Simultaneous cubic and quadratic diagonal equations in 12 prime variables

Abstract

The system of equations

$$\begin{aligned}&u_1p_1^2 + \cdots + u_sp_s^2 = 0,\\&v_1p_1^3 + \cdots + v_sp_s^3 = 0 \end{aligned}$$

has prime solutions $(p_1, \ldots , p_s)$ for $s \ge 12$, assuming that the system has solutions modulo each prime p. This is proved via the Hardy–Littlewood circle method, building on Wooley’s work on the corresponding system over the integers and recent results on Vinogradov’s mean value theorem. Additionally, a set of sufficient conditions for local solvability is given: If both equations are solvable modulo 2, the quadratic equation is solvable modulo 3, and for each prime p at least 7 of each of the $u_i$, $v_i$ are not zero modulo p, then the system has solutions modulo each prime p.

1 Introduction

Much work has been done in applying the Hardy–Littlewood circle method to find integral solutions to systems of simultaneous equations (see [2, 3, 10], and [12] for examples). In particular, recent progress on Vinogradov’s mean value theorem (see [1, 9]) has enabled progress on questions of this type. Here we consider the question of solving systems of equations with prime variables, generalizing the Waring–Goldbach problem in the same way existing work on integral solutions of systems of equations generalizes Waring’s problem. Following Wooley [12], we address here the simplest nontrivial case: one quadratic equation and one cubic equation. We find that under suitable local conditions, 12 variables will suffice for us to establish an eventually positive asymptotic formula guaranteeing solutions to the system of equations.

Consider a pair of equations of the form

$$\begin{aligned} \begin{aligned}&u_1p_1^2 + \cdots + u_sp_s^2 = 0,\\&v_1p_1^3 + \cdots + v_sp_s^3 = 0, \end{aligned} \end{aligned}$$

(1)

where $u_1, \ldots , u_s, v_1, \ldots , v_s$ are nonzero integer constants and $p_1, \ldots , p_s$ are variables restricted to prime values. We seek to prove the following theorem:

Theorem 1.1

If

• 1. the system (1) has a nontrivial real solution,

• 2. $s \ge 12$, and

• 3. for every prime p, the corresponding local system

  $$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_sx_s^2 \equiv 0 \pmod p,\\&v_1x_1^3 + \cdots + v_sx_s^3 \equiv 0 \pmod p \end{aligned} \end{aligned}$$

  (2)

  has a solution $(x_1, \ldots , x_s)$ with all $x_i \ne 0 \pmod p$,

then the system has a solution $(p_1, \ldots , p_s)$ with all $p_i$ prime. Moreover, if we let R(P) be the number of solutions $(p_1, \ldots , p_s)$ with each $p_i \le P$, each weighted by $(\log p_1)\ldots (\log p_s)$, then we have $R(P) \sim CP^{s-5}$ for some constant $C > 0$ uniformly over all choices of $u_1, \ldots , u_s, v_1, \ldots , v_s$.

In Sect. 9 we give a sufficient condition for (2) to be satisfied, giving us the explicit theorem

Theorem 1.2

Consider the system

$$\begin{aligned} \begin{aligned}&u_1p_1^2 + \cdots + u_sp_s^2 = U,\\&v_1p_1^3 + \cdots + v_sp_s^3 = V, \end{aligned} \end{aligned}$$

(3)

where $u_1, \ldots , u_s, v_1, \ldots , v_s$, are nonzero integer constants and U, V are integer constants. If

• 1. the system has a nontrivial real solution,

• 2. $s \ge 12$,

• 3. the quadratic form $u_1p_1^2 + \cdots + u_sp_s^2$ is indefinite,

• 4. $\displaystyle \sum \nolimits _{i=1}^{s} u_i \equiv U \pmod 2$ and $\displaystyle \sum \nolimits _{i=1}^s v_i \equiv V \pmod 2,$

• 5. $\displaystyle \sum \nolimits _{i=1}^s u_i \equiv U \pmod 3$, and

• 6. for each prime $p \ne 2$, at least 7 of each of the $u_i$ and the $v_i$ are not zero modulo p,

then the system has a solution $(p_1, \ldots , p_s)$ with all $p_i$ prime. Moreover, if we let R(P) be the nu mber of solutions $(p_1, \ldots , p_s)$, each weighted by $(\log p_1)\ldots (\log p_s)$, then we have $R(P) \sim CP^{s-5}$ where $C > 0$ uniformly over all choices of $u_1, \ldots , u_s, v_1, \ldots , v_s$, U, and V.

We use the Hardy–Littlewood circle method to prove these results. Section 2 performs the necessary setup for the application of the circle method: defining the relevant functions and the major arc/minor arc dissection. Section 3 consists of a number of preliminary lemmas, which are referenced throughout. Section 4 proves a Hua-type bound necessary for the minor arcs. Section 5 proves a Weyl-type bound on the minor arcs by means of Vaughan’s identity. Section 6 is the circle method reduction to the singular series and singular integral. Section 7 shows the convergence of the singular series and Sect. 8 shows that it is eventually positive, contingent on the local solvability of the system (3). Section 9 shows sufficient conditions for the solvability of the local system. This depends on a computer check of local solvability for a finite number of primes. Section 10 discusses several techniques which can be employed to improve the efficiency of this computation. Section 11 finishes the proof of Theorems 1.1 and 1.2. Appendix 1 contains the source code used to run the computations laid out in Sect. 10.

2 Notation and definitions

As is standard in the literature, we use $e(\alpha )$ to denote $e^{2\pi i \alpha }$. The letter p is assumed to refer to a prime wherever it is used, and $\varepsilon$ means a sufficiently small positive real number. The symbols $\Lambda$ and $\mu$ are the von Mangoldt and Möbius functions, respectively. Symbols in bold are tuples, with the corresponding symbol with a subscript denoting a component, i.e., $\mathbf {a} = (a_1, \ldots , a_k)$. The letter C is used to refer to a positive constant, with the value of C being allowed to change from line to line. We write $f(x) \ll g(x)$ for $f(x) = O(g(x))$, $f(x) \asymp g(x)$ if both $f(x) \ll g(x)$ and $g(x) \ll f(x)$ hold, and $f(x) \sim g(x)$ if $f(x)/g(x) \rightarrow 1$ as $x \rightarrow \infty$. When we refer to a solution of the system under study, we mean an ordered s-tuple of prime numbers $(p_1, \ldots , p_s)$ satisfying (1) or (3), depending on context.

Define the generating function

$$\begin{aligned} f_i(\alpha _2, \alpha _3) = \sum _{p \le P} (\log p) e(\alpha _2 u_i p^2 + \alpha _3 v_i p^3). \end{aligned}$$

(4)

Let $\mathcal {A}$ be the unit square $(\mathbb {R}/\mathbb {Z})^2$ and let $S_0$ be the set of solutions of the system (1). Then

$$\begin{aligned}&\int _{\mathcal {A}} \prod _{i=1}^s f_i(\alpha _2, \alpha _3) d\alpha _2 \, d\alpha _3 \nonumber \\&\quad = \int _{\mathcal {A}} \sum _{p_1, \ldots , p_s \le P} \prod _{i=1}^s \left( (\log p_i) e(\alpha _2 u_i p_i^2 + \alpha _3 v_i p_i^3) \right) d\alpha _2 \, d\alpha _3\nonumber \\&\quad = \sum _{\{p_1, \ldots , p_s\} \in S_0} \prod _{i=1}^s (\log p_i) = R(P) \end{aligned}$$

(5)

by orthogonality. Thus $R(P)>0$ if and only if there is a solution to the system (1).

We divide $\mathcal {A}$ into major and minor arcs. For any T with $1 \le T \le P$. and for all $q < T$, $1 \le a_2 \le q$, $1 \le a_3 \le q$, $(a_2, a_3, q) = 1$, let a typical major arc $\mathfrak {M}(a_2,a_3,q;T)$ consist of all $(\alpha _2, \alpha _3)$ such that

$$\begin{aligned} |\alpha _2-a_2/q| \le \frac{T}{qP^2} \qquad \text {and} \qquad |\alpha _3-a_3/q| \le \frac{T}{qP^3}. \end{aligned}$$

Let the major arcs $\mathfrak {M}(T)$ be the union of all such $\mathfrak {M}(a_2,a_3,q)$, and let the minor arcs $\mathfrak {m}(T)$ be the complement of $\mathfrak {M}(T)$ in $\mathcal {A}$.

We will use two distinct dissections in our argument: the primary dissection into $\mathfrak {M} = \mathfrak {M}(Q)$ and $\mathfrak {m} = \mathfrak {m}(Q)$ with $Q = (\log P)^A$, where A is a positive constant whose value will be fixed later, and a secondary dissection $\mathfrak {M}(R)$, $\mathfrak {m}(R)$ with $R = P^{\frac{1}{2}+\delta }$ for some sufficiently small positive $\delta$.

3 Preliminary lemmas

We begin by defining the necessary generating functions. Let

$$\begin{aligned} f(\pmb \alpha )= & {} \sum _{P< p \le 2P} e(\alpha _2p^2 + \alpha _3p^3),\nonumber \\ g(\pmb \alpha )= & {} \sum _{P < n \le 2P} e(\alpha _2n^2 + \alpha _3n^3),\nonumber \\ S(q,\pmb {a})= & {} \sum _{n=1}^q e\left( \frac{a_2n^2 + a_3n^3}{q}\right) , \end{aligned}$$

(6)

$$\begin{aligned} W(q,\pmb {a})= & {} \sum _{\begin{array}{c} n=1 \\ (n,q)=1 \end{array}}^q e\left( \frac{a_2n^2 + a_3n^3}{q}\right) ,\nonumber \\ v(\pmb \theta )= & {} \int _P^{2P} e(\theta _2x^2 + \theta _3x^3) dx, \end{aligned}$$

(7)

and for $\pmb \gamma \in \mathfrak {M}(R)$ let

$$\begin{aligned} V(\pmb \gamma ) = \frac{1}{q} S(q,\pmb {a}) v\left( \gamma _2 - \frac{a_2}{q}, \gamma _3 - \frac{a_3}{q}\right) . \end{aligned}$$

Lemma 3.1

We have the bounds

$$\begin{aligned} \int _{\mathcal {A}} |g(\pmb \alpha )|^{10} d\pmb \alpha \ll P^{\frac{31}{6}+\varepsilon } \end{aligned}$$

and

$$\begin{aligned} \int _{\mathcal {A}} |g(\pmb \alpha )|^{12} d\pmb \alpha \ll P^7. \end{aligned}$$

Proof

This is the relevant portion of Theorem 1.3 of [12]. $\square$

Lemma 3.2

We have the bounds

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^{10} d\pmb \alpha \ll P^{\frac{31}{6}+\varepsilon } d\pmb \alpha \end{aligned}$$

and

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^{12} d\pmb \alpha \ll P^7 d\pmb \alpha . \end{aligned}$$

Proof

For any positive integer k,

$$\begin{aligned} \int _{\mathcal {A}} |g(\pmb \alpha )|^{2k} d\pmb \alpha \end{aligned}$$

is the number of positive integer solutions to the system

$$\begin{aligned} x_1^2 + \cdots + x_k^2= & {} x_{k+1}^2 + \cdots + x_{2k}^2, \\ x_1^3 + \cdots + x_k^3= & {} x_{k+1}^3 + \cdots + x_{2k}^3 \end{aligned}$$

and

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^{2k} d\pmb \alpha \end{aligned}$$

is the number of prime solutions to the same system, so this lemma follows from Lemma 3.1. $\square$

Lemma 3.3

$$\begin{aligned} \sup _{\pmb \alpha \in \mathfrak {m}(R)} |g(\pmb \alpha )| \ll P^{\frac{5}{6} - \frac{\delta }{3} + \varepsilon }. \end{aligned}$$

Proof

This follows from Lemma 5.2 of [12]. $\square$

Lemma 3.4

$$\begin{aligned} v(\pmb \theta ) \ll \frac{P}{(1 + P^3|\theta _3|)^{1/2}}. \end{aligned}$$

Proof

If $|\theta _3| \le P^{-3}$, the result is immediate. Thus we assume $|\theta _3| > P^{-3}$. Let $K = (|\theta _3| P)^{\frac{1}{2}}$ and let $r(x) = \theta _2x^2+\theta _3x^3$. Then $r'(x) = 2\theta _2x + 3\theta _3x^2$ has at most one zero in [P, 2P]. Thus we can divide [P, 2P] into subsets $I_1$ and $I_2$ such that $|r'(x)| \ge K$ on $I_1$, where $I_1$ is the union of at most three intervals such that $r'(x)$ is monotonic on each, and $|r'(x)| \le K$ on $I_2$, where $I_2$ is the union of at most two intervals.

