This is the converse of the statement we proved above using a direct proof. From trying a few examples, this statement definitely appears to be true. So let's prove it.
A direct proof of this statement would require fixing an arbitrary \(n\) and assuming that \(n^2\) is even. But it is not at all clear how this would allow us to conclude anything about \(n\text{.}\) Just because \(n^2 = 2k\) does not in itself suggest how we could write \(n\) as a multiple of 2.
Try something else: write the contrapositive of the statement. We get, for all integers \(n\text{,}\) if \(n\) is odd then \(n^2\) is odd. This looks much more promising. Our proof will look something like this:
Let \(n\) be an arbitrary integer. Suppose that \(n\) is not even. This means that …. In other words …. But this is the same as saying …. Therefore \(n^2\) is not even.
Now we fill in the details:
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.