Read e-book online Primality Testing and Integer Factorization in Public-Key PDF

By Song Y. Yan

ISBN-10: 0387772677

ISBN-13: 9780387772677

Meant for complicated point scholars in desktop technological know-how and arithmetic, this key textual content, now in a new version, offers a survey of modern development in primality trying out and integer factorization, with implications for factoring dependent public key cryptography. For this up-to-date and revised variation, amazing new gains comprise a comparability of the Rabin-Miller probabilistic try out in RP, the Atkin-Morain elliptic curve try in ZPP and the AKS deterministic try.

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Extra info for Primality Testing and Integer Factorization in Public-Key Cryptography, 2nd ed. (Advances in Information Security)

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12. Let f (x) = 2x5 + x − 1 and p(x) = 3x2 + 2. Then 17 2 3 4 x − x + −1 + x 3 9 9 2x5 + x − 1 = (3x2 + 2) 2x5 + x − 1 = (3x2 + 2)(3x3 + 5x) + 5x + 6 in Q[x], in Z7 [x]. 4 (Euclid’s Algorithm for Polynomials). Let f and g be nonzero polynomials in F [x]. The Euclid’s algorithm for polynomials runs in exactly the same way as that for integers f = gq0 + r1 g r1 r2 = = = .. = = rn−2 rn−1 r1 q1 + r2 r2 q2 + r3 r3 q3 + r4 rn−1 qn−1 + rn rn qn + 0 Then, gcd(f, g) = rn . Moreover, if d(x) is the greatest common divisor of f (x) and g(x), then there are polynomials s(x) and t(x) such that d(x) = s(x)f (x) + t(x)g(x).

K. Proof. It is easy to see that gcd(a, b) = k i=1 lcm(a, b) = k i=1 pγi i , where γi is the lesser of αi and βi , pδi i , where δi is the greater of αi and βi . The result thus follows. 2. Suppose a and b are positive integers, then lcm(a, b) = ab . 42) Proof. Since γi + δi = αi + βi , it is now obvious that gcd(a, b) · lcm(a, b) = ab. The result thus follows. 6. Find gcd(240, 560) and lcm(240, 560). Since the prime factorizations of 240 and 560 are 240 = 24 · 3 · 5 = 24 · 31 · 51 · 70 , 560 = 24 · 5 · 7 = 24 · 30 · 51 · 71 , we get gcd(240, 560) = 2min(4,4) · 3min(1,0) · 5min(1,1) · 7min(0,1) = 24 · 30 · 51 · 70 = 80, lcm(240, 560) = 2max(4,4) · 3max(1,0) · 5max(1,1) · 7max(0,1) = 24 · 31 · 51 · 71 = 1680.

Otherwise, n can be factored into, say, ab, where a > 1 and b > 1. Apply the same argument to a and b: each is either a prime or a product of two numbers both > 1. The numbers other than primes involved in the expression for n are greater than 1 and decrease at every step; hence eventually all the numbers must be prime. Now we come to uniqueness. Suppose that the theorem is false and let n > 1 be the smallest number having more than one expression as the product of primes, say n = p1 p2 . . pr = q1 q2 .

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Primality Testing and Integer Factorization in Public-Key Cryptography, 2nd ed. (Advances in Information Security) by Song Y. Yan

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