2. On a chessboard, four rooks are placed in such a way that they attack all the white squares.

a) Provide an example of such an arrangement.

b) Determine the number of such arrangements

Zhenodarov R.G.
Solution. There are 32 white squares on a chessboard. If a rook is placed on a white square, it attacks 7 white squares; if it is placed on a black square, it attacks 8 squares. 1) Since there are only 4 rooks and 32 white squares, the rooks must be placed only on black squares. Consider the division of the chessboard into 4 equal vertical and 4 horizontal strips (each strip is the union of two adjacent lines).

2) If there is no rook on any strip (for definiteness, on a horizontal strip), then the 8 white squares of this strip must be attacked by 8 rooks vertically, which contradicts the problem's condition.
3) Therefore, there must be exactly 1 rook on each of the 4 horizontal and 4 vertical strips.

The horizontal and vertical strips intersect to form 16 cells of size $2 \times 2$ (2 black and 2 white squares). To determine the required placement of the rooks, it is necessary to select one square on each horizontal strip so that there is only one square on each vertical strip (see point 1 above). The total number of ways to choose such squares is $4 \times 3 \times 2 \times 1$. Considering the statements in points 1) and 2), within each cell, a rook can be placed in 2 ways, resulting in the number of placements being $2 \times 4$ !.

Answer. $2 \times 4$ !.

Recommendations for checking. 2 points for a correct example of rook placement;

2 points for noting that the rooks must be placed only on black squares;

4 points for proving that there must be exactly 1 rook on each of the 4 horizontal and 4 vertical strips.

3. Find all non-negative polynomials $g$ that satisfy the identity $g(x+y)=g(x)+g(y)+2 \sqrt{g(x) g(y)}$ for all non-negative $x$ and $y$.

Valeev N.F. Solution. If such a polynomial exists, then from the identity $g(x+y)=g(x)+g(y)+2 \sqrt{g(x) g(y)}$ for $y=x \geq 0$, we get $g(x+x)=g(x)+g(x)+2 \sqrt{g(x) g(x)}$ or $g(2 x)=4 g(x)$. We will look for polynomials $g$ in the form $g(x)=\sum_{j=0}^{n} a_{j} x^{j}$. Then from the identity $g(2 x)=4 \cdot g(x)$, it follows that
$g(2 x)-4 g(x) \equiv \sum_{j=0}^{n} a_{j}\left(2^{j}-4\right) x^{j} \equiv 0$ for all $x \geq 0$. This is possible only if $a_{j}\left(2^{j}-4\right)=0$. From this, we get that $a_{j}=0$ for all $j \neq 2$, and for $j=2$, $a_{j}$ can be any number. Thus, if the polynomial $g(x)$ exists, it must have the form $g(x)=a \cdot x^{2}$, where $a$ is any non-negative number. Let's check the identity $g(x+y)=g(x)+g(y)+2 \sqrt{g(x) g(y)}$ for $g(x)=a \cdot x^{2}$: for all non-negative $x$ and $y$, we have $a(x+y)^{2}=a x^{2}+a y^{2}+2 a \cdot x \cdot y=a x^{2}+a y^{2}+2|a| \cdot|x| y|=$.

$a x^{2}+a y^{2}+2 \sqrt{a x^{2} \cdot a y^{2}}=g(x)+g(y)+2 \sqrt{g(x) g(y)}$. Thus, $g(x)=a \cdot x^{2}$.

Answer. $g(x)=a \cdot x^{2}$, where $a \geq 0$.

Recommendations for checking. 2 points for only proving that the polynomial $g(x)=a \cdot x^{2}$ satisfies the identity.

2 points for obtaining an estimate for the degree of the polynomial $g$.