## Task 4 - 241234

Determine all real numbers $z$ with $1 \leq z \leq 5$ that satisfy the condition that the line with the equation $y=\frac{1}{3} x+z$ and the parabola with the equation $y=2 x^{2}$ have at least one intersection point with an integer abscissa.

For each number $z$ that satisfies this condition, provide - for the relevant line and parabola - the coordinates of all intersection points with an integer abscissa.
Let $x$ be the abscissa of an intersection point of the two curves. Then it holds that $2 x^{2}=\frac{1}{3} x+z$ or $x^{2}-\frac{1}{6} x-\frac{1}{2} z=0$, thus

$$
x=\frac{1}{12} \pm \sqrt{\frac{1}{144}+\frac{z}{2}}=\frac{1 \pm \sqrt{1+72 z}}{12}
$$

Case 1: It is $x=\frac{1+\sqrt{1+72 z}}{12} \in \mathbb{Z}$. Then, because

$$
0\frac{1-\sqrt{1+72 \cdot 1}}{12} \geq x=\frac{1-\sqrt{1+72 z}}{12} \geq \frac{1-\sqrt{1+72 \cdot 5}}{12}=\frac{1-19}{12}>\frac{-24}{12}=-2
$$

thus $x=-1$. Then $y=2 x^{2}=2$ and therefore $z=y-\frac{1}{3} x=\frac{7}{3}$.

Thus, there are exactly in the two cases $z=\frac{5}{3}$ and $z=\frac{7}{3}$ in the considered interval for $z$ intersection points with integer abscissas, namely in the first case the point (1|2) and in the second the point $(-1 \mid 2)$. (The other intersection points, if they exist, do not have integer abscissas, as they would have had to appear as solutions in the respective cases.)