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.**

## Comments

Very nice answer.