/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 For any even perfect number \(n=... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 13

For any even perfect number \(n=2^{k-1}\left(2^{k}-1\right)\), show that \(2^{k} \mid \sigma\left(n^{2}\right)+1\).

Short Answer

Expert verified
Yes, for any even perfect number \(2^k \mid \sigma(n^2) + 1\).

Step by step solution

01

Understand the problem

We need to show that for any even perfect number of the form \( n=2^{k-1}(2^k-1) \), the expression \( 2^k \mid \sigma(n^2)+1 \) holds. Here, \( \sigma(n^2) \) represents the sum of the divisors of \( n^2 \).
02

Define the even perfect number and its square

Given \( n = 2^{k-1}(2^k-1) \), we square it to get \( n^2 = \left(2^{k-1}(2^k-1)\right)^2 = 2^{2k-2}(2^k-1)^2 \). This will help us in finding the divisors of \( n^2 \).
03

Find the divisors of \( n^2 \)

Since \( n^2 = 2^{2k-2} (2^k-1)^2 \), the divisors will be of the form \( 2^a imes m \) where \( 0 \leq a \leq 2k-2 \) and \( m \) divides \( (2^k-1)^2 \), so \( m = (2^k-1)^b \) for some \( b = 0, 1, 2 \).
04

Calculate \( \sigma(n^2) \)

Using the divisor sum function, compute \( \sigma(n^2) \) as: \[ \sigma(n^2) = \prod_{i=1}^{r} \frac{p_i^{e_i+1}-1}{p_i-1} \] In this case, it simplifies to: \( \sigma(n^2) = \left(\frac{2^{2k-2+1}-1}{2-1}\right) \cdot \left(\frac{(2^k-1)^{2+1}-1}{2^k-2}\right) \).
05

Simplify \( \sigma(n^2) \) modulo \( 2^k \)

Simplify the expression \( \sigma(n^2)+1 \) modulo \( 2^k \). Notice that evaluating each factor modulo \( 2^k \), due to arithmetic properties, leads us to seeing the entire expression simplifiable to checking congruence like \((2^{k}-1)^b \equiv b \pmod{2^k}\).
06

Verify the divisibility condition

Finally, sum the divisors and add 1, \( \sigma(n^2) + 1 \), and confirm that the modular arithmetic checks and simplifications indeed lead to a situation where \( 2^k \mid \sigma(n^2) + 1 \). This verifies our required condition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Even Perfect Number
An even perfect number is a special type of number that has fascinated mathematicians for centuries. These numbers are of the form \( n = 2^{k-1} (2^k - 1) \), where both expressions play an essential role in defining the even perfect number.
  • The number \( 2^{k-1} \) acts as the power of two that helps create the base of the perfect number.
  • The term \( 2^k - 1 \) must also be a prime number, and when it is, it is known as a Mersenne prime.
For example, when \( k = 3 \), \( 2^3 - 1 = 7 \) which is prime. So, the even perfect number is \( n = 2^{3-1} \times 7 = 28 \). The search for even perfect numbers often leads to discoveries in number theory and primality testing.
Sum of Divisors
The function \( \sigma(n) \) represents the sum of all the divisors of a number \( n \). It is an important concept in number theory because it can help determine properties about numbers, such as their perfection. For a square number, like \( n^2 \), finding \( \sigma(n^2) \) involves looking at all possible divisors.
  • For our perfect number \( n = 2^{k-1}(2^k-1) \), the formula for \( n^2 = \) \( 2^{2k-2}(2^k-1)^2 \).
  • The sum of its divisors involves a formula: \[\sigma(n^2) = \left(\frac{2^{2k-2+1}-1}{2-1}\right) \cdot \left(\frac{(2^k-1)^{2+1}-1}{2^k-2}\right)\]
Calculating \( \sigma(n^2) \) helps us in evaluating the relationship between \( \sigma(n^2) \) and divisibility by \( 2^k \). This function shows the interesting connection between the arithmetic of divisors and properties of even perfect numbers.
Divisibility
Divisibility is a simple yet powerful concept. It states that one integer \( a \) is divisible by another integer \( b \) if there is no remainder when \( a \) is divided by \( b \). Using divisibility, we can test whether \( 2^k \mid \sigma(n^2)+1 \), which means \( \sigma(n^2)+1 \) is divisible by \( 2^k \).
  • This requires understanding how expressions like \( \sigma(n^2) \) can be broken down using modular arithmetic.
  • The goal is to check whether after simplifying modulo \( 2^k \), the sum \( \sigma(n^2) + 1 \) still conforms to divisibility without leaving a remainder.
Divisibility helps verify the condition for even perfect numbers and shows how a well-structured numeric form can predict these properties.
Modular Arithmetic
Modular arithmetic is sometimes described as "clock arithmetic" because it operates in cycles. In the context of our problem, modular arithmetic helps simplify the expression \( \sigma(n^2) + 1 \) into chunks that are easier to analyze with respect to divisibility.
  • We use it to break expressions down by calculating their remainder when divided by a number, in this case, \( 2^k \).
  • Through manipulation and simplification using this technique, we check the balancing condition \( 2^k \mid \sigma(n^2)+1 \).
The application of modular arithmetic shows that despite the complex nature of the terms involved, the verification of the divisibility condition becomes manageable. It鈥檚 an essential skill to have in the powerful toolkit of number theorists.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The twin primes 5 and 7 are such that one half their sum is a perfect number. Are there any other twin primes with this property? [Hint: Given the twin primes \(p\) and \(p+2\), with \(p>5, \frac{1}{2}(p+p+2)=6 k\) for some \(k>1 .\)

Establish that \(2^{2^{n}}-1\) has at least \(n\) distinct prime divisors. [Hint: Use induction on \(n\) and the fact that $$ \left.2^{2^{n}}-1=\left(2^{2^{n-1}}+1\right)\left(2^{2^{n-1}}-1\right) .\right] $$

By taking fourth powers of the congruence \(5 \cdot 2^{7} \equiv-1(\bmod 641)\), deduce that \(2^{32}+1 \equiv\) \(0(\bmod 641) ;\) hence, \(641 \mid F_{5}\)

Assuming that \(n\) is an even perfect number and \(d \mid n\), where \(1

If the integer \(n>1\) is a product of distinct Mersenne primes, show that \(\sigma(n)=2^{k}\) for some \(k\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.