A Course in Number Theory and Cryptography - download pdf or read online

By Neal Koblitz

ISBN-10: 0387942939

ISBN-13: 9780387942933

ISBN-10: 3540942939

ISBN-13: 9783540942931

The aim of this e-book is to introduce the reader to mathematics themes, either historic and smooth, which have been on the middle of curiosity in functions of quantity conception, fairly in cryptography. No heritage in algebra or quantity concept is thought, and the e-book starts with a dialogue of the elemental quantity concept that's wanted. The strategy taken is algorithmic, emphasizing estimates of the potency of the options that come up from the idea. a different characteristic is the inclusion of contemporary program of the speculation of elliptic curves. huge workouts and cautious solutions were integrated in the entire chapters. simply because quantity conception and cryptography are fast-moving fields, this new version includes colossal revisions and up-to-date references.

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Extra info for A Course in Number Theory and Cryptography

Example text

For example, if our plaintext and ciphertext message units are single letters from the 26-letter alphabet A-Z, then we can label the letters using the integers 0, 1, 2,. . " Thus, in place of A we write 0, in place of S we write 18, in place of X we write 23, and so on. As another example, if our message units are digraphs in the 27-letter alphabet consisting of A-Z and a blank, we might first let the blank have numerical equivalent 26 (one beyond Z), and then label the digraph whose two letters correspond to x, y E {0, 1, 2,.

Exercises 1. If we try to eliminate b' by subtracting the first two congruences, we arrive a t 459a' = 351 mod 729, which does not have a unique solution a' mod 729 (there are 27 solutions). We do better if we subtract the third congruence from the first, obtaining 437a' = 142 mod 729. To solve this, we must find the inverse of 437 modulo 729. By way of review of the Euclidean algorithm, let's go through that in detail: and then 2. = 362 - 437 mod 729. , a dirty joke), to encipher the letters (but not the blanks or punctuation) by a translation C P b mod 26.

The following are equivalent: --- 2x 3y = 1 mod 26, 7x + 89 2 mod 26; R to get a new vector (;:): - A-'B = (1714 11 lo) (i). (;) = (i:) mod 26. The matrix of the systems (b)-(c) does not have an inverse modulo 26, since its determinant is 14, which has a common factor of 2 with 26. , we can find the solution to the same congruence mod 13 and see if it gives a solution which works modulo 26. Modulo 13 we obtain (Z) (i) (i) (where (;) = in part (b) and (:) in part (c)). ) mod 13, respectively. Testing the possibilities modulo 26, we find that in part (b) there are no solutions, and in part (c) there are two solutions: x = 6, y = 7 and x = 19, y = 20.

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A Course in Number Theory and Cryptography by Neal Koblitz

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