Problem Solving (PS) | Re: x and y are integers such that 4x^2 – y^2 +4x + 4y – 3 = 0.

Another way to solve this question is as follows:

4x^2-y^2+4x+4y-3=0

(2x)^2-y^2+4x+4y-3=0

(2x+y)(2x-y)+4(x+y)=3

4(x+y/2)(x-y/2) + 4(x+y)=3

(x+y/2)(x-y/2) + (x+y) = 3/4

Since x and y are integers, (x+y) will always be an integer. RHS is a fraction. Thus, on LHS, y should always be odd so that there is a component of fraction. If y is even integer, RHS will completely be an integer.

y being positive or negative will have no effect on (x+y/2)(x-y/2).

y can be any odd number, and not necessarily
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