[proofplan] We prove equidistribution modulo $1$ by the Weyl criterion, reducing the theorem to the vanishing of exponential averages at every non-zero integer frequency. The core case is the case in which the leading coefficient is irrational; this is proved by induction on the degree using a finite-window van der Corput estimate, because each fixed difference $r(n+h)-r(n)$ has one lower degree and irrational leading coefficient. For a general polynomial, we take the highest irrational non-constant coefficient and split the average into residue classes modulo a common denominator of the rational higher-degree coefficients. On each residue class the rational higher-degree part contributes only a constant phase, and the remaining polynomial has irrational leading coefficient. [/proofplan]

[step:Convert equidistribution to exponential sum decay]Regard the polynomial in the statement as the map \begin{align*} p:\mathbb{R} &\to \mathbb{R} \\ t &\mapsto \alpha_d t^d+\alpha_{d-1}t^{d-1}+\cdots+\alpha_1t+\alpha_0 . \end{align*} Define \begin{align*} \mathrm e:\mathbb{R} &\to \mathbb{C} \\ t &\mapsto \exp(2\pi i t). \end{align*} For a real polynomial $r:\mathbb{R}\to\mathbb{R}$, an integer $k\in\mathbb{Z}\setminus\{0\}$, and a positive integer $N\in\mathbb{N}$, define \begin{align*} A_N(k,r):=\frac{1}{N}\sum_{n=0}^{N-1}\mathrm e(k r(n)). \end{align*} By the [Weyl Criterion](/page/Weyl%20Criterion), the sequence $(p(n))_{n=0}^{\infty}$ is [equidistributed modulo $1$](/page/Equidistribution%20Modulo%201) if and only if \begin{align*} \lim_{N\to\infty} A_N(k,p)=0 \end{align*} for every $k\in\mathbb{Z}\setminus\{0\}$. Thus it remains to prove this limit for each non-zero integer $k$.[/step]

[guided]Why reduce equidistribution to exponential sums in the first place? Equidistribution modulo $1$ is a statement about the fractional parts $p(n)-\lfloor p(n)\rfloor$ falling into intervals with the correct density. Manipulating fractional parts and indicator functions $\mathbb{1}_{[a,b)}$ directly is awkward — they are discontinuous, and counting them is not algebraically friendly. The standard remedy is to test with smooth periodic functions instead. The maps $t\mapsto\exp(2\pi i k t)$ for $k\in\mathbb{Z}$ are exactly the continuous group homomorphisms from $\mathbb{R}/\mathbb{Z}$ to the unit circle, and they separate points of $\mathbb{R}/\mathbb{Z}$. Weyl's criterion turns this observation into an iff statement. Formally, regard the polynomial in the statement as the map \begin{align*} p:\mathbb{R} &\to \mathbb{R} \\ t &\mapsto \alpha_d t^d+\alpha_{d-1}t^{d-1}+\cdots+\alpha_1t+\alpha_0 , \end{align*} so that $p|_{\mathbb{N}_0}$ is the sequence we wish to equidistribute. Define the basic character \begin{align*} \mathrm e:\mathbb{R} &\to \mathbb{C} \\ t &\mapsto \exp(2\pi i t), \end{align*} which is $1$-periodic, so $\mathrm e(t)=\mathrm e(t-\lfloor t\rfloor)$ depends only on the fractional part of $t$. For a real polynomial $r:\mathbb{R}\to\mathbb{R}$, an integer $k\in\mathbb{Z}\setminus\{0\}$, and a positive integer $N\in\mathbb{N}$, define the normalised exponential average \begin{align*} A_N(k,r):=\frac{1}{N}\sum_{n=0}^{N-1}\mathrm e(k r(n)). \end{align*} This is the empirical average of the character $\mathrm e(k\,\cdot\,)$ along the sequence $(r(n))_{n=0}^{N-1}$. We apply the [Weyl Criterion](/page/Weyl%20Criterion), which states: a real sequence $(x_n)_{n=0}^{\infty}$ is [equidistributed modulo $1$](/page/Equidistribution%20Modulo%201) if and only if $\lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}\mathrm e(kx_n)=0$ for every $k\in\mathbb{Z}\setminus\{0\}$. The hypothesis of the criterion is that $(x_n)$ is a real sequence, which is satisfied here because $p:\mathbb{N}_0\to\mathbb{R}$. Setting $x_n:=p(n)$ converts the conclusion of our theorem into \begin{align*} \lim_{N\to\infty}A_N(k,p)=0\quad\text{for every }k\in\mathbb{Z}\setminus\{0\}. \end{align*} This is the exponential-sum statement that the remainder of the proof will establish. The value $k=0$ is excluded because $A_N(0,p)=1$ for every $N$ — the constant character carries no information.[/guided]

[step:Derive a finite window van der Corput estimate for bounded sequences][claim:Finite window van der Corput estimate] Let $N,H\in\mathbb{N}$ satisfy $1\le H\le N$. Let $z:\{0,1,\ldots,N-1\}\to\mathbb{C}$ be a sequence satisfying $|z_n|\le 1$ for every $n\in\{0,1,\ldots,N-1\}$. Define \begin{align*} B_N&:=\frac{1}{N}\sum_{n=0}^{N-1}z_n,\\ C_{N,h}&:=\frac{1}{N}\sum_{n=0}^{N-1-h}z_{n+h}\overline{z_n} \end{align*} for $h\in\{1,\ldots,H-1\}$, where $\overline{w}$ denotes the complex conjugate of $w\in\mathbb{C}$. Then \begin{align*} |B_N|^2 \le \left(1+\frac{H-1}{N}\right) \left[ \frac{1}{H} + \frac{2}{H}\sum_{h=1}^{H-1}\left(1-\frac{h}{H}\right)\operatorname{Re} C_{N,h} \right], \end{align*} where $\operatorname{Re} w$ denotes the real part of $w\in\mathbb{C}$. [/claim] [proof] Define the extended sequence \begin{align*} \widetilde z:\mathbb{Z}&\to\mathbb{C}\\ m&\mapsto \begin{cases} z_m,&0\le m\le N-1,\\ 0,&\text{otherwise}. \end{cases} \end{align*} For $n\in\{-H+1,-H+2,\ldots,N-1\}$, define \begin{align*} W_n:=\sum_{j=0}^{H-1}\widetilde z_{n+j}. \end{align*} Each $z_m$ with $0\le m\le N-1$ occurs exactly $H$ times in the double sum $\sum_{n=-H+1}^{N-1}\sum_{j=0}^{H-1}\widetilde z_{n+j}$, namely for $j\in\{0,\ldots,H-1\}$ and $n=m-j$. Therefore \begin{align*} \sum_{n=-H+1}^{N-1}W_n = H\sum_{m=0}^{N-1}z_m. \end{align*} Applying the [Cauchy-Schwarz Inequality](/page/Cauchy-Schwarz%20Inequality) in the finite-dimensional complex inner product space $\mathbb{C}^{N+H-1}$ to the vectors $(1)_{n=-H+1}^{N-1}$ and $(W_n)_{n=-H+1}^{N-1}$ gives \begin{align*} H^2\left|\sum_{m=0}^{N-1}z_m\right|^2 \le (N+H-1)\sum_{n=-H+1}^{N-1}|W_n|^2. \end{align*} Expanding the squared modulus and grouping pairs of indices by their difference gives \begin{align*} \sum_{n=-H+1}^{N-1}|W_n|^2 &= \sum_{j=0}^{H-1}\sum_{\ell=0}^{H-1} \sum_{n=-H+1}^{N-1} \widetilde z_{n+j}\overline{\widetilde z_{n+\ell}}\\ &= H\sum_{m=0}^{N-1}|z_m|^2 + 2\operatorname{Re}\sum_{h=1}^{H-1}(H-h) \sum_{m=0}^{N-1-h}z_{m+h}\overline{z_m}. \end{align*} The