Problem 1806

For $k=1$ and $k=2$, the required polynomials are $f(x) = x+1$ and $f(x) = (x+1)(x+2)$, respectively. Let $k \ge 3$ and assume that such a polynomial $f(x)$ exists. For any prime number $p$, its degree in $\operatorname{lcm}(n+1, n+2, \dots, n+k)$ is $\max\{\alpha_1, \alpha_2, \dots, \alpha_{k-1}\}$, where $\alpha_i$ is the power of $p$ in the canonical representation of $n+i$, $i=1, \dots, k$. If this is, for example, $\alpha_s$, then it would be obtained if we take
$$\frac{(n+1)(n+2)\cdots(n+k)}{p^{\alpha_1}p^{\alpha_2}\cdots p^{\alpha_{s-1}}p^{\alpha_{s+1}}\cdots p^{\alpha_k}},$$
it being clear that the powers of $p$ in the denominator are divisors of $\prod_{1 \le i \ne s \le k} (s - i)$. Therefore
$$\operatorname{lcm}(n+1, n+2, \dots, n+k) = \frac{(n+1)(n+2)\dots(n+k)}{C_n}, \quad (1)$$
where $C_n$ is a divisor of $\prod_{1 \le i < j \le k} (j - i)$. Since $C_n$ can take a finite number of possible values, there will be a natural number $C$ such that for infinitely many $n$, $f(n) = \frac{(n+1)(n+2)\dots(n+k)}{C}$. So for infinitely many $x$, $f(x) = \frac{(x+1)(x+2)\dots(x+k)}{C}$, whence
$$f(x) = \frac{(x+1)(x+2)\dots(x+k)}{C}, \quad \forall x \in \mathbb{R}.$$
Therefore
$$\operatorname{lcm}(n+1, n+2, \dots, n+k) = \frac{(n+1)(n+2)\dots(n+k)}{C}, \quad \text{for all } n \in \mathbb{N}.$$
Let's assume this is possible. Let's choose a prime $p < k$ such that $p$ does not divide $k$. Let $n + k + 1 = p^m$ for sufficiently large $m$. From the above formula we have
$$\frac{\operatorname{lcm}(n+2, n+3, \dots, n+k+1)}{\operatorname{lcm}(n+1, n+2, \dots, n+k)} = \frac{n+k+1}{n+1}. \quad (2)$$
The degree of $p$ in the numerator of the left side is $m$ and in the denominator – at least 1, while the degree of $p$ in the numerator of the right side is $m$ and in the denominator – 0. We derive a contradiction! Therefore, the assumption is wrong and for $k \ge 3$ there does not exist a polynomial with the desired property. $\square$