4. (3 points) Find all pairs of geometric progressions such that their sum is also a geometric progression.
Answer: the denominators of these progressions must be the same and the sum of the first terms is not equal to 0.

Moreover, any progressions of two elements with non-zero sums of the first and second terms are suitable.

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_59517c05a28c86ab5adcg-3.jpg?height=309&width=538&top_left_y=1613&top_left_x=1270)

Let's call our progressions $a_{n}$ and $b_{n}$, and their denominators $p$ and $q$. Consider the first three terms of each progression. For the sum to also be a geometric progression, it is necessary that $(a_{1} + b_{1})(a_{1} p^{2} + b_{1} q^{2}) = (a_{1} p + b_{1} q)^{2}$.

Expanding the brackets and canceling out the same terms, we get $a_{1} b_{1} p^{2} + a_{1} b_{1} q^{2} = 2 a_{1} b_{1} p q$. Dividing by $a_{1} b_{1}$ and moving everything to the left side, we get $(p - q)^{2} = 0$. This is only possible when $p = q$. Additionally, it is required that $a_{n} + b_{n}$ is not equal to zero as the first term of the geometric progression.

5. (3 points) In how many ways can the figure shown in the diagram be colored by cells in blue, red, and white such that adjacent (i.e., having common sides) cells are colored differently?

Answer: $3 \cdot 48^{3}$.

Solution: The central cell can be colored in three ways. After that, we need to color four identical figures of six cells each with the same restriction on the color of the cell adjacent to the central cell (let's call it cell A).

Consider a $2 \times 2$ square. In such a square, there must be two cells of the same color, and they must be positioned diagonally. There are two ways to choose the diagonal, three ways to choose its color, and two ways to choose the color for each of the other cells, totaling 24 ways. However, squares painted in two colors have been counted twice, so there are 18 ways left.

Clearly, for each of the three colors, there are 6 ways such that the cell adjacent to not only the cells of the square has exactly this color. Let's call this cell cell B.

Cell A can be painted in two ways. For each of these ways, there are 6 ways to color the square such that cell B is painted in the same color. Then the cell between them can be painted in two ways, totaling $2 \cdot 6 \cdot 2 = 24$ options.

In addition, for each way to paint cell A, there are 12 ways to color the square such that cell B has a different color. In this case, the cell between them is painted uniquely, so we again get 24 ways. Therefore, there are 48 ways to color each such figure.

Thus, the answer to the problem is $3 \cdot 48^{3}$.