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\begin{center}
\vskip 1cm{\LARGE\bf 
A Note on Polynomial Sequences \\
\vskip .1in
Modulo Integers
}
\vskip 1cm
\large
Mohammad Javaheri\\ 
Department of Mathematics\\
School of Science\\
Siena College \\ 
Loudonville, NY 12211  \\
USA \\
\href{mailto:mjavaheri@siena.edu}{\tt mjavaheri@siena.edu} \\
\end{center}

\vskip .2 in

\begin{abstract}
We study the uniform distribution of the polynomial sequence
$\lambda(P)=(\lfloor P(k) \rfloor )_{k\geq 1}$ modulo integers, where
$P(x)$ is a polynomial with real coefficients. In the nonlinear case, we
show that $\lambda(P)$ is uniformly distributed in $\mathbb{Z}$ if and
only if $P(x)$ has at least one irrational coefficient other than the
constant term. In the case of even degree, we prove a stronger result:
$\lambda(P)$ intersects every congruence class modulo every integer if
and only if $P(x)$ has at least one irrational coefficient other than
the constant term.
\end{abstract}



\section{Introduction}


A sequence $(r_k)_{k\geq 1}$ of real numbers is said to be 
\textit{u.d.\ mod $1$} if for all $0\leq a&lt;b &lt;1$ we have
$$\lim_{N \rightarrow \infty}\dfrac{1}{N} \#\{k \in \{1,\ldots, N\}: a\leq \{r_k\} \leq b\}=b-a,$$
where $\{r_k\}$ denotes the fractional part of $r_k$. 
An integer sequence $(a_k)_{k\geq 1}$ is said to be 
\textit{u.d.\ mod an integer $m\geq 2$} if, for every integer $i$, one has
\begin{equation}\label{defud}
\lim_{N \rightarrow \infty}\dfrac{1}{N} \# \left \{ k \in \{1,\ldots, N\}: a_k \equiv \modd{i} {m} \right \} =\dfrac{1}{m}.
\end{equation}
A sequence is called \textit{u.d.\ in $\mathbb{Z}$} if it is u.d.\ mod $m$ for all $m\geq 2$ (or equivalently for all $m$ large enough). 
Given a sequence $(r_k)_{k\geq 1}$ of real numbers, if $(r_k/m)_{k\geq 1}$ is u.d.\ mod 1 for every $m\geq 2$, then $(\lfloor r_k \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$ \cite[Chap.\ 5]{kb}. Therefore, one can derive the following results on u.d.\ sequences in $\mathbb{Z}$ using existing results on u.d.\ sequences mod 1. 

\begin{example} \label{weylpol}
If $P(x)=\sum_{i=0}^n a_ix^i$ is a real polynomial with at least one irrational coefficient other than $a_0$, then $( \lfloor P(k)\rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$; see \cite[Chap.\ 5]{kb}. This result follows from the generalization of Weyl's distribution theorem which was proved by Weyl himself via his differencing method. Weyl's result was a generalization of Hardy and Littlewood's result on monomials \cite{HL1}. We prove the converse of this statement for nonlinear polynomials in Theorems \ref{nonlinear3} and \ref{nonlinear2}.
\end{example}

\begin{example}\label{bxalpha}
If $f(x)=\beta x^\alpha$, where $\alpha \in (1,\infty) \backslash \mathbb{N}$ and $\beta \in (0,1]$, then $( \lfloor f(k)\rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$. This follows from Weyl's criterion together with van der Corput inequalities \cite[Chap.\ 1]{kb}. 
\end{example}

\begin{example} \label{linearpol}
If $P(x)=\pm x+c$, $c\in \mathbb{R}$, then $( \lfloor P(k)\rfloor )_{k\geq 1}$ is clearly u.d.\ in $\mathbb{Z}$. Moreover, if $P(x)-P(0) \in \mathbb{Z}[x]$ and $( \lfloor P(k)\rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$, then $P(x)=\pm x +c$ for some $c\in \mathbb{R}$ \cite{niven}. 
\end{example}

\begin{example}
If $f(x)=\beta \alpha^x$ and $\beta&gt;0$, then the sequence $( \lfloor f(k)\rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$ for almost all $\alpha&gt;1$. This follows from Koksma's theorem \cite{Leveque}. 
\end{example}

