An Introduction to Cryptography (2nd Edition) (Discrete by Richard A. Mollin PDF

By Richard A. Mollin

ISBN-10: 1584886188

ISBN-13: 9781584886181

Carrying on with a bestselling culture, An creation to Cryptography, moment version presents an effective origin in cryptographic thoughts that includes all the needful history fabric on quantity idea and algorithmic complexity in addition to a historic examine the field.

With various additions and restructured fabric, this variation offers the tips in the back of cryptography and the functions of the topic. the 1st bankruptcy presents an intensive remedy of the maths essential to comprehend cryptography, together with quantity conception and complexity, whereas the second one bankruptcy discusses cryptographic basics, resembling ciphers, linear suggestions shift registers, modes of operation, and assaults. the subsequent a number of chapters speak about DES, AES, public-key cryptography, primality checking out, and numerous factoring tools, from classical to elliptical curves. the ultimate chapters are constructed from matters bearing on the net, akin to beautiful strong privateness (PGP), protocol layers, firewalls, and cookies, in addition to functions, together with login and community safeguard, viruses, clever playing cards, and biometrics. The e-book concludes with appendices on mathematical info, machine mathematics, the Rijndael S-Box, knapsack ciphers, the Silver-Pohlig-Hellman set of rules, the SHA-1 set of rules, radix-64 encoding, and quantum cryptography.

New to the second one Edition:
An introductory bankruptcy that gives additional info on mathematical evidence and complexity theory
Expanded and up-to-date workouts units, together with a few regimen exercises
More info on primality checking out and cryptanalysis

Accessible and logically equipped, An creation to Cryptography, moment version is the fundamental e-book at the basics of cryptography.

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Additional resources for An Introduction to Cryptography (2nd Edition) (Discrete Mathematics and Its Applications)

Example text

Some additional properties of congruences are given in the next result. 4 Let a, b, c ∈ Z, m, n ∈ N, and a ≡ b (mod n). Then each of the following holds. (a) am ≡ bm (mod mn). (b) am ≡ bm (mod n). (c) If m divides n, then a ≡ b (mod m). Proof. (a) Given that a ≡ b (mod n), a − b = kn for some integer k. Multiplying by m, we get (a − b)m = knm, so am − bm = (km)n, namely am ≡ bm (mod n). (b) Since n|(a − b), then n|(a − b)(am−1 + am−2 b + · · · + bm ) = am − bm . In other words, am ≡ bm (mod n).

5, the unique solution modulo n. One may wonder about the situation where the moduli are not relatively prime. D. 12 in his book t’ai-yen-lei-schu as follows. 75 on page 33, which are used in the following proof. 13 Generalized Chinese Remainder Theorem Let nj ∈ N, set = lcm(n1 , n2 , . . , nk ), and let rj ∈ Z be any integers for j = 1, 2, . . , k. Then the system of k simultaneous linear congruences given by x ≡ r1 (mod n1 ), x ≡ r2 (mod n2 ), .. x ≡ rk (mod nk ), © 2007 by Taylor & Francis Group, LLC 28 1.

Exactly one of the following holds. Prove it and provide counterexamples for the other two. (a) If a ≡ ±1 (mod p) for all primes p dividing n, then a2 ≡ 1 (mod n). (b) If a2 ≡ 1 (mod n), then a ≡ ±1 (mod p) for all primes p dividing n. (c) The congruence a ≡ ±1 (mod p) for all primes p dividing n holds if and only if a2 ≡ 1 (mod n). 64. Let a, b ∈ Z, n ∈ N. Prove that ax ≡ b (mod n) has a solution if and only if gcd(a, n) b. 65. 64, let g = gcd(a, n), and suppose that x = x0 is a solution of ax ≡ b (mod n).

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