Here’s the proof in that form: Proof (Direct): Assume that n is odd. Let n = 2k+1, k ∈ Z. n2−3 = (2k+1)2−3 = (4k2+4k+1)−3 = 4k2+4k−2 = 2(2k2+2k−1). As 2(2k2 + 2k − 1) is of the form 2j where j ∈ Z, we know that n2 − 3 is even. Therefore, n2 − 3 is even if n is odd. Proof (Direct): Assume that n is odd. Let n = 2k + 1, k ∈ Z.