Niven \cite{niven} showed that, given a nonlinear polynomial $P(x) \in \mathbb{Z}[x]$, there exist infinitely many integers $m$ such that $(P(k)  )_{k\geq 1}$ is not u.d.\ mod $m$. In this paper, our first goal is to extend this result to polynomials with rational coefficients in the following theorem:
\begin{theorem}\label{all2}
Let $P(x)$ be a polynomial with real coefficients. The sequence $(\lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$ if and only if $P(x)$ has an irrational coefficient other than the constant term or $P(x)=x/l+P(0)$ for a nonzero integer $l$. 
\end{theorem}
In the linear case, Theorem \ref{all2} follows from Theorem \ref{linear2}, and in the nonlinear case, it follows from Theorem \ref{nonlinear3} or Theorem \ref{nonlinear2}.

By adding the least integer operation to the arithmetic operations involved in defining polynomials, we obtain {\it generalized polynomials}. For example, $f(x)=\lfloor \lfloor a_1x^2+a_2 \rfloor x \rfloor + \lfloor a_3x+a_4 \rfloor x^2$ is a generalized polynomial. Hal\r{a}nd \cite{udgp} studied uniform distribution of generalized polynomials and showed that, under some conditions relating to the independence of coefficients of $f(x)$ over the rationals, the sequence $(f(k))_{k\geq 1}$ is u.d.\ mod 1.   
The second goal of this article is to study the range of the simplest generalized polynomials modulo integers, namely the range of $\lfloor P(x) \rfloor$ modulo integers, where $P(x)$ is a real polynomial. 
\begin{definition}\label{defcomplete}
We say a polynomial $P(x) \in \mathbb{R}[x]$ is {\it complete} modulo $m$ if, for every integer $n$, the equation $\lfloor P(x) \rfloor \equiv n$ (mod $m$) has a solution $x\in \mathbb{Z}$.  We say $P(x)$ is complete in $\mathbb{Z}$ if it is complete modulo every integer $m$ (or equivalently modulo all $m$ large enough).
\end{definition}
 
It follows from Example \ref{weylpol} that, if $P(x)$ has at least one irrational coefficient other than the constant term, then $P(x)$ is complete in $\mathbb{Z}$. 
The converse is not true in degree 1 (compare Theorems \ref{linear2} and \ref{linear3}). However, we will show in the following theorem that, at least in the even degree case, the converse is true. 

\begin{theorem}\label{gppmain}
Let $P(x)$ be an even-degree polynomial with real coefficients. Then the following statements are equivalent:
\begin{itemize}
\item[i.] $P(x)$ is complete modulo all primes large enough.
\item[ii.] $P(x)$ has an irrational coefficients other than the constant term. 
\item[iii.] The sequence $(\lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$.
\item[iv.] $P(x)$ is complete modulo all integers.
\end{itemize} 
\end{theorem}

We prove Theorem \ref{gppmain} in Section \ref{compevendeg}. Finally, in Section \ref{compmonom}, we consider polynomials of the form $P(x)=ax^n+c$, where $n&gt;1$ and $a, c \in \mathbb{R}$. In Theorem \ref{mon}, we show that $P(x)$ is complete modulo all primes large enough if and only if $a\notin \mathbb{Q}$. 

