Direct Proofs (4)

(4) If m is odd, then m^2 + 1 is even.

Proof. Assume that m is odd. Then m = 2k + 1 for some integer k. So
       
          m^2 + 1 = (2k +1)^2 + 1
                        = 4k^2 + 4k + 1 + 1
                        = 4k^2 + 4k +2
                        = 2(2k^2 + 2k + 1)
                        = 2q

where q = 2k^2 + 2k + 1 is an integer. So m^2 + 1 is even.

QED