(i) Let \(n\) be an even integer. Then \(n=2k\) for some integer \(k\). So \[n^2 = (2k)^2 = 4k^2 = 2 (2k^2) \] and since \(2k^2\) is an integer, \(n^2\) also satisfies the definition of an even integer.
(ii) To prove that \(n\) is even if \(n^2\) is even, we proceed by contraposition. Hence, suppose that \(n\) is not even and therefore odd. Then \(n=2k+1\) for some integer \(k\). So \[n^2=(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1\] and thus is also odd. Thus by contraposition, we have proved that if \(n^2\) is even, \(n\) is even.
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