First we consider $I_1$:

$$\begin{aligned} \int _{I_1} e(r(x)) dx = \int _{I_1} \frac{1}{2\pi i r'(x)} \frac{d}{dx}e(r(x)) dx, \end{aligned}$$

so, upon integrating by parts,

$$\begin{aligned} \int _{I_1} e(r(x)) dx = \frac{e(r(x))}{2\pi i r'(x)} \bigg |_{I_1} + \int _{I_1} \frac{r''(x)}{2\pi i r'(x)^2} e(r(x)) dx. \end{aligned}$$

The integral on the right is bounded by

$$\begin{aligned} \int _{I_1} \frac{|r''(x)|}{2\pi r'(x)^2} dx = \left| \int _{I_1} \frac{r''(x)}{2\pi r'(x)^2} dx \right| = \left| \frac{-1}{2\pi r'(x)} \bigg |_{I_1} \right| \ll \frac{1}{K}, \end{aligned}$$

since $r'(x)$ is monotonic on each interval in $I_1$. Thus

$$\begin{aligned} { \int _{I_1} e(r(x)) dx \ll \frac{e(r(x))}{2\pi i r'(x)} \bigg |_{I_1} + \frac{1}{K} \ll \frac{1}{K} \ll \frac{P}{(1+|\theta _3|P^3)^{1/2}}}. \end{aligned}$$

(8)

Next we consider $I_2$. Given an interval in $I_2$, let $x_0$ be one of its endpoints. Then for any x in $I_2$,

$$\begin{aligned} |x-x_0||2\theta _2 + 3\theta _3(x+x_0)| = |r'(x) - r'(x_0)| \le 2K. \end{aligned}$$

Moreover,

$$\begin{aligned} |2\theta _2+3\theta _3x_0| = \frac{|r'(x_0)|}{x_0} \le \frac{K}{x_0}. \end{aligned}$$

(9)

Applying the triangle identity to (9) yields

$$\begin{aligned} |3\theta _3x| - \frac{K}{x_0} \le |2\theta _2+3\theta _3(x+x_0)|. \end{aligned}$$

(10)

Also,

$$\begin{aligned} |3\theta _3x| - \frac{K}{x_0} \ge 3|\theta _3|P - \frac{K}{P} \ge 2|\theta _3|P. \end{aligned}$$

(11)

Combining (9), (10), and (11) yields

$$\begin{aligned} |x-x_0| \le \frac{2K}{2|\theta _3|P} = \frac{P}{(|\theta _3|P^3)^{1/2}}. \end{aligned}$$

Thus

$$\begin{aligned} \int _{I_2} e(r(x)) dx&\ll |e(r(x))| \left( {\text {meas}} (I_2)\right) \nonumber \\&\ll 2\max _{x\in I_2} |x-x_0| \nonumber \\&\ll \frac{P}{(1 + |\theta _3|P^3)^{1/2}}. \end{aligned}$$

(12)

Combining (8) and (12) now gives the desired result.

$\square$

Lemma 3.5

Let $t = 12-\delta$. Then

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^{t-1} d\pmb \alpha \ll P^{t-6+\frac{1+\delta }{12}+\varepsilon }. \end{aligned}$$

Proof

By Hölder’s inequality

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^{t-1} d\pmb \alpha \le \left( \int _{\mathcal {A}} |f(\pmb \alpha )|^{12} d\pmb \alpha \right) ^{\frac{t-11}{2}} \left( \int _{\mathcal {A}} |f(\pmb \alpha )|^{10} d\pmb \alpha \right) ^{\frac{13-t}{2}}. \end{aligned}$$

Applying Lemma 3.2 gives

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^{t-1} d\pmb \alpha \ll P^{\frac{7t-77}{2} + \frac{403-31t}{12} + \varepsilon } = P^{t-6+\frac{1+\delta }{12}+\varepsilon }. \end{aligned}$$

$\square$

Lemma 3.6

Let $R = P^{\frac{1}{2}+\delta }$ and let $\pmb \gamma \in \mathfrak {M}(R)$. Then

$$\begin{aligned} g(\pmb \gamma ) = V(\pmb \gamma ) + O\left( P^{\frac{5}{6} - \frac{\delta }{3}}\right) . \end{aligned}$$

This follows from Theorem 7.2 of [7].

Lemma 3.7

Let $\kappa (q)$ be the multiplicative function defined by

$$\begin{aligned} \kappa (p^j) = {\left\{ \begin{array}{ll} C p^{-1/2} &{} j=1, \\ C p^{-5/8} &{} j=2, \\ C p^{-j/4} &{} j>2. \end{array}\right. } \end{aligned}$$

Then there is a positive constant C such that

$$\begin{aligned} \max _{\begin{array}{c} \mathbf {a} \\ (q,a_2,a_3)=1 \end{array}}\frac{|S(q,\mathbf {a})|}{q} \le \kappa (q). \end{aligned}$$

Proof

The case $j=1$ follows from Theorem 2E of [6]. The cases with $j > 1$ follow from Theorem 7.1 of [7]. $\square$

Let

$$\begin{aligned} s_k(\mathbf {m}) = m_1^k + m_2^k + m_3^k - m_4^k - m_5^k - m_6^k. \end{aligned}$$

(13)

Lemma 3.8

Let $Q>0$ and let M(Q) be the number of solutions of the system

$$\begin{aligned}&s_2(\mathbf {m}) = 0, \\&s_1(\mathbf {m}) = 0 \end{aligned}$$

with all $m_j \le Q$. Then there is a positive constant C such that

$$\begin{aligned} M(Q) \sim C Q^3 \log Q. \end{aligned}$$

This is a result of Rogovskaya [5].

Lemma 3.9

If $(q, a_2, a_3)=1$, then

$$\begin{aligned} W(q, \mathbf {a}) \ll q^{\frac{1}{2}+\varepsilon }. \end{aligned}$$

In addition, if $(p,a_2,a_3)=1$, then

$$\begin{aligned} W(p, \mathbf {a}) \ll p^{\frac{1}{2}}. \end{aligned}$$

Proof

The case where $q = p$ follows from Theorem 2E of [6]. The case for general q follows from Lemma 8.5 of [4]. $\square$

4 Minor arc bounds

The primary purpose of this section is to prove the following theorem, which, together with the result of the next section, will provide the necessary minor arc bounds for our circle method approach.

Recall from the end of Sect. 2 that $\delta$ is a small positive number with $R = P^{\frac{1}{2}+\delta }$. Assume $\delta < 1$ and let $t = 12-\delta$.

Theorem 4.1

For $\delta$ sufficiently small,

$$\begin{aligned} \int _{\mathcal {A}} |f(\pmb \alpha )|^t d\pmb \alpha \ll P^{t-5}(\log P). \end{aligned}$$

Let

$$\begin{aligned} I_t(P) = \int _{\mathcal {A}} |f(\pmb \alpha )|^t d\pmb \alpha . \end{aligned}$$

Lemma 4.1

$$\begin{aligned} I_t(P)^2 \ll P^{2t-10} + P\mathop {\int _{\mathcal {A}} \int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |V(\pmb \alpha - \pmb \beta )||f(\pmb \alpha )|^{t-1}|f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta . \end{aligned}$$

Proof

$$\begin{aligned} I_t(P)= & {} \int _{\mathcal {A}} f(\pmb \alpha )f(-\pmb \alpha )|f(\pmb \alpha )|^{t-2} d\pmb \alpha \nonumber \\= & {} \sum _{P < p \le 2P}\int _{\mathcal {A}} e(\alpha _2p^2 + \alpha _3p^3) f(-\pmb \alpha )|f(\pmb \alpha )|^{t-2} d\pmb \alpha . \end{aligned}$$

(14)

Applying the Cauchy–Schwarz inequality to (14) yields

$$\begin{aligned} I_t(P)^2\ll & {} P\sum _{P < n \le 2P} \left| \int _{\mathcal {A}} e(\alpha _2n^2 + \alpha _3n^3) f(-\pmb \alpha )|f(\pmb \alpha )|^{t-2} d\pmb \alpha \right| ^2 \nonumber \\= & {} P \int _{\mathcal {A}}\int _{\mathcal {A}} g(\pmb \alpha -\pmb \beta ) f(-\pmb \alpha )|f(\pmb \alpha )|^{t-1} f(\pmb \beta )|f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta \nonumber \\\le & {} P \int _{\mathcal {A}}\int _{\mathcal {A}} |g(\pmb \alpha -\pmb \beta )| |f(\pmb \alpha )|^{t-1} |f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta . \end{aligned}$$

(15)

By Lemmas 3.3 and 3.5 and recalling that $t = 12-\delta$, we can bound the minor arc portion of (15):

$$\begin{aligned}&P \mathop {\int _{\mathcal {A}}\int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {m}(R)} |g(\pmb \alpha -\pmb \beta )| |f(\pmb \alpha )|^{t-1} |f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta \nonumber \\&\quad \ll P^{\frac{11}{6} - \frac{\delta }{3}+\varepsilon } \left( \int _{\mathcal {A}} |f(\pmb \alpha )|^{t-1}d\pmb \alpha \right) ^2 \nonumber \\&\quad \ll P^{2t-10-\frac{\delta }{6}+2\varepsilon } \ll P^{2t-10}. \end{aligned}$$

(16)

We now apply Lemma 3.6 to the major arc portion of (15).

$$\begin{aligned}&P \mathop {\int _{\mathcal {A}}\int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |g(\pmb \alpha -\pmb \beta )| |f(\pmb \alpha )|^{t-1} |f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta \nonumber \\&\quad = P \mathop {\int _{\mathcal {A}}\int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |V(\pmb \alpha -\pmb \beta )| |f(\pmb \alpha )|^{t-1} |f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta \nonumber \\&\qquad + O\left( P^{\frac{11}{6}-\frac{\delta }{3}} \left( \int _{\mathcal {A}} |f(\pmb \alpha )|^{t-1}d\pmb \alpha \right) ^2 \right) \nonumber \\&\quad = P \mathop {\int _{\mathcal {A}}\int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |V(\pmb \alpha -\pmb \beta )| |f(\pmb \alpha )|^{t-1} |f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta + O(P^{2t-10}). \end{aligned}$$

(17)

Combining (15), (16), and (17) yields the lemma. $\square$

Let $\pmb \gamma = \pmb \alpha - \pmb \beta$,

$$\begin{aligned} \lambda = \frac{t-6}{2} = 3 - \frac{\delta }{2} \end{aligned}$$

(18)

(note that $\lambda > 2$), and

$$\begin{aligned} J(\pmb \beta ) = \int _{\mathfrak {M}(R)} |V(\pmb \gamma )|^\lambda |f(\pmb \beta + \pmb \gamma )|^6 d\pmb \gamma . \end{aligned}$$

(19)

Lemma 4.2

$$\begin{aligned} I_t(P) \ll P^{t-5} + P^\lambda \sup _{\pmb \beta \in \mathcal {A}} J(\pmb \beta ). \end{aligned}$$

Proof

We begin by noting that

$$\begin{aligned} |V(\pmb \alpha -\pmb \beta )| |f(\pmb \alpha )|^{t-1} |f(\pmb \beta )|^{t-1} \end{aligned}$$

can be rewritten as

$$\begin{aligned} \begin{aligned}&\left( |V(\pmb \alpha -\pmb \beta )|^\lambda |f(\pmb \alpha )|^6 |f(\pmb \beta )|^t\right) ^{\frac{1}{2\lambda }} \\&\quad \times \left( |V(\pmb \alpha -\pmb \beta )|^\lambda |f(\pmb \beta )|^6 |f(\pmb \alpha )|^t\right) ^{\frac{1}{2\lambda }} \\&\quad \times \left( |f(\pmb \alpha )f(\pmb \beta )|^t\right) ^{1-\frac{1}{\lambda }}. \end{aligned} \end{aligned}$$

(20)

