[Quadratic equations. Vieta's theorem]

Can all roots of the equations $x^{2}-p x+q=0$ and $x^{2}-(p+1) x+q=0$ be integers if:
a) $q>0$

b) $q<0$ ?

#
a) For example, the roots of the equations $x^{2}-7 x+12=0$ and $x^{2}-8 x+12=0$ are integers (3 and 4, 2 and 6, respectively).

b) Each of these equations has roots of different signs. Let $x_{1}>0$ and $-x_{2}<0$, and $x_{3}>0$ and $x_{4}<0$ (the case when $x_{1}>x_{3}$ is considered similarly). Since all roots are integers, then $x_{3}-x_{1} \geq 1$ and $x_{2}-x_{4} \geq 1$. But $x_{1} - x_{2} = p$ and

$x_{3} - x_{4} = p + 1$. Then $1 = (p + 1) - p = (x_{3} - x_{4}) - (x_{1} - x_{2}) = (x_{3} - x_{1}) + (x_{2} - x_{4}) \geq 2$. Contradiction.

## Answer

a) They can; b) they cannot.

Vasya invented a new chess piece called "super-bishop". One "super-bishop" (denoted as A) attacks another (denoted as B) if they are on the same diagonal, there are no pieces between them, and the next cell along the diagonal after the "super-bishop" B is free. For example, in the diagram, piece a attacks piece b, but does not attack any of the pieces c, d, e, f, or g.

![](https://cdn.mathpix.com/cropped/2024_05_06_23a8a0dc60f629d15d84g-09.jpg?height=377&width=378&top_left_y=326&top_left_x=840)

What is the maximum number of "super-bishops" that can be placed on a chessboard such that each one is attacked by at least one other?

## Solution

An example is shown in the diagram.

![](https://cdn.mathpix.com/cropped/2024_05_06_23a8a0dc60f629d15d84g-09.jpg?height=380&width=378&top_left_y=976&top_left_x=840)

Estimate. We will prove that it is impossible to place more than 32 figures. First, note that "super-bishops" cannot be placed on the edge cells of the board (otherwise, they would not be attacked). We divide the inner $6 \times 6$ square into four $3 \times 3$ squares (see diagram). In each of these small squares, not all cells can be occupied, since otherwise, the "super-bishop" standing in the central cell of the small square would not be attacked. Therefore, no more than $6 \times 6 - 4 = 32$ figures can be placed.

## Answer

32.