(Integer) Variant of Hilbert’s irreducibility theorem Let $P\in{\mathbb Q}[X,Y]$ such that $P(x,.)$ has an integer root for any integer 
$x\in{\mathbb Z}$. Does it follow that $P$ has factors of the form
$Y-Q(X)$ for some $Q\in{\mathbb Q}[X]$, and does it follow also that if we denote
by $Q_1,Q_2, \ldots ,Q_r$ the polynomials that arise in this way in the factorisation of $P$, then for any $x\in{\mathbb Z}$ at least one of the $Q_i(x)$ is an integer.
 A: The strengthening of Hilbert’s irreducibility theorem  in  Cohen’s article (thanks to David Speyer for drawing attention to this article in his answer to this other MSE question) shows that the answer is YES to the first part of the question.
  The second part remains to be seen.
A: Rewrite the answer to include the first question (for the sake of completeness and for my own record). 
(I). Yes, $P(X,Y)$ has a factor of the form $Y-Q(X)$ with $Q(X)\in \mathbb Q[x]$. 
Decompose $P$ into 
$$P(X,Y)=(\prod_{1\le i\le r}(Y-Q_i(X))\prod_j P_j(X,Y)$$ with $\deg_Y(P_j)\ge 2$ (for the moment, $r$ could be zero). For $N$ big enough, by the strong version of Hilbert irreducibility theorem, the set of $k\in \mathbb Z$ such that $|k|\le N$ and one of the $P_j(k, Y)$ has a zero in $\mathbb Z$ (thus is reducible) has cardinality $\le c\sqrt{N}$ for some absolute constant $c$ 
(see the link given by Ewan in his answer, or this paper of M. Fried in a less general setting, but enough for our purpose). 
As $P(k, Y)$ has a root in $\mathbb Z$ for any $k=-N, -(N-1), ..., N$, taking $N>c^2/4$, we see that $r\ge 1$. This answers the first question.
(II). The answer is also yes. 
Write $Q_i(X)=F_i(X)/d$ with $F_i(X)\in \mathbb Z[X]$ and $d\in \mathbb N$. Let $k\in\mathbb Z$. Then 
$$Q_i(k)\in \mathbb Z\Longleftrightarrow \ d \mid F_i(k) \ \Longleftrightarrow F_i(k) \equiv 0 \mod d.$$ 
If $S_i\subset \mathbb Z$ is a set of representatives of the zeros of $F_i(X)$ in $\mathbb Z/d\mathbb Z$, then $Q_i(k)\in \mathbb Z$ for some $i\le r$ if and only if 
$$k\in \cup_{1\le i\le r} (S_i+d\mathbb Z)=(\cup_{1\le i\le r} S_i)+d\mathbb Z.$$ 
We saw above that the set of $k\in \mathbb Z$ such that $|k|\le N$ and one of the $P_j(k, Y)$ has a zero in $\mathbb Z$ has cardinality $\le c\sqrt{N}$. The cardinality of those $k$ such that one of the $Y-Q_i(k)$ has a zero in $\mathbb Z$ (equivalently $Q_i(k)\in\mathbb Z$) is about $2sN/d$, where $s$ is the cardinality of $\cup_i S_i$. As the union of these two types of $k$ has $2N+1$ elements, this forces $s=d$, hence $(\cup_i S_i)+d\mathbb Z=\mathbb Z$ and we are done.  
