By Jeffrey Hoffstein, Jill Pipher, Joseph H. Silverman
An advent to Mathematical Cryptography offers an creation to public key cryptography and underlying arithmetic that's required for the topic. all of the 8 chapters expands on a selected quarter of mathematical cryptography and offers an in depth checklist of exercises.
It is an appropriate textual content for complicated scholars in natural and utilized arithmetic and desktop technology, or the publication can be used as a self-study. This booklet additionally offers a self-contained therapy of mathematical cryptography for the reader with constrained mathematical background.
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Additional info for An Introduction to Mathematical Cryptography
8: Successive square powers of 3 modulo 1000 3 4 6 3218 = 32 · 32 · 32 · 32 · 32 7 ≡ 9 · 561 · 721 · 281 · 961 ≡ 489 (mod 1000). (mod 1000) We note that in computing the product 9 · 561 · 721 · 281 · 961, we may reduce modulo 1000 after each multiplication, so we never need to deal with very large numbers. We also observe that it has taken us only 11 multiplications to compute 3218 (mod 1000), a huge savings over the naive approach. And for larger exponents we would save even more. 19 goes by various names, including the Fast Powering Algorithm and the Square-and-Multiply Algorithm.
M − 1} and call Z/mZ the ring of integers modulo m. Note that whenever we perform an addition or multiplication in Z/mZ, we always divide the result by m and take the remainder in order to obtain an element in Z/mZ. 4 illustrates the ring Z/5Z by giving complete addition and multiplication tables modulo 5. 16. If you have studied ring theory, you will recognize that Z/mZ is the quotient ring of Z by the principal ideal mZ, and that the numbers 0, 1, . . , m − 1 are actually coset representatives for the congruence classes that comprise the elements of Z/mZ.
4. 20. Let p be a prime number, and suppose that p divides the product ab of two integers a and b. Then p divides at least one of a and b. More generally, if p divides a product of integers, say p | a1 a2 · · · an , then p divides at least one of the individual ai . Proof. Let g = gcd(a, p). Then g | p, so either g = 1 or g = p. If g = p, then p | a (since g | a), so we are done. 11 tells us that we can ﬁnd integers u and v satisfying au + pv = 1. We multiply both sides of the equation by b to get abu + pbv = b.