\section{u.d.\ polynomial sequences }\label{udpolyseq}

In this section, we determine all polynomials $P(x) \in \mathbb{R}[x]$ for which the sequence $( \lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$. Niven \cite[Thm.\ 3.1]{niven} showed that the sequence $( \lfloor \alpha k \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$ if and only if $\alpha$ is irrational or $\alpha=1/l$ for some nonzero integer $l$. By Example \ref{weylpol}, the sequence $( \lfloor \alpha k+ \beta \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$ for every irrational number $\alpha$. We will prove in Theorem \ref{linear2} that if the sequence $(\lfloor \alpha k+ \beta \rfloor )_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$, then either $\alpha$ is irrational or $\alpha=1/l$ for some nonzero integer $l$. First, we need a lemma.

\begin{lemma}\label{abm}
Let $a,b \in \mathbb{Z}$ such that $\gcd(a,b)=1$ and $b&gt;0$. Let $\beta \in \mathbb{R}$. Then the sequence $(\lfloor ak/b+\beta \rfloor )_{k\geq 1}$ is u.d.\ mod $m$ if and only if $\gcd(a,m)=1$. 
\end{lemma}

\begin{proof}
First, suppose that the sequence $(\lfloor ak/b+\beta \rfloor )_{k \geq 1}$ is u.d.\ mod $m$. Suppose that $d=\gcd(a,m)&gt;1$, and we derive a contradiction. Since we have assumed that $( \lfloor ak/b+\beta \rfloor )_{k\geq 1}$ is u.d.\ mod $m$, it follows that the sequence $(\lfloor ak/b+\beta \rfloor )_{k\geq 1}$ is u.d.\ mod $d$ \cite[Thm.\ 5.1]{niven}. One notes that the sequence $( \lfloor ak/b+\beta \rfloor )_{k\geq 1}$ modulo $d$ is periodic with period $b$. Therefore, if the number of solutions of $\lfloor ak/b+\beta \rfloor \equiv 0$ (mod $d$) with $1\leq k \leq b$ is given by $t$, then the number of solutions of $\lfloor ak/b+\beta \rfloor \equiv 0$ (mod $d$) with $1\leq k \leq sb$ is given by $st$, and so
$$\lim_{s \rightarrow \infty}\dfrac{1}{sb} \# \left \{ k \in \{1,\ldots, sb\}: \lfloor ak/b+\beta \rfloor  \equiv \modd{0} {d} \right \} =\lim_{s \rightarrow \infty}\dfrac{st}{sb}=\dfrac{t}{b}.$$
On the other hand, this limit must equal $1/d$ by the definition of u.d.\ mod $d$ (see equation \eqref{defud}). It follows that $t/b=1/d$, and so $d \mid b$. Since $d \mid a$ and $\gcd(a,b)=1$, we have a contradiction. 

For the converse, suppose that $\gcd(a,m)=1$. One notes that the sequence $(\lfloor ak/b+\beta \rfloor )_{k\geq 1}$ is periodic modulo $m$ with period $bm$. For each $0\leq i \leq m-1$, let $T_i$ denote the subset of elements $k \in \{1,\ldots, bm\}$ such that $\lfloor ak/b+\beta \rfloor \equiv i$ (mod $m$). We show that $|T_i|=b$ for all $0\leq i \leq m-1$. Fix $0 \leq i \leq m-1$, and let $T_i=\{t_1,\ldots, t_r\}$. For each $1\leq j \leq r$, we have 
$$\lfloor a(t_j+b)/b+\beta \rfloor  \equiv a+
\lfloor at_j/b+\beta \rfloor \equiv \modd{a+i} {m}.$$
In other words, the map $t_j \mapsto t_j+b$ is a one-to-one map from $T_i$ to $T_{a+i}$, where $t_j+b$ is computed modulo $bm$ and $a+i$ is computed modulo $m$. It follows that $|T_{a+i}| \geq |T_i|$, and so $|T_{qa+i}| \geq |T_i|$ for all $q \geq 0$, where $qa+i$ is computed modulo $bm$. Since $\gcd(a,m)=1$, we conclude that $|T_{i'}| \geq |T_i|$ for all $i,i'=0,\ldots, m-1$, and so $|T_i|=b$ for all $i=0,\ldots, m-1$. Thus, for $N=Qbm+R$, $0\leq R&lt;bm$, the number of solutions of $\lfloor ak/b+\beta \rfloor \equiv i$ (mod $m$) is between $Qb$ and $(Q+1)b$, which is sufficient to verify the definition of u.d.\ mod $m$ in \eqref{defud}. 
\end{proof}