diagonal contribution is bounded by $HN$ because $|z_m|\le 1$ for every $m$. Hence \begin{align*} \sum_{n=-H+1}^{N-1}|W_n|^2 \le HN + 2\operatorname{Re}\sum_{h=1}^{H-1}(H-h) \sum_{m=0}^{N-1-h}z_{m+h}\overline{z_m}. \end{align*} Dividing by $H^2N^2$ and using $B_N=N^{-1}\sum_{m=0}^{N-1}z_m$ and $C_{N,h}=N^{-1}\sum_{m=0}^{N-1-h}z_{m+h}\overline{z_m}$ gives \begin{align*} |B_N|^2 &\le \frac{N+H-1}{H^2N^2} \left[ HN + 2\operatorname{Re}\sum_{h=1}^{H-1}(H-h)NC_{N,h} \right]\\ &= \left(1+\frac{H-1}{N}\right) \left[ \frac{1}{H} + \frac{2}{H}\sum_{h=1}^{H-1}\left(1-\frac{h}{H}\right)\operatorname{Re} C_{N,h} \right]. \end{align*} This proves the claim. [/proof][/step]

[guided]The goal of this step is to bound a sum of size $N$ by an expression that involves only finitely many short-range correlations $C_{N,h}$ for $h<H$, together with an explicit error of size $1/H$. This is the engine that powers the induction in the next step: it converts the question of decay for an unknown sequence into a question about its $H-1$ shifted self-correlations. [claim:Finite window van der Corput estimate] Let $N,H\in\mathbb{N}$ satisfy $1\le H\le N$. Let $z:\{0,1,\ldots,N-1\}\to\mathbb{C}$ be a sequence satisfying $|z_n|\le 1$ for every $n\in\{0,1,\ldots,N-1\}$. Define \begin{align*} B_N&:=\frac{1}{N}\sum_{n=0}^{N-1}z_n,\\ C_{N,h}&:=\frac{1}{N}\sum_{n=0}^{N-1-h}z_{n+h}\overline{z_n} \end{align*} for $h\in\{1,\ldots,H-1\}$. Then \begin{align*} |B_N|^2 \le \left(1+\frac{H-1}{N}\right) \left[ \frac{1}{H} + \frac{2}{H}\sum_{h=1}^{H-1}\left(1-\frac{h}{H}\right)\operatorname{Re} C_{N,h} \right]. \end{align*} [/claim] [proof] Why do we shift and pad? The total $\sum_{m=0}^{N-1}z_m$ is a single sum of $N$ terms; if we instead write it as an average of $H$ overlapping length-$H$ window sums $W_n=\sum_{j=0}^{H-1}\widetilde z_{n+j}$, then $|W_n|^2$ naturally produces correlations $\widetilde z_{n+j}\overline{\widetilde z_{n+\ell}}$, and we can sort these by their lag $h=j-\ell$. Extend the sequence to all integers by setting it to zero outside $\{0,\ldots,N-1\}$: \begin{align*} \widetilde z:\mathbb{Z}&\to\mathbb{C}\\ m&\mapsto \begin{cases} z_m,&0\le m\le N-1,\\ 0,&\text{otherwise}. \end{cases} \end{align*} For $n\in\{-H+1,-H+2,\ldots,N-1\}$, define the window sum \begin{align*} W_n:=\sum_{j=0}^{H-1}\widetilde z_{n+j}. \end{align*} Each $z_m$ with $0\le m\le N-1$ appears in $W_n$ exactly when $n+j=m$ for some $j\in\{0,\ldots,H-1\}$, i.e.\ exactly $H$ times across $n\in\{-H+1,\ldots,N-1\}$. Hence \begin{align*} \sum_{n=-H+1}^{N-1}W_n=H\sum_{m=0}^{N-1}z_m. \end{align*} We now apply the [Cauchy-Schwarz Inequality](/page/Cauchy-Schwarz%20Inequality) in $\mathbb{C}^{N+H-1}$ — the index range has $N+H-1$ elements — to the vectors $(1)_{n=-H+1}^{N-1}$ and $(W_n)_{n=-H+1}^{N-1}$. Both are finite-dimensional, so the hypothesis of the inequality (vectors in an inner product space) is automatic: \begin{align*} H^2\left|\sum_{m=0}^{N-1}z_m\right|^2 \le (N+H-1)\sum_{n=-H+1}^{N-1}|W_n|^2. \end{align*} Why does this help? The