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Prime Numbers The Fundamental Theorem of Arithmetic

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Prime Numbers and RSA

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**Euclidean algorithm Eudoxus of Cnidus (about 375 BC),**

function gcd(a, b) if a = 0 return b while b ≠ 0 if a > b a := a − b else b := b − a return a

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**Euclid’s algorithm: GCD**

The greatest common divisor of M and N is the largest whole number that divides evenly into both M and N GCD (6 , 15 ) = 3 If GCD (M, N) = 1 then M and N are called relatively prime. Euclid’s algorithm: method to find GCD (M,N)

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**Euclid’s algorithm M and N whole numbers.**

Suppose M ≤ N. If N is divisible by M then GCD(M,N) = M. Otherwise, subtract from N the biggest multiple of M that is smaller than N. Call the remainder R. N=MK+R or R=MK-N. If Q divides into both M and N then Q divides into R. So: GCD(M,N) = GCD (M,R). Repeat until R divides into previous.

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**Example: GCD (105, 77) 77 does not divide 105.**

Subtract 1*77 from 105. Get R=28 28 does not divide into 77. Subtract 2*28 from 77. Get R=77-56=21 Subtract 21 from 28. Get 7. 7 divides into 21. Done. GCD (105, 77) = 7.

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**Example: GCD (105, 47) 47 does not divide 105.**

Subtract 2*47 from 105. Get R=11 11 does not divide into 47. Subtract 4*11 from 47. Get R=47-44=3 3 does not divide 11. Subtract 3*3 from 11. R=2 2 does not divide 3. Subtract 2 from 3. R=1 GCD (105, 47) = 1.

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Exercise: find GCD (1234,121)

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Relatively prime Two numbers M and N are called relatively prime if GCD(M,N)=1.

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Prime numbers A whole number is called prime if it is relatively prime to every smaller whole number.

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**Prime factorization theorem**

fundamental theorem of arithmetic : every natural number > 1 can be written as a unique product of prime numbers. Example: 6936=2 x 2 x 2 x 3 x 17 x 17 =2^3 x 3 x 17^2 No other way of writing 6936 as a product of prime powers practically proved by Euclid, first “ correct” proof in Disquisitiones Arithmeticae by Gauss.

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[GCD(6251,1473)=] [1] [3] [7] [11]

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**Large prime numbers Euclid: infinitely many prime numbers**

Proof: given a list of prime numbers, multiply all of them together and add one. Either the new number is prime or there is a smaller prime not in the list.

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Euclid’s proof Consider any finite set of primes. Multiply all of them together and add 1 (see Euclid number). Call this Q Dividing Q by any of these would give a remainder of 1. So Q is not divisible by any number in this list. Any non-prime can be decomposed into a product of primes, Either Q is prime itself, or there is a prime number in the decomposition of Q that is not in the original finite set of primes. Either way, there is at least one more prime that was not in the finite set we started with. This argument applies no matter what finite set we began with. So there are more primes than any given finite number.

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Infinitude of primes 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 … list of prime numbers

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**Testing for prime numbers**

Is 97 a prime number? How about 111? How about ?

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**Computers can factor small prime numbers**

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Primality test A primality test is an algorithm for determining whether an input number is prime. As of 2008, factorization is a computationally hard problem, whereas primality testing is comparatively easy. elliptic curve primality test O((log n)^6), Based on believed, but unproven properties 2002: Agrawal, Saxena and Kayal : AKS primality test, Õ((log n)^6) slower than probabilistic methods.

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RSA 200 RSA-200 = RSA-200 = ×

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How long to factor RSA 200? If a k-digit number is the product of two primes No known algorithm can factor in polynomial time, i.e., that can factor it in time O(k^p) for some constant p. There are algorithms faster than O((1+ε)^k) i.e., sub-exponential. For a quantum computer, Peter Shor discovered an algorithm in 1994 that solves it in polynomial time O(N^3) and O(N) memory. In 2001, the first 7-qubit quantum computer became the first to run Shor's algorithm. It factored the number 15 GNFS: O(2^(N^(1/3))

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**Racing Mathematicians**

German Federal Agency for Information Technology Security (BSI) team record for factorization of semiprimes ( RSA Factoring Challenge ) On May 9, 2005, factored RSA-200, a 663-bit number (200 decimal digits), using the general number field sieve. …later: RSA-640, a smaller number containing 193 decimal digits (640 bits), on November 4, 2005. Both factorizations required several months of computer time using the combined power of 80 AMD Opteron CPUs.

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**[The number 2*3*5*7*11+1=2311 is prime]**

True False

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**There are lots of primes known to man**

Prime number theorem: the number of primes less than or equal to N is on the order of N divided by log N.

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Largest known prime 243,112,609 − 1.

