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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 $m$.