Proof.

We proceed by contradiction. So suppose there are integers \(x\) and \(y\) such that \(x^2 = 4y + 2 = 2(2y + 1)\text{.}\) So \(x^2\) is even. We have seen that this implies that \(x\) is even. So \(x = 2k\) for some integer \(k\text{.}\) Then \(x^2 = 4k^2\text{.}\) This in turn gives \(2k^2 = (2y + 1)\text{.}\) But \(2k^2\) is even, and \(2y + 1\) is odd, so these cannot be equal. Thus we have a contradiction, so there must not be any integers \(x\) and \(y\) such that \(x^2 = 4y + 2\text{.}\)