Pepe wrote:
Is x² - y² even?
1) x + y is odd
2) x - y is odd
My answer would be D. But the OA is not D.
If x+y is odd I couldn't think for a possibility to get x² - y² even.
The same for x-y is odd. Again, I can't find a possibility to get x² - y² even.
So the answer has to be D, as for both statements alone we can say "NO", x² - y² is not even.
Did I miss something?
Source: Veritas Prep GMAT Simulator
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.
product of 2 nos (nos - because there is no possible combination for both x+y & x-y to be odd) will always be even
hence C...