Factoring integers, Producing primes and the RSA cryptosystem Harish-Chandra Research Institute

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1 RSA cryptosystem HRI, Allahabad, February, Factoring integers, Producing primes and the RSA cryptosystem Harish-Chandra Research Institute Allahabad (UP), INDIA February, 2005

2 RSA cryptosystem HRI, Allahabad, February,

3 RSA cryptosystem HRI, Allahabad, February, RSA 2048 =

4 RSA cryptosystem HRI, Allahabad, February, RSA 2048 = RSA 2048 is a 617 (decimal) digit number

5 RSA cryptosystem HRI, Allahabad, February, RSA 2048 = RSA 2048 is a 617 (decimal) digit number

6 RSA cryptosystem HRI, Allahabad, February, RSA 2048 =p q, p, q

7 RSA cryptosystem HRI, Allahabad, February, RSA 2048 =p q, p, q PROBLEM: Compute p and q

8 RSA cryptosystem HRI, Allahabad, February, RSA 2048 =p q, p, q PROBLEM: Compute p and q Price: US$ ( 87, 36, 000 Indian Rupee)!!

9 RSA cryptosystem HRI, Allahabad, February, RSA 2048 =p q, p, q PROBLEM: Compute p and q Price: US$ ( 87, 36, 000 Indian Rupee)!! Theorem. If a N! p 1 < p 2 < < p k primes s.t. a = p α 1 1 pα k k

10 RSA cryptosystem HRI, Allahabad, February, RSA 2048 =p q, p, q PROBLEM: Compute p and q Price: US$ ( 87, 36, 000 Indian Rupee)!! Theorem. If a N! p 1 < p 2 < < p k primes s.t. a = p α 1 1 pα k k Regrettably: RSAlabs believes that factoring in one year requires: number computers memory RSA Tb RSA , 000, Gb RSA ,000 4Gb.

11 RSA cryptosystem HRI, Allahabad, February,

12 RSA cryptosystem HRI, Allahabad, February, Challenge Number Prize ($US) RSA 576 $10,000 RSA 640 $20,000 RSA 704 $30,000 RSA 768 $50,000 RSA 896 $75,000 RSA 1024 $100,000 RSA 1536 $150,000 RSA 2048 $200,000

13 RSA cryptosystem HRI, Allahabad, February, Challenge Number Prize ($US) Status RSA 576 $10,000 Factored December 2003 RSA 640 $20,000 Not Factored RSA 704 $30,000 Not Factored RSA 768 $50,000 Not Factored RSA 896 $75,000 Not Factored RSA 1024 $100,000 Not Factored RSA 1536 $150,000 Not Factored RSA 2048 $200,000 Not Factored

14 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring

15 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene )

16 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler =

17 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler = Fermat, Gauss (Sieves - Tables)

18 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler = Fermat, Gauss (Sieves - Tables) 1880 Landry & Le Lasseur: =

19 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler = Fermat, Gauss (Sieves - Tables) 1880 Landry & Le Lasseur: = Pierre and Eugène Carissan (Factoring Machine)

20 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler = Fermat, Gauss (Sieves - Tables) 1880 Landry & Le Lasseur: = Pierre and Eugène Carissan (Factoring Machine) 1970 Morrison & Brillhart =

21 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler = Fermat, Gauss (Sieves - Tables) 1880 Landry & Le Lasseur: = Pierre and Eugène Carissan (Factoring Machine) 1970 Morrison & Brillhart = Quadratic Sieve QS (Pomerance) Number Fields Sieve NFS

22 RSA cryptosystem HRI, Allahabad, February, History of the Art of Factoring 220 BC Greeks (Eratosthenes of Cyrene ) 1730 Euler = Fermat, Gauss (Sieves - Tables) 1880 Landry & Le Lasseur: = Pierre and Eugène Carissan (Factoring Machine) 1970 Morrison & Brillhart = Quadratic Sieve QS (Pomerance) Number Fields Sieve NFS 1987 Elliptic curves factoring ECF (Lenstra)

