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Write $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $I_j={\int }_0^1|S{|}^j$. Character orthogonality gives $I_2=n$. For $I_4$, fix the difference $D=x^3-y^3$. If $D=0$, there are $n^2$ choices for the two equal pairs. If $D\ne 0$, any solution $z^3-w^3=D$ has $z-w\mid D$; fixing this divisor gives a quadratic equation for $w$, hence at most two choices. The subpower bound for the divisor function yields $I_4{\ll }_{\eta }{n}^{2+\eta }$.

The twice-differenced inequality in Weyl differencing reads $|S{|}^4\le 2n{\sum }_v{c}_{v}e(\theta v)$, with $c_v$ counting triples $(h,l,x)$ for which $v=3hl(2x+h+l)$ and all four shifted arguments are in $[1,n]$. Thus $c_0\ll n^2$, and $c_v{\ll }_{\eta }{n}^{\eta }$ for $v\ne 0$: the product $hl$ divides $v$, and the remaining equation determines $x$. The Fourier coefficients $r(v)$ of $|S{|}^4$ are nonnegative counts, with $r(0)=I_4$ and ${\sum }_{v}r(v)=n^4$. Multiply the inequality by $|S{|}^4$ and integrate. It follows that $I_8\ll n(n^2{I}_4+n^{\eta }{n}^4){\ll }_{\eta }{n}^{5+\eta }$, after choosing the divisor-bound exponents small enough.

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.

Put $S(\theta )={\sum }_{r=1}^{n}e(\theta r^3)$. Its modulus is the modulus of the Fourier transform in the question, independently of the sign convention. We use Hua's lemma in the cubic form ${\int }_0^1|S{|}^8{\ll }_{\delta }{n}^{5+\delta }$; the cubic eighth-moment proof by differencing establishes this from the subpower bound for the divisor function. Thereforeand $n\asymp N^{1/3}$ imply ${\sup }_m|S|{\gg }_{\delta }{n}^{1-\delta }$. Choose an actual $\theta \in m$ attaining at least half this bound; attainment of the supremum itself is unnecessary.

Fix $\epsilon >0$ and put ${\epsilon }_0=\min (\epsilon ,1/6)$. The Dirichlet approximation theorem with $Q=\lfloor N^{1-{\epsilon }_0}\rfloor$ gives coprime integers $a,q$ satisfyingThe cubic Weyl inequality, which is the quantitative equidistribution estimate needed here, saysCompare it with the lower bound. Choose $\delta ,\eta >0$ so small that $b=4(\delta +\eta )/3<{\epsilon }_0/2$. It follows thatBut $n^{-1}=O(N^{-1/3})$ and $q/n^3=O(N^{-{\epsilon }_0})$. Both are $o(N^{-b})$, so for large $N$ the first term must supply this lower bound. Consequently $q{\ll }_{\epsilon }{N}^b{\ll }_{\epsilon }{N}^{\epsilon }$, while the approximation above gives $\Vert q\theta \Vert \ll N^{{\epsilon }_0-1}\le N^{\epsilon -1}$. Enlarging the constants covers bounded $N$. ThusHere $\Vert \cdot \Vert$ is distance modulo one. A ninth moment of this size therefore forces at least one near-rational frequency, rather than merely a large pointwise sum somewhere outside $m$.

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.