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?

Question

Sahil Verma · Accepted Answer

We're given the equation: (x-y)^2 + 2y^2 = 27
Let's expand (x-y)^2:x^2 - 2xy + y^2 + 2y^2 = 27x^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 x&sup2; - 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 = 273y^2 - 2y - 26 = 0This is a quadratic in y, but it doesn't have integer solutions.If x = 3:9 - 6y + 3y^2 = 273y^2 - 6y - 18 = 0y^2 - 2y - 6 = 0(y-3)(y+1) = 0y = 3 or y = -1Both 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 = 273y^2 - 10y - 2 = 0
Therefore, x = 3 is a possible value for x that satisfies the equation.