Let

$$\begin{aligned} I_t^*(P) = \mathop {\int _{\mathcal {A}} \int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |V(\pmb \alpha - \pmb \beta )||f(\pmb \alpha )|^{t-1}|f(\pmb \beta )|^{t-1} d\pmb \alpha \, d\pmb \beta \end{aligned}$$

be the integral on the right in Lemma 4.1. Using (20) to apply Hölder’s inequality to $I_t^*(P)$, we obtain

$$\begin{aligned} I_t^*(P)\ll & {} \left( \mathop {\int _{\mathcal {A}}\int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |V(\pmb \alpha -\pmb \beta )|^\lambda |f(\pmb \alpha )|^6 |f(\pmb \beta )|^t d\pmb \alpha \pmb \beta \right) ^\frac{1}{\lambda } \nonumber \\&\times \left( \mathop {\int _{\mathcal {A}}\int _{\mathcal {A}}}_{\pmb \alpha -\pmb \beta \in \mathfrak {M}(R)} |f(\pmb \alpha )f(\pmb \beta )|^t d\pmb \alpha \pmb \beta \right) ^{1-\frac{1}{\lambda }} \end{aligned}$$

(21)

$$\begin{aligned}\le & {} I_t(P)^{2-\frac{1}{\lambda }} \left( \sup _{\pmb \beta \in \mathcal {A}} \int _{\mathfrak {M}(R)} |V(\pmb \gamma )|^\lambda |f(\pmb \beta + \pmb \gamma )|^6 d\pmb \gamma \right) ^\frac{1}{\lambda }. \end{aligned}$$

(22)

Applying (22) to Lemma 4.1, we have

$$\begin{aligned} I_t(P)^2 \ll P^{2t-10} + PI_t(P)^{2-\frac{1}{\lambda }}\left( \sup _{\pmb \beta \in \mathcal {A}} J(\pmb \beta ) \right) ^{\frac{1}{\lambda }}. \end{aligned}$$

Thus either $I_t(P) \ll P^{t-5}$ or

$$\begin{aligned} I_t(P) \ll P^\lambda \sup _{\pmb \beta \in \mathcal {A}} J(\pmb \beta ), \end{aligned}$$

which implies the desired result. $\square$

Lemma 4.3

Let N(q) be the number of solutions of the system

$$\begin{aligned} {\left\{ \begin{array}{ll} s_3(\mathbf{p}) \equiv 0 \pmod q, \\ s_2(\mathbf{p}) =0, \\ P < p_j \le 2P. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} J(\pmb \beta ) \ll P^{\lambda -3}\sum _{q \le R} \kappa (q)^\lambda q N(q). \end{aligned}$$

Proof

By (19) and the definition of $\mathfrak {M}(R)$,

$$\begin{aligned} J(\pmb \beta ) = \sum _{q \le R} \mathop {\sum _{a_2=1}^q \sum _{a_3=1}^q}_{(q,a_2,a_3)=1} \frac{|S(q,\mathbf {a})|^\lambda }{q^\lambda } \int _{-\frac{R}{qP^2}}^{\frac{R}{qP^2}} \int _{-\frac{R}{qP^3}}^{\frac{R}{qP^3}} |v(\pmb \theta )|^\lambda \left| f\left( \pmb \beta + \frac{\mathbf {a}}{q} + \pmb \theta \right) \right| ^6 d\pmb \theta . \end{aligned}$$

By Lemmas 3.4, 3.7, and the fact that for a given q, the intervals $\left[ \frac{a_2}{q} - \frac{R}{qP^2}, \frac{a_2}{q} + \frac{R}{qP^2}\right]$ are disjoint for distinct $a_2$,

$$\begin{aligned} J(\pmb \beta ) \le \sum _{q \le R} \int _{-\frac{R}{qP^3}}^{\frac{R}{qP^3}} \frac{\kappa (q)^\lambda P^\lambda }{(1 + P^3|\theta _3|)^{\lambda /2}} \sum _{a_3=1}^q \int _0^1 \left| f\left( \beta _2 + \phi , \beta _3 + \frac{a_3}{q} + \theta _3\right) \right| ^6 d\phi \, d\theta _3. \end{aligned}$$

(23)

We now examine the inner sum and integral.

$$\begin{aligned}&\sum _{a_3=1}^q \int _0^1 \left| f\left( \beta _2 + \phi , \beta _3 + \frac{a_3}{q} + \theta _3\right) \right| ^6 d\phi \\&\quad = \sum _{a_3=1}^q \int _0^1 \sum _{\begin{array}{c} \mathbf{p} \\ P< p_j \le 2P \end{array}} e\left( (\beta _2+\phi )s_2(\mathbf {p}) + \left( \beta _3 + \frac{a_3}{q} + \theta _3\right) s_3(\mathbf {p})\right) d\phi \\&\quad = \sum _{\begin{array}{c} \mathbf{p} \\ P < p_j \le 2P \end{array}} e(\beta _2s_2(\mathbf {p}) + (\beta _3+\theta _3)s_3(\mathbf {p})) \sum _{a_3=1}^q e\left( \frac{a_3}{q}s_3(\mathbf {p})\right) \int _0^1 e(\phi s_2(\mathbf {p}))d\phi . \end{aligned}$$

Now

$$\begin{aligned} \sum _{a_3=1}^q e\left( \frac{a_3}{q}s_3(\mathbf {p})\right) = {\left\{ \begin{array}{ll} 0 &{} s_3(\mathbf {p}) \not \equiv 0 \pmod q, \\ q &{} s_3(\mathbf {p}) \equiv 0 \pmod q \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \int _0^1 e(\phi s_2(\mathbf {p}))d\phi = {\left\{ \begin{array}{ll} 0 &{} s_2(\mathbf {p}) \ne 0, \\ 1 &{} s_2(\mathbf {p}) = 0, \end{array}\right. } \end{aligned}$$

so

$$\begin{aligned} \sum _{a_3=1}^q \int _0^1 \left| f\left( \beta _2 + \phi , \beta _3 + \frac{a_3}{q} + \theta _3\right) \right| ^6 d\phi \ll qN(q). \end{aligned}$$

(24)

Substituting (24) into (23) yields

$$\begin{aligned} J(\pmb \beta ) \ll \sum _{q \le R} \kappa (q)^\lambda qN(q) \int _{-\frac{R}{qP^3}}^{\frac{R}{qP^3}} \frac{P^\lambda }{(1 + P^3|\theta _3|)^{\lambda /2}} d\theta _3. \end{aligned}$$

Since $\lambda > 2$, this becomes

$$\begin{aligned} J(\pmb \beta ) \ll P^{\lambda -3}\sum _{q \le R} \kappa (q)^\lambda qN(q). \end{aligned}$$

$\square$

Lemma 4.4

Let $N_1(q)$ be the number of solutions to the system

$$\begin{aligned} {\left\{ \begin{array}{ll} s_3(\mathbf{p}) \equiv 0 \pmod q, \\ s_2(\mathbf{p}) = 0, \\ P < p_j \le 2P \quad p_j \not \mid q. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} qN(q) \ll q(\log q)^6 + qN_1(q). \end{aligned}$$

Proof

First, note that

$$\begin{aligned} qN(q) = \sum _{a_3=1}^q \int _0^1 \left| f\left( x, \frac{a_3}{q}\right) \right| ^6 dx. \end{aligned}$$

Let

$$\begin{aligned} f_|(\pmb \alpha ) = \sum _{\begin{array}{c} P \le p < 2P \\ p|q \end{array}} e(\alpha _2p^2 + \alpha _3p^3) \end{aligned}$$

and

$$\begin{aligned} f_{\not \mid }(\pmb \alpha ) = \sum _{\begin{array}{c} P \le p < 2P \\ p\not \mid q \end{array}} e(\alpha _2p^2 + \alpha _3p^3). \end{aligned}$$

Thus

$$\begin{aligned} f(\pmb \alpha ) = f_|(\pmb \alpha ) + f_{\not \mid }(\pmb \alpha ). \end{aligned}$$

Since $|f_|(\pmb \alpha )| \ll \log q$,

$$\begin{aligned} |f(\pmb \alpha )|^6 \ll (\log q)^6 + |f_{\not \mid }(\pmb \alpha )|^6. \end{aligned}$$

Now

$$\begin{aligned} \sum _{a_3=1}^q\int _0^1 \left| f_{\not \mid }\left( x, \frac{a_3}{q}\right) \right| ^6 dx = qN_1(q), \end{aligned}$$

so

$$\begin{aligned} qN(q) \ll q(\log q)^6 + qN_1(q). \end{aligned}$$

$\square$

Lemma 4.5

Let $N_2(q)$ be the number of solutions of the system

$$\begin{aligned} {\left\{ \begin{array}{ll} s_3(\mathbf{r}) \equiv 0 \pmod q, \\ s_2(q\mathbf{m} + \mathbf{r}) =0, \\ 1 \le r_j \le q (q,r_j)=1, \\ \frac{P-r_j}{q} < m_j \le \frac{2P - r_j}{q}. \end{array}\right. } \end{aligned}$$

(25)

Then

$$\begin{aligned} N_1(q) \le N_2(q). \end{aligned}$$

Proof

We classify the solutions $\mathbf {p}$ counted by $N_1(q)$ according to the residue class $r_j$ of each $p_j$ modulo q, and let $m_j = \frac{p_j-r_j}{q}$. Thus

$$\begin{aligned} 0 = s_2(q\mathbf {m} + \mathbf {r}) \equiv s_2(\mathbf {r}) \pmod q, \end{aligned}$$

so $N_1(q) \le N_2(q)$. $\square$

Lemma 4.6

Let $N_3(q)$ be the number of solutions of the system

$$\begin{aligned} {\left\{ \begin{array}{ll} s_3(\mathbf {r}) \equiv 0 \pmod q, \\ s_2(\mathbf {r}) \equiv 0 \pmod q, \\ 1 \le r_j \le q \quad (q,r_j)=1. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} N_2(q) \ll N_3(q)P^4q^{-5}(\log P)\left( \frac{q^2}{P} + 1\right) . \end{aligned}$$

Proof

Let $\mathbf {r}$, $\mathbf {m}$ be a solution counted in $N_2(q)$, i.e., let $\mathbf {r}$, $\mathbf {m}$ satisfy (25). Expanding the third equation of (25) gives

$$\begin{aligned} q^2s_2(\mathbf {m}) + 2q(r_1m_1 + r_2m_2 + r_3m_3 - r_4m_4 - r_5m_5 - r_6m_6) + s_2(\mathbf {r}) = 0. \end{aligned}$$

Since $s_2(\mathbf {r}) \equiv 0 \pmod q$ by the second equation of (25), this can be rewritten as

$$\begin{aligned} qs_2(\mathbf {m}) + 2(r_1m_1 + r_2m_2 + r_3m_3 - r_4m_4 - r_5m_5 - r_6m_6) + \frac{s_2(\mathbf {r})}{q} = 0 \end{aligned}$$

with each term remaining integer-valued. For a fixed $\mathbf {r}$, define

$$\begin{aligned} H_j(\alpha ) = \sum _{\frac{P-r_j}{q} < m \le \frac{2P - r_j}{q}} e\left( \alpha (qm^2 + 2r_jm)\right) . \end{aligned}$$

(26)

Thus the number of $\mathbf {m}$ satisfying (25) for a given $\mathbf {r}$ is

$$\begin{aligned} \int _0^1 H_1(\alpha )H_2(\alpha )H_3(\alpha )H_4(-\alpha )H_5(-\alpha )H_6(-\alpha ) e\left( \frac{s_2(\mathbf {r})}{q} \alpha \right) d\alpha . \end{aligned}$$

By Hölder’s inequality this is

$$\begin{aligned} \le \prod _{j=1}^6 \left( \int _0^1 |H_j(\alpha )|^6 d\alpha \right) ^{\frac{1}{6}}. \end{aligned}$$

The integral

$$\begin{aligned} \int _0^1 |H_j(\alpha )|^6 d\alpha \end{aligned}$$

counts the number of solutions of

$$\begin{aligned} qs_2(\mathbf {m}) + 2r_j s_1(\mathbf {m}) = 0. \end{aligned}$$

(27)

Let $s_2(\mathbf {m}) = u$ and $s_1(\mathbf {m}) = v$. Then (27) becomes $qu + 2r_j v = 0$. For any solution, we have $|v| \le \frac{6P}{q}$, and since $(q,r_j)=1$, $v = \frac{v'q}{(q,2)}$. Thus the number of choices for $v'$ is $\le 1 + 24P/q^2$, and u is determined by $v'$.