23 RSA cryptosystem HRI, Allahabad, February, Carissan s ancient Factoring Machine

24 RSA cryptosystem HRI, Allahabad, February, Carissan s ancient Factoring Machine Figure 1: Conservatoire Nationale des Arts et Métiers in Paris

25 RSA cryptosystem HRI, Allahabad, February, Carissan s ancient Factoring Machine Figure 1: Conservatoire Nationale des Arts et Métiers in Paris shallit/papers/carissan.html

26 RSA cryptosystem HRI, Allahabad, February, Figure 2: Lieutenant Eugène Carissan

27 RSA cryptosystem HRI, Allahabad, February, Figure 2: Lieutenant Eugène Carissan = minutes = minutes = minutes

28 RSA cryptosystem HRI, Allahabad, February, Contemporary Factoring

29 RSA cryptosystem HRI, Allahabad, February, Contemporary Factoring ❶ 1994, Quadratic Sieve (QS): (8 months, 600 voluntaries, 20 countries) D.Atkins, M. Graff, A. Lenstra, P. Leyland RSA 129 = = =

30 RSA cryptosystem HRI, Allahabad, February, Contemporary Factoring ❶ 1994, Quadratic Sieve (QS): (8 months, 600 voluntaries, 20 countries) D.Atkins, M. Graff, A. Lenstra, P. Leyland RSA 129 = = = ❷ (February ), Number Fields Sieve (NFS): (160 Sun, 4 months) RSA 155 = = =

31 RSA cryptosystem HRI, Allahabad, February, Contemporary Factoring ❶ 1994, Quadratic Sieve (QS): (8 months, 600 voluntaries, 20 countries) D.Atkins, M. Graff, A. Lenstra, P. Leyland RSA 129 = = = ❷ (February ), Number Fields Sieve (NFS): (160 Sun, 4 months) RSA 155 = = = ❸ (December 3, 2003) (NFS): J. Franke et al. (174 decimal digits) RSA 576 = = =

32 RSA cryptosystem HRI, Allahabad, February, Contemporary Factoring ❶ 1994, Quadratic Sieve (QS): (8 months, 600 voluntaries, 20 countries) D.Atkins, M. Graff, A. Lenstra, P. Leyland RSA 129 = = = ❷ (February ), Number Fields Sieve (NFS): (160 Sun, 4 months) RSA 155 = = = ❸ (December 3, 2003) (NFS): J. Franke et al. (174 decimal digits) RSA 576 = = = ❹ Elliptic curves factoring: introduced by da H. Lenstra. suitable to find prime factors with 50 digits (small)

33 RSA cryptosystem HRI, Allahabad, February, Contemporary Factoring ❶ 1994, Quadratic Sieve (QS): (8 months, 600 voluntaries, 20 countries) D.Atkins, M. Graff, A. Lenstra, P. Leyland RSA 129 = = = ❷ (February ), Number Fields Sieve (NFS): (160 Sun, 4 months) RSA 155 = = = ❸ (December 3, 2003) (NFS): J. Franke et al. (174 decimal digits) RSA 576 = = = ❹ Elliptic curves factoring: introduced by da H. Lenstra. suitable to find prime factors with 50 digits (small)

34 RSA cryptosystem HRI, Allahabad, February, All: sub exponential running time

35 RSA cryptosystem HRI, Allahabad, February, RSA Adi Shamir, Ron L. Rivest, Leonard Adleman (1978)

36 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem

37 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998)

38 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it

39 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it A (Alice) B (Bob) C (Charles)

40 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it ❶ ❷ ❸ ❹ A (Alice) B (Bob) C (Charles)

41 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it A (Alice) B (Bob) C (Charles) ❶ Key generation Bob has to do it ❷ ❸ ❹

42 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it A (Alice) B (Bob) C (Charles) ❶ Key generation ❷ Encryption Bob has to do it Alice has to do it ❸ ❹

43 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it A (Alice) B (Bob) C (Charles) ❶ Key generation ❷ Encryption ❸ Decryption Bob has to do it Alice has to do it Bob has to do it ❹

44 RSA cryptosystem HRI, Allahabad, February, The RSA cryptosystem 1978 R. L. Rivest, A. Shamir, L. Adleman (Patent expired in 1998) Problem: Alice wants to send the message P to Bob so that Charles cannot read it A (Alice) B (Bob) C (Charles) ❶ Key generation ❷ Encryption ❸ Decryption ❹ Attack Bob has to do it Alice has to do it Bob has to do it Charles would like to do it

45 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation

46 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation

47 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q )

48 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1)

49 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t.

