Let $n$ be a positive integer not less than $4$. Show that there exists a polynomial with integral coefficients $$f(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+\cdots + a_1 x + a_0$$
such that for any positive integer $m$ and any $k \ge 2$ distinct integers $r_1$, $r_2$, $\cdots$, $r_k$, it always hold that $f(m)\ne f(r_1)f(r_2)\cdots f(r_k)$.
Let's construct a polynomial such that for any integer $u$, it always hold $f(u)\equiv 2\pmod{4}$. If so, because $k\ge 2$, we always have $f(r_1)f(r_2)\cdots f(r_k)\equiv 0\pmod{4}$ but $f(m)\equiv 2\pmod{4}$, therefore they can never equal.
There exist many candidates, the polynomial below is one of them $$f(x)=(x+1)(x+2)\cdots (x+n) + 2$$
It is clearly a $n$-degree polynomial with integer coefficients and the coefficient to $x^n$ is $1$. Meanwhile, because $n\ge 4$, $(x+1)(x+2)\cdot (x+n)$ is a product of $n\ge 4$ consecutive number thus must be a multiple of $4$. This means $f(x)\equiv 2\pmod{4}$.