Let

$$\begin{aligned} h(\pmb \alpha ) = \sum _{\frac{P-r_j}{q} < m \le \frac{2P - r_j}{q}} e(\alpha _1 m + \alpha _2 m^2). \end{aligned}$$

For fixed pair u, v, the number of choices of $\mathbf {m}$ is

$$\begin{aligned} \int _{\mathcal {A}} |h(\pmb \alpha )|^6 e(-\alpha _1v - \alpha _2u) d\pmb \alpha \le \int _{\mathcal {A}} |h(\pmb \alpha )|^6 d\pmb \alpha . \end{aligned}$$

But this is the number of solutions of the system

$$\begin{aligned} \begin{aligned}&s_2(\mathbf {m}) = 0, \\&s_1(\mathbf {m}) = 0, \end{aligned} \end{aligned}$$

so by Lemma 3.8,

$$\begin{aligned} \int _{\mathcal {A}} |h(\pmb \alpha )|^6 d\pmb \alpha \ll \left( \frac{P}{q}\right) ^3 \log P. \end{aligned}$$

So, given $\mathbf {r}$ satisfying the first two equations of (25) and $(q,r_j)=1$, the number of solutions to the third equation of (25) is

$$\begin{aligned} \ll \left( 1 + \frac{P}{q^2}\right) \frac{P^3}{q^3} \log P = P^4q^{-5}\left( 1 + \frac{q^2}{P}\right) \log P. \end{aligned}$$

Thus

$$\begin{aligned} N_2(q) \ll N_3(q)P^4q^{-5}(\log P)\left( 1 + \frac{q^2}{P}\right) . \end{aligned}$$

$\square$

Lemma 4.7

Let $N_3(q)$ be as defined in Lemma 4.6 above. Then there exists a positive constant C such that

$$\begin{aligned} N_3(q) \ll q^4 \prod _{p | q}\left( 1 + \frac{C}{p}\right) . \end{aligned}$$

Proof

We begin by observing that $N_3(q)$ is a multiplicative function, and that by orthogonality,

$$\begin{aligned} N_3(p^k) = p^{-2k} \sum _{b_2=1}^{p^k} \sum _{b_3=1}^{p^k} |W(p^k, b_2, b_3)|^6. \end{aligned}$$

Sorting the terms of this sum by the value of $(p^k, b_2, b_3) = p^{k-j}$, where $0 \le j \le k$, gives

$$\begin{aligned} N_3(p^k) = p^{-2k} \sum _{j=0}^k \mathop {\sum _{a_2=1}^{p^j} \sum _{a_3=1}^{p^j}}_{(p^j,a_2,a_3)=1} |W(p^k, p^{k-j}a_2, p^{k-j}a_3)|^6. \end{aligned}$$

If $j=0$, then

$$\begin{aligned} W(p^k, p^{k-j}a_2, p^{k-j}a_3) = \phi (p^k) = p^k(1-1/p) \end{aligned}$$

and if $j > 0$, then

$$\begin{aligned} W(p^k, p^{k-j}a_2, p^{k-j}a_3) = p^{k-j}W(p^j,a_2,a_3). \end{aligned}$$

Thus

$$\begin{aligned} N_3(p^k) = p^{4k}(1-1/p)^6 + p^{4k} \sum _{j=1}^k \mathop {\sum _{a_2=1}^{p^j} \sum _{a_3=1}^{p^j}}_{(p^j,a_2,a_3)=1} p^{-6j} |W(p^j,a_2,a_3)|^6. \end{aligned}$$

By Lemma 3.9,

$$\begin{aligned} \sum _{\begin{array}{c} \mathbf {a} \\ (p,a_2,a_3)=1 \end{array}} p^{-6} |W(p,a_2,a_3)|^6 \ll p^{-1}, \end{aligned}$$

and for $j \ge 2$,

$$\begin{aligned} \sum _{\begin{array}{c} \mathbf {a} \\ (p^j,a_2,a_3)=1 \end{array}} p^{-6j} |W(p^j,a_2,a_3)|^6 \ll p^{-4j+6j/2+j\varepsilon } \ll p^{-j + j\varepsilon }. \end{aligned}$$

Thus

$$\begin{aligned} N_3(p^k) \le p^{4k}\left( 1 + \frac{C}{p}\right) \end{aligned}$$

and the lemma follows by multiplicativity. $\square$

Proof of Theorem 4.1

By Lemma 4.2,

$$\begin{aligned} I_t(P) \ll P^{t-5} + P^\lambda \sup _{\pmb \beta \in \mathcal {A}} J(\pmb \beta ). \end{aligned}$$

Bounding $J(\pmb \beta )$ with Lemma 4.3 yields

$$\begin{aligned} I_t(P) \ll P^{t-5} + P^{2\lambda -3}\sum _{q \le R}\kappa (q)^\lambda q N(q). \end{aligned}$$

(28)

Lemmas 4.4, 4.5, and 4.6 successively bound N(q) in terms of $N_1(q)$, then $N_2(q)$, then $N_3(q)$, and Lemma 4.7 bounds $N_3(q)$. Collecting these bounds and applying them to (28) gives

$$\begin{aligned} I_t(P)\ll & {} P^{t-5} + P^{2\lambda + 1}(\log P)\sum _{q\le R}\kappa (q)^\lambda \left( P^{-4}q(\log q)^6 \right. \\&\left. +\left( \frac{q^2}{P}+1\right) \prod _{p|q}\left( 1 + \frac{C}{p}\right) \right) . \end{aligned}$$

Since $q \le R = P^{\frac{1}{2}+\delta }$,

$$\begin{aligned} P^{-4}q(\log q)^6 \ll P^{-3} \ll 1, \end{aligned}$$

and

$$\begin{aligned} \frac{q^2}{P} \le q^{\frac{4\delta }{1+2\delta }}, \end{aligned}$$

so we have

$$\begin{aligned} I_t(P) \ll P^{t-5} + P^{2\lambda + 1}(\log P)\sum _{q\le R}\kappa (q)^\lambda q^{\frac{4\delta }{1+2\delta }} \prod _{p|q}\left( 1 + \frac{C}{p}\right) . \end{aligned}$$

(29)

We now desire a bound on

$$\begin{aligned} \sum _{q\le R}\kappa (q)^\lambda q^{\frac{4\delta }{1+2\delta }} \prod _{p|q}\left( 1 + \frac{C}{p}\right) . \end{aligned}$$

Since $\kappa$ is multiplicative, it suffices to bound

$$\begin{aligned} \prod _{p\le R}\left( 1 + \sum _{j=1}^{\infty }\kappa (p^j)^\lambda p^{j\frac{4\delta }{1+2\delta }}\right) . \end{aligned}$$

We have

$$\begin{aligned} \sum _{j=1}^{\infty }\kappa (p^j)^\lambda p^{j\frac{4\delta }{1+2\delta }}\ll & {} p^{-5/4} + p^{-3/2} + \sum _{j=3}^{\infty } p^{-\frac{2}{3}j} \\\ll & {} p^{-5/4}. \end{aligned}$$

Thus

$$\begin{aligned} \prod _{p\le R}\left( 1 + \left( 1+\frac{C}{p}\right) \sum _{j=1}^\infty \kappa (p^j)^\lambda p^{j\frac{4\delta }{1+2\delta }}\right) \ll \prod _{p\le R}(1 + Cp^{-5/4}) \ll 1, \end{aligned}$$

which implies that

$$\begin{aligned} \sum _{q\le R}\kappa (q)^\lambda q^{\frac{4\delta }{1+2\delta }} \prod _{p|q}\left( 1 + \frac{C}{p}\right) \ll 1. \end{aligned}$$

(30)

Applying (30) to (29) yields

$$\begin{aligned} I_t(P) \ll P^{t-5} + P^{2\lambda +1}(\log P), \end{aligned}$$

which, upon applying the definition of $\lambda$ in (18), is

$$\begin{aligned} I_t(P) \ll P^{t-5}(\log P). \end{aligned}$$

$\square$

5 A pointwise minor arc bound sensitive to multiple coefficients

In this section, we will work with a narrower set of minor arcs $\mathfrak {m}(Q)$, where $Q = (\log P)^A$. Henceforth $\mathfrak {m}$ will be assumed to mean $\mathfrak {m}(Q)$ rather than $\mathfrak {m}(R)$ unless otherwise specified. Let $\pmb \alpha = (\alpha _1, \ldots , \alpha _k)$ and let

$$\begin{aligned} F_k(\pmb \alpha ) = \sum _{n \le P} \Lambda (n) e(\alpha _1 n + \alpha _2 n^2 + \cdots + \alpha _k n^k). \end{aligned}$$

This section consists of the proof of the following theorem and corollary:

Theorem 5.1

For $D > 0$, where $D = D(A)$ can be made arbitrarily large by increasing A, if $(\alpha _2, \alpha _3) \in \mathfrak {m}(Q)$, then

$$\begin{aligned} \sup _{\pmb \alpha \in \mathfrak {m}(Q)}F_3(\pmb \alpha ) \ll P(\log P)^{-D}. \end{aligned}$$

Corollary 5.1

For each i, $1\le i \le s$,

$$\begin{aligned} \sup _{(\alpha _2, \alpha _3)\in \mathfrak {m}(Q)} f_i(\alpha _2, \alpha _3) \ll P(\log P)^{-D}. \end{aligned}$$

Proof

Take $(0 ,u_i\alpha _2, v_i\alpha _3)$ as the argument of $F_3$ in Theorem 5.1 and sum over the dyadic intervals, noting that multiplying $\alpha _2$ and $\alpha _3$ by the integer coefficients $u_i$ and $v_i$ does not move them out of $\mathfrak {m}(Q)$, and that there are trivially $\ll P^{1/2}\log P$ prime powers $\le P$ which contribute $\ll P^{1/2}(\log P)^2$ to the sum. $\square$

We begin by citing some known results on Vinogradov’s mean value theorem. Let

$$\begin{aligned} J_{s,k}(P) = \int _{[0,1)^k} |F_k(\pmb \alpha )|^{2s} d\pmb \alpha . \end{aligned}$$

We cite the bound

$$\begin{aligned} J_{3,2}(P) \ll P^3 \log P \end{aligned}$$

(31)

from [5] (cf. [7] chap. 7 exercise 2) and for $s > 6$

$$\begin{aligned} J_{s,3}(P) \ll P^{2s-6} \end{aligned}$$

(32)

from equation (7) of [1].

Let $X = (\log P)^B$ for some $B > 0$ to be fixed later. For brevity, we let $h(n) := e(\alpha _1n + \alpha _2n^2 + \alpha _3n^3)$. Then

$$\begin{aligned} F_3(\pmb \alpha ) = \sum _{n \le P} \Lambda (n)h(n). \end{aligned}$$

Applying Vaughan’s identity [8] to this sum yields

$$\begin{aligned} \sum _{n \le P} \Lambda (n)h(n) = S_1 + S_2 + S_3 + S_4, \end{aligned}$$

(33)

where

$$\begin{aligned} S_1= & {} \sum _{n \le X} \Lambda (n)h(n), \\ S_2= & {} \sum _{n \le P} \left( \sum _{\begin{array}{c} kl=n \\ k \le X \end{array}} \mu (k) \log l \right) h(n), \\ S_3= & {} \sum _{n \le P} \sum _{\begin{array}{c} kl=n \\ k \le X^2 \end{array}} \left( \sum _{\begin{array}{c} m,n \\ mn=k \\ m \le X, n \le X \end{array}} \Lambda (m) \mu (n) \right) h(n), \\ S_4= & {} \sum _{n \le P} \left( \sum _{\begin{array}{c} kl=n \\ k>X, l>X \end{array}} a(k)b(l) \right) h(n), \end{aligned}$$

with

$$\begin{aligned} a(k)= & {} \sum _{\begin{array}{c} l | k \\ l>X \end{array}} \Lambda (l), \\ b(l)= & {} {\left\{ \begin{array}{ll} \mu (l), &{} l>X \\ 0, &{} l \le X. \end{array}\right. } \end{aligned}$$

This now enables us to bound each of the sums $S_1$, $S_2$, $S_3$, $S_4$ individually to obtain the desired bound on $F_3(\pmb \alpha )$. The bounds on these four sums constitute Lemmas 5.1–5.4.

