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Write $e(t)=e^{2\pi it}$ and $N=n+1$. A convenient form of the quadratic Weyl inequality is the following: if $q\ge 1$, $(a,q)=1$ and $|\alpha -a/q|\le q^{-2}$, thenThe implied constant is absolute. In particular the familiar square-root saving is $|S|\ll (N^2/q+N+q{)}^{1/2}\log (2q{)}^{1/2}$. We prove all estimates used in this statement.

First, the finite geometric series identity giveswhere $\Vert \theta \Vert$ denotes distance to the nearest integer and the second bound is interpreted as infinite when that distance is zero. Indeed the sum is $(1-e(L\theta ))/(1-e(\theta ))$, and $|1-e(\theta )|=2|\sin \pi \theta |\ge 4\Vert \theta \Vert$. This proves the exponential geometric sum bound.

Expand the squared quadratic exponential sum by the difference $h$ between its indices. Since $(x+h{)}^2-x^2=2hx+h^2$, the diagonal and the paired off-diagonals giveFor $q\ge 16$, partition the $h$ into consecutive blocks of length at most $L=\lfloor q/8\rfloor$. Within one block, two distinct indices have difference $0<|d|<q/2$. Since the denominator of $2a/q$ after reduction is at least $q/2$, $\Vert 2ad/q\Vert \ge 1/q$. ThereforeThus the points $2\alpha h$ in a block are separated on the circle by at least $3/(4q)$.

For any such separated set, at most an absolute number of points lie in each pair of arcs whose distances from zero are between $j/(4q)$ and $(j+1)/(4q)$. The closest few contribute $O(N)$ to the capped reciprocal sum; the remaining arcs contribute $O(q/j)$ for $1\le j\le 4q$. Summing the harmonic series, using ${\sum }_{j\le m}{j}^{-1}\le 1+\log m$, proves a block bound $O(N+q\log (2q))$. This is the required separated reciprocal-distance sum estimate, with its proof supplied here. There are $O(N/q+1)$ blocks, soFor $q<16$, the trivial bound $|S|\le N$ already implies this estimate after increasing the absolute constant. This completes the proof, including the small denominators and the possible common factor of $2$ and $q$.

Write $e(t)=e^{2\pi it}$. We first prove the needed cancellation of quadratic exponential sums, rather than invoke equidistribution without proof. The number $\sqrt{2}$ is irrational: writing it as a reduced fraction would make its numerator even from $u^2=2v^2$, and then its denominator even as well, a contradiction. Thus $2hr\sqrt{2}$ is irrational for nonzero integers $h,r$.

Fix $h\ne 0$ and set $z_n=e(h\sqrt{2}{n}^2)$ for $1\le n\le L$, extending it by zero otherwise. For integers $L\ge H\ge 1$, each term appears exactly $H$ times in the shifted sum, soFor any complex numbers $w_1,\dots ,w_J$, nonnegativity of ${\sum }_j|w_j-J^{-1}{\sum }_i{w}_i{|}^2$ gives $|{\sum }_j{w}_j{|}^2\le J{\sum }_j|w_j{|}^2$. Apply this special Cauchy-Schwarz inequality to the displayed outer sum, and then expand the inner squared modulus. This givesFor fixed $H$ and large $L$, each correlation is the geometric serieswhose absolute value is at most $2/|1-e(2hr\sqrt{2})|$, independently of $L$. This follows by multiplying the geometric sum by $1-e(2hr\sqrt{2})$; its denominator is nonzero by irrationality. Divide the inequality by $L^2$ and let $L\to \infty$, keeping $H$ fixed. It gives a limit superior at most $1/H$. Taking arbitrarily large $H$ provesThis is the quadratic instance of the Van der Corput inequality for finite scalar sequences, with its proof supplied here.

To turn cancellation into the particular approximation, put $\epsilon =10^{-8}$ and choose an integer $M>2/{\sin }^2(\pi \epsilon )$. The Fejér kernel is the explicit finite trigonometric polynomialThe expansion follows by counting pairs of indices with difference $h$. Its constant coefficient is one, so the cancellation just proved givesIf $\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}\ge \epsilon$, the geometric-series formula givesFor all sufficiently large $L$, the displayed average exceeds $1/2$. Consequently at least one $n\in \{1,\dots ,L\}$ satisfiesIt is a positive integer, as required. Every auxiliary cancellation and kernel identity used in this existence proof has been derived; a smooth cutoff or an unproved Weyl criterion is unnecessary.