50 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1

51 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1 Note. One could take e = 3 and p q 2 mod 3

52 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1 Note. One could take e = 3 and p q 2 mod 3 Experts recommend e =

53 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1 Note. One could take e = 3 and p q 2 mod 3 Experts recommend e = He computes arithmetic inverse d of e modulo ϕ(m)

54 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1 Note. One could take e = 3 and p q 2 mod 3 Experts recommend e = He computes arithmetic inverse d of e modulo ϕ(m) (i.e. d N (unique ϕ(m)) s.t. e d 1 (mod ϕ(m)))

55 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1 Note. One could take e = 3 and p q 2 mod 3 Experts recommend e = He computes arithmetic inverse d of e modulo ϕ(m) (i.e. d N (unique ϕ(m)) s.t. e d 1 (mod ϕ(m))) Publishes (M, e) public key and hides secret key d

56 RSA cryptosystem HRI, Allahabad, February, Bob: Key generation He chooses randomly p and q primes (p, q ) He computes M = p q, ϕ(m) = (p 1) (q 1) He chooses an integer e s.t. 0 e ϕ(m) and gcd(e, ϕ(m)) = 1 Note. One could take e = 3 and p q 2 mod 3 Experts recommend e = He computes arithmetic inverse d of e modulo ϕ(m) (i.e. d N (unique ϕ(m)) s.t. e d 1 (mod ϕ(m))) Publishes (M, e) public key and hides secret key d Problem: How does Bob do all this?- We will go came back to it!

57 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption

58 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption Represent the message P as an element of Z/MZ

59 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption Represent the message P as an element of Z/MZ (for example) A 1 B 2 C 3... Z 26 AA 27...

60 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption Represent the message P as an element of Z/MZ (for example) A 1 B 2 C 3... Z 26 AA Sukumar = Note. Better if texts are not too short. Otherwise one performs some padding

61 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption Represent the message P as an element of Z/MZ (for example) A 1 B 2 C 3... Z 26 AA Sukumar = Note. Better if texts are not too short. Otherwise one performs some padding C = E(P) = P e (mod M)

62 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption Represent the message P as an element of Z/MZ (for example) A 1 B 2 C 3... Z 26 AA Sukumar = Note. Better if texts are not too short. Otherwise one performs some padding C = E(P) = P e (mod M) Example: p = , q = , M = , e = = 65537, P = Sukumar:

63 RSA cryptosystem HRI, Allahabad, February, Alice: Encryption Represent the message P as an element of Z/MZ (for example) A 1 B 2 C 3... Z 26 AA Sukumar = Note. Better if texts are not too short. Otherwise one performs some padding C = E(P) = P e (mod M) Example: p = , q = , M = , e = = 65537, P = Sukumar: E(Sukumar) = (mod ) = = C = JGEBNBAUYTCOFJ

64 RSA cryptosystem HRI, Allahabad, February, Bob: Decryption

65 RSA cryptosystem HRI, Allahabad, February, Bob: Decryption P = D(C) = C d (mod M)

66 RSA cryptosystem HRI, Allahabad, February, Bob: Decryption P = D(C) = C d (mod M) Note. Bob decrypts because he is the only one that knows d.

67 RSA cryptosystem HRI, Allahabad, February, Bob: Decryption P = D(C) = C d (mod M) Note. Bob decrypts because he is the only one that knows d. Theorem. (Euler) If a, m N, gcd(a, m) = 1, a ϕ(m) 1 (mod m). If n 1 n 2 mod ϕ(m) then a n 1 a n 2 mod m.