\begin{theorem}\label{linear2}
Let $\alpha, \beta \in \mathbb{R}$. Then the sequence $(\lfloor \alpha k + \beta \rfloor)_{k\geq 1}$ is u.d.\ in $\mathbb{Z}$ if and only if $\alpha$ is irrational or $\alpha=1/l$ for some nonzero integer $l$. 
\end{theorem}

\begin{proof}
If $\alpha$ is irrational, then the claim follows from Example \ref{weylpol} \cite[Thm.\ 3.2]{niven}. If $\alpha=1/l$ for some nonzero integer $l$, then the sequence $(\lfloor k/l+\beta \rfloor)_{k\geq 1}$ is u.d.\ mod $m$ for every $m$ by Lemma \ref{abm}. Thus, suppose that $\alpha=a/b$ for integers $a,b$ with $\gcd(a,b)=1$, $|a|&gt;1$, and $b&gt;0$. It follows from Lemma \ref{abm} that the sequence $( \lfloor ak/b+\beta \rfloor )_{k\geq 1}$ is not u.d.\ mod $|a|$, hence it is not u.d.\ in $\mathbb{Z}$. 
\end{proof}

Next, we discuss nonlinear polynomials. Niven \cite[Thm.\ 4.1]{niven} proved that if $P(x)$ is a nonlinear polynomial with integer coefficients, then there exist infinitely many integers $m$ such that $(P(k))_{k\geq 1}$ is not u.d.\ mod $m$. We prove generalizations of this statement in the next two theorems. 

\begin{theorem}\label{nonlinear3}
Let $P(x)$ be a nonlinear polynomial with real coefficients. If $P(x)$ has no irrational coefficients other than the constant term, then there exists infinitely many mutually coprime integers $m$ such that $(\lfloor P(k) \rfloor )_{k \geq 1}$ is not u.d.\ mod $m$. 
\end{theorem}

\begin{proof} Let $P(x)=\sum_{i=0}^n a_ix^i$ 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 $s_i$, $1\leq i \leq n$, and let $Q(x)=N(P(x)-P(0)) \in \mathbb{Z}[x]$. Choose an integer $a$ such that $Q^\prime(a)$ has an arbitrarily large prime factor $p&gt;6N$ (this can be done, since $Q'(x)$ is a nonconstant polynomial). We define $f(x)=Q(x)-Q(a)$. Then $f(a) \equiv 0$ (mod $p^2$) and $f^\prime(a) \equiv 0$ (mod $p$). It follows from Hensel's Lemma \cite[Thm.\ 3.4.1]{Gouvea} that $f(a+kp) \equiv f(a) \equiv 0$ (mod $p^2$) for all integer values of $k$. In particular, the equation $Q(x) \equiv Q(a)$ (mod $p^2$) has at least $p$ solutions for $x\in \{1,\ldots, p^2\}$. It follows that, given an integer $s\geq 1$, we have $|T|\geq sp$, where $T$ denotes the set of solutions $x \in \{1,\ldots, sp^2\}$ of $Q(x) \equiv Q(a)$ (mod $p^2$).

