4. Two vertices of a square lie on the parabola $y=x^{2}$, and one of its sides lies on the line $y=2 x-22$. What is the area of the square?
Answer: 180 or 980.

Solution. It is easy to check that the parabola and the line from the condition of the problem do not intersect.

The line $l$, parallel to the line $2 x-y-22=0$, is given by the equation of the form $2 x-y+b=0$, where $b$ is some constant. The distance between these lines $d=\frac{|b+22|}{\sqrt{5}}$ (this expression can be obtained in different ways). Let's find the distance between the points $M_{1}\left(x_{1}, y_{1}\right)$ and $M_{2}\left(x_{1}, y_{2}\right)$, where the line $l$ intersects the parabola $y=x^{2}$:

$$
\begin{gathered}
y=2 x+b=x^{2} ; \quad x_{1,2}=1 \pm \sqrt{1+b} ; \quad x_{1}-x_{2}=2 \sqrt{b+1} \\
y_{1}-y_{2}=2\left(x_{1}-x_{2}\right)=4 \sqrt{b+1} ; \quad M_{1} M_{2}^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=20(b+1)
\end{gathered}
$$

By equating the squares of the distances between adjacent vertices of the square and between its opposite sides, we get an equation in terms of the parameter $b$

$$
20(b+1)=\frac{(b+22)^{2}}{5}
$$

From here, $b=8$ or $b=48$. In this case, the area of the square $d^{2}=180$ or $d^{2}=980$.

Evaluation. 13 points for a complete solution.