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In, Bruedern and Wooley mention the following fact on the bottom of page 6: Let $$\Phi(\alpha) = \sum_{h\le 6H}\sum_{P<x\le 2P}e(\alpha h(3x^2 + 3xh + h^2))$$ They then assert that $$\int_{0}^1\left|\Phi(\alpha)\right|^4d\alpha\ll H^3P^{2 + \epsilon}$$ for all $\epsilon>0$

However, I don't have access to the paper mentioned in which this fact is proven. Can someone show me a proof of this? It is supposed to be along the lines of the proof of Hua's Lemma, but I can't see how this is the case.

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Google Scholar gives this publicly available pdf for the cited paper. – Dougal Aug 27 '14 at 6:27
Dougal's link is valuable of course, but it's amazing to me that a reference to the paper gets more upvotes here than a response by the author himself. – Lucia Aug 28 '14 at 3:05
up vote 7 down vote accepted

First, the paper is Bruedern and Wooley!

Next ... by Cauchy's inequality, one has $$|\Phi(\alpha)|^2\le 6H\Psi(\alpha),$$ say, where $$\Psi(\alpha)=\sum_{h\le 6H}\left| \sum_{P<x\le 2P}e(\alpha h(3x^2+3xh+h^2))\right|^2.$$ Thus the mean value in question is bounded above by $6H$ times $$\int_0^1\Psi(\alpha)|\Phi(\alpha)|^2\, d\alpha ,$$ and by orthogonality that counts the number of solutions of the Diophantine equation $$3h(x_1-x_2)(x_1+x_2+h)=g_1(3y_1^2+3y_1g_1+g_1^2)-g_2(3y_2^2+3y_2g_2+g_2^2),$$ with $h,g_1,g_2\le 6H$ and $P<x_1,x_2,y_1,y_2\le 2P$. When $x_1=x_2$ and $h$ is free, then given $g_2$ and $y_2$, one finds that $g_1$ and $y_1$ are determined by a divisor estimate. So the number of diagonal solutions is $\ll HP\cdot (HP)^{1+\epsilon}=(HP)^{2+\epsilon}$. Meanwhile, when $x_1\ne x_2$, the right hand side here is also non-zero. Then given a fixed choice for $g_1,g_2,y_1,y_2$, one finds that $x_1-x_2$ and $h$ are determined by a divisor function estimate. There are consequently $\ll (HP)^2\cdot (HP)^\epsilon$ such solutions. Combine these estimates, throw back in the factor $6H$ that arose from applying Cauchy's inequality, and one gets the estimate claimed.

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Inventiones papers up to 1996 are publicly available here.

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