Is x² - y² even

Note that we are not told that \(x\) and \(y\) are integers.

Is x^2-y^2 even?

(1) x + y is odd --> now, if \(x=0.5\) and \(y=0.5\) then \(x^2-y^2=(x-y)(x+y)=0=even\) but \(x=2\) and \(y=1\) then \(x^2-y^2=(x-y)(x+y)=3=odd\). Two different answers, not sufficient.

(2) x - y is odd --> now, if \(x=1.5\) and \(y=0.5\) then \(x^2-y^2=(x-y)(x+y)=2=even\) but \(x=2\) and \(y=1\) then \(x^2-y^2=(x-y)(x+y)=3=odd\). Two different answers, not sufficient.

(1)+(2) \(x+y=odd\) and \(x-y=odd\) --> \(x^2-y^2=(x-y)(x+y)=odd*odd=odd\). Sufficient.

Answer: C.

P.S. There is one more thing: \(x+y=odd\) doesn't mean that \(x^2-y^2=integer\), for example if \(x=0.7\) and \(y=0.3\) --> \(x^2-y^2=(x-y)(x+y)=0.4\neq{integer}\);
Similarly \(x-y=odd\) doesn't mean that \(x^2-y^2=integer\), for example if \(x=1.3\) and \(y=0.3\) --> \(x^2-y^2=(x-y)(x+y)=1.6\neq{integer}\).

Hope it's clear.