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Put $e(t)=e^{2\pi it}$ and let $P(n)=\theta n^k+\cdots$ be a real polynomial of degree $k\ge 2$. A standard form of Weyl inequality is: if $(a,q)=1$ and $|\theta -a/q|\le q^{-2}$, then for every $\eta >0$The implied constant is uniform in all lower coefficients. For degree one the finite geometric series bound is $|\sum e(\theta n+\beta )|\le \min (N,(2\Vert \theta \Vert {)}^{-1})$, where $\Vert t\Vert$ is distance to the nearest integer.

Here is the Weyl differencing proof. Extend $z_n=e(P(n))$ by zero outside $[1,N]$. Expanding $|\sum z_n{|}^2$ by differences of indices gives a sum of correlations ${\sum }_n{z}_{n+h}\overline{z_n}$. Repeated Cauchy-Schwarz inequality and this correlation identity yield, after $k-1$ steps, with $p=2^{k-1}$,The interval $I(\mathbf{h})$ ensures that every shifted argument remains in $[1,N]$. For completeness, the induction is the following: at stage $j$, the factor is $N^{2^j-j-1}$ and there is a sum over $j$ shifts. Squaring, Cauchy-Schwarz over those $O(N^j)$ shifts supplies $N^j$; expanding the square of each remaining interval sum introduces the next shift. The new factor is $N^{2^{j+1}-j-2}$, as asserted for stage $j+1$.

The final differenced phase is linear, of slope $k!\theta h_1\cdots h_{k-1}$. Zero products contribute $O_k(N^{k-1})$ to the sum. For nonzero products the finite geometric series bound applies. Grouping by $r=|h_1\cdots h_{k-1}|\le N^{k-1}$, the subpower bound for the divisor function bounds the number of representations by $O_{k,\rho }(N^{\rho })$ for any chosen $\rho >0$. The divisor input is elementary: for $t=k-1$, the number of ordered factorizations of $r$ is ${\prod }_{p^v\parallel r}\left(\genfrac{}{}{0ex}{}{v+t-1}{t-1}\right)$. For all sufficiently large primes this is bounded factorwise by $p^{ϵv}$, and for the finitely many smaller primes the polynomial in $v$ is bounded by a constant times $p^{ϵv}$. Multiplication gives $O_{t,ϵ}(r^{ϵ})$. ThusThe reduced denominator of $k!a/q$ is between $q/k!$ and $q$. Its approximation error is bounded by a constant depending on $k$ times the inverse square of that denominator. Split the $r$-range into blocks of length a sufficiently small fixed multiple of $q$. Within a block the values $k!\theta r$ are separated modulo one by a constant multiple of $1/q$: rational differences have spacing at least $1/q$, while the approximation error in a block difference is smaller than half that spacing. At most a bounded number of values occupy each distance interval of width $1/q$. Summing reciprocals of their ordered distances proves the reciprocal fractional-part sum near a rational estimateFor $q>N^k$, the proposed inequality follows from $|S|\le N$. Otherwise the last display and the differencing estimate giveChoose $\rho$ sufficiently small, absorb the logarithm into a further arbitrarily small power of $N$, and take $p$th roots. This proves the stated Weyl inequality.

For the recurrence assertion, a rational $a=u/v$ gives an integer $a{m}^2$ with $m=v$. For irrational $a$, Weyl differencing implies that $a{m}^2$ is an equidistributed sequence modulo one. One way to avoid a rate issue for unusually good rational approximations is to keep a fixed differencing range $H$: the Van der Corput inequality for finite scalar sequences gives, for each nonzero integer $j$, a limsup bound ${\mathrm{limsup}}_N|N^{-1}{\sum }_{m\le N}e(ja{m}^2){|}^2\ll 1/H$, because every fixed nonzero difference has irrational linear slope and its geometric average tends to zero. Let $H\to \infty$ and apply the Weyl criterion. Hence some $m$ always enters the open $ϵ$-neighborhood of zero.

Finally, $U_m=\{a\in \mathbb{R}/\mathbb{Z}:\Vert a{m}^2\Vert <ϵ\}$ are open sets covering the compact circle group. A finite subcover has a largest index $N$. That single $N$ works for every real $a$. This is uniform square recurrence on the circle; the compactness step gives existence, not an explicit quantitative bound.