2. Determine all polynomials $P_{n}(x)$ of the form

$$
P_{n}(x)=n!x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+(-1)^{n} n(n+1)
$$

with integer coefficients for which the roots $x_{i}, i=1,2, \ldots, n$ satisfy

$$
x_{i} \in[i, i+1], \text { for } i=1,2, \ldots, n
$$
Solution. For $n=1$, the only polynomial of the given form is $P_{1}(x)=x-2$ and it satisfies the conditions of the problem.

For $n=2$, we need to find an integer $a_{1}$ such that

$$
P_{2}(x)=2 x^{2}+a_{1} x+6
$$

satisfies the conditions of the problem. Then for the roots of this polynomial, we have

$$
1 \leq x_{1} \leq 2 \leq x_{2} \leq 3, \quad x_{1}+x_{2}=-\frac{a_{1}}{2}, \quad x_{1} x_{2}=3
$$

From these inequalities, we get $0 \leq x_{2}-x_{1} \leq 2$, so

$$
0 \leq\left(x_{2}-x_{1}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\frac{1}{4} a_{1}^{2}-12 \leq 4 \text {, i.e. } 48 \leq a_{1}^{2} \leq 64
$$

and also $a_{1}<0$. Since $a_{1}$ is an integer, from the last inequalities it follows that $a_{1}=-7$ or $a_{1}=-8$. It is easy to check that for these values of $a_{1}$, the conditions of the problem are satisfied.

We will prove that for $n \geq 3$ there is no polynomial for which the conditions of the problem are satisfied. Let

$$
\begin{aligned}
P_{n}(x) & =n!x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+(-1)^{n} n(n+1) \\
& =n!\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)
\end{aligned}
$$

Then

$$
(-1)^{n} n(n+1)=n!(-1)^{n} x_{1} x_{2} \ldots x_{n}
$$

so from $x_{i} \in[i, i+1]$, for $i=1,2, \ldots, n$ it follows

$$
n!\leq x_{1} x_{2} \ldots x_{n}=\frac{(n+1)!}{(n-1)!} \leq \frac{n+1}{2}<n<n!
$$

which is a contradiction.

Thus, the desired polynomials are $x-2,2 x^{2}-7 x+6$ and $2 x^{2}-8 x+6$.