Lemma 5.1

$$\begin{aligned} S_1 \ll X. \end{aligned}$$

(34)

Proof

Since $|h(n)| \ll 1$,

$$\begin{aligned} S_1 = \sum _{n \le X} \Lambda (n)h(n) \ll \sum _{m \le X} \Lambda (n) \ll X, \end{aligned}$$

where the last bound is a classical result of Chebyshev. $\square$

Lemma 5.2

$$\begin{aligned} S_3 \ll P(\log P)^{B - A/12 + 4}. \end{aligned}$$

Proof

$$\begin{aligned} S_3 = \sum _{n \le P} \sum _{\begin{array}{c} kl=n \\ k \le X^2 \end{array}} \left( \sum _{\begin{array}{c} m_1,m_2 \\ m_1m_2=k \\ m_1 \le X, m_2 \le X \end{array}} \Lambda (m_1) \mu (m_2) \right) h(m_2). \end{aligned}$$

(35)

Let

$$\begin{aligned} c_3(k) := \sum _{\begin{array}{c} m_1,m_2 \\ m_1m_2=k \\ m_1 \le X, m_2 \le X \end{array}} \Lambda (m_1) \mu (m_2) \end{aligned}$$

and note for future reference that

$$\begin{aligned} |c_3(k)| \le \sum _{m|k}\Lambda (m) = \log k. \end{aligned}$$

Interchanging the order of summation in (35) yields

$$\begin{aligned} S_3= & {} \sum _{k \le X^2} c_3(k) \sum _{l \le P/k} h(kl) \nonumber \\= & {} \sum _{k \le X^2} c_3(k) \sum _{l \le P/k} e(\alpha _1 kl + \alpha _2 k^2 l^2 + \alpha _3 k^3 l^3). \end{aligned}$$

(36)

We now use Dirichlet’s theorem on Diophantine approximation to obtain integers $b_j$, $q_j$ for $j \in \{2,3\}$ such that $(b_j, q_j)=1$,

$$\begin{aligned} \left| \alpha _j k^j - \frac{b_j}{q_j}\right|\le & {} \frac{(\log (P/k))^{A/2}}{q_j(P/k)^j}, \nonumber \\ q_j\le & {} \frac{(P/k)^j}{(\log (P/k))^{A/2}}. \end{aligned}$$

(37)

Assume for contradiction that $q_j \le (\log (P/k))^{A/2}$ for both $j=2$ and $j=3$ and rewrite (37) as

$$\begin{aligned} \left| \alpha _j - \frac{b_j}{k^j q_j}\right| \le \frac{(\log (P/k))^{A/2}}{q_j P^j}. \end{aligned}$$

Let $b_j' = b_j/(k^j,b_j)$, $q_j' = k^j q_j/(k^j,b_j)$. Then

$$\begin{aligned} \left| \alpha _j - \frac{b_j}{k^j q_j}\right| \le \frac{(\log (P/k))^{A/2}}{q_j' P^j}, \end{aligned}$$

$(b_j',q_j')=1$, and $q_j' \le (\log (P/k))^{A/2}$ for $j \in \{2,3\}$. Let $q = \text {lcm}(q_2',q_3')$ and $a_j = b_j'q/q_j$. Then $(a_2,a_3,q)=1$, $q \le (\log (P/k))^A$, and

$$\begin{aligned} \left| \alpha _j - \frac{a_j}{q}\right| \le \frac{(\log (P/k))^A}{qP^j}. \end{aligned}$$

This implies that $(\alpha _2, \alpha _3) \in \mathfrak {M}(Q)$. However, we have $(\alpha _2, \alpha _3) \in \mathfrak {m}(Q)$, which is the desired contradiction, so we may assume that $q_j > (\log (P/k))^{A/2}$ for at least one $j_0 \in \{2,3\}$.

Define $\alpha _j' = \alpha _j k^j$ and note that (37) becomes

$$\begin{aligned} \left| \alpha _j k^j - \frac{b_j}{q_j}\right| \le \frac{(\log (P/k))^{A/2}}{q_j(P/k)^j}. \end{aligned}$$

(38)

We now need a bound on

$$\begin{aligned} H(\pmb \alpha ', P/k) := \sum _{l \le P/k} e(\alpha _1' l + \alpha _2'l^2 + \alpha _3'l^3). \end{aligned}$$

By Theorem 5.2 of [7], using the Diophantine approximation of (38), we have

$$\begin{aligned} H(\pmb \alpha ', P/k) \ll (\log P) \left( J_{3,2}(2P/k) \left( \frac{P}{k}\right) ^3 \left( \frac{1}{q_{j_0}'} + \frac{k}{P} + \frac{q_{j_0}'k^{j_0}}{P^{j_0}} \right) \right) ^{1/6}. \end{aligned}$$

Now by (31), we have $J_{3,2}(P) \ll P^3(\log P)$, so

$$\begin{aligned} H(\pmb \alpha ', P/k) \ll \frac{P}{k} (\log P)^2 \prod _{j=1}^3 \left( \frac{1}{q_j'} + \frac{k}{P} + \frac{q_j'k^j}{P^j}\right) ^{1/6}. \end{aligned}$$

(39)

Now $\frac{k}{P} \ll P^{-1/2}$, and $\frac{q_j'k^j}{P^j} \ll (\log P)^{2jB-A}$, $1/q_{j_0}' \ll (\log P)^{2Bj_0 - A/2}$, so

$$\begin{aligned} \frac{1}{q_{j_0}'} + \frac{k}{P} + \frac{q_{j_0}'k^{j_0}}{P^{j_0}} \ll (\log P)^{2Bj_0 - A/2}, \end{aligned}$$

(40)

assuming $2Bj_0 - A/2 < 0$.

Applying the bound of (40) to (39) yields

$$\begin{aligned} H(\pmb \alpha ', P/k)\ll & {} \frac{P}{k} (\log P)^2 (\log P)^{(2Bj_0 - A/2)/12} \nonumber \\\ll & {} \frac{P}{k} (\log P)^{B - A/12 + 2}, \end{aligned}$$

(41)

since $j_0 \le 3$.

Substituting the bound of (41) into (36), we obtain

$$\begin{aligned} S_3\ll & {} \sum _{k \le X^2} (\log k) \frac{P}{k} (\log P)^{B - A/12 + 2} \\\ll & {} P(\log P)^{B - A/12 + 4}. \end{aligned}$$

$\square$

Lemma 5.3

$$\begin{aligned} S_2 \ll P(\log P)^{B-A/12+4}. \end{aligned}$$

Proof

$$\begin{aligned} S_2= & {} \sum _{n \le P} \left( \sum _{\begin{array}{c} kl=n \\ k \le X \end{array}} \mu (k) \log l \right) h(n) \nonumber \\= & {} \sum _{k \le X} \mu (k) \sum _{l< P/k} h(kl) \int _1^l \frac{dt}{t}\nonumber \\= & {} \sum _{k \le X} \mu (k) \int _1^{P/k} \sum _{l< P/k} h(kl) \frac{dt}{t}\nonumber \\= & {} \int _1^{P/k} \left( \sum _{k \le X} \mu (k) \sum _{l < P/k} h(kl) \right) \frac{dt}{t}. \end{aligned}$$

(42)

Now by (41),

$$\begin{aligned} \sum _{l < P/k} h(kl) = H(\pmb \alpha ', P/k) \ll \frac{P}{k} (\log P)^{B - A/12 + 2}. \end{aligned}$$

Substituting this into (42) yields

$$\begin{aligned} S_2\ll & {} \int _1^{P/k} \sum _{k \le X} \frac{P}{k} (\log P)^{B - A/12 + 2} \frac{dt}{t} \\\ll & {} P(\log P)^{B - A/12 + 2} \left( \sum _{k \le X} \frac{\mu (k)}{k}\right) \int _1^{P/k} \frac{dt}{t} \\\ll & {} P(\log P)^{2 + B - A/12} (\log X) (\log P/k) \\\ll & {} P(\log P)^{4 + B - A/12}. \end{aligned}$$

$\square$

Lemma 5.4

$$\begin{aligned} S_4 \ll P(\log P)^{4-\min \{A,B\}/(4b^2)} \end{aligned}$$

Proof

We begin by splitting $S_4$ into dyadic ranges. Let $\mathcal {M} = \{X2^m : 0 \le k, 2^m \le P/X^2\}$. Then

$$\begin{aligned} S_4 = \sum _{M \in \mathcal {M}} S_4(M), \end{aligned}$$

(43)

where

$$\begin{aligned} S_4(M) = \sum _{M < k \le 2M} \sum _{l \le P/k} a(k)b(l)h(kl). \end{aligned}$$

Our goal is now to replace the sum over the range $l \le P/k$ with one over the range $l \le P/M$. We begin by considering the integral

$$\begin{aligned} I(x) := \int _\mathbb {R} \frac{\sin (2\pi Rt)}{\pi t}e(-xt)dt, \end{aligned}$$

where $R>0$ is a constant. Computing the integral via the residue theorem gives

$$\begin{aligned} I(x) = {\left\{ \begin{array}{ll} 1, &{} |x|<R, \\ 0, &{} |x|>R.\end{array}\right. } \end{aligned}$$

Now for $x \ne R$, $t \ge 1$,

$$\begin{aligned} \int _{|t|>T} \frac{\sin (2\pi Rt)}{\pi t}e(-xt)dt = \int _{|t|>T} \frac{e\big ((R-x)t\big )-e\big (-(R+x)t\big )}{2\pi it} dt. \end{aligned}$$

(44)

Integrating the right-hand side of (44) by parts gives

$$\begin{aligned} \int _{|t|>T} \frac{\sin (2\pi Rt)}{\pi t}e(-xt)dt \ll \frac{1}{T|R-x|} + \frac{1}{T|R+x|} + \frac{1}{T^3} \ll \frac{1}{T\big |R-|x|\big |}. \end{aligned}$$

Thus we can rewrite I(x) as an integral over $[-T,T]$ with an acceptable error term:

$$\begin{aligned} I(x) = \int _{-T}^T \frac{\sin (2\pi Rt)}{\pi t}e(-xt)dt + O\left( \frac{1}{T\big |R-|x|\big |}\right) . \end{aligned}$$

We now take $R = \log (\lfloor P\rfloor + \frac{1}{2})$, $x = \log (kl)$, giving us

$$\begin{aligned} S_4(M)= & {} \sum _{M< k \le 2M} \sum _{l \le P/M} a(k)b(l)h(kl)I(\log (kl)) \\= & {} \int _{-T}^T \sum _{M < k \le 2M} \sum _{l \le P/M} \frac{a(k)b(l)}{(kl)^{2\pi it}} h(kl) \frac{\sin (2\pi Rt)}{\pi t} dt + O\left( \frac{P^2\log P}{T}\right) . \end{aligned}$$

Now

$$\begin{aligned} \frac{\sin (2\pi Rt)}{\pi t} \ll \frac{1}{\pi t} \ll \frac{1}{|t|} \end{aligned}$$

and

$$\begin{aligned} \frac{\sin (2\pi Rt)}{\pi t} \ll \frac{2\pi Rt}{\pi t} \ll R, \end{aligned}$$

so

$$\begin{aligned} \frac{\sin (2\pi Rt)}{\pi t} \ll \min (R, 1/|t|). \end{aligned}$$

Take $T = P^3$, $a(k,t) = a(k)k^{-2\pi it}$, $b(l,t) = b(l)l^{-2\pi it}$, and let

$$\begin{aligned} S_4(M, t) = \sum _{M < k \le 2M} \sum _{l \le P/k} a(k, t)b(l, t)h(kl). \end{aligned}$$

(45)

Then

$$\begin{aligned} S_4(M)\ll & {} \sup _{|t|<T}|S_4(M,t)| \int _{-T}^T \frac{\sin (2\pi Rt)}{\pi t} dt \\\ll & {} 1 + (\log P)\sup _{|t|<T}|S_4(M,t)|. \end{aligned}$$

We now consider $S_4(M,t)$. Let $b>6$. By Hölder’s inequality

$$\begin{aligned} S_4(M,t)^{2b} \ll \left( \sum _{M< k \le 2M} |a(k,t)|^\frac{2b}{2b-1} \right) ^{2b-1} \sum _{M < k \le 2M} \left| \sum _{l \le P/M} b(l,t)h(kl) \right| ^{2b}. \end{aligned}$$

(46)

Now $|a(k,t)| = |a(k)| \le \log k \ll \log M \ll \log P$, so

$$\begin{aligned} S_4(M,t)^{2b}\ll & {} \left( M(\log P)^\frac{2b}{2b-1}\right) ^{2b-1} \sum _{M< k \le 2M} \left| \sum _{l \le P/M} b(l,t)h(kl) \right| ^{2b} \nonumber \\\ll & {} (\log P)^{2b} M^{2b-1} \sum _{M < k \le 2M} \left| \sum _{l \le P/M} b(l,t)h(kl) \right| ^{2b}. \end{aligned}$$

(47)

Expanding the 2b-th power in (47) yields

$$\begin{aligned}&\left| \sum _{l \le P/M} b(l,t)h(kl) \right| ^{2b} \nonumber \\&\quad = \sum _{\begin{array}{c} \mathbf{l} \\ l_j \le P/M \end{array}} \left( \prod _{i=1}^b b(l_i,t) \prod _{i=b+1}^{2b} \overline{b(l_i,t)} \right) \nonumber \\&\qquad \times \, e\left( \alpha _1ks_1(\mathbf{l}) + \alpha _2 k^2 s_2(\mathbf{l}) + \alpha _3 k^3 s_3(\mathbf{l})\right) , \end{aligned}$$

