Every prime larger than 3 is either of form 6k+1, or 6k+5; the other four possibilities are either divisible by 2 or by 3 (or by both). Now (6k+1)² − 1 = 6k(6k+2) = 12k(3k+1) and at least one of k and 3k+1 must be even. Also (6k+5)² − 1 = (6k+4)(6k+6) = 12(3k+2)(k+1) and at least one of 3k+2 and k+1 must be even.