We show that the sequence $(\lfloor P(k) \rfloor )_{k\geq 1}$ is not u.d.\ mod $m=p^{2}$. On the contrary, suppose that $(\lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ mod $p^2$. It follows from the definition that for each $0 \leq t &lt; p^2$,
$$\lim_{s \rightarrow \infty}\dfrac{1}{sp^2} |S_t|=\dfrac{1}{p^2},$$
where $S_t$ is the set of $x\in  \{1,\ldots, sp^2\}$ such that $\lfloor P(x) \rfloor  \equiv t$ (mod $p^2$). In particular, for $s$ large enough, one has
\begin{equation}\label{lims2}
\dfrac{1}{sp^2}|S_t|\leq \dfrac{2}{p^2}
\end{equation}
for all $0\leq t &lt;p^2$. If $x\in T$, then $Q(x)=Q(a)+\alpha(x)\cdot p^2$ for some $\alpha(x)\in \mathbb{Z}$. It follows that 
$$\lfloor P(x)-P(0) \rfloor =\left \lfloor \dfrac{Q(x)}{N} \right \rfloor =\left \lfloor \dfrac{Q(a)+\alpha(x) \cdot p^2}{N} \right \rfloor.$$
We note that $\lfloor (Q(a)+(\beta+N) p^2)/N \rfloor \equiv \lfloor (Q(a)+\beta p^2)/N \rfloor$ (mod $p^2$) for every $\beta \in \mathbb{Z}$, hence there are at most $N$ congruence classes modulo $p^2$ among the values $\lfloor P(x)-P(0) \rfloor$. Since $|T| \geq sp&gt;6sN$, 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 $|S|&gt;6s$. 

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,
 $$\dfrac{1}{sp^2}|S_{t_{-1}}|+\dfrac{1}{sp^2}|S_{t_0}|+\dfrac{1}{sp^2}|S_{t_1}| \geq \dfrac{1}{sp^2} \left |S_{t_{-1}} \cup S_{t_0} \cup S_{t_1} \right | \geq \dfrac{1}{sp^2} \left |S \right | &gt;\dfrac{6}{p^2},$$
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 $0\leq j &lt;N$. Choose $M$ large enough so that the polynomials $f_j(x)=P_j(x)+M$, $0\leq j&lt;N$, are all irreducible over $\mathbb{Q}[x]$ (the existence of $M$ follows from Hilbert's irreducibility theorem \cite[Chap.\ 9]{HIT}). Let $f(x)=f_0(x) \cdots f_{N-1}(x)$ and
  $$R=\prod_{0 \leq i&lt;j &lt;N} \Res(f_i,f_j) \in \mathbb{Z},$$
 where $\Res(f_i,f_j)$ is the resultant of polynomials $f_i$ and $f_j$. Since $f_i$ and $f_j$ as irreducible polynomials in $\mathbb{Q}[x]$ have no common zeros for all $0 \leq i&lt;j&lt;N$, we must have $\Res(f_i,f_j) \neq 0$, and so $R\neq 0$. Therefore, for any prime $p&gt;p_0$, $\Res(f_i, f_j) \not \equiv 0$ (mod $p$), where $p_0$ is the greatest prime factor of $R$. In other words, for any prime $p&gt;p_0$, the polynomials $f_j$, $0\leq j&lt;N$, have no common zeros modulo $p$. By the Chebotarev density theorem \cite{FriedJarden,LenstraStevenhagen}, there exist infinitely many primes $p$ such that $f(x)$ splits completely into $nN$ linear factors modulo $p$. It follows that there exists arbitrarily large $p&gt;p_0$ such that $f(x)$ has $nN$ distinct zeros modulo $p$. Therefore, the number of solutions of $N \lfloor P(Nt+j) \rfloor \equiv -M$ (mod $p$) is at least $nN&gt;N$. This is in contradiction with equation \eqref{countN}, and the claim follows.
\end{proof}

