If x and y are integers and if (x-y)^2 + 2*y^2 = 27, which of the following could be the value of x?

We're given the equation: (x-y)^2 + 2y^2 = 27

Let's expand (x-y)^2:
x^2 - 2xy + y^2 + 2y^2 = 27
x^2 - 2xy + 3y^2 = 27

Now, since x and y are integers, x^2 - 2xy + 3y^2 must equal 27.

Let's consider some properties of this equation:
- The left side is always positive (or zero) for real x and y.
- 27 is odd, so x^2 - 2xy must be odd (since 3y^2 is always even for integer y).
- For x² - 2xy to be odd, x must be odd.

Let's try some odd values for x and see if we can find integer solutions for y:
If x = 1:
1 - 2y + 3y^2 = 27
3y^2 - 2y - 26 = 0
This is a quadratic in y, but it doesn't have integer solutions.
If x = 3:
9 - 6y + 3y^2 = 27
3y^2 - 6y - 18 = 0
y^2 - 2y - 6 = 0
(y-3)(y+1) = 0
y = 3 or y = -1
Both of these work! When y = 3, left side = 9 - 18 + 27 = 18, and when y = -1, left side = 9 + 6 + 3 = 18.
If x = 5:
25 - 10y + 3y^2 = 27
3y^2 - 10y - 2 = 0

Therefore, x = 3 is a possible value for x that satisfies the equation.