of Public key = {3, 187} and ciphertext, C = 156, how to recover the plaintext, P ? n The decryption is P = Cd mod 187 P = 156 d mod 187 Actually there are infinite possibility of d that we can try to recover the plaintext. Such condition make RSA very secure because the it is computationally expensive to do brute-force analysis (trying all the possible key) n

How secure is your RSA ? In 1977, RSA inventors offered a $100 reward for the return of a plaintext sentences, an event they predicted might not occur for some 40 quadrillion years with the public key length of 129 decimal digits or around 428 bits n In April 1994, a group working over the Internet claimed the prize after only eight months of work n