Chapter 9.3, Problem 21E

(a)

To determine

A value of n greater than 2

Answer to Problem 21E

  $\begin{array}{l}2n=10\\ n^2-1=24\\ n^2+1=26\end{array}$

Explanation of Solution

Given:

The given expressions are:

  $2n,n^2-1,n^2+1$

Calculation:

Let the value of n = 5

Therefore, the terms will be

  $\begin{array}{l}2n=2\left(5\right)=10\\ n^2-1={\left(5\right)}^2-1=25-1=24\\ n^2+1={\left(5\right)}^2+1=25+1=26\end{array}$

(b)

To determine

To verify: The numbers obtained form a Pythagorean triple

Explanation of Solution

Given:

The given expressions are:

  $2n,n^2-1,n^2+1$

Calculation:

Let the value of n = 5

Therefore, the terms will be

  $\begin{array}{l}2n=2\left(5\right)=10\\ n^2-1={\left(5\right)}^2-1=25-1=24\\ n^2+1={\left(5\right)}^2+1=25+1=26\end{array}$

Hence,

  $\begin{array}{l}{c}^2=a^2+b^2\\ {\left(26\right)}^2={\left(10\right)}^2+{\left(24\right)}^2\\ 676=100+576\\ 676=676\end{array}$