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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.