6sN$, it follows that there exists an integer $r$ such that the equation $\lfloor P(x)-P(0) \rfloor \equiv r$ (mod $p^2$) has more than $6s$ solutions in the set $\{1,\ldots, sp^2\}$. Let $S$ be the set of $x\in T$ such that $\lfloor P(x)-P(0) \rfloor \equiv r$ (mod $p^2$). In particular $|S|>6s$.
For every $x\in \mathbb{Z}$, we have $\lfloor P(x) \rfloor = \lfloor P(x)-P(0) \rfloor+\lfloor P(0) \rfloor +u$ for some $u \in \{-1,0,1\}$, and so $ S \subseteq S_{t_{-1}} \cup S_{t_0} \cup S_{t_1}$, where $t_u=r+\lfloor P(0) \rfloor +u$ (computed modulo $p^2$). Therefore,
$$\dfrac{1}{sp^2}|S_{t_{-1}}|+\dfrac{1}{sp^2}|S_{t_0}|+\dfrac{1}{sp^2}|S_{t_1}| \geq \dfrac{1}{sp^2} \left |S_{t_{-1}} \cup S_{t_0} \cup S_{t_1} \right | \geq \dfrac{1}{sp^2} \left |S \right | >\dfrac{6}{p^2},$$
which contradicts the inequality \eqref{lims2} as $s \rightarrow \infty$.
\end{proof}
We now prove a statement that is stronger than the statement of Theorem \ref{nonlinear3}.
\begin{theorem}\label{nonlinear2}
Let $P(x)$ be a nonlinear polynomial with real coefficients. If the sequence $(\lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ mod all primes large enough, then $P(x)$ has at least one irrational coefficient other than the constant term.
\end{theorem}
\begin{proof}
Suppose on the contrary that $P(x)=\sum_{i=0}^n a_ix^i$ such that $a_i=r_i/s_i \in \mathbb{Q}$ with $\gcd(r_i,s_i)=1$ for all $1\leq i \leq n$. Let $N$ be the least common multiple of $s_i$, $1\leq i \leq n$. Since the sequence $(\lfloor P(k) \rfloor )_{k\geq 1}$ is assumed to be u.d.\ mod all primes large enough, the sequence $( N \lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ mod all primes $p$ large enough. The value $N \lfloor P(k) \rfloor$ is periodic modulo $p$ with period $Np$. Therefore, it follows from the uniform distribution of $( N \lfloor P(k) \rfloor )_{k\geq 1}$ modulo $p$ that, with ${\mathcal U}=\{0,\ldots, p-1\} \times \{0,\ldots, N-1\}$, we have
\begin{equation}\label{countN}
\#\{(t,j) \in {\mathcal U}: N \lfloor P(Nt+j) \rfloor \equiv \modd{i} {p} \}=N,
\end{equation}
for every integer $i$. Let $P_j(x)=N \lfloor P(Nx+j) \rfloor \in \mathbb{Z}[x]$ for $0\leq j **|a|$ and $tj+i-s$ has the same sign as $a$. Then, write $tj+i-s=aq+u$, where $q\geq 1$ and $u \in \{0,\ldots, |a|-1\}$. Since there exists an integer $0 \leq l N$ and an integer $m$ such that $Q(x)+A_j \not \equiv m$ (mod $p$) for all $x\in \mathbb{Z}$ and $j \in \{0,\ldots, N-1\}$. We claim that $P(x)$ is not complete modulo $p$. On the contrary, suppose there exists an integer $x$ such that $\lfloor P(x)\rfloor \equiv K$ (mod $p$), where $K$ is such that $NK \equiv m$ (mod $p$). But then, writing $x=Nk+j$ with $0\leq j **