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If $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$, the displayed estimate holds. Twice applying Weyl differencing bounds $|S{|}^4$ by $O(n^3+n{\sum }_{0<|h|,|l|<n}\min (n,\Vert 6\theta hl{\Vert }^{-1}))$. Group by $6hl$ and use the subpower bound for the divisor function. The reciprocal fractional-part sum near a rational bounds the resulting sum by $O(\log (2q)(n^3/q+n^2+n+q))$. Take fourth roots, absorbing logarithms into $n^{\eta }$ when $q\le n^3$; for larger $q$ the trivial bound suffices. Leiden lecture notes, Section 7.2 develop this cubic differencing estimate.

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.

The rational approximation below is taken in lowest terms, $(a,q)=1$, with $1\le q\le Q$, and $X>0$. This is necessary for denominator-sensitive exponential sum estimates and for the final Ramanujan sum to equal $\mu (q)$. If $0<X<1$, the defining sum over $a\le X$ is empty and $S_4=0$; hence we may assume $1\le X<\sqrt{N}$.

Set $c_u={\sum }_{a\mid u,a\le X}\mu (a)$, so $|c_u|\le \tau (u)$, and write the given Vaughan identity term as the Type II sumUse a dyadic decomposition into $u\in (U,2U]$, $d\in (D,2D]$, retaining $ud\le N$. Empty blocks may be discarded; every remaining block has $U,D\ge X$, $UD<N$, and there are $O({\log }^2(2N))$ blocks. Denote a block by $T$. The Cauchy-Schwarz inequality givesBy the allowed divisor-square summatory bound, the first factor is $O(U{\log }^3(2N))$. Expand the second factor. The diagonal $d_1=d_2$ contributes $O(UD{\log }^2(2N))$, using $0\le \Lambda (d)\le \log (2N)$. For a pair $d_1\ne d_2$, the allowed $u$ form the intervalThus the exponential geometric sum bound applies to this interval even though the original block has a hyperbolic cutoff. For each difference $h=|d_1-d_2|\le D$, there are $O(D)$ ordered pairs, giving an upper boundfor the expanded second factor, with an absolute constant $C$ accommodating integer endpoints. The supplied reciprocal fractional-part sum near a rational, with $M\asymp D$ and $R\asymp U$, bounds the last sum byOne may also derive this directly from the separated reciprocal-distance sum by splitting $h$ into blocks of length at most $q/2$; the approximation error between two block points is at most $1/(2q)$, and the rational rotations are separated by $1/q$.

Combining the estimates yieldsSince $UD<N$ and $U,D\ge X$, both $U,D\le N/X$. Taking square roots provesSumming the dyadic blocks establishes the rational-phase Type II estimate and, in particular, the requested bound with an explicit logarithmic power:

The Siegel–Walfisz theorem states that for fixed $A,C>0$, uniformly for $q\le (\log N{)}^A$ and $(b,q)=1$,The constant is allowed to be ineffective. Apply this with $C=A+B+2$ to the reduced residue classes. Their contribution to the prime exponential sum issince there are at most $q\le (\log N{)}^A$ residue classes. To evaluate the Ramanujan sum, the Möbius divisor-sum identity gives $1_{(b,q)=1}={\sum }_{d\mid (b,q)}\mu (d)$. Then the finite geometric series givesWhen $(a,q)=1$, only $m=1$ remains, giving $c_q(a)=\mu (q)$. The omitted nonunits contribute only prime powers $p^j$ with $p\mid q$. For each such prime, the sum of their Von Mangoldt function weights is at most $\log N$, so their total is at most $\omega (q)\log N=O_A(\log N\log \log (3N))$, absorbed by the stated error. This proves the major-arc value of the prime exponential sum:The original PDF does not state coprimality in this last request. Without it the asserted main term is genuinely false: $a=q=2$ makes $e(an/q)=1$ and hence ${\hat{\Lambda }}_N(a/q)=\psi (N)\sim N$, whereas $\mu (2)/\phi (2)=-1$. For an arbitrary numerator the correct main term is $N{c}_q(a)/\phi (q)$, with $c_q(a)$ given by the divisor formula above. Reducing the fraction before applying the boxed result is the equivalent repair.

Write $\alpha =a/q+\beta$, where $|\beta |\le q^{-2}$. Split the $m$-interval into $O(M/q+1)$ consecutive blocks of length at most $q/2$. If distinct integers $m,n$ lie in one block, then $0<|m-n|<q/2$. Since $(a,q)=1$,whereas $|\beta (m-n)|\le 1/(2q)$. Hence $\Vert \alpha m-\alpha n\Vert \ge 1/(2q)$.

The points $\alpha m$ in one block are therefore $1/(2q)$-separated modulo one. Order them by distance from the nearest integer. Apart from a bounded number of endpoints, the $j$th closest point has distance $\gg j/q$, and consequentlyMultiplying by $O(M/q+1)$ givesThis is the reciprocal fractional-part sum near a rational estimate.

Write $M_0=CR{T}^2$ and split the last sum in part d according as ${\tau }_4(m)<X$ or ${\tau }_4(m)\ge X$. For the first part, the reciprocal fractional-part sum near a rational givesFor the second part, use the supplied second-moment estimate and truncation of a divisor weight by its second moment:Substitute $M_0\asymp R{T}^2$, take fourth roots, and factor out $R^{1/2}{T}^{1/2}$. The four terms from the bounded-weight estimate become, after harmless enlargement by $(\log q)(\log RT{)}^{O(1)}$,with the smaller $XR$ term absorbed by $X{M}_0$. The large-weight part contributes $X^{-1/4}$. Finally, the two diagonal terms from part d contribute $R^{-1/4}$ and $T^{-1/4}$; since $X\ge 1$, the first is absorbed by $X^{1/4}{R}^{-1/4}$. Therefore

Suppose $|a_u|\le \tau (u)$, $|b_d|\le \log (2N)$, $u,d>X\ge 1$, $ud\le N$, and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$. A dyadic decomposition reduces $S=\sum a_u{b}_{d}e(\theta ud)$ to blocks $u\asymp U,d\asymp D$. The Cauchy-Schwarz inequality, divisor-square summatory bound and the reciprocal fractional-part sum near a rational giveThe condition $ud\le N$ makes the allowed $u$ values for each pair $d_1,d_2$ an interval, so the exponential geometric sum bound applies despite the hyperbolic cutoff. Since $UD\le N$ and $U,D\gg X$, take square roots and sum the $O({\log }^2(2N))$ blocks to get the displayed estimate.

For a real degree-$k$ polynomial with leading coefficient $\theta$, $k\ge 2$, and coprime integers $a,q$ satisfying $|\theta -a/q|\le q^{-2}$, the displayed bound applies to its exponential sum over $1\le n\le N$. Repeated Weyl differencing reduces the phase to slope $k!\theta h_1\cdots h_{k-1}$. A geometric-sum estimate, the subpower bound for the divisor function and the reciprocal fractional-part sum near a rational complete the estimate. The arbitrary positive exponent $\eta$ absorbs divisor bounds and logarithms.