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**An example gcd(1071,1029) =gcd(1029,42) (42= 1071 mod 1029**

=21: since b=0, we return a

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**Run time of Euclidean (O(N^2)). Red: fast, blue: slow**

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RSA

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**The RSA encryption algorithm**

N=PQ (product of two primes) Φ(N) = (P-1)(Q-1) Encryption key: 1<E<φ(N) such that GCD(E , φ) = 1 Decryption key: D such that DE ≡ 1 mod (φ) M< φ

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**Encryption/Decryption**

C=MD mod (N) R=CE mod (N) CLAIM: R=M (the original message)

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**Short digression: modular arithmetic**

A ≡ B mod (C) Means that B is the remainder when C is divided into A For example, 13 ≡ 1 mod (12) If it is 3:30 now then in 13 hours it will be 4:30. Shorthand: B=mod(A,C) Arithmetic: mod (MN, C)=mod(mod(M,C) x mod(N,C), C))

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**Laws of exponents Regular exponents Modular exponents (ab)c = ac bc**

mod((ab)c ,m)= mod(mod (ac ,m) mod (bc ,m), m) ab+c=ab ac mod(ab+c,m)= mod(mod (ab ,m) mod (ac ,m), m) (ab)c = abc mod((mod (ab ,m))c, m)

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**Examples mod((13)25 ,12)= mod((mod (13 ,12))25, 12)=**

mod(225, 12)= mod(mod(24, 12)6 x mod(2,12), 12)= mod(mod(4, 12)5 x2, 12)=mod(8 , 12)=8

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**Proof of RSA C=ME mod (N) R=CDmod (N) = (MD mod (N))E mod (N) =**

(MDE mod (N)) DE ≡ 1 mod (N) (M1 mod (N)) =M (since M< N) Fermat’s little theorem: aP-1 ≡ 1 mod (P)

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**Plaintext to numbers - A B C D E F G H I J K L M N O P Q R S T U V W X**

Y Z 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

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**Plaintext message numerical version Kill Bill**

Note: M<φ so may need to send message in pieces, e.g. one letter at a time

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Example N = 5 x 7 =35 Φ=4x6=24 E=11 then GCD (E, Φ)=1 D=11 then DxE=121=5x24+1 So DxE≡1 mod 24 In this case decryption is just the inverse of encryption because E=D. Generally note true.

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- A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 28 27 33 32 30 31

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EXERCISE Use the cipher table above to decrypt the following ciphertext into plain text:

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Solution: Flattery will get you nowhere, but don't stop trying

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**Simple; multiply numbers**

Difficult: factor numbers. example x 99991= (easy) M= = A xB Find A and B (difficult) Computer: check primes up to square-root (roughly 38000).

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**How long to factor large products?**

what if the number to be factored is not ten digits, but rather 400 digits? square-root : 200 digits. lifetime of universe: approx. 10^{18} seconds. If computer could test one trillion factorizations per second, in the lifetime of the universe it could check 10^{30} possibilities. But there are 10^{200} possibilities.

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RSA outline find two huge prime numbers, p and q (about 200 digits) (private key), N=pq public key. Baby prime example p=23 and q=41 pq = (23)(41) = 943, the ``public key’’. E: encryption key E is relatively prime to (p-1)(q-1) = (22)(40) = 880, E=7 is ok as public key To encode the message number M=35. : C = M^e (mod N) = 35^7 (mod 943) = (mod 943) = 545 The number C=545 is the encoding of M that is sent. D: decryption key ed = 1 { mod} (p-1)(q-1): D=503 works, since 7*503 = 3521 = 4(880) + 1

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Why decoding is “easy” Must calculate C^D (mod N) = 545^{503} (mod 943). 503 = so 545^{503} = 545^{ } = 545^{256}545^{128}… 545^1. (mod 943) all the exponents that are powers of 2. For example, 545^2 ( mod 943) = 545*545 = (mod 943) = 923 Square again: 545^4 (mod 943) = (545^2)^2 ( mod} 943) = 923*923 = (mod 943) = 400, and so on. We obtain the following table: 545^1 (mod 943) = 545 545^2 (mod} 943) =…= 18 545^{256} (mod 943) = 324 So …545^{503} (mod 943) = 324*18*215*795*857*400*923*545 (mod 943) = 35=M !

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**Memory efficient modular exponentiation: Montgomery reduction**

Memory-efficient method Makes use of: given two integers a and b, the following two equations are equivalent: C = ab mod m C = ( (a mod m)x (bmod m) ) mod The algorithm is as follows: 1. Set c = 1, e' = 0. 2. Increase e' by 1. 3. Set c =b mod m 4. If e' <e, goto step 2. Else, c contains the correct solution to c = b^e mod m in every pass through step 3, c = b^(e’) mod m holds true. In summary, this algorithm basically counts up e' by ones until e' reaches e, doing a multiply by b and the modulo operation each time it adds one

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**The example b = 4, e = 13, and m = 497 is presented again**

