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By Kraft, James S.; Washington, Lawrence C
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 particular Factorization A Puzzle Irrationality. Read more...
summary: IntroductionDiophantine EquationsModular ArithmeticPrimes and the Distribution of PrimesCryptographyDivisibilityDivisibilityEuclid's Theorem Euclid's unique facts 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 detailed FactorizationPreliminary effects the basic Theorem of mathematics Euclid and the basic Theorem of ArithmeticChapter Highlights difficulties functions of detailed Factorization A Puzzle Irrationality
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Additional info for An Introduction to Number Theory with Cryptography
9) implies that (a/d) | (b/d)(y0 − v). 13 implies that (a/d) divides (y0 − v). By definition, this means that there is an integer t with a y0 − v = t . 9), we get a b a (u − x0 ) = t . 11) d , we have a b u − x0 = t d or b u = x0 + t. 12), we have b u = x0 + t d and a v = y0 − t. 4), we have completed the proof. 4), we first verify that d | c. If it doesn’t, we’re done since there are no solutions. If it does, we divide both sides by d to get a new equation ax+by =c and in this equation, gcd(a , b ) = 1.
The last line tells us that 36 = 456 · (−1) + 123 · 4. Moving to the third row of our gcd calculation, we see that 15 = 87−2·36 =(3rd row) −2·(4th row) in our row and column language. This becomes x y 456 1 0 123 0 1 87 1 −3 36 −1 4 15 3 −11 (3rd row) − 2·(4th row). We continue in this way and end when we have 3 = gcd(456, 123) in the first column: x y 456 1 0 123 0 1 87 1 −3 36 −1 4 15 3 −11 6 −7 26 (4th row) − 2·(5th row) 3 17 −63 (5th row) − 2·(6th row). This tells us that 3 = 456 · 17 + 123 · (−63).
Let q be the largest integer less than or equal to a/b, so q ≤ a/b < q + 1. 18 Chapter 1 Divisibility Multiplying by b yields bq ≤ a < bq + b, which implies that 0 ≤ a − bq < b. Let r = a − bq. Then 0 ≤ r < b. Since a = bq + r, we have proved that the desired q and r exist. It remains to prove that q and r are unique. If a = bq + r = bq1 + r1 with 0 ≤ r, r1 < b then b(q − q1 ) = r1 − r. Since the left-hand side of this equation is a multiple of b, so is r1 − r. Because 0 ≤ r, r1 < b, we must have −b < r1 − r < b.
An Introduction to Number Theory with Cryptography by Kraft, James S.; Washington, Lawrence C