Write $e(t)=\exp (2\pi it)$ and use $\Vert t\Vert$ for distance to the nearest integer. The Dirichlet approximation theorem, in its pigeonhole form, says that for every real $\alpha$ and integer $Q\ge 1$ there are coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$. Indeed put the $Q+1$ fractional parts of $0,\alpha ,\dots ,Q\alpha$ into $Q$ equal intervals; two lie in the same interval. Their difference gives $|q\alpha -a|\le 1/Q$, and reducing the fraction preserves the claimed bound. This includes rational $\alpha$.

For $t\ge s$, the finite geometric series has length $L=t-s+1$. When $\theta \notin \mathbb{Z}$,The last inequality uses the concavity of sine on $[0,\pi /2]$. The triangle inequality also gives $L$. At integer $\theta$ the sum has absolute value $L$, interpreting the reciprocal as infinity. Thus the precise exponential geometric sum bound isThe interval must be nonempty; for $t<s$ its sum is zero and its length should be defined as $\max (0,t-s+1)$, rather than using a potentially negative bound.

For the separated points, assume $Q\ge 2$ and $\delta >0$. An arc of length $2r\le 1$ contains at most $1+2r/\delta$ of the points, by ordering them along the arc. Points with $\Vert {\theta }_i\Vert \le 1/Q$ contribute $O(Q+{\delta }^{-1})$. In the dyadic annulus $2^j/Q<\Vert {\theta }_i\Vert \le 2^{j+1}/Q$, each summand is $O(Q{2}^{-j})$ and there are at most $1+2^{j+2}/(Q\delta )$ points. There are $O(\log (2Q))$ nonempty annuli, giving the stronger separated reciprocal-distance sumIf there is only one point the bound is immediate; otherwise $\delta \le 1/2$. The printed $\log Q$ requires a lower bound away from $Q=1$; $\log (2Q)$ avoids that endpoint defect.

A useful version of the quadratic Weyl inequality is the following, with $(a,q)=1$, $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$:For $N\ge 2$, expand the square of this quadratic exponential sum by $h=n-m$:The exponential geometric sum bound therefore gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. For $q\ge 8$, partition these $h$ into blocks of length at most $\lfloor q/4\rfloor$. Two different points in one block have difference $d$ with $0<|d|<q/4$. The reduced denominator of $2a/q$ is at least $q/2$, so $\Vert 2ad/q\Vert \ge 1/q$, whileThus the rotations $2\alpha h$ are separated by at least $1/(2q)$ in each block. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, and henceFor $q<8$ or $N<2$, the trivial bound supplies the same assertion with an absolute constant. The same proof applies to ${\sum }_{n=1}^{N}e(\alpha n^2)$, whose difference sums also have interval endpoints.

We finally prove a polynomial rate of quantitative quadratic recurrence, without using the optional finite-group assumption. For $0<\epsilon \le 1/2$, use the periodic triangular bump $b(t)=(1-\Vert t\Vert /\epsilon {)}_{+}$. It is ${\epsilon }^{-1}$ times the periodic convolution of two interval indicators, and therefore has the nonnegative Fourier coefficientsIts Fourier series is absolutely convergent, ${\sum }_r\hat{b}(r)=b(0)=1$, and the coefficient tail is ${\sum }_{|r|>R}\hat{b}(r)\ll 1/(\epsilon R)$. Suppose that $\Vert \alpha n^2\Vert \ge \epsilon$ for every $1\le n\le N$. Then ${\sum }_{n}b(\alpha n^2)=0$. Choose $R=\lceil C{\epsilon }^{-2}\rceil$ with a sufficiently large absolute $C$, so that the contribution of frequencies $|r|>R$ has absolute value at most $N\epsilon /2$. The zero frequency contributes $N\epsilon$. Since all coefficients are nonnegative and their sum is one, some $1\le r\le R$ must satisfyApply the Dirichlet approximation theorem to $r\alpha$ with $Q=N$, obtaining a reduced $a/q$, $q\le N$. The quadratic Weyl inequality then impliesSet $\epsilon =N^{-1/8}$. For sufficiently large $N$, the $N\log (2N)$ term is absorbed on the left, giving $q\ll {\epsilon }^{-2}\log (2N)$. The integer $n=rq$ consequently satisfiesMoreover the approximating integer is $arq$, andThis contradicts the assumed avoidance. Increasing the absolute constant handles the finitely many smaller $N$, uniformly for every real $\alpha$. Thus

There is a minor range issue in the printed bound: for $\epsilon >1$ and positive $C$, the upper bound is below one. We prove the intended result for $0<\epsilon \le 1$; a valid formulation for every $\epsilon >0$ replaces the upper bound by $\max \{1,{\epsilon }^{-C}\}$. In fact the argument below gives $C=32$. All auxiliary estimates are proved here.