The example b = 4, e = 13, and m = 497 is presented again. The algorithm passes through step 3 thirteen times: * e' = 1. c = (1 * 4) mod 497 = 4 mod 497 = 4. * e' = 2. c = (4 * 4) mod 497 = 16 mod 497 = 16. * e' = 3. c = (16 * 4) mod 497 = 64 mod 497 = 64. * e' = 4. c = (64 * 4) mod 497 = 256 mod 497 = 256. * e' = 5. c = (256 * 4) mod 497 = 1024 mod 497 = 30. * e' = 6. c = (30 * 4) mod 497 = 120 mod 497 = 120. * e' = 7. c = (120 * 4) mod 497 = 480 mod 497 = 480. * e' = 8. c = (480 * 4) mod 497 = 1920 mod 497 = 429. * e' = 9. c = (429 * 4) mod 497 = 1716 mod 497 = 225. * e' = 10. c = (225 * 4) mod 497 = 900 mod 497 = 403. * e' = 11. c = (403 * 4) mod 497 = 1612 mod 497 = 121. * e' = 12. c = (121 * 4) mod 497 = 484 mod 497 = 484. * e' = 13. c = (484 * 4) mod 497 = 1936 mod 497 = 445. The final answer for c is therefore 445 requires O(e) multiplications to complete. the computation time decreases by a factor of at least O(e) in this method.

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**An efficient method: the right-to-left binary algorithm**

third method: combines previous method and “exponentiation by squaring” convert exponent e to binary notation. That is, e can be written as: e = a_0 x 1 + a_1 x 2^1 + a_2 x 2^2 +… +a_N x 2^N “length” of e is N bits. a_i = 0 or 1 b^e = (b^0)^(a_0) x (b^1)^(a_1) x…x (b^N)^(a_N) So C= (b^0)^(a_0) x (b^1)^(a_1) x…x (b^N)^(a_N) mod M

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RSA history Algorithm described in 1977 by Ron Rivest, Adi Shamir, and Leonard Adleman at MIT; Clifford Cocks, a British mathematician working for the UK intelligence agency GCHQ, described an equivalent system in an internal document in His discovery, however, was not revealed until 1997 due to its top-secret classification, and Rivest, Shamir, and Adleman devised RSA independently of Cocks' work. MIT was granted U.S. Patent 4,405,829 for a "Cryptographic communications system and method" that used the algorithm in The patent would have expired in 2003, but was released to the public domain by RSA on 21 September 2000.

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Pretty Good Privacy

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**PGP Based on “public key” cryptography**

Binds public key to user name or address Authentication: “digital signature” used to verify identity of sender integrity checking: used to detect whether a message has been altered since it was completed Encryption: based on RSA/DSA Decryption based on public key Web of trust: third party vetting

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**Zimmerman, 1992 As time goes on…**

you accumulate keys from other “trusted” parties. Others each choose their own trusted parties. everyone gradually accumulates and distributes with their key certifying signatures from others Expectation: anyone receiving it will trust at least one or two of the signatures. emergence of a decentralized fault-tolerant web of confidence for all public keys.

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**Any agency wanting to read PGP messages would probably use easier means than standard cryptanalysis,**

e.g. rubber-hose cryptanalysis or black-bag cryptanalysis i.e. installing some form of trojan horse or keystroke logging software/hardware on the target computer to capture encrypted keyrings and their passwords. The FBI has used this attack against PGP. such vulnerabilities apply to any encryption software.

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**Criminal investigation of Zimmerman**

PGP encryption found its way outside the US. Cryptosystems using keys > 40 bits were considered munitions by US export regulations; PGP keys >= 128 bits. Feb 1993: Zimmermann targeted by US Govt for "munitions export without a license". Penalties … were substantial. Zimmermann challenged these regulations in a curious way. Published PGP source code as hardback book (MIT Press) To build: buy the $60 book, scan pages using an OCR program, GNU C Compiler. PGP would thus be available anywhere in the world. Export of munitions restricted; export of books is protected ( First Amendment). US export regulations regarding cryptography remain in force, but were liberalized substantially …. PGP … can be exported internationally except to 7 specific countries and a named list of groups and individuals.

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**Mathematical Cryptography**

William S. Jevons,” The Principles of Science: A Treatise on Logic and Scientific Method,” (1890s) observed many situations where 'direct' operation is “easy,” but ‘inverse' operation ‘hard’. Example: encryption is easy; decryption is hard. Jevons Ch 7: multiplication of integers is easy, finding (prime) factors of product is hard. Jevons anticipated RSA Algorithm for public key cryptography

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**The future of cryptography?**

As of 2005, the largest number factored by a general-purpose factoring algorithm was 663 bits long (see RSA-200), using a state-of-the-art distributed implementation. RSA keys are typically 1024–2048 bits long. Some experts believe that 1024-bit keys may become breakable in the near term (though this is disputed); few see any way that 4096-bit keys could be broken in the foreseeable future. Therefore, it is generally presumed that RSA is secure if n is sufficiently large. If n is 256 bits or shorter, it can be factored in a few hours on a personal computer, using software already freely available. Keys of 512 bits (or less) have been shown to be practically breakable in 1999 when RSA-155 was factored by using several hundred computers. A theoretical hardware device named TWIRL and described by Shamir and Tromer in 2003 called into question the security of 1024 bit keys. It is currently recommended that n be at least 2048 bits long.

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Is RSA safe? In 1994, Peter Shor published Shor's algorithm, showing that a quantum computer could in principle perform the factorization in polynomial time. However, quantum computation is still in the early stages of development and may never prove to be practical.

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