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.
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.