**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.

Answered: 13 Prove that n=2²°² is a square. 2K+1…

Math

Advanced Math

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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Question

Expert Solution

Step 1

(.) Given ,

$n = 2^{2 k + 1}$

(.). Euler's $ϕ -$ function $ϕ (n)$ counts the number of positive integers less than $n$ and relatively prime to $n$ .

i.e. $ϕ (n) = \{x : 1 \leq x < n a n d g c d (x, n) = 1\}$

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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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