11.2. A line with a positive slope passes through the point (0, 2020) and intersects the parabola $y=x^{2}$ at two points with integer coordinates. What values can the slope take? List all possible options and explain why there are no others.
Answer: $81,192,399,501,1008,2019$

## Solution:

The equation of the line is $y=kx+b$. Since the line passes through the point $(0,2020)$, then $b=2020$.

The points of intersection of the parabola and the line are the integer and distinct roots of the equation $x^{2}=$ $kx+2020$, or $x^{2}-kx-2020=0$. According to Vieta's theorem,

$$
\left\{\begin{array}{c}
x_{1}+x_{2}=k \\
x_{1} x_{2}=-2020
\end{array}\right.
$$

Since the roots are integers, they are divisors of the number -2020. Factorize -2020. Clearly, the factors have different signs. Note that the larger root in absolute value is positive, and the smaller one in absolute value is negative (since $k>0$): $-2020=-1 \cdot 2020=-2 \cdot 1010$ $=-4 \cdot 505=-5 \cdot 404=-10 \cdot 202=-20 \cdot 101$.

For each option, write down $k: (-1+2020)=2019; (-2+1010)=1008; (-4+505)=501; (-5+404)=$ $399; (-10+202)=192; (-20+101)=81$. In total, there are six different options.

## Criteria:

7 points - correct answer and complete correct solution;

for each lost solution, deduct 1 point;

only answer - 0 points.