(48)

where

$$\begin{aligned} s_j(\mathbf{l}) = l_1^j + \cdots + l_b^j - l_{b+1}^j - \cdots - l_{2b}^j. \end{aligned}$$

Collecting terms in (48) by values of $s_j$ yields

$$\begin{aligned} \left| \sum _{l \le P/M} b(l,t)h(kl) \right| ^{2b} = \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^j \end{array}} R_1(\mathbf {v}) e(\alpha _1kv_1 + \alpha _2k^2v_2 + \alpha _3k^3v_3), \end{aligned}$$

(49)

where

$$\begin{aligned} R_1(\mathbf {v}) = \sum _{\begin{array}{c} \mathbf {l} \\ l_j \le P/M \\ \mathbf {s}(\mathbf {l}) = \mathbf {v} \end{array}} \prod _{i=1}^b b(l_i,t) \prod _{i=b+1}^{2b} \overline{b(l_i,t)} \ll J_{b,3}(P/M) \ll (P/M)^{2b-6} \end{aligned}$$

by (32). Substituting (49) into (47) yields

$$\begin{aligned} S_4(M,t)^{2b}\ll & {} (\log P)^{2b}M^{2b-1} \nonumber \\&\times \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} R_1(\mathbf {v}) \sum _{M< k \le 2M} e(\alpha _1kv_1 \nonumber \\&+ \alpha _2k^2v_2 + \alpha _3k^3v_3) \nonumber \\\ll & {} (\log P)^{2b}M^5 P^{2b-6} \nonumber \\&\times \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} \sum _{M < k \le 2M} \nonumber \\&e(\alpha _1kv_1+\alpha _2k^2v_2 + \alpha _3k^3v_3). \end{aligned}$$

(50)

We now repeat the procedure followed from (46) to (50). By Hölder’s inequality

$$\begin{aligned} S_4(M,t)|^{4b^2}\ll & {} \left( (\log P)^{2b} M^5 P^{2b-6} \right) ^{2b} \left( \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} 1^\frac{2b}{2b-1} \right) ^{2b-1} \nonumber \\&\times \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} \left| \sum _{M < k \le 2M} e(\alpha _1kv_1 + \alpha _2k^2v_2 + \alpha _3k^3v_3) \right| ^{2b} \end{aligned}$$

(51)

$$\begin{aligned}\ll & {} (\log P)^{4b}M^{10b}P^{4b^2-12b} \left( b^3P^6M^{-6} \right) ^{2b-1}\nonumber \\&\times \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} \left| \sum _{M< k \le 2M} e(\alpha _1kv_1 + \alpha _2k^2v_2 + \alpha _3k^3v_3) \right| ^{2b} \nonumber \\\ll & {} (\log P)^{4b^2} M^{6-2b} P^{4b^2-6} \nonumber \\&\times \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} \left| \sum _{M < k \le 2M} e(\alpha _1kv_1 + \alpha _2k^2v_2 + \alpha _3k^3v_3) \right| ^{2b}. \end{aligned}$$

(52)

We expand the 2b-th power in (52) and collect like terms. Thus

$$\begin{aligned}&\left| \sum _{M< k \le 2M} e(\alpha _1kv_1 + \alpha _2k^2v_2 + \alpha _3k^3v_3) \right| ^{2b} \nonumber \\&\quad = \sum _{\begin{array}{c} \mathbf {k} \\ M < k_j \le 2M \end{array}} e(\alpha _1s_1(\mathbf {k})v_1 + \alpha _2s_2(\mathbf {k})v_2 + \alpha _3s_3(\mathbf {k})v_3) \nonumber \\&\quad = \sum _{\begin{array}{c} \mathbf {u} \\ |u_j| \le b2^jM^j \end{array}} R_2(\mathbf {u}) e(\alpha _1u_1v_1 + \alpha _2u_2v_2 + \alpha _3u_3v_3), \end{aligned}$$

(53)

where

$$\begin{aligned} R_2(\mathbf {u}) = \sum _{\begin{array}{c} \mathbf {k} \\ M < k_j \le 2M \\ \mathbf {s}(\mathbf {k}) = \mathbf {u} \end{array}} 1 \ll J_{b,3}(2M) \ll M^{2b-6} \end{aligned}$$

by (32). Substituting (53) into (52), we obtain

$$\begin{aligned} S_4(M,t)^{4b^2}&\ll (\log P)^{4b^2} P^{4b^2-6}\\&\quad \times \sum _{\begin{array}{c} \mathbf {u} \\ |u_j| \le b2^jM^j \end{array}} \left| \sum _{\begin{array}{c} \mathbf {v} \\ |v_j| \le bP^jM^{-j} \end{array}} e(\alpha _1u_1v_1 + \alpha _2u_2v_2 + \alpha _3u_3v_3) \right| . \end{aligned}$$

Summing over each of the $v_j$ gives

$$\begin{aligned} S_4(M,t)^{4b^2} \ll (\log P)^{4b^2} P^{4b^2-6} \sum _{\begin{array}{c} \mathbf {u} \\ |u_j| \le b2^jM^j \end{array}} \prod _{j=1}^3 \min \left( \frac{P^j}{M^j},\frac{1}{\Vert \alpha _ju_j\Vert }\right) . \end{aligned}$$

Applying Lemma 2.2 of [7] yields

$$\begin{aligned} S_4(M,t)^{4b^2} \ll (\log P)^{4b^2+3} P^{4b^2} \prod _{j=1}^3 \left( \frac{1}{q_j} + \frac{1}{M^j} + \frac{M^j}{P^j} + \frac{q_j}{P^j}\right) . \end{aligned}$$

(54)

Combining (54) with (43) and (45), we obtain

$$\begin{aligned} S_4 \ll P(\log P)^4 \prod _{j=1}^3 \left( \frac{1}{q_j} + \frac{1}{X^j} + \frac{q_j}{P^j}\right) ^{1/(4b^2)}. \end{aligned}$$

Recalling that $q_j > (\log P)^A$ for some j and $X = (\log P)^B$, this is

$$\begin{aligned} S_4 \ll P (\log P)^{4-\min (A,B)/(4b^2)} \end{aligned}$$

(55)

for $b > 6$. $\square$

Proof of Theorem 5.1

Using the Vaughan’s identity breakdown of (33) and the estimates for the $S_i$ found in Lemmas 5.1, 5.2, 5.3, and 5.4, we have

$$\begin{aligned} F_3(\pmb \alpha )= & {} S_1 + S_2 + S_3 + S_4 \\\ll & {} (\log P)^B + P(\log P)^{B - A/12 + 4} + P(\log P)^{B-A/12+4} \\&+P (\log P)^{4-\min (A,B)/(4b^2)}. \end{aligned}$$

So, taking $B > 4b^2D(D+4)$ and $A > 12(B + D + 4)$ for some $D > 0$ yields

$$\begin{aligned} F_3(\pmb \alpha ) \ll P(\log P)^{-D} \end{aligned}$$

uniformly in $\pmb \alpha$. $\square$

6 Major arc approximations

On a typical major arc $\mathfrak {M}(a_2,a_3,q)$, let $\alpha _2 = \frac{a_2}{q}+\theta$, $\alpha _3 = \frac{a_3}{q}+\omega$, with $\theta < \frac{Q}{qP^2}$, $\omega < \frac{Q}{qP^3}$, and $q < Q$. For ease of notation, let $\frac{Q}{qP^2} = \Theta$, $\frac{Q}{qP^3} = \Omega$. Let

$$\begin{aligned} W_i(q,a_2,a_3)= & {} \sum _{\begin{array}{c} r=1 \\ (r,q)=1 \end{array}}^q e\left( \frac{a_2u_ir^2 + a_3v_ir^3}{q}\right) , \\ f_i^*(\alpha _2, \alpha _3)= & {} \frac{1}{\phi (q)} W_i(q,a_2,a_3) \int _0^P e(\theta u_i x^2 + \omega v_i x^3) dx, \\ T_i(x,a_2,a_3)= & {} \sum _{p \le x}(\log p) e\left( \frac{a_2u_ip^2 + a_3v_ip^3}{q}\right) , \end{aligned}$$

and for $x > \sqrt{P}$,

$$\begin{aligned} T_i^\dagger (x,a_2,a_3) = \sum _{\sqrt{P} < p \le x}(\log p) e\left( \frac{a_2u_ip^2 + a_3v_ip^3}{q}\right) . \end{aligned}$$

We begin with preliminary bounds on $T_i(x,a_2,a_3)$ and $T_i^\dagger (q,a_2,a_3)$.

Lemma 6.1

$$\begin{aligned} T_i(x,a_2,a_3) = \frac{x}{\phi (q)}W_i(q,a_2,a_3) + O(x\exp (-C(\log x)^{1/2})). \end{aligned}$$

Proof

The exponential function $e((a_2u_ip^2 + a_3v_ip^3)/q)$ is only sensitive to the residue class of p modulo q, so

$$\begin{aligned} T_i(x,a_2,a_3)= & {} \sum _{\begin{array}{c} r=1 \\ (r,q)=1 \end{array}}^q\sum _{\begin{array}{c} p \le x \\ p \equiv r\, (\text {mod } q) \end{array}} (\log p) e\left( \frac{a_2u_ir^2 + a_3v_ir^3}{q}\right) + O(q^\varepsilon \log q) \\= & {} \sum _{\begin{array}{c} r = 1\\ (r,q)=1 \end{array}}^q \left( e\left( \frac{a_2u_ir^2 + a_3v_ir^3}{q}\right) \sum _{\begin{array}{c} p \le x\\ p \equiv r \, (\text {mod } q) \end{array}} \log p \right) + O(q^\varepsilon \log q). \end{aligned}$$

Now by the Siegel–Walfisz theorem we have that

$$\begin{aligned} \sum _{\begin{array}{c} p \le x\\ p \equiv r \, (\text {mod } q) \end{array}} \log p = \frac{x}{\phi (q)} + O(x\exp (-C(\log x)^{1/2})), \end{aligned}$$

(56)

so

$$\begin{aligned} T_i(x,a_2,a_3)= & {} \sum _{\begin{array}{c} r = 1\\ (r,q)=1 \end{array}}^q \left( e\left( \frac{a_2u_ir^2 + a_3v_ir^3}{q}\right) \left( \frac{x}{\phi (q)} + O(x\exp (-C(\log x)^{1/2}))\right) \right) \\= & {} \frac{x}{\phi (q)} W_i(q,a_2,a_3) + W_i(q,a_2,a_3)\left( O(x\exp (-C(\log x)^{1/2}))\right) \\= & {} \frac{x}{\phi (q)}W_i(q,a_2,a_3) + O(x\exp (-C(\log x)^{1/2})). \end{aligned}$$

$\square$

Corollary 6.1

For $x > \sqrt{P}$,

$$\begin{aligned} T_i^\dagger (x,a_2,a_3) = \frac{x}{\phi (q)}W_i(q,a_2,a_3) + O(x\exp (-C(\log x)^{1/2})). \end{aligned}$$

Proof

$$\begin{aligned} T_i^\dagger (x,a_2,a_3)= & {} T_i(x,a_2,a_3) - T_i(\sqrt{P},a_2,a_3) \\= & {} \frac{x}{\phi (q)}W_i(q,a_2,a_3) + O(x\exp (-C(\log x)^{1/2})) \\&+O(P^{1/2}\exp (-C(\log P)^{1/2})\\= & {} \frac{x}{\phi (q)}W_i(q,a_2,a_3) + O(x\exp (-C(\log x)^{1/2})). \end{aligned}$$

$\square$

Lemma 6.2

On $\mathfrak {M}(q,a_2,a_3)$,

$$\begin{aligned} f_i(\alpha _2, \alpha _3) = f_i^*(\alpha _2, \alpha _3) + O(P\exp (-C(\log P)^{1/2})) \end{aligned}$$

for some positive constant C.

