## Task 6 - 110936

Determine all ordered pairs $(x, y)$ of integers $x, y$ that are solutions to the following equation!

$$
2 x^{2}-2 x y-5 x-y+19=0
$$
By the substitution $y=z-3$ with $z \in \mathbb{Z}$, the equation transforms into

$$
2 x^{2}-2 x z+6 x-5 x-z+3+19=0 \quad \text { i.e. } \quad 2 x^{2}-2 x z+x-z=-22
$$

or $(2 x+1)(x-z)=-22$.

Thus, $2 x+1$ is an integer divisor of -22. Since this term is also odd, the following four cases arise:

1. Case: $2 x+1=1$ and $x-z=-22$. Then $x=0, z=22$ and $y=19$.
2. Case: $2 x+1=-1$ and $x-z=22$. Then $x=-1, z=-23$ and $y=-26$.
3. Case: $2 x+1=11$ and $x-z=-2$. Then $x=5, z=7$ and $y=4$.
4. Case: $2 x+1=-11$ and $x-z=2$. Then $x=-6, z=-8$ and $y=-11$.

Verification confirms all results. The equation is thus solved exactly by the integer pairs $(x, y) \in\{(-6,-11),(-1,-26),(0,19),(5,4)\}$.