68 RSA cryptosystem HRI, Allahabad, February, Bob: Decryption P = D(C) = C d (mod M) Note. Bob decrypts because he is the only one that knows d. Theorem. (Euler) If a, m N, gcd(a, m) = 1, a ϕ(m) 1 (mod m). If n 1 n 2 mod ϕ(m) then a n 1 a n 2 mod m. Therefore (ed 1 mod ϕ(m)) D(E(P)) = P ed P mod M

69 RSA cryptosystem HRI, Allahabad, February, Bob: Decryption P = D(C) = C d (mod M) Note. Bob decrypts because he is the only one that knows d. Therefore (ed 1 mod ϕ(m)) Theorem. (Euler) If a, m N, gcd(a, m) = 1, a ϕ(m) 1 (mod m). If n 1 n 2 mod ϕ(m) then a n 1 a n 2 mod m. D(E(P)) = P ed P mod M Example(cont.):d = mod ϕ( ) = D(JGEBNBAUYTCOFJ) = (mod ) = Sukumar

70 RSA cryptosystem HRI, Allahabad, February, RSA at work

71 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm

72 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c?

73 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod )

74 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod )

75 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod ) Compute the binary expansion b = [log 2 b] j=0 ɛ j 2 j

76 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod ) Compute the binary expansion b = [log 2 b] j=0 ɛ j 2 j =

77 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod ) Compute the binary expansion b = [log 2 b] j=0 ɛ j 2 j = Compute recursively a 2j mod c, j = 1,..., [log 2 b]:

78 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod ) Compute the binary expansion b = [log 2 b] j=0 ɛ j 2 j = Compute recursively a 2j mod c, j = 1,..., [log 2 b]: ( 2 a 2j mod c = a 2j 1 mod c) mod c

79 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod ) Compute the binary expansion b = [log 2 b] j=0 ɛ j 2 j = Compute recursively a 2j mod c, j = 1,..., [log 2 b]: ( 2 a 2j mod c = a 2j 1 mod c) mod c Multiply the a 2j mod c with ɛ j = 1

80 RSA cryptosystem HRI, Allahabad, February, Repeated squaring algorithm Problem: How does one compute a b mod c? (mod ) Compute the binary expansion b = [log 2 b] j=0 ɛ j 2 j = Compute recursively a 2j mod c, j = 1,..., [log 2 b]: ( 2 a 2j mod c = a 2j 1 mod c) mod c Multiply the a 2j mod c with ɛ j = 1 ) a b mod c = mod c ( [log2 b] j=0,ɛ j =1 a2j mod c

81 RSA cryptosystem HRI, Allahabad, February, #{oper. in Z/cZ to compute a b mod c} 2 log 2 b

82 RSA cryptosystem HRI, Allahabad, February, #{oper. in Z/cZ to compute a b mod c} 2 log 2 b JGEBNBAUYTCOFJ is decrypted with 131 operations in Z/ Z

83 RSA cryptosystem HRI, Allahabad, February, #{oper. in Z/cZ to compute a b mod c} 2 log 2 b JGEBNBAUYTCOFJ is decrypted with 131 operations in Z/ Z Pseudo code: e c (a, b) = a b mod c

84 RSA cryptosystem HRI, Allahabad, February, #{oper. in Z/cZ to compute a b mod c} 2 log 2 b JGEBNBAUYTCOFJ is decrypted with 131 operations in Z/ Z Pseudo code: e c (a, b) = a b mod c e c (a, b) = if b = 1 then a mod c if 2 b then e c (a, b 2 )2 mod c else a e c (a, b 1 2 )2 mod c

85 RSA cryptosystem HRI, Allahabad, February, #{oper. in Z/cZ to compute a b mod c} 2 log 2 b JGEBNBAUYTCOFJ is decrypted with 131 operations in Z/ Z Pseudo code: e c (a, b) = a b mod c e c (a, b) = if b = 1 then a mod c if 2 b then e c (a, b 2 )2 mod c else a e c (a, b 1 2 )2 mod c To encrypt with e = , only 17 operations in Z/MZ are enough

86 RSA cryptosystem HRI, Allahabad, February, Key generation

87 RSA cryptosystem HRI, Allahabad, February, Key generation Problem. Produce a random prime p Probabilistic algorithm (type Las Vegas) 1. Let p = Random( ) 2. If isprime(p)=1 then Output=p else goto 1

88 RSA cryptosystem HRI, Allahabad, February, Key generation Problem. Produce a random prime p Probabilistic algorithm (type Las Vegas) 1. Let p = Random( ) 2. If isprime(p)=1 then Output=p else goto 1 subproblems:

89 RSA cryptosystem HRI, Allahabad, February, Key generation Problem. Produce a random prime p Probabilistic algorithm (type Las Vegas) 1. Let p = Random( ) 2. If isprime(p)=1 then Output=p else goto 1 subproblems: A. How many iterations are necessary? (i.e. how are primes distributes?)