Proof

First, we isolate the range $(\sqrt{P}, P]$, bounding the remainder immediately.

$$\begin{aligned}&|f_i(\alpha _2, \alpha _3) - f_i^*(\alpha _2, \alpha _3)| \\&\quad = \left| \sum _{p \le P} (\log p)e(\alpha _2 u_i p^2 + \alpha _3 v_i p^3) \right. \\&\qquad \left. - \frac{1}{\phi (q)}W_i(q,a_2,a_3)\int _0^P e(\theta u_i x^2 + \omega v_i x^3) dx \right| \\&\quad = \left| \sum _{\sqrt{P} < p \le P} (\log p)e(\alpha _2 u_i p^2 + \alpha _3 v_i p^3) \right. \\&\qquad \left. -\frac{1}{\phi (q)}W_i(q,a_2,a_3)\int _{\sqrt{P}}^P e(\theta u_i x^2 + \omega v_i x^3) dx \right| \\&\qquad + O(P^{1/2}\log P). \end{aligned}$$

Now

$$\begin{aligned}&\left| \sum _{\sqrt{P}< p \le P} (\log p)e(\alpha _2 u_i p^2 + \alpha _3 v_i p^3) \right. \nonumber \\&\qquad \left. - \frac{1}{\phi (q)}W_i(q,a_2,a_3)\int _{\sqrt{P}}^P e(\theta u_i x^2 + \omega v_i x^3) dx \right| \nonumber \\&\quad = \Bigg | W_i(q,a_2,a_3) \sum _{\begin{array}{c} \sqrt{P} < p \le P \\ p \equiv r \, (\text {mod } q) \end{array}} (\log p)e(\theta u_i p^2 + \omega v_i p^3)\nonumber \\&\qquad - \frac{1}{\phi (q)}W_i(q,a_2,a_3)\int _{\sqrt{P}}^P e(\theta u_i x^2 + \omega v_i x^3) dx \Bigg | \end{aligned}$$

(57)

$$\begin{aligned}&\quad = \sum _{\sqrt{P} < m \le P} \bigg [ (\log m)e\left( \frac{a_2u_im^2+a_3v_im^3}{q}\right) \mathbb {1}_\mathcal {P}\nonumber \\&\qquad - \frac{1}{\phi (q)} W_i(q,a_2,a_3) \bigg ] e(\theta u_i m^2 + \omega v_i m^3) + O(|\omega | P^{5/2}), \end{aligned}$$

(58)

where $\mathbb {1}_\mathcal {P}$ is the indicator function of the primes.

We now apply Abel summation to (58), with the term in square brackets serving as the coefficient. This yields that

$$\begin{aligned}&|f_i(\alpha _2, \alpha _3) - f_i^*(\alpha _2, \alpha _3)| \nonumber \\&\quad = e(\theta u_i P^2 + \omega v_i P^3) \bigg ( T_i(x,a_2,a_3) - \frac{1}{\phi (q)}\sum _{\sqrt{P}< m \le P}W_i(q,a_2,a_3)\bigg ) \nonumber \\&\qquad -\int _{\sqrt{P}}^P 2\pi i(2\theta u_i x + 3\omega v_i x^2) \bigg (T_i(x,a_2,a_3) - \frac{1}{\phi (q)}\sum _{\sqrt{P} < m \le x}W_i(q,a_2,a_3) \bigg ) dx \nonumber \\&\qquad + O(|\omega | P^{5/2}). \end{aligned}$$

(59)

Now Corollary 6.1 gives that for $x > \sqrt{P}$,

$$\begin{aligned} T_i^\dagger (x,a_2,a_3) - \frac{1}{\phi (q)}\sum _{\sqrt{P} < m \le x}W_i(q,a_2,a_3) \ll x\exp (-C(\log x)^{1/2}), \end{aligned}$$

so

$$\begin{aligned}&|f_i(\alpha _2, \alpha _3) - f_i^*(\alpha _2, \alpha _3)| \\&\quad = e(\theta u_i P^2 + \omega v_i P^3) \left( O(\phi (q) P \exp (-C(\log P)^{1/2}))\right) \\&\qquad - 2\pi i\int _0^P (2\theta u_i x + 3\omega v_i x^2) \left( O(\phi (q) x \exp (-C(\log P)^{1/2})) \right) dx\\&\qquad + O(P^{1/2}\log P) + O(|\omega | P^{5/2}) \\&\quad \ll (1 + |\theta |P^2 + |\omega |P^3) \phi (q) P \exp (-C(\log P)^{1/2}) \\&\quad \ll (\log P)^A \frac{\phi (q)}{q} P \exp (-C(\log P)^{1/2}) \\&\quad \ll P \exp (-C(\log P)^{1/2}). \end{aligned}$$

$\square$

For clarity of notation, let

$$\begin{aligned} A(q)= & {} \mathop {\sum _{a_2=1}^q \sum _{a_3=1}^q}_{(a_2,a_3,q)=1} \frac{1}{\phi (q)^s} \prod _{i=1}^s W_i(q,a_2,a_3), \\ \mathfrak {S}(Q)= & {} \sum _{q<Q} A(q), \\ J(Q)= & {} \int _{|\theta |<Q/P^2} \int _{|\omega |<Q/P^3} \prod _{i=1}^s \int _0^P e(\theta u_i x^2 + \omega v_i x^3) dx \, d\omega \, d\theta . \end{aligned}$$

We are now able to state the primary lemma of this section:

Lemma 6.3

For some $E > 0$,

$$\begin{aligned} R(P) = \mathfrak {S}(Q)J(Q) + O(P^{s-5}(\log P)^{-E}). \end{aligned}$$

Proof

We first introduce variant major arcs whose length is independent of q:

$$\begin{aligned} \mathfrak {B}(q, \mathbf {r}, Q) = \left\{ (a_2,a_3) : |\alpha _2 - a_2/q|< \frac{Q}{P^2}, |\alpha _3 - a_3/q| < \frac{Q}{P^3}\right\} \end{aligned}$$

for $1 \le Q \le P$, $q < Q$, $1 \le a_2 \le q$, $1 \le a_3 \le q$, and $(a_2, a_3, q) = 1$. Let $\mathfrak {B}$ be the union of all such $\mathfrak {B}(q, \mathbf {r}, Q)$ and note that $\mathfrak {M} \subseteq \mathfrak {B}$ and thus $\mathfrak {B} \setminus \mathfrak {M} \subseteq \mathfrak {m}$.

It follows immediately from Lemma 6.2 that

$$\begin{aligned} \left| \prod _{i=1}^s f_i(\alpha _2, \alpha _3) - \prod _{i=1}^s f_i^*(\alpha _2, \alpha _3)\right| \ll P^s\exp (-C(\log P)^{1/2}). \end{aligned}$$

(60)

Summing (60) over all arcs in $\mathfrak {B}$ gives

$$\begin{aligned}&\int _\mathfrak {B} \left| \prod _{i=1}^s f_i(\alpha _2, \alpha _3) - \prod _{i=1}^s f_i^*(\alpha _2, \alpha _3)\right| d\alpha _2 \, d\alpha _3 \nonumber \\&\quad = \sum _{q<Q} \mathop {\sum _{a_2=1}^q \sum _{a_3=1}^q}_{(a_2,a_3,q)=1} \int _{\mathfrak {B}(a_2,a_3,q)} \left| \prod _{i=1}^s f_i(\alpha _2, \alpha _3) - \prod _{i=1}^s f_i^*(\alpha _2, \alpha _3)\right| d\alpha _2 \, d\alpha _3 \nonumber \\&\quad \ll \sum _{q<Q} \mathop {\sum _{a_2=1}^q \sum _{a_3=1}^q}_{(a_2,a_3,q)=1} \int _{-Q/P^2}^{Q/P^2} \int _{-Q/P^3}^{Q/P^3} P^s\exp (-C(\log P)^{1/2}) d\alpha _2 \, d\alpha _3 \nonumber \\&\quad \ll Q^3P^{s-5}\exp (-C(\log P)^{1/2}). \end{aligned}$$

(61)

We now wish to compute

$$\begin{aligned}&\int _\mathfrak {B} \prod _{i=1}^s f_i^*(\alpha _2, \alpha _3) d\alpha _2 d\alpha _3 \nonumber \\&\quad = \sum _{q<Q} \mathop {\sum _{a_2=1}^q \sum _{a_3=1}^q}_{(a_2,a_3,q)=1} \prod _{i=1}^s \frac{1}{\phi (q)} W_i(q,a_2,a_3) \nonumber \\&\qquad \times \int _{-Q/P^2}^{Q/P^2} \int _{-Q/P^3}^{Q/P^3} \int _0^P \nonumber \\&\qquad e(\theta u_i x^2 + \omega v_i x^3) dx \, d\theta \, d\omega \nonumber \\&\qquad + O(P^{s-5}(\log P)^{-E}) \nonumber \\&\quad = \mathfrak {S}(Q) J(Q) + O(P^{s-5}(\log P)^{-E}). \end{aligned}$$

(62)

Combining (61) and (62) yields the bound

$$\begin{aligned} \int _\mathfrak {B} \prod _{i=1}^s f_i(\alpha _2, \alpha _3) = \mathfrak {S}(Q) J(Q) + O(P^{s-5} (\log P)^{-E}). \end{aligned}$$

(63)

Combining Theorem 4.1 and Corollary 5.1 yields the minor arc bound

$$\begin{aligned} \int _\mathfrak {m} \prod _{i=1}^s f_i(\alpha _2, \alpha _3) \ll P^{s-5} (\log P)^{-E}, \end{aligned}$$

(64)

and moreover, since $\mathcal {A}\setminus \mathfrak {B} \subseteq \mathfrak {m}$, by Corollary 5.1 and Theorem 4.1 we have

$$\begin{aligned} \int _{\mathcal {A}\setminus \mathfrak {B}} \prod _{i=1}^s f_i(\alpha _2, \alpha _3) \ll P^{s-5} (\log P)^{-E}. \end{aligned}$$

(65)

Now by (5), (63), and (65) we have

$$\begin{aligned} R(P) = \mathfrak {S}(Q)J(Q) + O(P^{s-5} (\log P)^{-E}). \end{aligned}$$

(66)

$\square$

7 Convergence of the singular series

Lemma 7.1

Let $(q_1,q_2)=1$. Then

$$\begin{aligned} W_i(q_1q_2,a_2,a_3) = W_i(q_2,a_2q_1,a_3q_1^2)W_i(q_1,a_2q_2,a_3q_2^2). \end{aligned}$$

Proof

Each residue class r modulo $q_1q_2$ with $(r, q_1q_2) = 1$ is uniquely represented as $cq_1+dq_2$ with $1 \le c \le q_2$, $(c, q_2) = 1$, $1 \le d \le q_1$, and $(d, q_1) = 1$. Also, $cq_1$, $dq_2$ run over all residue classes modulo $q_2$, $q_1$ with $(cq_1, q_2) = 1$, $(dq_2, q_1) = 1$, respectively. Thus

$$\begin{aligned}&W_i(q_1q_2,a_2,a_3) = \sum _{\begin{array}{c} c=1 \\ (c,q_2)=1 \end{array}}^{q_2} \sum _{\begin{array}{c} d=1 \\ (d,q_1)=1 \end{array}}^{q_1} e\left( \frac{a_2u_i(cq_1+dq_2)^2 + a_3v_i(cq_1+dq_2)^3}{q_1q_2} \right) \\&\quad = \sum _{\begin{array}{c} c=1 \\ (c,q_2)=1 \end{array}}^{q_2} \sum _{\begin{array}{c} d=1 \\ (d,q_1)=1 \end{array}}^{q_1} e\left( \frac{a_2u_ic^2q_1+a_3v_ic^3q_1^2}{q_2} \right) e\left( \frac{a_2u_id^2q_2 + a_3v_id^3q_2^2}{q_1} \right) \\&\quad \qquad \qquad \qquad \qquad = W_i(q_2,a_2q_1,a_3q_1^2)W_i(q_1,a_2q_2,a_3q_2^2). \end{aligned}$$

$\square$

Lemma 7.2

A(q) is multiplicative.