\section{Complete even-degree polynomials}\label{compevendeg}
Let $P(x)$ be a polynomial such that $P(x)-P(0) \in \mathbb{Q}[x]$. Since the sequence $(\lfloor P(k) \rfloor)_{k\geq 1}$ modulo any integer $m$ is periodic,
it follows from Definition \ref{defcomplete} that the polynomial $P(x)$ is complete modulo $m$ if and only if 
\begin{equation}\label{defgpp}
\lim_{N \rightarrow \infty}\dfrac{1}{N} \# \left \{ k \in \{1,\ldots, N\}: \lfloor P(k) \rfloor \equiv \modd{i} {m} \right \} &gt;0,
\end{equation}
for every integer $i$. Condition \eqref{defgpp} is weaker than condition \eqref{defud}. Therefore, if $(\lfloor P(k) \rfloor )_{k\geq 1}$ is u.d.\ mod $m$, then $P(x)$ is complete modulo $m$. The converse is not true for linear polynomials as shown by the following theorem in comparison with Theorem \ref{linear2}. 
\begin{theorem}\label{linear3}
The linear polynomial $P(x)=\alpha x+ \beta$ is complete in $\mathbb{Z}$ if and only if either $| \alpha | \in (0,1]$ or $\alpha$ is irrational. 
\end{theorem}

\begin{proof}
If $\alpha$ is irrational, then the claim follows from Theorem \ref{linear2}. Thus, suppose $\alpha=a/b$ where $a,b$ are coprime integers, $a\neq 0$, and $b&gt;0$. Suppose that $P(x)$ is complete in $\mathbb{Z}$, and so the set $\{\lfloor \alpha k + \beta \rfloor: k\geq 1\}$ contains the numbers $0,\ldots, |a|-1$ modulo $|a|$. Let $k=bq+l$ where $0\leq l &lt;b$. Then
$$\left \lfloor \dfrac{a}{b}(bq+l) + \beta \right \rfloor=aq+\left \lfloor \dfrac{al}{b}+\beta \right \rfloor.$$
Therefore, the $b$ numbers $\lfloor al/b +\beta \rfloor,~0\leq l&lt;b$, must contain the numbers $0, \ldots, |a|-1$  modulo $|a|$. In particular $b \geq |a|$ and so $\alpha \in [-1,0) \cup (0,1]$. 

For the converse, suppose $b\geq |a|$. Then the numbers $al/b +\beta,~0\leq l \leq b$ are apart by $|a/b| \leq 1$, and they stretch from $\beta $ to $a+\beta$. Therefore, the numbers $\lfloor al/b+\beta \rfloor$, $0\leq l &lt;b$, include $|a|$ consecutive integers, say $s,\ldots, s+|a|-1$. Given any $i,j \in \mathbb{Z}$, we show that there exists an integer $x$ such that $\lfloor ax/b+\beta \rfloor \equiv i$ (mod $j$). We choose $t \in \mathbb{Z}$ such that $|tj+i-s|&gt;|a|$ 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 $0 \leq l &lt;b$ such that $\lfloor al/b+\beta \rfloor =s+u$, with $x=bq+l$, we have $\lfloor ax/b+\beta \rfloor =aq + \lfloor al/b+\beta \rfloor  \equiv aq+s+u \equiv i$ (mod $j$). It follows that $P(x)$ is complete in $\mathbb{Z}$, and the proof is completed. 
\end{proof}

To prove Theorem \ref{gppmain}, we need the following two lemmas.

\begin{lemma}\label{primeRp}
Let $R(x)$ be a polynomial with integer coefficients and no real zeros. Then there exist infinitely many primes $p$ such that $R(x)$ has no zeros modulo $p$.
\end{lemma}

\begin{proof}
Suppose on the contrary that $R(x)$ has a zero modulo all primes large enough. It follows from the Chebotarev density theorem \cite{FriedJarden,LenstraStevenhagen} that every element of the Galois group of the splitting field of $R(x)$ has a fixed point in the action on the zeros. In particular, complex conjugation must have a fixed point on the set of the zeros of $R(x)$, which contradicts our assumption that $R(x)$ has no real zeros.
\end{proof}

\begin{lemma}\label{Qa}
Let $Q(x)$ be a polynomial of even degree with integer coefficients, and let $A_0,\ldots, A_{N-1} \in \mathbb{Z}$. Then, there exist an arbitrarily large prime $p$ and an integer $m$ such that $Q(x)+A_i \not \equiv m$ (mod $p$) for all $x\in \mathbb{Z}$ and $i \in \{0,\ldots, N-1\}$. 
\end{lemma}