For $0<\epsilon \le 1/2$, let $m=\lceil {\epsilon }^{-2}\rceil$, and use the Fejér kernelExpanding the square proves the identity and nonnegativity. The finite geometric series formula and $|\sin \pi t|\ge 2\Vert t\Vert$ give $F_m(t)\le (4m\Vert t{\Vert }^2{)}^{-1}$ off the integers. In particular $F_m(t)\le 1/4$ whenever $\Vert t\Vert \ge \epsilon$.

Suppose, for a contradiction, that no $1\le n\le N$ has $\Vert \alpha n^2\Vert <\epsilon$. Summing the Fejér kernel along the quadratic sequence and separating its constant term givesThe coefficients on the right sum to $m-1$. Also $|S_{-k}|=|S_k|$. Hence some $1\le k<m$ has $|S_k|\ge \eta N$, where $\eta =1/(2m)$.

We need only an elementary Van der Corput inequality for finite scalar sequences. For $|z_n|\le 1$, extended by zero outside $[N]$, each term of $S={\sum }_n{z}_n$ occurs in exactly $H$ windows of length $H$. Applying the Cauchy-Schwarz inequality to the window sums and expanding their squares gives, for $H\le N$,Consequently $|S{|}^2\le 2N^2/H+(4N/H){\sum }_{h=1}^{H-1}|C_h|$. Take $H=\lceil 8{\eta }^{-2}\rceil$. If $|S|\ge \eta N$, some $1\le h<H$ must have $|C_h|\ge {\eta }^{2}N/8$; otherwise the displayed upper bound is less than $3{\eta }^2{N}^2/4$.

For the large quadratic exponential sum just found, put $z_n=e(k\alpha n^2)$. Its quadratic exponential sum has multiplicative derivativeThus $C_h$ is a finite geometric series. Its absolute value is at most $1/(2\Vert 2kh\alpha \Vert )$, unless that distance is zero, in which case the desired estimate is automatic. It follows thatSet $n_{\ast }=2kh$. The distance to the nearest integer satisfies $\Vert \ell x\Vert \le \ell \Vert x\Vert$ for a positive integer $\ell$, by multiplying a nearest integer to $x$. ThereforeThis proves the needed quantitative quadratic recurrence once $N$ is chosen polynomially in ${\epsilon }^{-1}$.

For explicit bookkeeping, $m\le 2{\epsilon }^{-2}$ and $H\le 33m^2$. Choose $N=\lfloor {\epsilon }^{-32}\rfloor$. For $0<\epsilon \le 1/2$ we have $N>33792{\epsilon }^{-11}\ge 8mH/({\eta }^2\epsilon )$ and $N\ge H$, while $2mH\le 528{\epsilon }^{-6}\le N$. Hence $n_{\ast }\le N$ and $\Vert \alpha n_{\ast }^2\Vert <\epsilon$, contradicting our supposition. The estimates have substantial slack even at $\epsilon =1/2$.

For $1/2<\epsilon \le 1$, simply take $n=1$, since $\Vert \alpha \Vert \le 1/2<\epsilon$ and $1\le {\epsilon }^{-32}$. For the intended range, the conclusion isFor $\epsilon >1$, the same choice $n=1$ proves the corrected all-range version.

For coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$, expand the squared quadratic exponential sum by its difference $h$. The exponential geometric sum bound gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. In each block of length at most $q/4$, the rotations $2\alpha h$ are separated by at least $1/(2q)$: the reduced denominator of $2a/q$ is at least $q/2$, and the approximation error between two block points is at most $1/(2q)$. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, giving the displayed inequality. For bounded $q$ the trivial estimate supplies the same conclusion.

For a universal constant $C>0$ and $0<\epsilon \le 1$, every real $\alpha$ admits $1\le n\le {\epsilon }^{-C}$ with distance to the nearest integer of $\alpha n^2$ less than $\epsilon$. A Fejér kernel detects failure of recurrence as a large quadratic exponential sum. The Van der Corput inequality for finite scalar sequences then produces a short linear near-return, whose suitable multiple gives the quadratic return. To include $\epsilon >1$, use the bound $\max \{1,{\epsilon }^{-C}\}$.