odd+odd=even odd+even=odd, even +even =even
odd*odd=odd, odd*even=even, even*even=even
(odd)^{2}=odd, (even)^{2}= even
x+y is even, hence, either x and y are both even or x and y are both odd. Thus x can be either even or odd. a) and b) are both wrong.
When x and y are both even, xy is even; when both x and y are odd, xy is odd. Hence, xy can be either even or odd. e) is wrong
Since x+y is even, (x+y)^{2} is also even. (x+y)+x+z is odd. Hence, even+(x+z)=odd . x+z must be odd. Hence when z is even, x must be odd. c) is correct.
Since x and y must be either both odd or both even, and we also know when z is even, x must be odd. Hence,when z is even, y must be odd too. xy is odd. d) is wrong.
10/27/2012

Linghong C.
Comments
Very nice answer.