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Exercise 5. (a) Use Proposition 3.5.20 to prove the following: if k> integer m with 0 ≤ m < N then there exists an integer will need the fact that N = 1. 3 0 and k is relatively prime to N, then for any such that (k)= m. (To do this problem, you (b) Show that I found in the previous exercise can be chosen between 0 and N - 1. (c) Show that if k> 0 is relatively prime to N, then the complex numbers (k), j = 0,... N - 1 are exactly the Nth roots of unity in rearranged order.(Hint: show that according to the previous exercise, all N of the Nth roots of unity are included somewhere among these N numbers. So all N numbers are accounted for.)

Exercise 5. (a) Use Proposition 3.5.20 to prove the following: if k> integer m with 0 ≤ m < N then there exists an integer will need the fact that N = 1. 3 0 and k is relatively prime to N, then for any such that (k)= m. (To do this problem, you (b) Show that I found in the previous exercise can be chosen between 0 and N - 1. (c) Show that if k> 0 is relatively prime to N, then the complex numbers (k), j = 0,... N - 1 are exactly the Nth roots of unity in rearranged order.(Hint: show that according to the previous exercise, all N of the Nth roots of unity are included somewhere among these N numbers. So all N numbers are accounted for.)

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.2: Integral Domains And Fields
Problem 12E
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Please do Exercise 5 part A-D and please show step by step and explain

From Proposition 3.5.20 in the book we have: "Given a modular equation kx = m (mod N), where k, m, N are integers. Then the equation has an integer solution for x if and only if m is an integer multiple of the greatest common divisor of k and N." Exercise 5. (a) Use Proposition 3.5.20 to prove the following: if k> 0 and k is relatively prime to N, then for any integer m with 0 <m<N then there exists an integer & such that (k)= m. (To do this problem, you will need the fact that N = 1. 3 (b) Show that I found in the previous exercise can be chosen between 0 and N - 1. (c) Show that if k> 0 is relatively prime to N, then the complex numbers (k), j = 0,... N - 1 are exactly the Nth roots of unity in rearranged order.(Hint: show that according to the previous exercise, all N of the Nth roots of unity are included somewhere among these N numbers. So all N numbers are accounted for.) N-1 (d) Show that if k is relatively prime to N, then the sum e2ijk/N is equal to 0. j=0 N'-1 (e) Now suppose 0 <k < N is not relatively prime to N. Then show that there is a number N' which rijk/N = 0. Show also that Σe2rijk/N = 0 by breaking this sum of N terms into N/N' sums, and show that each of these sums is equal to zero. divides N such that j=0
Transcribed Image Text:From Proposition 3.5.20 in the book we have: "Given a modular equation kx = m (mod N), where k, m, N are integers. Then the equation has an integer solution for x if and only if m is an integer multiple of the greatest common divisor of k and N." Exercise 5. (a) Use Proposition 3.5.20 to prove the following: if k> 0 and k is relatively prime to N, then for any integer m with 0 <m<N then there exists an integer & such that (k)= m. (To do this problem, you will need the fact that N = 1. 3 (b) Show that I found in the previous exercise can be chosen between 0 and N - 1. (c) Show that if k> 0 is relatively prime to N, then the complex numbers (k), j = 0,... N - 1 are exactly the Nth roots of unity in rearranged order.(Hint: show that according to the previous exercise, all N of the Nth roots of unity are included somewhere among these N numbers. So all N numbers are accounted for.) N-1 (d) Show that if k is relatively prime to N, then the sum e2ijk/N is equal to 0. j=0 N'-1 (e) Now suppose 0 <k < N is not relatively prime to N. Then show that there is a number N' which rijk/N = 0. Show also that Σe2rijk/N = 0 by breaking this sum of N terms into N/N' sums, and show that each of these sums is equal to zero. divides N such that j=0
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