Any nonzero polynomial can be suitably modified to become everywhere nonzero on ${\mathbb Z}^n$? If we have a "graded" sequence of polynomials $g_1\in {\mathbb C}(x_1),g_2\in {\mathbb C}(x_1,x_2),g_3\in {\mathbb C}(x_1,x_2,x_3) \ldots, g_n\in {\mathbb C}(x_1,x_2,\ldots,x_n)$ such that $g_i$ is non-constant
in $x_i$ for each $i$, I call the map 
$f : {\mathbb C}^n \to {\mathbb C}^n$ defined by
$$
f(x_1,x_2,\ldots,x_n)=(g_1(x_1),g_2(x_1,x_2),\ldots,g_n(x_1,x_2,\ldots,x_n))
$$
a flag transformation. For any polynomial 
$P\in {\mathbb C}(x_1,x_2,\ldots,x_n)$, one can compose
$P$ with $f$ to obtain another polynomial which I denote by $P\circ f$ :
$$
P\circ f(x_1,x_2,\ldots,x_n)=
P(g_1(x_1),g_2(x_1,x_2),\ldots,g_n(x_1,x_2,\ldots,x_n)) 
$$
Given a nonzero polynomial $P$, can we always find a flag transformation
$f$ such that $P\circ f$ has no zeroes on ${\mathbb Z}^n$ ?
This is clearly true for $n=1$, because in that case the zeroes of $P$ are bounded in some interval $[-M,M]$ and we can take $g_1(x_1)=M(1+x_1^2)$.
 A: We'll prove that it is always possible.
Take the polynomial $P$ and write it as
$$  \sum_r x_n^r P_r(x_1, \dots, x_{n-1}) $$
for polynomials $P_0, \dots, P_k$ not all zero.  The linear dependences between the $P_r$ in $\mathbb{C}[x_1, \dots, x_{n-1}]$ form a linear subspace of $\mathbb{C}^{k+1}$, whose intersection with the set $\{(1, t, \dots, t^k): t \in \mathbb{R}\}$ is finite.  In particular, we can choose $a_n \in \mathbb{R}$ such that no solution $t$ lies in $\mathbb{Z}+a_n$.  Applying the flag transformation $x_n \mapsto x_n   + a_n$, we get that $P(x_1, \dots, x_{n-1}, m)\neq 0$ in $\mathbb{C}[x_1, \dots, x_{n-1}]$ for all $m\in \mathbb{Z}$.
Now for each fixed value of $m$ we can apply the same argument to get a finite number of values of $t$ for which $P(x_1, \dots, x_{n-2}, t, m)=0$ in $\mathbb{C}[x_1, \dots, x_{n-2}]$.  Taking the union over $m$, we have a countable collection of 'bad' $t$ values.  We can now apply a translation in $x_{n-1}$ to make these bad values non-integers.
Keep going iteratively to get a flag transformation (in fact, a translation) which pushes all zeros of $P$ off the integer lattice.
