13 Prove that n=2²°² is a square. 2K+1 where K = 1 (n)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:**Question 3: Prove that \( n = 2^{2k+1} \), where \( k \geq 1 \), \(\Phi(n)\) is a square.**
This problem asks you to demonstrate that for the given expression \( n = 2^{2k+1} \), the Euler's totient function \(\Phi(n)\) results in a perfect square, provided that \( k \) is greater than or equal to 1.
### Solution Outline:
1. **Understand the Totient Function**: Euler's totient function \(\Phi(n)\) calculates the number of integers up to \( n \) that are relatively prime to \( n \).
2. **Apply Euler's Formula**: For \( n = 2^{2k+1} \), \(\Phi(n) = 2^{2k+1} \times (1 - \frac{1}{2}) = 2^{2k+1-1}\).
3. **Verify It's a Square**: Show that \( 2^{2k} \) results in a perfect square, being \( (2^k)^2 \).
4. **Conclude**: Since \( 2k \) is an integer, \( 2^{2k} \) is indeed a perfect square.
This solution illustrates how to apply number theory concepts to demonstrate properties of exponential functions under constraints.
Expert Solution

Step 1
(.) Given ,
(.). Euler's function counts the number of positive integers less than and relatively prime to .
i.e.
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Follow-up Questions
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Follow-up Question
n does not equal to 22K+1. It should b 2^2k+1.
Solution
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