3. Let $f(x)$ be a polynomial with integer coefficients, for which

$$
f(0)=23, f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=\ldots=f\left(x_{n}\right)=2014
$$

for some distinct $x_{1}, x_{2}, \ldots, x_{n}$. Find the maximum value of $n$.
Solution. Define $g(x)=f(x)-2014$. If $f\left(x_{i}\right)=2014$, then $x_{i}$ is a zero of $g(x)$, so $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) q(x)$ for some polynomial $q(x)$ with integer coefficients, where $n$ is the maximum number of $x_{i}$. Substituting $x=0$ into $g(x)$, we get

$$
g(0)=-1991=-11 \cdot 181=\prod_{i=1}^{n}\left(x_{i}\right) q(0)
$$

Since 11 and 181 are prime numbers, it follows that -1991 can be written as a product of at most 4 different factors, that is, since

$$
-1991=-1 \cdot 1 \cdot 11 \cdot 181
$$

we get that $n \leq 4$. It remains to give an example when $n=4$. Let

$$
g(x)=(x-1)(x+1)(x+11)(x+181), q(x)=1
$$

Then the polynomial

$$
f(x)=(x-1)(x+1)(x+11)(x+181)+2014
$$

satisfies the conditions of the problem.