We will prove the contrapositive. Let \(n\) be an arbitrary integer. Suppose that \(n\) is not even, and thus odd. Then \(n= 2k+1\) for some integer \(k\text{.}\) Now \(n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\text{.}\) Since \(2k^2 + 2k\) is an integer, we see that \(n^2\) is odd and therefore not even.
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