(e) Now suppose 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please do part E and please show step by step and explain
![From Proposition 3.5.20 in the book we have: “Given a modular equation \( kx \equiv m \, (\text{mod} \, N) \), 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 \( 0 \leq m < N \) there exists an integer \( \ell \) such that \( (\zeta^k)^\ell = \zeta^m \). (To do this problem, you will need the fact that \( \zeta^N = 1 \).)
(b) Show that \( \ell \) 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 \( (\zeta^k)^j, j = 0, \ldots, N - 1 \) are exactly the \( N \)th roots of unity in rearranged order. *(Hint: show that according to the previous exercise, all \( N \) of the \( N \)th roots of unity are included somewhere among these \( N \) numbers. So all \( N \) numbers are accounted for.)*
(d) Show that if \( k \) is relatively prime to \( N \), then the sum
\[
\sum_{j=0}^{N-1} e^{2\pi i jk/N}
\]
is equal to 0.
(e) Now suppose \( 0 < k < N \) is not relatively prime to \( N \). Then show that there is a number \( N' \) which divides \( N \) such that
\[
\sum_{j=0}^{N-1} e^{2\pi i jk/N} = 0
\]
Show also that
\[
\sum_{j=0}^{N'/N} e^{2\pi i jk/N} = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F5e60d98a-23a1-40dc-89b0-cb1812d712a5%2Ftgev2qa_processed.png&w=3840&q=75)
Transcribed Image Text:From Proposition 3.5.20 in the book we have: “Given a modular equation \( kx \equiv m \, (\text{mod} \, N) \), 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 \( 0 \leq m < N \) there exists an integer \( \ell \) such that \( (\zeta^k)^\ell = \zeta^m \). (To do this problem, you will need the fact that \( \zeta^N = 1 \).)
(b) Show that \( \ell \) 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 \( (\zeta^k)^j, j = 0, \ldots, N - 1 \) are exactly the \( N \)th roots of unity in rearranged order. *(Hint: show that according to the previous exercise, all \( N \) of the \( N \)th roots of unity are included somewhere among these \( N \) numbers. So all \( N \) numbers are accounted for.)*
(d) Show that if \( k \) is relatively prime to \( N \), then the sum
\[
\sum_{j=0}^{N-1} e^{2\pi i jk/N}
\]
is equal to 0.
(e) Now suppose \( 0 < k < N \) is not relatively prime to \( N \). Then show that there is a number \( N' \) which divides \( N \) such that
\[
\sum_{j=0}^{N-1} e^{2\pi i jk/N} = 0
\]
Show also that
\[
\sum_{j=0}^{N'/N} e^{2\pi i jk/N} = 0
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