Let x,y∈Z. Prove that if x and y have the same parity, then x^2+xy is even

Answer 1

either both x and y are even or both are odd

either x=2m and y=2n or not

if x=2m and y = 2n then x^2 +xy = (2m)^2 + (2m)(2n) = 4m^2 +4mn = 4(m^2 +mn) = 2[2(m^2+mn)] which is a multiple of 2 and therefore even

or x and y are both odd and

x=2m+1 and y=2n+1

then x^2 +xy = (2m+1)^2 + (2m+1)(2n+1) = 4m^2 + 4m + 1 + 4mn+2m+2n+1 = 4m^2 +6m + 4mn +2n + 2 =

2[2m^2 +3m +2mn +n +1] which is a multiple of 2 and even