# Why should $\sum_{m=1}^N e(\alpha m^3)$ be big for some $\alpha?$

I'm going through a "circle method" proof of the fact that every large enough natural number $$n$$ is the sum of nine cubes. At some point a lot of control over the function $$f(\alpha)=\sum_{m=1}^N e(\alpha m^3)$$ is needed. Here $$N=\lfloor n^{1/3}\rfloor$$ and $$e(z)=e^{2\pi i z}$$. If we first study the function $$g(\alpha)=\sum_{m=1}^N e(\alpha m)=\frac{e(\alpha(N+1))-e(\alpha)}{e(\alpha)-1}$$ we find that it takes bigger values when $$\alpha$$ is close to an integer. The author then says that "a similar analysis shows that $$f$$ is bigger whenever $$\alpha$$ is close to a rational number with small denominator". He/She uses the following major arcs $$\mathcal{U}=\left[\frac{1}{n^{1-\frac{1}{300}}},1+\frac{1}{n^{1-\frac{1}{300}}}\right]$$ $$\mathfrak{M}(a,q)=\left\lbrace \alpha\in\mathcal{U} \mid \left|\alpha-\frac{a}{q}\right|\le\frac{1}{n^{1-\frac{1}{300}}}\right\rbrace\quad a,q\in\mathbb{N}\quad (a,q)=1$$ $$\mathfrak{M}=\bigcup_{1\le q\le n^{1/300}}\bigcup_{\substack{1\le a\le q-1 \\ (a,q)=1}}\mathfrak{M}(a,q)$$ and then says it can be proved that $$\int_\mathfrak{M}f(\alpha)^9e(-\alpha n)\,d\alpha$$ is big (something like $$cn^2$$). I'm completely lost here. How does one prove that? Is there an easy reference on Weyl sums and lower bounds? The author works out the bounds for the integral over the minor arcs but not over the major arcs.

Can someone either work it out here or point to a reference where it is worked out in detail?

Thanks!!

Edit: The only (almost trivial) progress I've been able to make is $$\vert\mathfrak{M}\vert=\sum_{1\le q\le n^{1/300}}\sum_{\substack{1\le a\le q-1 \\ (a,q)=1}}\vert\mathfrak{M}(a,q)\vert\le 2\frac{\varphi(q)}{n^{1-\frac{2}{300}}}\le\frac{\varphi(n^{\frac{1}{300}})}{n^{1-\frac{2}{300}}}<2\frac{n^{\frac{1}{300}}}{n^{1-\frac{2}{300}}}\to 0$$ when $$n\to\infty$$. Also $$f\left(\frac{a}{q}\right)=\sum_{m=1}^N e\left(\frac{a}{q}m^3\right)=\sum_{r=1}^q\sum_{\substack{1\le k\le N \\ k\,\equiv_q\,r}}e\left(\frac{a}{q}k^3\right)=\frac{N}{q}\sum_{r=1}^q e\left(\frac{a}{q}r^3\right)$$ The last equality may not be true but $$N/q$$ is an increasingly good approximation (asymptotically true) of how many integers $$1\le k\le N$$ are in each residue class modulo $$q$$. Maybe this last sum, from $$r=1$$ to $$q$$ is easier to estimate. It shouldn't be difficult to get an asymptotic expression for $$\vert\mathfrak{M}\vert$$ and argue that $$f(\alpha)\approx f(a/q)$$ in $$\mathfrak{M}(a,q)$$ so that $$\int_{\mathfrak{M}(a,q)}f(\alpha)^9e(-\alpha n)\, d\alpha\approx\int_{\mathfrak{M}(a,q)}f\left(\frac{a}{q}\right)^9e(-\alpha n)\, d\alpha=\vert\mathfrak{M}(a,q)\vert f\left(\frac{a}{q}\right)^9e(-\alpha n)$$

Better than an answer with a solution: I found the entire set of papers containing the whole proof (two weeks ago but couldn't make this edit earlier) and they come from a number theory course from the University of Leiden

• I would suggest Vaughan The Hardy-Littlewood Method (pg 18-19 Thm 2.2-2.3 give a general estimate of this type for the general form of your integral ($\int_\mathfrak{M}f(\alpha)^se(-\alpha n)\,d\alpha, f(\alpha)= \sum_m e(m\alpha^k)$ for arbitrary $k \ge 2,s > 2^k$ with a main term of $n^{s/k-1}$ which is precisely $n^2$ for $s=9, \alpha=3$ showing one needs power $9$ for cubes ($9>2^3=8$!) btw cambridge.org/us/academic/subjects/mathematics/number-theory/… – Conrad Jul 8 '20 at 14:07
• @Conrad is it available for free somewhere? – augustoperez Jul 8 '20 at 22:33
• @augustoperez isn't the answer to the title question "take $\alpha = 0$"? have you analyzed what happens near $\alpha = 0$? – mathworker21 Jul 10 '20 at 0:42
• @augustoperez have u tried something like this: $|\sum_{m=1}^N e(\frac{am^3}{q})^2 = \sum_{m,n=1}^N e(\frac{am^3}{q})e(\frac{-an^3}{q}) = \sum_{n,h=1}^N e(\frac{a(n+h)^3}{q})e(\frac{-an^3}{q}) = \sum_{n,h=1}^N e(\frac{3an^2h+3anh^2+ah^3}{q})$ (where second equality is not exactly true but is almost true if $q$ is small) then sum over $n$ and use formula for quadratic gauss sum? – mathworker21 Jul 10 '20 at 19:02
• (1) think more. (2) you definitely can find exact values for quadratic gauss sums (did you try wikipedia) – mathworker21 Jul 10 '20 at 19:19