/*! 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 15 Prove that a positive integer \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Prove that a positive integer \(a>1\) is a square if and only if in the canonical form of \(a\) all the exponents of the primes are even integers.

Short Answer

Expert verified
A positive integer is a square if and only if all exponents in its canonical form are even integers.

Step by step solution

01

Understand the Canonical Form

The canonical form of a positive integer essentially breaks down the number into a product of prime factors, each raised to some power, i.e., \( a = p_1^{e_1} p_2^{e_2} imes ext{...} imes p_n^{e_n} \) where \( p_1, p_2, ext{...}, p_n \) are prime numbers and \( e_1, e_2, ext{...}, e_n \) are their respective exponents.
02

Define a Perfect Square

An integer is a perfect square if there exists another integer \( b \) such that \( a = b^2 \). This means every exponent in the prime factorization of \( a \) should be even because \( b = p_1^{f_1} p_2^{f_2} imes ext{...} \) implies \( a = (p_1^{f_1})^2 = p_1^{2f_1} \) and so on.
03

Show that Even Exponents Imply a Square

Assume all the exponents \( e_1, e_2, ext{...}, e_n \) are even. Then \( e_i = 2k_i \) for some integer \( k_i \). The expression \( a = p_1^{2k_1} p_2^{2k_2} imes ext{...} \) can be rewritten as \( a = (p_1^{k_1} p_2^{k_2} imes ext{...})^2 \), which proves \( a \) is a square because it's clear that \( a = b^2 \) for \( b = p_1^{k_1} p_2^{k_2} imes ext{...} \).
04

Prove the Converse (Square Implies Even Exponents)

Assume \( a \) is a square, so \( a = b^2 \) for some integer \( b \). Given \( b = p_1^{f_1} p_2^{f_2} imes ext{...} \), \( a = b^2 = p_1^{2f_1} p_2^{2f_2} imes ext{...} \). This means that each exponent \( e_i \) in \( a \)'s canonical form is twice the exponent in \( b \)'s form, hence even.
05

Conclusion

Since both conditions hold (even exponents imply a square, and a square implies even exponents), we have shown that a positive integer \( a > 1 \) is a square if and only if all the exponents in its canonical form are even integers.

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Key Concepts

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

Prime Factorization
Prime factorization is the process of expressing a number as the product of its prime numbers. Imagine breaking down an integer, like 60, into the building blocks that form it. These building blocks are called prime numbers. For example, the prime factorization of 60 is given by:
  • 60 = 2 \( \times \) 2 \( \times \) 3 \( \times \) 5
Here, each of the numbers 2, 3, and 5 is a prime number, which means they are only divisible by 1 and themselves. When we write the prime factorization, it’s helpful to use exponents to show how many times each prime factor appears.
  • 60 = 2^2 \( \times \) 3^1 \( \times \) 5^1
The exponents tell us the power or the number of times each prime factor is used. Breaking down a number into its primes helps us understand the number better and is essential in many areas of mathematics, such as finding the canonical form of an integer.
Perfect Squares
A perfect square is an integer that can be expressed as the square of another integer. Think of perfect squares as numbers that result from an integer being multiplied by itself. For example, let's consider several numbers:
  • 1 is a perfect square because it is \( 1 \times 1 = 1^2 \)
  • 4 is a perfect square because it is \( 2 \times 2 = 2^2 \)
  • 9 is a perfect square because it is \( 3 \times 3 = 3^2 \)
To determine whether a number is a perfect square using prime factorization, simply check if every prime factor appears with an even exponent. For instance, the number 36 can be broken down as:
  • 36 = 2^2 \( \times \) 3^2
The exponents are both even, confirming 36 is indeed a perfect square, as it equals \((2 \times 3)^2 = 6^2\). Every perfect square must adhere to this rule because of how multiplication works.
Even Exponents
When exploring even exponents, the focus is on the powers to which prime factors are raised in the canonical form of integers. An exponent is an even number if it can be expressed as \( 2n \), where \( n \) is an integer. This means it can be divided evenly by 2. For example:
  • 4, 6, 8 are all even exponents because they can be rewritten as \( 2\times 2, 3\times 2, 4\times 2 \), respectively.
Even exponents have a neat property: they play a critical role in perfect squares. If a number's prime factorization has even exponents for all prime numbers involved, it suggests that the number is a perfect square. For instance, in the number 144, we see:
  • 144 = 2^4 \( \times \) 3^2
Both exponents are even, thereby confirming that 144 is a perfect square (\( 12^2 \)). This property is what connects even exponents to perfect squares directly.
Integer Exponents
In the context of prime factorization, integer exponents tell us how many times a prime number divides another number. Every integer exponent can either be odd or even. This distinction is vital for determining whether a number is a perfect square. Let's explore the basics.
  • An integer exponent indicates multiplication of a number by itself a certain number of times. For example, \( 7^3 \) means multiplying 7 by itself three times, i.e., \( 7 \times 7 \times 7 \).
  • The rules of exponents, such as \( a^m \times a^n = a^{m+n} \), help in simplifying and evaluating expressions.
Understanding integer exponents allows us to utilize them in identifying the properties of numbers, especially within prime factorization. If a number is truly a perfect square, each prime factor in its prime factorization will have an even integer exponent.

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Most popular questions from this chapter

Modify Euclid's proof that there are infinitely many primes by assuming the existence of a largest prime \(p\) and using the integer \(N=p !+1\) to arrive at a contradiction.

An integer is said to be square-free if it is not divisible by the square of any integer greater than \(1 .\) Prove the following: (a) An integer \(n>1\) is square-free if and only if \(n\) can be factored into a product of distinct primes. (b) Every integer \(n>1\) is the product of a square-free integer and a perfect square. [Hint: If \(n=p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{s}^{k_{3}}\) is the canonical factorization of \(n\), then write \(k_{i}=\) \(2 q_{i}+r_{i}\) where \(r_{i}=0\) or 1 according as \(k_{i}\) is even or odd.]

A conjecture of Lagrange (1775) asserts that every odd integer greater than 5 can be written as a sum \(p_{1}+2 p_{2}\), where \(p_{1}, p_{2}\) are both primes. Confirm this for all odd integers through 75 .

It has been conjectured that every even integer can be written as the difference of two consecutive primes in infinitely many ways. For example, $$ 6=29-23=137-131=599-593=1019-1013=\cdots $$ Express the integer 10 as the difference of two consecutive primes in 15 ways.

Let the sequence of primes, with 1 adjoined, be denoted by \(p_{0}=1, p_{1}=2, p_{2}=3\), \(p_{3}=5, \ldots .\) For each \(n \geq 1\), it is known that there exists a suitable choice of coefficients \(\epsilon_{k}=\pm 1\) such that $$ p_{2 n}=p_{2 n-1}+\sum_{k=0}^{2 n-2} \epsilon_{k} p_{k} \quad p_{2 n+1}=2 p_{2 n}+\sum_{k=0}^{2 n-1} \epsilon_{k} p_{k} $$ To illustrate: $$ 13=1+2-3-5+7+11 $$ and $$ 17=1+2-3-5+7-11+2 \cdot 13 $$

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