Chapter 12.4, Problem 26AYU

In Problems 23-27, prove each statement.

$a+b$ is a factor of $a^{2n\text{ + 1}}-b^{2n\text{ + 1}}$.

To determine

To prove: The given statement $a + b$ is a factor of $a^{2n + 1} + b^{2n + 1}$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 26AYU

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.

Explanation of Solution

Given:

Statements says the series $a + b$ is a factor of $a^{2n + 1} + b^{2n + 1}$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$a + b$ is a factor of $a^{2n + 1} + b^{2n + 1}$   -----(1)

Step 1: Show that statement (1) is true for the natural number $n = 1$.

That is $a^{2\left(1\right)+1}+b^{2\left(1\right)+1}=a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right)$. Here both the terms contains $a + b$. $a + b$ is a factor of $a^{2n + 1} + b^{2n + 1}$ when $n = 1$. Hence the statement is true for the natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $a + b$ is a factor of $a^{2k + 1} + b^{2k + 1}$   -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove tha $a + b \text{is a factor of} a^{2k + 3} + b^{2k + 3}$

Consider $a^{2k + 3} + b^{2k + 3}$

$a^{2k + 3} + b^{2k + 3} = a^2\left(a^{2k + 1} + b^{2k + 1}\right)-b^{2k + 1}\left(a^2-b^2\right)$

In here, the first term is a multiple of $\left(a^{2k + 1} + b^{2k + 1}\right)$, which has a factor $a + b$ from equation (2) and the second term is a multiple of $\left(a^2 - b^2\right)$, which obviously $\left(a^2 - b^2\right) = \left(a + b\right)\left(a - b\right)$ has the factor $a + b$. Therefore $a + b$ is a factor of $\left(a^{2k + 1} + b^{2k + 1}\right)$

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.