Chapter 12: Q 26. (page 830)

In Problem, prove each statement.
$a + b$ is a factor of $a^{2 n + 1} + b^{2 n + 1}$

Short Answer

Expert verified

The given statement is shown.

Step by step solution

Step 1. Given information

The given expression is $a + b$ .

Step 2. Prove the given statement.

First we consider the statement for $n = 1$ .

$\begin{array}{rcl} a^{2.1 + 1} + b^{2.1 + 1} & = & a^{3} + b^{3} \\ = & (a + b) (a^{2} + b^{2} - a b) \end{array}$

which is divisible by $a + b$

Hence, condition holds true.

Now we assume the statement is true for some $k$

Now we consider,

$\begin{array}{rcl} a^{2 (k + 1) + 1} + b^{2 (k + 1) + 1} & = & a^{2 k + 3} + b^{2 k + 3} \\ = & a^{2 k + 1} \cdot a^{2} + b^{2 k + 1} \cdot b^{2} \\ = & (a^{2 k + 1} + b^{2 k + 1} - b^{2 k + 1}) \cdot a^{2} + b^{2 k + 1} b^{2} \\ = & (a^{2 k + 1} + b^{2 k + 1}) a^{2} - a^{2} b^{2 k + 1} + b^{2 k + 1} b^{2} \\ = & (a^{2 k + 1} + b^{2 k + 1}) a^{2} + b^{2 k + 1} (b^{2} - a^{2}) \end{array}$

Since $(a^{2 k + 1} + b^{2 k + 1})$ is divisible by $a + b$ and $b^{2 k + 1} (b^{2} - a^{2})$ is also divisible by $a + b$ , we say that condition also holds true.

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In Problem, prove each statement.
$a + b$ is a factor of $a^{2 n + 1} + b^{2 n + 1}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Prove the given statement.

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In Problem, prove each statement.a+bis a factor ofa2n+1+b2n+1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Prove the given statement.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Mechanics Maths

Geometry

Pure Maths

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Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Problem, prove each statement.
$a + b$ is a factor of $a^{2 n + 1} + b^{2 n + 1}$