$P \in \mathbb C[X]$ such that $P(\mathbb R\setminus\mathbb Q) \subset \mathbb R\setminus\mathbb Q$ Which polynomials $P(X)$ in $\mathbb C[X]$ satisfy
$5) \ P(\mathbb R\setminus\mathbb Q) \subset \mathbb R\setminus\mathbb Q$?
The followings have already been solved: 
$1) \ P(\mathbb C) \subset \mathbb R$?
$2) \ P(\mathbb U) \subset \mathbb U$ with $\mathbb U$ being the unit circle $\big\{z\in\mathbb{Z}\,\big|\,|z|=1\big\}$?
$2') \ P(\mathbb U) = \mathbb U$?
$3) \ P(\mathbb Z) \subset \mathbb Z$?
$4) \ P(\mathbb Q) \subset \mathbb Q$?
$4') \ P(\mathbb Q) = \mathbb Q$?
 A: (4) is the set of polynomials whose coefficients are all rational.
Note that we can get the coefficients of $P$ as rational linear combinations of the values of $P$ on the integers, using finite differences.
A: This is a solution to (4').  We claim that the only possible polynomials $P(X)\in \mathbb{C}[X]$ with the property that $P(\mathbb{Q})=\mathbb{Q}$ are of the form $P(X)=a\,X+b$ where $a$ and $b$ are rational numbers with $a\neq 0$.
To verify this, it suffices to prove the claim when $P(X)$ is monic with integral coefficients (why?).  Suppose on the contrary that there exists a monic polynomial $P(X)\in\mathbb{Z}[X]$ of degree at least $2$ with the required property.  Clearly, $P(X)$ must have an odd degree.  Thus, for integers $n$ with sufficiently large $|n|$, the equation $P\left(x_n\right)=n$ has a unique real solution $x_n$.
Suppose from now on that $n$ is an integer with sufficiently large $|n|$.  Now, the assumption that $P(\mathbb{Q})=\mathbb{Q}$ means that each solution $x_n$ must be rational.  As $x_n$ must also be an algebraic integer, we conclude that $x_n$ must be an integer.  However, it can easily be shown that $$P(k+1)-P(k)\to \pm\infty\text{ as }k\to \pm\infty$$ (this is where we need the assumption that the degree of $P(X)$ is at least $2$).

For (2) and (2'), consider a holomorphic function $f:\mathbb{C}\to\mathbb{C}$ such that $f(\mathbb{U})\subseteq \mathbb{U}$.  Write $\mathbb{D}$ for the open unit disc $\Big\{z\in\mathbb{C}\,\boldsymbol{\big|}\,|z|<1\Big\}$, so that $\mathbb{U}=\partial\mathbb{D}$.  I shall prove that $$f(z)=u\;\sigma_1(z)\,\sigma_2(z)\,\cdots\,\sigma_n(z)$$ for some integer $n\geq 0$, for some $u\in\mathbb{U}$, and for some (holomorphic) automorphisms $\sigma_1,\sigma_2,\ldots,\sigma_n\in\text{Aut}(\mathbb{D})$ of $\mathbb{D}$.
By the Maximum Modulus Principle, the maximum value of $|f|$ on $\bar{\mathbb{D}}$ is attained on $\mathbb{U}$, whence $f|_{\bar{\mathbb{D}}}$ maps $\bar{\mathbb{D}}$ to $\bar{\mathbb{D}}$ holomorphically.  If $f$ has no zeros inside $\mathbb{D}$, then again using the Maximum Modulus Principle, the minimum value of $|f|$, restricted to $\bar{\mathbb{D}}$, is attained on $\mathbb{U}$.  Consequently, $f$ maps the open disc $\mathbb{D}$ to $\mathbb{U}$.  By the Open Mapping Theorem (for holomorphic functions), either $f$ is constant or $f(\mathbb{D})$ is an open subset of $\mathbb{C}$.  Since any subset of $\mathbb{U}$ is not an open subset of $\mathbb{C}$, we conclude that $f$ is constant.  That is, $f(z)=u$ for some $u\in\mathbb{U}$.
From now on, we assume that $f$ has zeros inside $\mathbb{D}$.  Then, $f$ has finitely many zeros $w_1,w_2,\ldots,w_n$ in $\mathbb{D}$ (counting multiplicities).  There exists an automorphism $\tau_1$ of $\mathbb{D}$ that sends $0$ to $w_1$.  Let $f_1:=f\circ \tau_1$.  That is, $f_1(z)=z\,g_1(z)$ for some holomorphic function $g_1:\mathbb{C}\to\mathbb{C}$.  Now, suppose we have a function $g_k$ along with $\tau_1,\tau_2,\ldots,\tau_k\in\text{Aut}(\mathbb{D})$.  If $k<n$, then we can find an automorphism $\tau_{k+1}$ of $\mathbb{D}$ that sends $0$ to $\left(\tau_k^{-1}\circ\tau_{k-1}^{-1}\circ\cdots\circ\tau_1^{-1}\right)\left(w_{k+1}\right)$.  Set $f_{k+1}:=g_k\circ\tau_{k+1}$.  Then, $f_{k+1}(z)=z\,g_{k+1}(z)$ for some holomorphic function $g_{k+1}:\mathbb{C}\to\mathbb{C}$.  This process terminates when $k=n$.
From the paragraph above, we obtain a holomorphic function $g_n:\mathbb{C}\to\mathbb{C}$ without zeros in $\mathbb{D}$ such that $g_n$ maps $\bar{\mathbb{D}}$ to itself.  This means $g_n(z)=u$ for some constant $u\in\mathbb{U}$.  Write $\sigma_k:=\tau_k^{-1}\circ\tau_{k-1}^{-1}\circ\cdots\circ\tau_1^{-1}$ for $k=1,2,\ldots,n$.  Then,
$$f(z)=u\,\prod_{i=1}^n\,\sigma_i(z)\,,$$
as required.
Now, if $P(X)\in\mathbb{C}[X]$ is a polynomial satisfying $P(\mathbb{U})\subseteq\mathbb{U}$, then by the work above, we can easily see that $P(X)=u\,X^n$ for some integers $n\geq 0$ and $u\in\mathbb{U}$.  Furthermore, if $P(\mathbb{U})=\mathbb{U}$, then $n>0$ must hold.

For (5), we fist show that, if $P(X)$ is nonconstant, then $P(X)\in\mathbb{Q}[X]$.  It is obvious that such a nonconstant $P(X)$ must satisfy $P(X)\in\mathbb{R}[X]$.  Therefore, $P(\mathbb{R})$ contains an open interval $I \neq \emptyset$.  This interval $I$ has infinitely many rational numbers, and they must be mapped via $P$ from an infinite set of rational numbers.  This proves that $P(X)\in\mathbb{Q}[X]$.
As before, we may assume that $P(X)\in\mathbb{Z}[X]$ is monic.  If $P(X)$ has a degree $d>1$, then we can use a similar argument to (4') to yield a contradiction (since $P(\mathbb{Z})$ must contain all sufficiently large positive integers).  Thus, $P(X)$ is linear.  
In conclusion, there are two types of polynomials satisfying the condition (5).  The first class consists of constant polynomials $P(X)=t$, where $t\in \mathbb{R}\setminus\mathbb{Q}$.  The second class consists of polynomials $P(X)=a\,X+b$, where $a$ and $b$ are rational numbers with $a\neq 0$.
