1. The problem statement, all variables and given/known data For n ∈ Z+ prove by contrapositive that if 2n3 + 3n2 + 4n + 5 is odd then n is even. 2. Relevant equations 3. The attempt at a solution If n is odd then ( 2n³ + 3n² + 4n + 5 ) is even. By definition, a number n is odd if n = 2k + 1 for some integer k. A number n is even if n = 2k for some integer k. If x and y are two integers for which x + y is even, then by definition x and y have the same parity. It follows that two integers will have the same parity if they are both odd or both even. 2(2k + 1)³ + 3(2k + 1)² + 4(2k + 1) + 5 We see here that 2(2k + 1)³ can be written as 2n where n = (2k+1)³. Thus we conclude 2(2k + 1)³ is an even number. By the same logic, 4(2k + 1) can be written as 2n where n = 2(2k+1). I get stuck after this... How can i prove 3(2k + 1)² is odd so that i can add it to 5 and get even parity?