1. For the quadratic equation $x^{2} + p x + q = 0$, the coefficients $p$ and $q$ were increased by one. This operation was repeated four times. Provide an example of such an initial equation so that each of the five resulting equations has integer roots.
1. It is not difficult to check that the roots of the equation $x^{2}+3 x+2=0$ are -1 and -2. After increasing the coefficients by one, we get the equation $x^{2}+4 x+3=0$ with roots -1 and -3, then the equation with roots -1 and -4, then - with roots -1 and -5, and finally, the equation with roots -1 and -6.

To verify the above, one would have to solve 5 quadratic equations. Instead, one can use Vieta's theorem. All our equations have the form

$$
x^{2}+(p+1) x+p=0
$$

where $p$ is an integer. Then, by the converse of Vieta's theorem, the roots of this equation will be the integers -1 and $-p$:

$$
(-1)+(-p)=-(p+1), \quad(-1)(-p)=p
$$

Comments. $1^{\circ}$. Compare with problem 2 for 9th grade.

$2^{\circ}$. Any equation with one root -1 and the other an integer will do. When the coefficients are increased by one, the root -1 will remain, and the second root will decrease by one.

$3^{\circ}$. The fact that the root -1 remains can also be understood as follows: increasing the coefficients $p$ and $q$ by one means adding $x+1$ to the original trinomial. In this case, the value of the trinomial at the point -1 will not change. Therefore, if -1 was a root, it will remain a root.

$4^{\circ}$. Let's try to explain how one might guess that one of the roots should be -1. Let $x_{1}$ and $x_{2}$ be the roots. Then (by Vieta's theorem)

$$
x_{1}+x_{2}=-p, \quad x_{1} x_{2}=q
$$

This means that when $p$ and $q$ are increased by one, the sum of the roots decreases by one. The simplest way to achieve this is to keep $x_{1}$ unchanged and decrease $x_{2}$ by 1:

$$
x_{1}+\left(x_{2}-1\right)=-(p+1)
$$

Let's see what happens to the product of the roots:

$$
x_{1}\left(x_{2}-1\right)=x_{1} x_{2}-x_{1} .
$$

Thus, the product of the roots will decrease by $x_{1}$, but it should increase by 1. Therefore, we need to take $x_{1}=-1$.