## Task B-2.2.

How many integers $x$ are there for which the value of the expression

$$
\frac{\left(x-1^{2}\right)\left(x-3^{2}\right) \ldots\left(x-(2 k-1)^{2}\right) \ldots\left(x-2019^{2}\right)}{\left(x-2^{2}\right)\left(x-4^{2}\right) \ldots\left(x-(2 k)^{2}\right) \ldots\left(x-2020^{2}\right)}
$$

is negative?
## Solution.

The sign of the given expression changes at the zeros of the numerator and denominator as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_1c7eac828d9f4d45279fg-07.jpg?height=132&width=1060&top_left_y=2367&top_left_x=498)

Using this representation (or sign tables), we conclude that the value of the expression is positive for all $x2020^{2}$, and on the intervals determined by the zeros of the numerator and denominator $1^{2}, 2^{2}, 3^{2}, \ldots, 2019^{2}, 2020^{2}$, the sign will be negative on the intervals

$$
\left\langle 1^{2}, 2^{2}\right\rangle,\left\langle 3^{2}, 4^{2}\right\rangle, \ldots,\left\langle(2 k-1)^{2},(2 k)^{2}\right\rangle, \ldots,\left\langle 2019^{2}, 2020^{2}\right\rangle
$$

The number of integers within these intervals is

$$
\begin{aligned}
& \left(2^{2}-1^{2}-1\right)+\left(4^{2}-3^{2}-1\right)+\cdots+\left((2 k)^{2}-(2 k-1)^{2}-1\right)+\cdots+\left(2020^{2}-2019^{2}-1\right) \\
& =\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+\cdots+\left((2 k)^{2}-(2 k-1)^{2}\right)+\cdots+\left(2020^{2}-2019^{2}\right)-1010 \\
& =(2-1)(2+1)+(4-3)(4+3)+\cdots+ \\
& \quad((2 k)-(2 k-1))((2 k)+(2 k-1))+\cdots+(2020-2019)(2020+2019)-1010 \\
& =1+2+3+4+\cdots+2020-1010=\frac{2020 \cdot(1+2020)}{2}-1010=2040200 .
\end{aligned}
$$