By Kraft, James S.; Washington, Lawrence C

ISBN-10: 1482214423

ISBN-13: 9781482214420

ISBN-10: 1931962022

ISBN-13: 9781931962025

IntroductionDiophantine EquationsModular ArithmeticPrimes and the Distribution of PrimesCryptographyDivisibilityDivisibilityEuclid's Theorem Euclid's unique evidence The Sieve of Eratosthenes The department set of rules the best universal Divisor The Euclidean set of rules different BasesLinear Diophantine EquationsThe Postage Stamp challenge Fermat and Mersenne Numbers bankruptcy Highlights difficulties specific FactorizationPreliminaryRead more...

summary: IntroductionDiophantine EquationsModular ArithmeticPrimes and the Distribution of PrimesCryptographyDivisibilityDivisibilityEuclid's Theorem Euclid's unique evidence The Sieve of Eratosthenes The department set of rules the best universal Divisor The Euclidean set of rules different BasesLinear Diophantine EquationsThe Postage Stamp challenge Fermat and Mersenne Numbers bankruptcy Highlights difficulties designated FactorizationPreliminary effects the basic Theorem of mathematics Euclid and the elemental Theorem of ArithmeticChapter Highlights difficulties functions of distinct Factorization A Puzzle Irrationality

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**Example text**

Before we get to the main result of this section, we look at two more examples that will help us understand the general situation. First, consider 6x − 9y = 20. Notice that 3 must divide the lefthand side but 3 is not a divisor of the right-hand side. This tells us that this equation can never have an integer solution. To make things notationally simpler, let d = gcd(a, b). We then see that in order for ax + by = c to have a solution, we must have d | c. Now let’s look at an example where this does occur, say 6x + 9y = 21.

An ) = a1 x1 + a2 x2 + · · · + an xn . Proof. We’ll use mathematical induction (see Appendix A). 11, the result is true for n = 2. Assume that it is true for n = k. 3) for some integers y1 , y2 , . . , yk . But gcd(a1 , a2 , . . , ak+1 ) = gcd(gcd(a1 , a2 , . . , ak ), ak+1 ) = gcd(a1 , a2 , . . 11. 3) into this equation yields gcd(a1 , a2 , . . ,ak+1 ) = (a1 y1 + a2 y2 + · · · + ak yk )x + ak+1 y = a1 (xy1 ) + a2 (xy2 ) + · · · + ak (xyk ) + ak+1 yk+1 , which is the desired result, with xi = xyi for 1 ≤ i ≤ k and xk+1 = y.

7. If n is composite, then n has a prime factor √ p ≤ n. Proof. Since n is composite, we can write n = ab with 1 < a ≤ b < n. Then a2 ≤ ab = n, √ so a ≤ n. Let p be a prime number dividing a. Then p ≤ a ≤ √ n. 07, so we could stop after crossing out the multiples of 7. 6). Why is the process called a sieve? In our example, the multiples of the primes 2, 3, 5, 7 created a net. The numbers that fell through this net are the prime numbers. CHECK YOUR UNDERSTANDING 4. Use the Sieve of Eratosthenes to compute the prime numbers less than 20.

### An Introduction to Number Theory with Cryptography by Kraft, James S.; Washington, Lawrence C

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