Proof.
Let \(n\) be an arbitrary integer. Suppose \(n\) is even. Then \(n = 2k\) for some integer \(k\text{.}\) Now \(n^2 = (2k)^2 = 4k^2 = 2(2k^2)\text{.}\) Since \(2k^2\) is an integer, \(n^2\) is even.
The format of the proof will be this: Let \(n\) be an arbitrary integer. Assume that \(n\) is even. Explain explain explain. Therefore \(n^2\) is even.
To fill in the details, we will basically just explain what it means for \(n\) to be even, and then see what that means for \(n^2\text{.}\) Here is a complete proof.
Let \(n\) be an arbitrary integer. Suppose \(n\) is even. Then \(n = 2k\) for some integer \(k\text{.}\) Now \(n^2 = (2k)^2 = 4k^2 = 2(2k^2)\text{.}\) Since \(2k^2\) is an integer, \(n^2\) is even.