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the 20 1, 2, 3 Factor the number 211 – 1 by Fermat's factorization method. -

the 20 1, 2, 3 Factor the number 211 – 1 by Fermat's factorization method. -

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.3: Quadratic Equations
Problem 47E
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ELEMENTARY PROBLEMS 5.4 Use Fermat’s method to factor each of the following numbers: (a) 2279. (b) 10541. (c) 340663 [Hint: The smallest square just exceeding 340663 is 584².] 2. Prove that a perfect square must end in one of the following pairs of digits: 00, 01 o4 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96. [Hint: Because x² = (50 + x)² (mod 100) and x = (50 – x)² (mod 100), it suffices examine the final digits of x² for the 26 values x = Factor the number 211 – 1 by Fermat's factorization method. 4. In 1647, Mersenne noted that when a number can be written as a sum of two relative prime squares in two distinct ways, it is composite and can be factored as follows = a² - 0, 1, 2, ..., 25.] - t+ z = -9 + zv = u (ac+ bd)(ac – bd) %D - Use this result to factor the numbers 493 = 182 + 132 = 22² + 3² and 38025 = 1682 + 992 = 156² + 1172 5.)Employ the generalized Fermat method to factor each of the following numbers: (a) 2911 [Hint: 1382 = 672 (mod 2911).] (b) 4573 [Hint: 1772 = 922 (mod 4573).] (c) 6923 [Hint: 2082 = 932 (mod 6923).] 6. Factor 13561 with the help of the congruences 233² = 32 . 5 (mod 13561) 12812 = 2* . 5 (mod 13561) and 7. (a) Factor the number 4537 by searching for x such that x² - k. 4537 is the product of small prime powers. (b) Use the procedure indicated in part (a) to factor 14429. [Hint: 1202 – 14429 = -29 and 30032 – 625 · 14429 = –116.1 8. Use Kraitchik's method to factor the number 20437.
Transcribed Image Text:ELEMENTARY PROBLEMS 5.4 Use Fermat’s method to factor each of the following numbers: (a) 2279. (b) 10541. (c) 340663 [Hint: The smallest square just exceeding 340663 is 584².] 2. Prove that a perfect square must end in one of the following pairs of digits: 00, 01 o4 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96. [Hint: Because x² = (50 + x)² (mod 100) and x = (50 – x)² (mod 100), it suffices examine the final digits of x² for the 26 values x = Factor the number 211 – 1 by Fermat's factorization method. 4. In 1647, Mersenne noted that when a number can be written as a sum of two relative prime squares in two distinct ways, it is composite and can be factored as follows = a² - 0, 1, 2, ..., 25.] - t+ z = -9 + zv = u (ac+ bd)(ac – bd) %D - Use this result to factor the numbers 493 = 182 + 132 = 22² + 3² and 38025 = 1682 + 992 = 156² + 1172 5.)Employ the generalized Fermat method to factor each of the following numbers: (a) 2911 [Hint: 1382 = 672 (mod 2911).] (b) 4573 [Hint: 1772 = 922 (mod 4573).] (c) 6923 [Hint: 2082 = 932 (mod 6923).] 6. Factor 13561 with the help of the congruences 233² = 32 . 5 (mod 13561) 12812 = 2* . 5 (mod 13561) and 7. (a) Factor the number 4537 by searching for x such that x² - k. 4537 is the product of small prime powers. (b) Use the procedure indicated in part (a) to factor 14429. [Hint: 1202 – 14429 = -29 and 30032 – 625 · 14429 = –116.1 8. Use Kraitchik's method to factor the number 20437.
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