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(a) If \(\phi(n) \mid n-1\), prove that \(n\) is a square-free integer. [Hint: Assume that \(n\) has the prime factorization \(n=p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{r}^{k_{r}}\), where \(k_{1} \geq 2\). Then \(p_{1} \mid \phi(n)\), whence \(p_{1} \mid n-1\), which leads to a contradiction.] (b) Show that if \(n=2^{k}\) or \(2^{k} 3^{j}\), with \(k\) and \(j\) positive integers, then \(\phi(n) \mid n\).

Short Answer

Expert verified
If \(\phi(n) \mid n-1\), \(n\) must be square-free as otherwise a prime \(p\) would divide \(n-1\), creating a contradiction.

Step by step solution

01

Understand the Problem

We want to prove that if \(\phi(n) \,|\, n-1\), then \(n\) is square-free. Square-free means that \(n\) does not have any prime factors raised to a power greater than 1.
02

Analyze Prime Factorization

Assume \(n\) has the prime factorization \(n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}\) with one of the \(k_i \geq 2\), say \(k_1 \geq 2\). Our goal is to show that this leads to a contradiction.
03

Apply Euler's Totient Formula

Recall that Euler's Totient Function \(\phi(n) = n \left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)\cdots\left(1 - \frac{1}{p_r}\right)\). If \(k_1 \geq 2\), then the contribution to \(\phi(n)\) from \(p_1\) is \(p_1^{k_1-1}(p_1-1)\), meaning \(p_1 \mid \phi(n)\).
04

Use the Divisibility Condition

Since \(\phi(n) \mid n - 1\), it follows from our assumption that \(p_1 \mid n-1\). This implies \(n \equiv 1 \pmod{p_1}\), leading to a contradiction since \(n = p_1^{k_1} (\text{other terms})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Euler's Totient Function
Euler's Totient Function, denoted as \( \phi(n) \), is a fundamental concept in number theory, representing the count of integers less than \( n \) that are coprime to \( n \). In simpler terms, it tells us how many numbers up to \( n \) are not divisible by \( n\)'s prime factors.

For any positive integer \( n \), expressed with its prime factorization as \( n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} \), Euler's Totient Function is calculated using the formula:

\[ \phi(n) = n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_r}\right) \]

Here's how it works:
  • The function multiplies \( n \) by the term \( 1 - \frac{1}{p_i} \) for each distinct prime factor \( p_i \).
  • This deduction accounts for the multiples of each prime factor, effectively removing them from the count of coprime numbers up to \( n \).
Euler's Totient Function plays a key role in several cryptographic algorithms and is crucial for problems involving divisibility conditions, especially when analyzing the properties of numbers like square-free integers.
Prime Factorization
Prime factorization involves breaking down a composite number into prime numbers that multiply together to form the original number. Each prime is raised to a power that indicates how many times it appears in the factorization.

Consider a number \( n \) with its prime factorization represented as:

\[ n = p_1^{k_1} \times p_2^{k_2} \times \cdots \times p_r^{k_r} \]

Here, each \( p_i \) is a prime number, and \( k_i \) is the exponent showing how many times the prime \( p_i \) divides \( n \).

  • A number is considered square-free if all \( k_i = 1 \), meaning no prime factor is repeated in its factorization.
  • If any \( k_i \) is greater than 1, the number is not square-free as it has a squared factor.
Understanding prime factorization is essential for determining number properties, particularly distinguishing between square-free and non-square-free integers. This forms the basis for proving properties related to factors, such as demonstrating that \( n \) is square-free when Euler's Totient Function divides \( n - 1 \).
Divisibility Condition
Divisibility conditions are rules that help determine when one integer divides another without leaving a remainder. These conditions are crucial in number theory and form the foundation of many mathematical proofs.

In the problem context, if Euler's Totient Function \( \phi(n) \) divides \( n-1 \), it implies a fundamental relationship between \( n \) and its divisors.

Consider an integer \( n \) with its transformation \( n - 1 \). If all prime factors, especially those contributing to \( \phi(n) \), also divide \( n - 1 \), this sets up a path to deducing other properties of \( n \).

  • The requirement \( \phi(n) \mid n - 1 \) often indicates structural properties in \( n \)'s composition, such as it being square-free.
  • This is because any repeated prime factors in \( n \)'s factorization contradict the existence of \( \phi(n) \mid n - 1 \) as shown by the derived equivalence \( n \equiv 1 \pmod{p_1} \).
This condition facilitates proofs about integer properties and helps to unlock constraints in problem set-ups, leading to deeper insights into the nature of numbers.

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