If x > 1 is (x+y)(x-y) an even integer?

#1:
1)\(x = 5, y = 5\)
\(x + y = 10\), even
\((x+y)*(x-y) = 0\), even integer
2)\(x = 5 + \sqrt{2}, y = 5 - \sqrt{2}\)
\((x+y)*(x-y) = 10*2*\sqrt{2}\), not even integer
#1 - insufficient

#2:
Since X is bigger than 1 the only way we can get prime number as \(2*x\) is if X is pretty much a half of prime number, for example x = 3,5 (7), x =2,5(5),x = 8,5 (17) etc. Its worth mentioning that the number will always look like "a,5" with 5 being the only possible decimal and a - combination of units and tenth digits if necessary.
Now, with just #2 we have no idea about value of y, so with x = 3,5 we can either choose y = 3,5 which would give us 0, even integer, or y = 3, which would give us \(6,5*0,5 = 3,25\) - not even integer
#2 - insufficient

#1 + #2.
Using the information we accumulated from #2 and taking into account #1 we can conclude that y is also a number which looks like "b,5" with 5 being the only possible decimal. In this case whether you subtract y from x or add y to x you will get an integer value. "a,5" - "b,5" = "a-b" and "a,5" + "b,5" = "a + b + 1". For instance, x = 5,5, y = 3,5: (x+y)*(x-y) = 9*2 = 18. To figure out if we get only even integer as a result lets generalise our numbers:
x = a + 0,5
y = b + 0,5
a, b - integers bigger than 1
(x+y)*(x-y) = (a+b+1)*(a-b)
1)a - even, b - odd: even*odd = even - good
2)a - even, b - even: odd*even = even - good
3)a - odd, b - odd: odd*even = even - good
That makes #1 and #2 combined sufficient to answer our question, thus answer C