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Chapter 4: Problem 9

Show that \(a^{m}+1\) is composite if \(a\) and \(m\) are integers greater than 1 and \(m\) is odd. [Hint: Show that \(x+1\) is a factor of the polynomial \(x^{m}+1\) if \(m\) is odd. \(]\)

Short Answer

Expert verified
\(a^m + 1\) is composite if \(a\) and \(m\) are greater than 1 and \(m\) is odd because it can be factorized as \((a + 1)(a^{m-1} - a^{m-2} + ... + a - 1)\).

Step by step solution

01

- Set Up the Problem

Given that we need to prove that \(a^m + 1\) is composite if \(a\) and \(m\) are integers greater than 1, and \(m\) is odd.
02

- Use the Hint

According to the hint, we need to show that \(x + 1\) is a factor of the polynomial \(x^m + 1\) if \(m\) is odd.
03

- Substitute and Factorize

Let \(x = a\). Then the polynomial becomes \(a^m + 1\). Since \(m\) is odd, we can write \(m = 2k + 1\) for some integer \(k\). Thus, we factorize \(x^m + 1\) as follows: \[ \[\begin{equation} x^m + 1 = x^{2k+1} + 1 = (x + 1)(x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) \end{equation}\] \]
04

- Verify the Factorization

We can verify that \(x^{2k+1} + 1\) is factored as \( (x + 1)(x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) \) by multiplying these two factors. For example, \( x(x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) + (x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) \) will result in the cancellation of all terms except \( x^{2k+1} + 1 \).
05

- Conclude Composite Nature

Since \(x + 1\) is a factor of \(x^m + 1\) and both \(a\) and \(m\) are integers greater than 1, \(a^m + 1\) must be composite because it can be factored into a product of two non-trivial polynomials. Therefore, \(a^m + 1\) is composite if \(a\) and \(m\) are integers greater than 1 and \(m\) is odd.

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Key Concepts

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

Integer Factorization
Integer factorization is the process of breaking down a composite number into smaller non-trivial divisors, which when multiplied together give the original number. For example, the number 12 can be factorized into 2, 2, and 3 because 2 \( \times \) 2 \( \times \) 3 equals 12.

Understanding integer factorization is crucial for solving problems related to composite numbers. It allows us to express a number as a product of smaller integers, which is useful in a variety of mathematical contexts, including proofs and cryptographic algorithms.

In the exercise, we see integer factorization at play when we take the polynomial \( x^m + 1 \) and factorize it into \( (x + 1)(x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) \). This shows that \( x^m + 1 \) is composite because it can be expressed as a product of two non-trivial factors.
Polynomial Factorization
Polynomial factorization involves expressing a polynomial as a product of its factors. This is similar to integer factorization but applied to polynomials instead of integers.

In our exercise, we need to factorize the polynomial \( x^m + 1 \) where \( m \) is odd. The hint suggests that we use a specific factorization technique showing that \( x + 1 \) is a factor of \( x^m + 1 \) if \( m \) is odd.

Specifically, we substitute \( x = a \) and consider \( a^m + 1 \), which becomes composite as a result of the factorization:

Selecting an odd \( m \), let's say \( m = 2k + 1 \), we can rewrite and factor it as:

\[ x^m + 1 = x^{2k+1} + 1 = (x + 1)(x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) \]

This factorization helps us to see that \( x^m + 1 \) is not a prime polynomial and is composite because it can be broken down into the product of simpler polynomials.
Odd Exponents
An odd exponent means the power to which a number is raised is odd. For example, in the expression \( x^3 \), 3 is an odd exponent.

Working with odd exponents is critical in our problem because they significantly influence how the polynomial \( x^m + 1 \) behaves and can be factorized. Importantly, if \( m \) is odd, the polynomial \( x^m + 1 \) can be broken down using polynomial factorization.

In the given exercise, if \( m = 2k + 1 \) (which is clearly odd), it enables us to factorize \( x^m + 1 \) into two parts:

\( x^m + 1 = x^{2k+1} + 1 = (x + 1)(x^{2k} - x^{2k-1} + x^{2k-2} - ... + x - 1) \)

This factorization wouldn't work the same way if \( m \) were even because the pattern and potential factors differ.

Thus, the property of having an odd exponent is key to our proof that \( a^m + 1 \) is composite when \( a \) and \( m \) are integers greater than 1, and \( m \) is odd.

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Most popular questions from this chapter

Show that if \(p\) is an odd prime, then there are exactly \((p-1) / 2\) quadratic residues of \(p\) among the integers \(1,2, \ldots, p-1\) If \(p\) is an odd prime and \(a\) is an integer not divisible by \(p\) , the Legendre symbol \(\left(\frac{a}{p}\right)\) is defined to be 1 if \(a\) is a quadratic residue of \(p\) and \(-1\) otherwise.

Prove that the set of positive rational numbers is countable by showing that the function \(K\) is a one-to- one correspondence between the set of positive rational numbers and the set of positive integers if \(K(m / n)=p_{1}^{2 a_{1}} p_{2}^{2 a_{2}} \cdots \cdots p_{s}^{2 a_{s}} q_{1}^{2 b_{1}-1} q_{2}^{2 b_{2}-1} \ldots \cdots q_{t}^{2 b_{t}-1}\) where gcd \((m, n)=1\) and the prime-power factorizations of \(m\) and \(n\) are \(m=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots \cdot p_{s}^{a_{s}}\) and \(n=q_{1}^{b_{1}} q_{2}^{b_{2}} \cdots q_{t}^{b_{t}}\)

Write an algorithm in pseudocode for generating a sequence of pseudorandom numbers using a linear congruential generator. The middle-square method for generating pseudorandom numbers begins with an \(n\) -digit integer. This number is squared, initial zeros are appended to ensure that the result has 2\(n\) digits, and its middle \(n\) digits are used to form the next number in the sequence. This process is repeated to generate additional terms.

Show that the integer \(m\) with one's complement representation \(\left(a_{n-1} a_{n-2} \ldots a_{1} a_{0}\right)\) can be found using the equation \(m=-a_{n-1}\left(2^{n-1}-1\right)+a_{n-2} 2^{n-2}+\cdots+a_{1} \cdot 2+a_{0}\).

To break a Vigen猫re cipher by recovering a plaintext message from the ciphertext message without having the key, the first step is to figure out the length of the key string. The second step is to figure out each character of the key string by deter- mining the corresponding shift. Exercises 21 and 22 deal with these two aspects. Once the length of the key string of a Vig猫nere cipher is known, explain how to determine each of its characters. Assume that the plaintext is long enough so that the frequency of its letters is reasonably close to the frequency of letters in typical English text.
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