Let $R$ be a commutative ring and let $f\in R[x_1,x_2,\cdots,x_{n-1}],n\geq 2$ be a polynomial.
Definition: We say $f$ is $n$-severable over $R$ if there exists a partition (of set) $$R=\coprod_{i=1}^n R_i , \qquad R_i\neq \varnothing ,$$ such that for any sequence $\mathbf{a}=(a_{i_1},\cdots,a_{i_{n-1}}), a_{i_j}\in R_{i_j}$, $f(\mathbf{a})$ lies in $R_{i_n}$. Here $(i_1,\cdots,i_n)$ is a rearrangement of $(1,\cdots,n)$.
Let me give a somewhat trivial example. Put $R=\Bbb{Z}$ and $n=2$, one can easily show that $f(x)=x+1$ is $2$-severable with partition $\Bbb{Z}=$ {odd numbers} $\cup$ {even numbers}. On the other hand, $f(x)=2x+1$ is not $2$-severable, since it possesses a fixed point $-1$. In general, one has following result, whose proof is straightforward.
Claim: For any $R$ and $f\in R[x]$, $f$ is $2$-severable if and only if there is no periodic element in $R$ of odd period under the iteration $f$.
To be honest, I haven't tried much beyond the above examples, and I feel that it is hopeless to obtain an explicit criterion for the severability of a general polynomial. So to narrow down the question, here is what I mainly interest in:
Question: Is there any $n$-severable polynomial $f$ over $\Bbb{Q}$ with $n\geq3$?
For the case $n=3$, I only calculated a few linear functions $f=ax_1+bx_2$ and didn't find any satisfied one yet. Also, for any $n\geq 2$, there is an $n$-severable polynomial $n(n-1)/2-(x_1+\cdots+x_{n-1})$ over $\Bbb{Z}/n\Bbb{Z}$, so it induces a collection of $n$-severable polynomials over $\Bbb{Z}$, but I've no idea whether any of them is severable over $\Bbb{Q}$.
I also posted it on MSE, but did not receive any answer yet. Any advise or guidance would be appreciated.
