5. Prove that the equation ø(n) = ¢(n+ 2) is satisfied by n = 2(2p – 1) whenever p and 2p – 1 are both odd primes. %3D -
5. Prove that the equation ø(n) = ¢(n+ 2) is satisfied by n = 2(2p – 1) whenever p and 2p – 1 are both odd primes. %3D -
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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![### Problems 7.2
1. **Calculate**
- \(\phi(1001)\), \(\phi(5040)\), and \(\phi(36,000)\).
2. **Verify**
- Show that the equality \(\phi(n) = \phi(n+1) = \phi(n+2)\) holds when \(n = 5186\).
3. **Simultaneous Integer Properties**
- Show that the integers \(n = 3^k \cdot 568\) and \(n = 3^k \cdot 638\), where \(k \geq 0\), satisfy simultaneously:
\[ \tau(m) = \tau(n), \quad \sigma(m) = \sigma(n), \quad \text{and} \quad \phi(m) = \phi(n) \]
4. **Establish Assertions**
- (a) If \(n\) is an odd integer, then \(\phi(2n) = \phi(n)\).
- (b) If \(n\) is an even integer, then \(\phi(2n) = 2\phi(n)\).
- (c) \(\phi(3n) = 3\phi(n)\) if and only if \(3 \mid n\).
- (d) \(\phi(3n) = 2\phi(n)\) if and only if \(3 \nmid n\).
- (e) \(\phi(n) = n/2\) if and only if \(n = 2^k\) for some \(k \geq 1\).
*[Hint: Write \(n = 2^k N\), where \(N\) is odd, and use the condition \(\phi(n) = n/2\) to show that \(N = 1\).]*
5. **Equation Proof**
- Prove that the equation \(\phi(n) = \phi(n+2)\) is satisfied by \(n = 2(2p - 1)\) whenever \(p\) and \(2p - 1\) are both odd primes.
6. **Perfect Square Investigation**
- Show that there are infinitely many integers \(n\) for which \(\phi(n)\) is a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fadf70025-15e9-4d93-968e-9d86049b61bd%2F5d05c6a3-0cc6-44c8-b0ea-3366b8d90b70%2Fdncfqp.jpeg&w=3840&q=75)
Transcribed Image Text:### Problems 7.2
1. **Calculate**
- \(\phi(1001)\), \(\phi(5040)\), and \(\phi(36,000)\).
2. **Verify**
- Show that the equality \(\phi(n) = \phi(n+1) = \phi(n+2)\) holds when \(n = 5186\).
3. **Simultaneous Integer Properties**
- Show that the integers \(n = 3^k \cdot 568\) and \(n = 3^k \cdot 638\), where \(k \geq 0\), satisfy simultaneously:
\[ \tau(m) = \tau(n), \quad \sigma(m) = \sigma(n), \quad \text{and} \quad \phi(m) = \phi(n) \]
4. **Establish Assertions**
- (a) If \(n\) is an odd integer, then \(\phi(2n) = \phi(n)\).
- (b) If \(n\) is an even integer, then \(\phi(2n) = 2\phi(n)\).
- (c) \(\phi(3n) = 3\phi(n)\) if and only if \(3 \mid n\).
- (d) \(\phi(3n) = 2\phi(n)\) if and only if \(3 \nmid n\).
- (e) \(\phi(n) = n/2\) if and only if \(n = 2^k\) for some \(k \geq 1\).
*[Hint: Write \(n = 2^k N\), where \(N\) is odd, and use the condition \(\phi(n) = n/2\) to show that \(N = 1\).]*
5. **Equation Proof**
- Prove that the equation \(\phi(n) = \phi(n+2)\) is satisfied by \(n = 2(2p - 1)\) whenever \(p\) and \(2p - 1\) are both odd primes.
6. **Perfect Square Investigation**
- Show that there are infinitely many integers \(n\) for which \(\phi(n)\) is a
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