left-hand side is what we want to bound. The right-hand side now contains only $|W_n|^2$, which we expand and group by lag. Setting $h:=j-\ell$ for $j,\ell\in\{0,\ldots,H-1\}$: \begin{align*} \sum_{n=-H+1}^{N-1}|W_n|^2 &= \sum_{j=0}^{H-1}\sum_{\ell=0}^{H-1} \sum_{n=-H+1}^{N-1} \widetilde z_{n+j}\overline{\widetilde z_{n+\ell}}\\ &= H\sum_{m=0}^{N-1}|z_m|^2 + 2\operatorname{Re}\sum_{h=1}^{H-1}(H-h) \sum_{m=0}^{N-1-h}z_{m+h}\overline{z_m}. \end{align*} Here the $j=\ell$ diagonal contributes $H$ copies of $\sum_m|z_m|^2$, and each off-diagonal lag $h\in\{1,\ldots,H-1\}$ is achieved by exactly $H-h$ pairs $(j,\ell)$ with $j-\ell=h$, paired with its conjugate. The hypothesis $|z_m|\le 1$ gives $\sum_m|z_m|^2\le N$, so the diagonal is at most $HN$. Substituting, dividing by $H^2N^2$, and using the definitions of $B_N$ and $C_{N,h}$: \begin{align*} |B_N|^2 &\le \frac{N+H-1}{H^2N^2} \left[ HN+2\operatorname{Re}\sum_{h=1}^{H-1}(H-h)NC_{N,h} \right]\\ &= \left(1+\frac{H-1}{N}\right) \left[ \frac{1}{H}+\frac{2}{H}\sum_{h=1}^{H-1}\left(1-\frac{h}{H}\right)\operatorname{Re} C_{N,h} \right]. \end{align*} The $1/H$ term is the unavoidable contribution we pay for shrinking the window. The remaining sum is the genuine correlation content; if the correlations decay, the right-hand side goes to $1/H$, which is arbitrarily small. [/proof][/guided]

[step:Prove exponential sum decay when the leading coefficient is irrational]For $D\in\mathbb{N}$, let $\mathsf{L}(D)$ denote the following assertion: if \begin{align*} r:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto \beta_Dt^D+\beta_{D-1}t^{D-1}+\cdots+\beta_1t+\beta_0 \end{align*} is a real polynomial of degree $D$ with $\beta_D\notin\mathbb{Q}$, then \begin{align*} \lim_{N\to\infty}A_N(k,r)=0 \end{align*} for every $k\in\mathbb{Z}\setminus\{0\}$. First prove $\mathsf{L}(1)$. Let $r(t)=\beta_1t+\beta_0$ with $\beta_1\notin\mathbb{Q}$. For $k\in\mathbb{Z}\setminus\{0\}$, the number $\mathrm e(k\beta_1)$ is not equal to $1$, because $\mathrm e(k\beta_1)=1$ would imply $k\beta_1\in\mathbb{Z}$ and hence $\beta_1\in\mathbb{Q}$. The finite geometric series formula gives \begin{align*} A_N(k,r) &= \frac{\mathrm e(k\beta_0)}{N} \sum_{n=0}^{N-1}\mathrm e(k\beta_1 n)\\ &= \frac{\mathrm e(k\beta_0)}{N} \cdot \frac{1-\mathrm e(k\beta_1N)}{1-\mathrm e(k\beta_1)}. \end{align*} Since $|\mathrm e(k\beta_0)|=1$ and $|1-\mathrm e(k\beta_1N)|\le 2$, \begin{align*} |A_N(k,r)| \le \frac{2}{N|1-\mathrm e(k\beta_1)|} \to 0. \end{align*} Thus $\mathsf{L}(1)$ holds. Now assume $D\ge 2$ and $\mathsf{L}(D-1)$ holds. Let $r:\mathbb{R}\to\mathbb{R}$ be a degree $D$ polynomial with irrational leading coefficient $\beta_D$, and fix $k\in\mathbb{Z}\setminus\{0\}$. For each $h\in\mathbb{N}$, define the difference polynomial \begin{align*} q_h:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto r(t+h)-r(t). \end{align*} By the [Binomial Theorem](/page/Binomial%20Theorem), the coefficient of $t^{D-1}$ in $q_h$ is $Dh\beta_D$. Since $Dh$ is a non-zero integer and $\beta_D\notin\mathbb{Q}$, the coefficient $Dh\beta_D$ is irrational. Thus $q_h$ has degree $D-1$ with irrational leading coefficient. For $N>h$, define $z_n:=\mathrm e(k r(n))$ for $n\in\{0,\ldots,N-1\}$. Then $|z_n|=1$, and the correlation in the finite-window van der Corput estimate satisfies \begin{align*} C_{N,h} &= \frac{1}{N}\sum_{n=0}^{N-1-h}z_{n+h}\overline{z_n}\\ &= \frac{1}{N}\sum_{n=0}^{N-1-h} \mathrm e(k(r(n+h)-r(n)))\\ &= \frac{N-h}{N}A_{N-h}(k,q_h). \end{align*} By $\mathsf{L}(D-1)$ applied to $q_h$, we have $A_{N-h}(k,q_h)\to 0$ as $N\to\infty$. Hence $C_{N,h}\to 0$ for each fixed $h\in\mathbb{N}$. Fix $H\in\mathbb{N}$. Applying the finite-window van der Corput estimate for all $N\ge H$ gives \begin{align*} |A_N(k,r)|^2 \le \left(1+\frac{H-1}{N}\right) \left[ \frac{1}{H} + \frac{2}{H}\sum_{h=1}^{H-1}\left(1-\frac{h}{H}\right)\operatorname{Re} C_{N,h} \right]. \end{align*} Taking $\limsup_{N\to\infty}$ and using $C_{N,h}\to 0$ for each of the finitely many $h\in\{1,\ldots,H-1\}$ yields \begin{align*} 0\le \limsup_{N\to\infty}|A_N(k,r)|^2\le \frac{1}{H}. \end{align*} Since $H\in\mathbb{N}$ is arbitrary, letting $H\to\infty$ gives \begin{align*} \lim_{N\to\infty}|A_N(k,r)|^2=0. \end{align*} Therefore $A_N(k,r)\to 0$, and $\mathsf{L}(D)$ holds. By induction, $\mathsf{L}(D)$ holds for every $D\in\mathbb{N}$.[/step]

[guided]The strategy is induction on the degree $D$. The base case $D=1$ is a finite geometric series — exactly the case where the standard equidistribution proof for $n\alpha$ works. The inductive step uses the van der Corput estimate to reduce decay for a degree-$D$ polynomial to decay for the degree-$(D-1)$ difference polynomials $q_h(t)=r(t+h)-r(t)$. The key arithmetic input is that the leading coefficient of $q_h$ is still irrational, so we can apply the inductive hypothesis. For $D\in\mathbb{N}$, let $\mathsf{L}(D)$ be the assertion: for every real polynomial \begin{align*} r:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto\beta_Dt^D+\beta_{D-1}t^{D-1}+\cdots+\beta_1t+\beta_0 \end{align*} of degree $D$ with $\beta_D\notin\mathbb{Q}$, and every $k\in\mathbb{Z}\setminus\{0\}$, $\lim_{N\to\infty}A_N(k,r)=0$. \emph{Base case $\mathsf{L}(1)$.} Let $r(t)=\beta_1t+\beta_0$ with $\beta_1\notin\mathbb{Q}$, and fix $k\in\mathbb{Z}\setminus\{0\}$. We first verify that $\mathrm e(k\beta_1)\ne 1$: if $\mathrm e(k\beta_1)=\exp(2\pi i k\beta_1)=1$ then $k\beta_1\in\mathbb{Z}$, which would force $\beta_1=(k\beta_1)/k\in\mathbb{Q}$, contradicting irrationality. Therefore the denominator in the geometric series formula is non-zero, and \begin{align*} A_N(k,r) =\frac{\mathrm e(k\beta_0)}{N}\sum_{n=0}^{N-1}\mathrm e(k\beta_1)^n =\frac{\mathrm e(k\beta_0)}{N}\cdot\frac{1-\mathrm e(k\beta_1N)}{1-\mathrm e(k\beta_1)}. \end{align*} The constant phase $\mathrm e(k\beta_0)$ has modulus $1$, and $|1-\mathrm e(k\beta_1N)|\le 2$ by the triangle inequality. Hence $|A_N(k,r)|\le 2/(N|1-\mathrm e(k\beta_1)|)\to 0$. \emph{Inductive step.