\begin{proof}
Choose $M \in \mathbb{Z}$ so that $Q(x)+M+A_i$ has no real zeros for all $0\leq i&lt;N$. We let 
$$R(x)=(Q(x)+M+A_0)\cdots (Q(x)+M+A_{N-1}).$$
Then $R(x)$ has no real zeros. By Lemma \ref{primeRp}, there exists an arbitrarily large prime $p$ such that $R(x) \not \equiv 0$ (mod $p$) for all $x \in \mathbb{Z}$. It follows that $Q(x)+A_i \neq -M$ for all $x\in \mathbb{Z}$ and $i\in \{0,\ldots, N-1\}$. 
\end{proof}

We are now ready to prove Theorem \ref{gppmain}. 
\begin{proof} 
Let $P(x)=\sum_{i=0}^n a_ix^i$ 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 $s_i$, $1\leq i \leq n$. One has
$$\lfloor P(Nk+j) \rfloor =\lfloor P(j) \rfloor +\sum_{i=1}^n \dfrac{r_i}{s_i}((Nk+j)^i-j^i).$$
And so
\begin{align}\nonumber
N\lfloor P(Nk+j) \rfloor &amp;=N\lfloor P(j) \rfloor +N\sum_{i=1}^n \dfrac{r_i}{s_i}((Nk+j)^i-j^i). \\ \nonumber
&amp;= N\lfloor P(j) \rfloor -\sum_{i=1}^n r_i \dfrac{N}{s_i}j^i +Q(Nk+j) \\ \label{np}
&amp;= A_j +Q(Nk+j),
\end{align}
 where $Q(x)=N(P(x)-P(0)) \in \mathbb{Z}[x]$ and $A_j \in \mathbb{Z}$ depending on $j$ and $P(x)$, $0\leq j &lt;N$. By Lemma \ref{Qa}, there exist an arbitrarily large prime $p&gt;N$ 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 $0\leq j &lt;N$, we have $Q(Nk+j)+A_j \equiv N\lfloor P(Nk+j) \rfloor \equiv NK \equiv m$ (mod $p$). This is a contradiction, and so (i) implies (ii). The implication (ii) $\Rightarrow$ (iii) was discussed in Example \ref{weylpol}. The implications (iii) $\Rightarrow$ (iv) and (iv) $\Rightarrow$ (i) are straightforward, and the proof of Theorem \ref{gppmain} is completed. \end{proof}

\section{Complete monomials}\label{compmonom}

Let $p$ be a prime and $n$ be a positive integer that divides $p-1$. An $n$th power \emph{character} modulo $p$ is any homomorphism $\chi: \mathbb{Z}_p^* \rightarrow \mathbb{C}$ that is onto the group of $n$th roots of unity. By a theorem of A. Brauer \cite{RJ}, given $n,l \geq 1$, there exists a constant $z(n,l)$ such that for every prime $p&gt;z(n,l)$ and any $n$th power character $\chi$ modulo $p$, there exists an integer $t$ such that 
\begin{equation}\label{xi}
\chi(t)=\chi(t+1)=\cdots =\chi(t+l-1).
\end{equation}

A number $x$ is an $n$th power residue modulo $p$, if there exists $y$ such that $x \equiv y^n$ (mod $p$). If $\chi$ is an $n$th power character modulo $p$ and $x$ is an $n$th power residue modulo $p$, then $\chi(x)=\chi(y^n)=(\chi(y))^n=1$. Therefore, to show that a number $z$ is not an $n$th power residue modulo $p$, it is sufficient to find an $n$th power character modulo $p$ such that $\chi(z) \neq 1$. We use this fact in the proof of the following lemma. 

