Chapter 1.1, Problem 53E

Solution

To prove: If $n$is a positive integer, then $n^2+2n$ is either odd or a multiple of 4.

Given Information:

A positive integer $n$ .

Concept Used:

If $n$ is odd then it must be of the form $n=2k+1$ , where $k$ is a positive integer.

If $n$ is even then it must be of the form $n=2k$ , where $k$ is a positive integer.

Proof:

Let $n$ be a positive integer.

Then $n$ can be either even or odd.

Case 1: Let $n$ be odd.

If $n$ is odd then it must be of the form $n=2k+1$ , where $k$ is a positive integer.

Substitute $n=2k+1$ into $n^2+2n$ and simplify as follows.

  $\begin{array}{c}{n}^2+2n={\left(2k+1\right)}^2+2\left(2k+1\right)\\ =4k^2+1+4k+4k+2\\ =4k^2+8k+2+1\\ =2\left(2k^2+4k+1\right)+1\\ =2m+1\left(2k^2+4k+1=m\right)\end{array}$

Thus, if $n$ is odd then $n^2+2n$ is also odd.

Case 2: Let $n$ be even.

If $n$ is even then it must be of the form $n=2k$ , where $k$ is a positive integer.

Substitute $n=2k$ into $n^2+2n$ and simplify as follows.

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

Thus, if $n$ is even then $n^2+2n$ is a multiple of 4.

Therefore, from case 1 and 2 it follows that if $n$is a positive integer, then $n^2+2n$ is either odd or a multiple of 4.