2. Given a positive integer $n \geq 2$, let the polynomial $f(x)=x^{n}+b_{n-1} x^{n-1}+\ldots+b_{0}$, where $b_{i} \in \mathbb{R}, i=0,1,2, \ldots, n-1$. If there exists a non-constant infinite positive integer geometric sequence $\left\{a_{k}\right\}$ such that for any positive integer $k$, there exists a positive integer $x_{k}$ such that $a_{k}=f\left(x_{k}\right)$, find all possible $f(x)$.
【Answer】 $f(x)=(x+a)^{n}$, where $a$ is any integer.
Obviously, $2^{nc}$ can all be represented as the function value of $f$, and they form a geometric sequence with a common ratio of $2^{n}$. (3 points)
【Proof】The common ratio of the geometric sequence $\left\{a_{k}\right\}$ is clearly a positive rational number. First, we explain that it must be a positive integer; otherwise, let the common ratio $q=\frac{s}{t}$, $a_{k+1}=a_{1} \frac{s^{k}}{t^{k}}$ is a positive integer, thus $t^{k} \mid a_{1}$, which is impossible when $k$ is sufficiently large, so only $t=1$. (6 points) Since the geometric sequence is non-constant, hence $q>1$.

Obviously, there exists a sufficiently large $K$ such that $f(x)$ is strictly increasing on $(K, \infty)$.
For sufficiently large $a_{k}>K$, consider $a_{k}=f\left(x_{k}\right), a_{k+n}=f\left(x_{k+n}\right)$, and $a_{k+n}=a_{k} q^{n}$.
Thus, $f\left(x_{n+k}\right)=q^{n} f\left(x_{k}\right)$. (9 points)
At this point, its $n$-th term coefficient is $q^{n}$, and the $(n-1)$-th term coefficient is $b_{n-1} q^{n}$.
Now consider $f(q x+r)$, it is still a polynomial, with the $n$-th term coefficient being $q^{n}$, and the $(n-1)$-th term coefficient being $n q^{n-1} r+b_{n-1} q^{n-1}$. We take $r_{0}=\frac{q-1}{n} b_{n-1}$, then the $n$-th term coefficient and $(n-1)$-th term coefficient of $f\left(q x+r_{0}\right)$ are equal to those of $q^{n} f(x)$. (12 points) Therefore, we can know that $q^{n} f(x)-f\left(q x+r_{0}-1\right)$ and $f\left(q x+r_{0}+1\right)-q^{n} f(x)$ are both $(n-1)$-th degree polynomials with positive leading coefficients. So when $x$ is sufficiently large, there is $f\left(q x+r_{0}-1\right)1$, hence $\operatorname{Im}(\alpha)=0$. Therefore, all roots of $f(x)$ are real. (18 points)

Now let $f(x)=\left(x+x_{1}\right)\left(x+x_{2}\right) \cdots\left(x+x_{n}\right)\left(x_{1} \leq x_{2} \leq \cdots \leq x_{n}\right)$.
Then $q^{n} f(x)=\left(q x+q x_{1}\right)\left(q x+q x_{2}\right) \cdots\left(q x+q x_{n}\right), \quad\left(q x_{1} \leq q x_{2} \leq \cdots \leq q x_{n}\right)$.
$$
f\left(q x+r_{0}\right)=A\left(q x+r_{0}+x_{1}\right)\left(q x+r_{0}+x_{2}\right) \cdots\left(q x+r_{0}+x_{n}\right), \quad\left(r_{0}+x_{1} \leq r_{0}+x_{2} \leq \cdots \leq r_{0}+x_{n}\right)
$$

Since $q^{n} f(x)=f\left(q x+r_{0}\right)$, it follows that $q x_{1}=x_{1}+r_{0}, q x_{2}=x_{2}+r_{0}, \cdots q x_{n}=x_{n}+r_{0}$.
That is, $x_{k}=\frac{r_{0}}{q-1}(1 \leq k \leq n)$, thus $f(x)=(x+a)^{n}, a=\frac{r_{0}}{q-1}$ is a rational number.
Then by $a_{k}=\left(x_{k}+a\right)^{n}$, we know that $a$ is an integer, and the conclusion holds. (21 points)