} Assume $D\ge 2$ and $\mathsf{L}(D-1)$. Let $r$ be a degree-$D$ polynomial with $\beta_D\notin\mathbb{Q}$, and fix $k\in\mathbb{Z}\setminus\{0\}$. Define for each $h\in\mathbb{N}$ the \emph{difference polynomial} \begin{align*} q_h:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto r(t+h)-r(t). \end{align*} Why do we look at $q_h$? Because van der Corput will involve correlations $z_{n+h}\overline{z_n}=\mathrm e(k(r(n+h)-r(n)))=\mathrm e(kq_h(n))$, so $C_{N,h}$ is essentially $A_N$ for the polynomial $q_h$. To apply the inductive hypothesis, we need $q_h$ to be a degree-$(D-1)$ polynomial with irrational leading coefficient. The [Binomial Theorem](/page/Binomial%20Theorem) gives, for the leading coefficient $\beta_D t^D$, \begin{align*} \beta_D(t+h)^D-\beta_D t^D=\sum_{j=0}^{D-1}\binom{D}{j}\beta_D h^{D-j}t^j, \end{align*} whose leading $t^{D-1}$ term is $D h\beta_D$. Lower-order terms in $r$ contribute only to $t^j$ for $j\le D-2$. So the leading coefficient of $q_h$ is $Dh\beta_D$. Since $D\in\mathbb{N}$ and $h\in\mathbb{N}$, $Dh$ is a non-zero integer; if $Dh\beta_D$ were rational then $\beta_D=(Dh\beta_D)/(Dh)$ would be rational, contradiction. So $q_h$ has degree $D-1$ and irrational leading coefficient — exactly the hypothesis of $\mathsf{L}(D-1)$. Now connect this to van der Corput. Set $z_n:=\mathrm e(kr(n))$ for $n\in\{0,\ldots,N-1\}$; then $|z_n|=1$, so the claim of the previous step applies. For $h\in\{1,\ldots,H-1\}$ and $N>h$, \begin{align*} C_{N,h} =\frac{1}{N}\sum_{n=0}^{N-1-h}\mathrm e(k(r(n+h)-r(n))) =\frac{1}{N}\sum_{n=0}^{N-1-h}\mathrm e(kq_h(n)) =\frac{N-h}{N}A_{N-h}(k,q_h). \end{align*} By $\mathsf{L}(D-1)$ applied to $q_h$, we have $A_{N-h}(k,q_h)\to 0$ as $N\to\infty$, and $(N-h)/N\to 1$, so $C_{N,h}\to 0$ for each fixed $h$. Fix any $H\in\mathbb{N}$ and apply the finite-window van der Corput estimate from the previous step. For all $N\ge H$, \begin{align*} |A_N(k,r)|^2 \le\left(1+\frac{H-1}{N}\right)\!\left[\frac{1}{H}+\frac{2}{H}\sum_{h=1}^{H-1}\!\left(1-\frac{h}{H}\right)\operatorname{Re}C_{N,h}\right]. \end{align*} Take $\limsup_{N\to\infty}$: $(H-1)/N\to 0$, and for each of the finitely many $h\in\{1,\ldots,H-1\}$ we have $\operatorname{Re}C_{N,h}\to 0$, so the bracketed term tends to $1/H$. Hence \begin{align*} \limsup_{N\to\infty}|A_N(k,r)|^2\le\frac{1}{H}. \end{align*} This bound holds for every $H\in\mathbb{N}$, so letting $H\to\infty$ gives $\limsup_{N\to\infty}|A_N(k,r)|^2=0$. Combined with the immediate lower bound $0\le|A_N(k,r)|^2$, this proves $|A_N(k,r)|^2\to 0$ and therefore $A_N(k,r)\to 0$. Hence $\mathsf{L}(D)$ holds, completing the induction.[/guided]

[step:Split into residue classes to reduce the general case to the leading irrational case]Let \begin{align*} m:=\max\{j\in\{1,\ldots,d\}:\alpha_j\notin\mathbb{Q}\}. \end{align*} Then $\alpha_m\notin\mathbb{Q}$ and $\alpha_\ell\in\mathbb{Q}$ for every $\ell\in\{m+1,\ldots,d\}$. Define the rational higher-degree part \begin{align*} p_{\mathrm{rat}}:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto \sum_{\ell=m+1}^{d}\alpha_\ell t^\ell, \end{align*} with the convention that the empty sum is $0$ when $m=d$. Choose $Q\in\mathbb{N}$ such that $Q\alpha_\ell\in\mathbb{Z}$ for every $\ell\in\{m+1,\ldots,d\}$; when $m=d$, take $Q=1$. For each residue $a\in\{0,1,\ldots,Q-1\}$, define \begin{align*} s_a:\mathbb{R}&\to\mathbb{R}\\ u&\mapsto \sum_{\ell=0}^{m}\alpha_\ell(Qu+a)^\ell. \end{align*} The polynomial $s_a$ has degree $m$, and its leading coefficient is $\alpha_m Q^m$, which is irrational because $\alpha_m\notin\mathbb{Q}$ and $Q^m\in\mathbb{N}$. For $u\in\mathbb{N}_0:=\{0,1,2,\ldots\}$ and $\ell\in\{m+1,\ldots,d\}$, the integer $(Qu+a)^\ell-a^\ell$ is divisible by $Q$. Hence \begin{align*} \alpha_\ell\bigl((Qu+a)^\ell-a^\ell\bigr)\in\mathbb{Z}, \end{align*} because $Q\alpha_\ell\in\mathbb{Z}$. Summing over $\ell$ gives \begin{align*} p_{\mathrm{rat}}(Qu+a)-p_{\mathrm{rat}}(a)\in\mathbb{Z}. \end{align*} Therefore, for every $k\in\mathbb{Z}\setminus\{0\}$, \begin{align*} \mathrm e(kp(Qu+a)) &= \mathrm e(kp_{\mathrm{rat}}(Qu+a))\, \mathrm e(ks_a(u))\\ &= \mathrm e(kp_{\mathrm{rat}}(a))\, \mathrm e(ks_a(u)). \end{align*} For $N\in\mathbb{N}$ and $a\in\{0,\ldots,Q-1\}$, define \begin{align*} M_{N,a}:=\#\{u\in\mathbb{N}_0:Qu+a\le N-1\}. \end{align*} For all $N\ge Q$, each $M_{N,a}$ is positive, and the residue classes modulo $Q$ decompose the sum: \begin{align*} A_N(k,p) &= \sum_{a=0}^{Q-1} \frac{1}{N} \sum_{u=0}^{M_{N,a}-1} \mathrm e(kp(Qu+a))\\ &= \sum_{a=0}^{Q-1} \frac{M_{N,a}}{N}\, \mathrm e(kp_{\mathrm{rat}}(a))\, A_{M_{N,a}}(k,s_a). \end{align*} For each fixed $a$, we have $M_{N,a}\to\infty$ as $N\to\infty$. Since $s_a$ has irrational leading coefficient, the assertion $\mathsf{L}(m)$ proved in the previous step applies to $s_a$, giving \begin{align*} A_{M_{N,a}}(k,s_a)\to 0. \end{align*} The sum over $a\in\{0,\ldots,Q-1\}$ is finite and $0\le M_{N,a}/N\le 1$, so \begin{align*} \lim_{N\to\infty}A_N(k,p)=0. \end{align*}[/step]

[guided]The previous step handled the case of irrational \emph{leading} coefficient. But the hypothesis of the theorem only requires \emph{some} non-constant coefficient $\alpha_j$ ($j\ge 1$) to be irrational. If only $\alpha_2$ is irrational while $\alpha_3,\ldots,\alpha_d$ are rational, the leading coefficient $\alpha_d$ is rational, so the previous step does not apply directly. The idea is to push the rational higher-degree contributions into an integer (which $\mathrm e$ does not see) by working along arithmetic progressions $n=Qu+a$ for a fixed common denominator $Q$. Define the highest irrational index: \begin{align*} m:=\max\{j\in\{1,\ldots,d\}:\alpha_j\notin\mathbb{Q}\}. \end{align*} This maximum exists because the hypothesis asserts the set on the right is non-empty. By construction $\alpha_m\notin\mathbb{Q}$ and every higher-index coefficient $\alpha_\ell$ for $\ell>m$ is rational. Split the polynomial accordingly into a rational high part and an irrational-leading low part: \begin{align*} p_{\mathrm{rat}}:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto\sum_{\ell=m+1}^{d}\alpha_\ell t^\ell, \end{align*} with $p_{\mathrm{rat}}\equiv 0$ when $m=d$ (empty sum). Choose $Q\in\mathbb{N}$ with $Q\alpha_\ell\in\mathbb{Z}$ for every $\ell\in\{m+1,\ldots,d\}$: take $Q$ to be the least common multiple of the denominators of the (finitely many) rational numbers $\alpha_{m+1},\ldots,\alpha_d$ in lowest terms, and set $Q=1$ if $m=d$. For each residue $a\in\{0,1,\ldots,Q-1\}$, define the restriction polynomial \begin{align*} s_a:\mathbb{R}&\to\mathbb{R}\\ u&\mapsto\sum_{\ell=0}^{m}\alpha_\ell(Qu+a)^\ell. \end{align*} The polynomial $s_a$ has degree $m$ because $\alpha_m\ne 0$, and expanding $(Qu+a)^m$ via the [Binomial Theorem](/page/Binomial%20Theorem) shows the coefficient of $u^m$ is $\alpha_m Q^m$. Since $\alpha_m\notin\mathbb{Q}$ and $Q^m\in\mathbb{N}$, $\alpha_mQ^m\notin\mathbb{Q}$ — the inductive hypothesis $\mathsf{L}(m)$ from the previous step applies to $s_a$. Why does $p_{\mathrm{rat}}$ disappear modulo $1$? For $u\in\mathbb{N}_0$, $a\in\{0,\ldots,Q-1\}$, and $\ell\in\{m+1,\ldots,d\}$, the [Binomial Theorem](/page/Binomial%20Theorem) gives \begin{align*} (Qu+a)^\ell-a^\ell=\sum_{j=1}^{\ell}\binom{\ell}{j}(Qu)^j a^{\ell-j}=Q\sum_{j=1}^{\ell}\binom{\ell}{j}Q^{j-1}u^j a^{\ell-j}\in Q\mathbb{Z}. \end{align*} Multiplying by $\alpha_\ell$ and using $Q\alpha_\ell\in\mathbb{Z}$: \begin{align*} \alpha_\ell\bigl((Qu+a)^\ell-a^\ell\bigr)\in\mathbb{Z}. \end{align*} Summing over $\ell\in\{m+1,\ldots,d\}$: \begin{align*} p_{\mathrm{rat}}(Qu+a)-p_{\mathrm{rat}}(a)\in\mathbb{Z}. \end{align*} Since $\mathrm e$ is identically $1$ on $\mathbb{Z}$, this contributes a constant phase only: \begin{align*} \mathrm e(kp(Qu+a)) &=\mathrm e(kp_{\mathrm{rat}}(Qu+a))\,\mathrm e(ks_a(u))\\ &=\mathrm e(kp_{\mathrm{rat}}(a))\,\mathrm e(ks_a(u)). \end{align*} We now reassemble the average. For $N\in\mathbb{N}$ and $a\in\{0,\ldots,Q-1\}$, let $M_{N,a}:=\#\{u\in\mathbb{N}_0:Qu+a\le N-1\}$ count the elements of residue class $a$ in $\{0,\ldots,N-1\}$. Each $n\in\{0,\ldots,N-1\}$ lies in exactly one residue class modulo $Q$, so the residue classes partition $\{0,\ldots,N-1\}$. Splitting $A_N(k,p)$ accordingly and substituting: \begin{align*} A_N(k,p) &=\sum_{a=0}^{Q-1}\frac{1}{N}\sum_{u=0}^{M_{N,a}-1}\mathrm e(kp(Qu+a))\\ &=\sum_{a=0}^{Q-1}\frac{M_{N,a}}{N}\mathrm e(kp_{\mathrm{rat}}(a))A_{M_{N,a}}(k,s_a). \end{align*} Finally, take $N\to\infty$. For each fixed $a$, $M_{N,a}/N\to 1/Q$ and in particular $M_{N,a}\to\infty$. The polynomial $s_a$ has degree $m\ge 1$ and irrational leading coefficient, so $\mathsf{L}(m)$ from the previous step gives $A_{M_{N,a}}(k,s_a)\to 0$. Each summand has modulus at most $|M_{N,a}/N|\cdot|A_{M_{N,a}}(k,s_a)|\le|A_{M_{N,a}}(k,s_a)|\to 0$, and there are only $Q$ summands, so $A_N(k,p)\to 0$.[/guided]

[step:Apply the Weyl criterion to finish the proof] We have proved that for every $k\in\mathbb{Z}\setminus\{0\}$, \begin{align*} \lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}\mathrm e(kp(n))=0. \end{align*} By the [Weyl Criterion](/page/Weyl%20Criterion), the sequence $(p(n))_{n=0}^{\infty}$ is [equidistributed modulo $1$](/page/Equidistribution%20Modulo%201). This is the desired conclusion. [/step]