solve the equation in Z

Question

Solve the equation over $\textbf{Z}$ :

2$x^2$ - 2$xy$ - 5$x$ - $y$ + 19 = 0

I tried to obtain some $(A+B)^2$ terms, but I didn't make it.

Thanks for your time!

Answer 1

$2x^2 - 2xy - 5x - y + 19 = 0 \to y = \dfrac{2x^2 - 5x + 19}{2x + 1} = x - 3 + \dfrac{22}{2x + 1}$. So $2x + 1 | 22$, and $2x + 1 = 1, -1 , 11, -11$. we can then find $y$ for each value of $x$ listed.

Answer 2

The discriminant of $2x^2-(2y+5)x+(19-y)$ as a quadratic in $x$ necessarily needs to be a perfect square. Thus we need values for $y$ s.t. $(2y+5)^2-8(19-y)$ is a perfect square. Now $$(2y+5)^2-8(19-y)=s^2 \iff (2y+7)^2-176 = s^2$$ so we need integer solutions for $$(2y+7+s)(2y+7-s)=2^4 \cdot 11$$

Consider the possible (finite number of) factorisations, solve for $y, x$ and exclude non-integer cases.