\begin{lemma}\label{last}
For any positive integer $l$, there exist infinitely many primes $p$ such that all of the numbers $t,t+1,\ldots, t+l-1$ are $n$th power non-residues modulo $p$ for some positive integer $t$. 
\end{lemma}

\begin{proof}
We can assume, without loss of generality, that $n$ is prime and $l\geq 4$. By a result of Mills \cite[Thm.\ 3]{Mills}, for every $m\geq 1$, there exist infinitely many primes $p$ with an $n$th power character $\chi$ modulo $p$ such that 
$$\chi(2)\neq 1, ~\forall 2 \leq i\leq m:~\chi(p_i)=1,$$
where $p_i$ is the $i$th prime. Let $t$ be defined by \eqref{xi}. We can choose $t&gt;1$ by adding multiples of $p$ if necessary. Choose an integer $m$ large enough so that $p_m&gt;t+l-1$. Choose $i \in \{0,\ldots, l-1\}$ such that $t+i-1=2(2d+1)$ for some integer $d$. Then
$$\chi(2(2d+1))=\chi(2)\chi(2d+1) \neq 1.$$
It then follows from equation \eqref{xi} that $\chi(t)=\chi(t+1)=\cdots=\chi(t+l-1) \neq 1$ i.e., none of the values $t,t+1,\ldots, t+l-1$, are $n$th power residues modulo $p$. 
\end{proof}

\begin{theorem}\label{mon}
Let $P(x)=ax^n+c$, where $a\in \mathbb{Q}$ and $c\in \mathbb{R}$. If $n&gt;1$, then $P(x)$ is not complete modulo all primes large enough, hence $(\lfloor P(k) \rfloor )_{k\geq 1}$ is not u.d.\ in $\mathbb{Z}$.  
\end{theorem}

\begin{proof}
Let $a=M/N$, where $M$ and $N$ are integers and $M&gt;0$. Let $Q(x)=Mx^n$ and $A_0,\ldots, A_{N-1}$ be given by equation \eqref{np}. On the contrary, suppose $P(x)$ is complete modulo all primes $p$ large enough. By Lemma \ref{last}, for $l=1+\max_i M^{n-1}A_i - \min_i M^{n-1}A_i$, there exists an arbitrarily large prime $p&gt;|MN|$ and an integer $t$ such that $t+j$ is not an $n$th power residue modulo $p$ for any $0\leq j &lt;l$. 

Let $K=t+\max_i M^{n-1}A_i$, and choose $L$ such that $M^{n-1}L \equiv K$ (mod $p$). Since $P(x)$ is complete modulo $p$, there exists an integer $x$ such that $N \lfloor P(x) \rfloor \equiv L$ (mod $p$). Writing $x=Nk+j$ with $0\leq j&lt;N$, we have
$$M^{n-1}(Q(x)+A_j) \equiv M^{n-1}N \lfloor P(x) \rfloor \equiv M^{n-1}L \equiv \modd{K} {p}.$$
Since $t \leq K-M^{n-1}A_j&lt;t+l$ and $K-M^{n-1}A_{j} \equiv M^{n-1}Q(x) \equiv (Mx)^n$ (mod $p$) is an $n$th power residue modulo $p$, we have a contradiction, and the claim follows.
\end{proof}

\begin{remark}
In light of the proofs of Theorems \ref{gppmain} and \ref{mon}, one can generalize Theorem \ref{gppmain} to all nonlinear polynomials if the following statement is true: Given a nonlinear polynomial $P(x)$ with integer coefficients and a positive integer $l$, there exist an arbitrarily large prime $p$ and a positive integer $k$ such that $P(x) \not \equiv k+i$ (mod $p$) for all $i=0,\ldots, l-1$. 
\end{remark}

\section{Acknowledgment}

I would like to thank Siena College for providing a sabbatical leave during which this work was completed. I would also like to thank the referee for many useful comments and suggestions.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11K36.

\noindent \emph{Keywords: } 
uniform distribution, polynomial sequence.

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\vspace*{+.1in}
\noindent
Received September 11 2018;
revised version received  December 16 2018.
Published in {\it Journal of Integer Sequences}, December 17 2018.

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