The problem asks us to prove that if $n$ is an even integer, then $n^2$ is even, using a direct proof.

1. Problem Description

The problem asks us to prove that if $n$ is an even integer, then $n^2$ is even, using a direct proof.

2. Solution Steps

A direct proof starts by assuming the hypothesis is true and then showing the conclusion is true.

* Assume $n$ is an even integer.

This means that $n$ can be written as $2k$ for some integer $k$.

$n = 2k$

* Now, we want to show that $n^2$ is also even.

$n^2 = (2k)^2$

* Squaring the right side, we get:

$n^2 = 4k^2$

* We can rewrite this as:

$n^2 = 2(2k^2)$

* Since $k$ is an integer, $2k^2$ is also an integer. Let $m = 2k^2$.

$n^2 = 2m$

* Since $n^2$ can be written as $2$ times an integer $m$, $n^2$ is an even integer.

3. Final Answer

Therefore, if $n$ is an even integer, then $n^2$ is even.