90 RSA cryptosystem HRI, Allahabad, February, Key generation Problem. Produce a random prime p Probabilistic algorithm (type Las Vegas) 1. Let p = Random( ) 2. If isprime(p)=1 then Output=p else goto 1 subproblems: A. How many iterations are necessary? (i.e. how are primes distributes?) B. How does one check if p is prime? (i.e. how does one compute isprime(p)?) Primality test

91 RSA cryptosystem HRI, Allahabad, February, Key generation Problem. Produce a random prime p Probabilistic algorithm (type Las Vegas) 1. Let p = Random( ) 2. If isprime(p)=1 then Output=p else goto 1 subproblems: A. How many iterations are necessary? (i.e. how are primes distributes?) B. How does one check if p is prime? (i.e. how does one compute isprime(p)?) Primality test False Metropolitan Legend: Check primality is equivalent to factoring

92 RSA cryptosystem HRI, Allahabad, February, A. Distribution of prime numbers

93 RSA cryptosystem HRI, Allahabad, February, A. Distribution of prime numbers π(x) = #{p x t. c. p is prime}

94 RSA cryptosystem HRI, Allahabad, February, A. Distribution of prime numbers π(x) = #{p x t. c. p is prime} Theorem. (Hadamard - de la vallee Pussen ) π(x) x log x

95 RSA cryptosystem HRI, Allahabad, February, A. Distribution of prime numbers Quantitative version: π(x) = #{p x t. c. p is prime} Theorem. (Hadamard - de la vallee Pussen ) π(x) x log x Theorem. (Rosser - Schoenfeld) if x 67 x log x 1/2 < π(x) < x log x 3/2

96 RSA cryptosystem HRI, Allahabad, February, A. Distribution of prime numbers Quantitative version: Therefore π(x) = #{p x t. c. p is prime} Theorem. (Hadamard - de la vallee Pussen ) π(x) x log x Theorem. (Rosser - Schoenfeld) if x 67 x log x 1/2 < π(x) < x log x 3/ < P rob (Random( ) = prime <

97 RSA cryptosystem HRI, Allahabad, February, If P k is the probability that among k random numbers there is a prime one, then

98 RSA cryptosystem HRI, Allahabad, February, If P k is the probability that among k random numbers there is a prime one, then P k = 1 ( ) k 1 π(10100 )

99 RSA cryptosystem HRI, Allahabad, February, If P k is the probability that among k random numbers there is a prime one, then P k = 1 ( ) k 1 π(10100 ) Therefore < P 250 <

100 RSA cryptosystem HRI, Allahabad, February, If P k is the probability that among k random numbers there is a prime one, then P k = 1 ( ) k 1 π(10100 ) Therefore < P 250 < To speed up the process: One can consider only odd random numbers not divisible by 3 nor by 5.

101 RSA cryptosystem HRI, Allahabad, February, If P k is the probability that among k random numbers there is a prime one, then P k = 1 ( ) k 1 π(10100 ) Therefore < P 250 < To speed up the process: One can consider only odd random numbers not divisible by 3 nor by 5. Let Ψ(x, 30) = # {n x s.t. gcd(n, 30) = 1}

102 RSA cryptosystem HRI, Allahabad, February, To speed up the process: One can consider only odd random numbers not divisible by 3 nor by 5.

103 RSA cryptosystem HRI, Allahabad, February, To speed up the process: One can consider only odd random numbers not divisible by 3 nor by 5. Let Ψ(x, 30) = # {n x s.t. gcd(n, 30) = 1} then

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