Chapter 5.4, Problem 49WE

To determine

To show: the square of any odd integer is odd.

Explanation of Solution

Concept Used:

  $a^n\cdot a^m=a^{n+m}$

Calculation:

It is known that for any integer n , $2n$ is an even integer and $2n+1$ is an odd integer.

Consider the square of odd integer $2n+1$ ,

  $\begin{array}{c}{\left(2n+1\right)}^2=\left(2n+1\right)\left(2n+1\right)\\ =2n\left(2n\right)+2n\left(1\right)+\left(1\right)2n+1\left(1\right)\\ =4n^{1+1}+2n+2n+1\end{array}$

  $\begin{array}{c}=4n^2+4n+1\\ =2\left(2n^2+2n\right)+1\end{array}$

Now, since n is an integer, so $k=2n^2+2n$ is also an integer.

Then, ${\left(2n+1\right)}^2=2k+1$ for some integer $k=2n^2+2n$

Therefore ${\left(2n+1\right)}^2=2k+1$ is odd.