Below is a proof of a result.Proof We consider two cases.Case 1. a and b are even. Then

Chapter 3, Problem 3.37

(choose chapter or problem)

Below is a proof of a result.Proof We consider two cases.Case 1. a and b are even. Then a = 2r and b = 2s for integers r and s. Thusa2 b2 = (2r)2 (2s)2 = 4r 2 4s2 = 2(2r 2 2s2).Since 2r 2 2s2 is an integer, a2 b2 is even.Case 2. a and b are odd. Then a = 2r + 1 and b = 2s + 1 for integers r and s. Thusa2 b2 = (2r + 1)2 (2s + 1)2 = (4r 2 + 4r + 1) (4s2 + 4s + 1)= 4r 2 + 4r 4s2 4s = 2(2r 2 + 2r 2s2 2s).Since 2r 2 + 2r 2s2 2s is an integer, a2 b2 is even.Which of the following is proved?(1) Let a, b Z. Then a and b are of the same parity if and only if a2 b2 is even.(2) Let a, b Z. Then a2 b2 is even.(3) Let a, b Z. If a and b are of the same parity, then a2 b2 is even.(4) Let a, b Z. If a2 b2 is even, then a and b are of the same parity.