## Task A-2.1.

Determine all pairs $(a, b)$ of integers such that the area of the triangle whose vertices are the points where the parabola $y=x^{2}+a x+b$ intersects the coordinate axes is 3.
## Solution.

Let $A$ and $B$ be the intersections of the parabola with the $x$-axis, and $C$ be the intersection with the $y$-axis.

![](https://cdn.mathpix.com/cropped/2024_05_30_659ee8c29a87e0ecc7d7g-07.jpg?height=810&width=666&top_left_y=1077&top_left_x=615)

Then $A=\left(x_{1}, 0\right), B=\left(x_{2}, 0\right)$, where $x_{1}, x_{2}$ are the roots of the function $f(x)=x^{2}+a x+b$ and $|A B|=\left|x_{1}-x_{2}\right|$.

By Viète's formulas, we have

$$
\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=a^{2}-4 b, \quad \text { i.e. } \quad|A B|=\sqrt{a^{2}-4 b} . \quad 2 \text { points }
$$

The point $C$ has coordinates $(0, b)$, so the length of the height of triangle $A B C$ is $|b|$.

The area of triangle $A B C$ is

$$
3=\frac{\sqrt{a^{2}-4 b} \cdot|b|}{2}
$$

from which it follows that $\left(a^{2}-4 b\right) \cdot b^{2}=36$.

Since $b^{2}$ divides the number 36, it follows that $b^{2} \in\{1,4,9,36\}$.

We consider four cases:

1. If $b^{2}=1$, then $a^{2}-4 b=36$. From $b=-1$ we get $a^{2}=40$, and from $b=1$ we get $a^{2}=32$, so there is no solution in this case.

1 point

2. If $b^{2}=4$, then $a^{2}-4 b=9$. From $b=-2$ we get $a^{2}=1$, and from $b=2$ we get $a^{2}=17$, so the solutions in this case are $(-1,-2)$ and $(1,-2)$.

1 point

3. If $b^{2}=9$, then $a^{2}-4 b=4$. From $b=-3$ we get $a^{2}=-8$, and from $b=3$ we get $a^{2}=16$, so the solutions in this case are $(-4,3)$ and $(4,3)$.

1 point

4. If $b^{2}=36$, then $a^{2}-4 b=1$. From $b=-6$ we get $a^{2}=-23$, and from $b=6$ we get $a^{2}=25$, so the solutions in this case are $(-5,6)$ and $(5,6)$.