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P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness Month Emily List Wittenberg University s07.elist@wittenberg.edu

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April 2006: Mathematics Awareness Month Mathematics and Internet Security

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Definitions P: yes or no decision problems that can be solved by an algorithm that runs in polynomial time. Polynomial time: the number of steps needed to solve a problem can be expressed as a function. Where x is the size of the input and n is a constant.

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Whats so great about polynomial time? Running time of algorithm t(n) Maximum size solvable in 1 second nN 0 =100 million 100 N 0 1000 N 0 100nN 1 =1 million100 N 1 1000 N 1 n2n2 N 2 =10,00010 N 2 31.6 N 2 n3n3 N 3 =4644.64 N 3 10 N 3 2n2n N 4 =26N 4 +6.64N 4 +9.97 Current computer 100 times faster 1000 times faster Ramachandran, Vijaya. P versus NP

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NP: a problem that can be verified using an algorithm that runs in polynomial time IMPORTANT: This does not mean not polynomial time Definitions Continued

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What would a solution to P = NP? look like? or P NP P

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Why is P vs NP important? Clay Mathematics Institute: $1,000,000 prize Internet security implications Public Key Encryption Whitfield Diffie and Martin Hellman, 1976 RSA public-key cryptosystem Ronald Rivest, Adi Shamir, and Leonard Aldeman, 1977

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RSA Encryption Uses a function that is NP but not known to be P to encrypt information. Fermats Little Theorem: Let a and p be integers such that p is prime and gcd(a, p) =1, then

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Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof.

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Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m.

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Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m. Then by Fermats little theorem: (m (p-1) ) (q-1)k 1 (m e ) f m (p-1)(q-1)k m m (mod p)

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Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m. Then by Fermats little theorem: (m (p-1) ) (q-1)k 1 (m e ) f m (p-1)(q-1)k m m (mod p) Similarly, (m e ) f m (p-1)(q-1)k m m (mod q).

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Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m. Then by Fermats little theorem: (m (p-1) ) (q-1)k 1 (m e ) f m (p-1)(q-1)k m m (mod p) Similarly, (m e ) f m (p-1)(q-1)k m m (mod q). Therefore, by the Chinese Remainder Theorem we have (m e ) f (mod n) m.

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RSA Example Necessary InformationWhere is comes fromWhat it is for this example p,qprimep=67 q=89 npq5963 Φ(n)Number of integers less than n that are relatively prime to n. (p-1)(q-1) 5808 e,fe,f >1 such thate = 37 f = 157 We want to encrypt the number 17: x e (mod n) 17 16 (mod 5963) 5064 To decrypt: 5064 f (mod 5963) 5064 157 17

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Why is RSA secure? Its nearly impossible to find f without the factors of n. Since we do not have an algorithm that runs in polynomial time to find factorizations, finding the factors n is nearly impossible.

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Is this number prime, if so what are its factors? 2039568783564019774057658669290345772 8019399331434826309477264645328306272 2701277632936616063144088173312372882 6771238795387094001583065673383282791 5449969836607190676644003707421711780 5690872792848149112022286332144876183 3763265120835748216479339929612499173 1983621930427428024380310401500056379 0123

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1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

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1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

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1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

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1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

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Does the Sieve of Eratosthenes run in polynomial time? NO. Why not? For a number with N digits, the number of steps the sieve needs is [10 N ] 1/2 which is exponential.

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Primes is in P In 2002, Manindra Agrawal, Neeraj Kayal and Nitin Saxena came up with an algorithm that runs in polynomial and give the primality of a number. This algorithm is beautiful Carl Pomerance The proof is simple, elegant and beautiful R. Balasubramanian

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AKS Algorithm From PRIMES is in P

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Explanation of AKS Lemma 2.1 Let a be an integer, n is a natural number, n > 2 and gcd(a,n)=1. Then n is prime iff (X+ a) n X n +a(mod n). Proof. By the binomial theorem: the coefficient of x i in ((X+a) n –(X n +a) is a n-i.

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Explanation of AKS Lemma 2.1 Let a be an integer, n is a natural number, n > 2 and gcd(a,n)=1. Then n is prime iff (X+ a) n X n +a(mod n). Proof. By the binomial theorem: the coefficient of x i in ((X+a) n –(X n +a) is a n-i. Suppose n is prime. Then 0 (mod n) and hence all of the coefficients are zero.

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Explanation of AKS Lemma 2.1 Let a be an integer, n is a natural number, n > 2 and gcd(a,n)=1. Then n is prime iff (X+ a) n X n +a(mod n). Proof. By the binomial theorem: the coefficient of x i in ((X+a) n –(X n +a) is a n-i. Suppose n is prime. Then 0 (mod n) and hence all of the coefficients are zero. Suppose n is composite. Consider a prime q that is a factor of n and let q k divide n, but q k+1 does not. Then q k does not divide and gcd( a n-q, q k ) =1 Hence, the coefficient of X q is not zero (mod n). Therefore (X+a) n X n +a (mod n).

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Does AKS ruin RSA? NO!! Why not? AKS does not factor a number, it only tells us if it is prime or not. RSA is secure as long as we dont have an algorithm that can factor in polynomial time.

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Acknowledgements Manindra Agrawal, Neeraj Kayal, and Nitin Saxena. PRIMES is in P. (http://www.cse.iitk.ac.in/news/primality_v3.ps), Februaruy 2003.http://www.cse.iitk.ac.in/news/primality_v3.ps P vs NP Problem. Clay Mathematics Institute, (http://www.claymath.org/millennium/P_vs_NP/) Ramachandran, Vijaya. P versus NP. University of Texas Lectures on the Millennium Prize Problems, May 2001. (http://www.claymath.org/video/) Stewart, Ian. Ian Stewart on Minesweeper. Clay Mathematics Institute, (http://www.claymath.org/Popular_Lectures/Minesweeper) Kaliski, Burt. The Mathematics of the RSA Public-Key Cryptosystem. RSA Laboratories. Polynomial time. Wikipedia, (http://en.wikipedia.org/wiki/Polynomial _time)

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