Proof

Let $(q_1,q_2)=1$. Then

$$\begin{aligned} A(q_1q_2) = \mathop {\sum _{a_2=1}^{q_1q_2}\sum _{a_3=1}^{q_1q_2}}_{(a_2,a_3,q_1q_2)=1} \frac{1}{\phi (q_1q_2)^s} \prod _{i=1}^s W_i(q_1q_2,a_2,a_3). \end{aligned}$$

Now $a_2$ and $a_3$ can be represented by $b_1q_2+b_2q_1$ and $c_1q_2+c_2q_1$, respectively, with $1 \le b_1,c_1 \le q_1$, $1 \le b_2,c_2 \le q_2$. So we can rewrite our sum as

$$\begin{aligned} A(q_1q_2) = \mathop {\sum _{b_1=1}^{q_1}\sum _{c_1=1}^{q_1}}_{(b_1,c_1,q_1)=1} \mathop {\sum _{b_2=1}^{q_2}\sum _{c_2=1}^{q_2}}_{(b_2,c_2,q_2)=1} \frac{1}{\phi (q_1q_2)^s} \prod _{i=1}^s W_i(q_2,b_2q_1^2,c_2q_1^3) W_i(q_1,b_1q_2^2,c_1q_2^3). \end{aligned}$$

Now since $(q_1,q_2)=1$, $(c_2,b_2,q_1)=1$, and $(b_1,c_1,q_2)=1$, we have that $b_2q_1^2,c_2q_1^3,b_1q_2^2,c_1q_2^3$ run through complete sets of residue classes modulo $q_2,q_2,q_1,q_1$, respectively. Thus

$$\begin{aligned} A(q_1q_2)= & {} \mathop {\sum _{b_1=1}^{q_2}\sum _{c_1=1}^{q_2}}_{(b_1,b_1,q_2)=1} \mathop {\sum _{b_2=1}^{q_1}\sum _{c_2=1}^{q_1}}_{(b_2,c_2,q_1)=1} \frac{1}{\phi (q_1q_2)^s} \prod _{i=1}^s W_i(q_2,b_2,c_2) W_i(q_1,b_1,c_1) \\= & {} A(q_1)A(q_2). \end{aligned}$$

$\square$

Let $\mathfrak {S}$ be the completed singular series

$$\begin{aligned} \mathfrak {S} = \sum _{q=1}^\infty A(q). \end{aligned}$$

Since A(q) is multiplicative,

$$\begin{aligned} \mathfrak {S} = \prod _p \left( 1 + \sum _{k=1}^\infty A(p^k)\right) . \end{aligned}$$

(67)

Lemma 7.3

$\mathfrak {S}$ converges absolutely.

Proof

$$\begin{aligned} A(p^k) = \mathop {\sum _{a_2=1}^{p^k}\sum _{a_3=1}^{p^k}}_{(a_2,a_3,p^k)=1} \frac{1}{\phi (p^k)^s} \prod _{i=1}^s W_i(p^k,a_2,a_3). \end{aligned}$$

By Lemma 3.9 and the fact that there are $\ll p^{2k}$ choices for the pair a, b, we have

$$\begin{aligned} A(p^k)\ll & {} p^{2k} \phi (p^k)^{-s} ((p^k)^{\frac{1}{2}+\varepsilon })^s \\\ll & {} (p^k)^{2-\frac{1}{2}s+\varepsilon }. \end{aligned}$$

Since $s \ge 7$, we have

$$\begin{aligned} A(p^k) \ll (p^k)^{-\frac{3}{2}+\varepsilon }. \end{aligned}$$

(68)

Thus

$$\begin{aligned} \sum _{k=1}^\infty A(p^k) \ll \sum _{k=1}^\infty (p^k)^{-\frac{3}{2}+\varepsilon } = \frac{p^{-3/2+\varepsilon }}{1-p^{-3/2+\varepsilon }} \ll p^{-3/2+\varepsilon }. \end{aligned}$$

Then

$$\begin{aligned} \sum _p \sum _{k=1}^\infty A(p^k) \ll \sum _p p^{-3/2+\varepsilon } \end{aligned}$$

converges, so

$$\begin{aligned} \mathfrak {S} = \prod _p \left( 1 + \sum _{k=1}^\infty A(p^k)\right) \end{aligned}$$

converges. $\square$

8 Positivity of the singular series

To show that R(P) is eventually positive, we now need to show that $\mathfrak {S}$ is positive.

Lemma 8.1

There exists $R > 0$ such that

$$\begin{aligned} \frac{1}{2} < \prod _{p \ge R} \left( 1 + \sum _{k=1}^\infty A(p^k)\right) . \end{aligned}$$

Proof

By (68), we have $A(p^k) \ll (p^k)^{-3/2+\varepsilon } \ll (p^k)^{-1/4}$. Choose C, R such that $A(p^k) \le Cp^{-5/4}< Cp^{-1/4} < \frac{1}{8}$ for all $p \ge R-1$. Then

$$\begin{aligned}&\prod _{p\ge R}\left( 1-Cp^{-5/4}\right) \ge 1 - \sum _{p\ge R} Cp^{-5/4} \\&\quad \ge 1 - C\int _{R-1}^\infty x^{-5/4} dx = 1 - 4C(R-1)^{-1/4} \ge \frac{1}{2}. \end{aligned}$$

$\square$

We now need only show that for $p \le R$, $1 + \sum _{k=1}^\infty A(p^k) > 0$. For $1 \le t \le s$, define $M_t(q)$ to be the number of solutions $(x_1, \ldots , x_s)$ to the simultaneous congruences

$$\begin{aligned} \begin{aligned}&\sum _{i=1}^t u_ix_i^2 \equiv 0 \pmod q, \\&\sum _{i=1}^t v_ix_i^3 \equiv 0 \pmod q \end{aligned} \end{aligned}$$

with $(x_i,q)=1$ for all i.

Lemma 8.2

For any positive integer q,

$$\begin{aligned} M_s(q) = \frac{\phi (q)^s}{q^2} \sum _{d|q} A(d). \end{aligned}$$

Proof

$$\begin{aligned} M_s(q)= & {} \frac{1}{q^2} \sum _{r_2=1}^q \sum _{r_3=1}^q \sum _{\begin{array}{c} x_1=1 \\ (x_1,q)=1 \end{array}}^q \cdots \\&\quad \times \sum _{\begin{array}{c} x_s=1 \\ (x_s,q)=1 \end{array}}^q e\left( \frac{r_2(u_1x_1^2 + \cdots + u_sx_s^2) + r_3(v_1x_1^3 + \cdots + v_sx_s^3)}{q}\right) \\= & {} \frac{1}{q^2} \sum _{r_2=1}^q \sum _{r_3=1}^q \prod _{i=1}^s \sum _{\begin{array}{c} x_i=1 \\ (x_1,q)=1 \end{array}}^q e\left( \frac{r_2u_ix_i^2 + r_3v_ix_i^3}{q}\right) . \end{aligned}$$

Let $d = \frac{q}{(r_2,r_3,q)}$, $a_1 = \frac{r_2}{(r_2,r_3,q)}$, and $a_2 = \frac{r_3}{(r_2,r_3,q)}$. Then, rearranging according to the value of d, we have

$$\begin{aligned} M_s(q)= & {} \frac{1}{q^2} \sum _{d|q} \mathop {\sum _{a_2=1}^d \sum _{a_3=1}^d}_{(a_2,a_3,d)=1} \prod _{i=1}^s \frac{\phi (q)}{\phi (d)} \sum _{\begin{array}{c} x_i=1 \\ (x_i,d)=1 \end{array}}^d e\left( \frac{a_2u_ix_i^2 + a_3v_ix_i^3}{d}\right) \\= & {} \frac{\phi (q)^s}{q^2} \sum _{d|q} A(d). \end{aligned}$$

$\square$

Lemma 8.3

For positive integers $t, \gamma$ with $t > \gamma$,

$$\begin{aligned} M_s(p^t) \ge M_s(p^\gamma )p^{(t-\gamma )(s-2)}. \end{aligned}$$

Proof

This is [11], Lemma 6.7, with the added observation that (in that paper’s notation)

$$\begin{aligned} \max \{|b_1-a_1|_p, |b_2-a_2|_p\} \le p^{-\gamma } \Rightarrow p^{\gamma } | (b_1-a_1), (b_2-a_2). \end{aligned}$$

So if $a_1,b_1 \not \equiv 0$ (mod p), then $a_2,b_2 \not \equiv 0$ (mod p). Thus the argument lifts solutions over reduced residue classes modulo $p^\gamma$ to solutions over reduced residue classes modulo $p^t$, so it applies here without modification. $\square$

Theorem 8.1

If for every prime p there exists a positive integer $\gamma$ such that $M_s(p^\gamma ) > 0$, then $\mathfrak {S} > 0$.

Proof

By Lemma 8.2,

$$\begin{aligned} 1 + \sum _{k=1}^\infty A(p^k)= & {} \lim _{t\rightarrow \infty }\frac{p^{2t}}{\phi (p^t)^s}M(p^t) \\\ge & {} \lim _{t\rightarrow \infty }p^{(2-s)t}M(p^t). \end{aligned}$$

By Lemma 8.3, for some positive integer $\gamma$,

$$\begin{aligned} 1 + \sum _{k=1}^\infty A(p^k)\ge & {} \lim _{t\rightarrow \infty }p^{(2-s)t}M(p^\gamma )p^{(t-\gamma )(s-2)} \nonumber \\\ge & {} \lim _{t\rightarrow \infty } p^{(-\gamma )(s-2)} > 0. \end{aligned}$$

(69)

The lemma now follows from (67), Lemma 8.1, and (69). $\square$

In Sects. 9 and 10 we prove that, under the conditions of Theorem 1.2, for every p there exists a positive integer $\gamma$ such that $M(p^\gamma )>0$.

9 Solvability of the local problem

We now consider the local system

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_sx_s^2 \equiv 0 \pmod p,\\&v_1x_1^3 + \cdots + v_sx_s^3 \equiv 0 \pmod p \end{aligned} \end{aligned}$$

(70)

with $x_i \ne 0$ in $\mathbb {Z}/p\mathbb {Z}$.

We will prove the following result:

Theorem 9.1

The system

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_sx_s^2 \equiv U \pmod p,\\&v_1x_1^3 + \cdots + v_sx_s^3 \equiv V \pmod p \end{aligned} \end{aligned}$$

(71)

has a solution $(x_1, \ldots , x_s)$ with all $x_i \ne 0$ modulo every prime p if

• 1. $\displaystyle \sum _{i=1}^s u_i \equiv U \pmod 2$ and $\displaystyle \sum _{i=1}^s v_i \equiv V \pmod 2,$

• 2. $\displaystyle \sum _{i=1}^s u_i \equiv U \pmod 3,$ and

• 3. for each prime p at least 7 of the $u_i$, $v_i$ are not zero modulo p.

Observe that if the system

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_tx_t^2 \equiv U \pmod p,\\&v_1x_1^3 + \cdots + v_tx_t^3 \equiv V \pmod p \end{aligned} \end{aligned}$$

(72)

has a solution for all $u_1, \ldots , u_t, v_1, \ldots , v_t \ne 0$, then so does the system

$$\begin{aligned} \begin{aligned}&u_{i_1}x_{i_1}^2 + \cdots + u_{i_t}x_{i_t}^2 \equiv U \pmod p,\\&v_{j_1}x_{j_1}^3 + \cdots + v_{j_t}x_{j_t}^3 \equiv V \pmod p \end{aligned} \end{aligned}$$

(73)

for any $\{i_1, \ldots , i_t\}$, $\{j_1, \ldots , j_t\} \subset \{1, \ldots , s\}$. Also observe that the conditions of Theorem 9.1 guarantee solvability modulo $p=2$ and $p=3$: $p=2$ is immediate and for $p=3$, the condition guarantees that the quadratic equation is satisfied and each term $v_ix_i^3$ of the cubic equation can be independently set to 1 or $-1$, allowing us to set $v_1x_1^3 = V$ if $V \not \equiv 0 \pmod 3$ and partition the remainder of $\{1, \ldots , t\}$ into groups of 2 and 3, which can be zeroed by setting them to $\{1, -1\}$ and $\{1, 1, 1\}$.

Thus we have reduced Theorem 9.1 to this lemma:

Lemma 9.1

For all $u_i$, $v_i \ne 0 \pmod p$, $p \ge 5$, $t \ge 7$, U, V, there exist $\{x_1, \ldots , x_s\}$ with $x_i \ne 0 \pmod p$ such that

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_tx_t^2&\equiv U \pmod p,\\&v_1x_1^3 + \cdots + v_tx_t^3&\equiv V \pmod p. \end{aligned} \end{aligned}$$

(74)

Lemma 9.2

Suppose $p > 3$, and that a and b are not both equal to p. Then $|W_i(p,a_2,a_3)| \le 2\sqrt{p}+1.$

Proof

Corollary 2F of [6] gives

$$\begin{aligned} \left| \sum _{\begin{array}{c} r=0 \end{array}}^{p-1} e\left( \frac{a_2u_ir^2+a_3v_ir^3}{p}\right) \right| \le 2p^{1/2}. \end{aligned}$$

Now

$$\begin{aligned} |W_i(p,a_2,a_3)|= & {} \left| \sum _{\begin{array}{c} r=1 \end{array}}^{p-1} e\left( \frac{a_2u_ir^2+a_3v_ir^3}{p}\right) \right| \\\le & {} \left| \sum _{\begin{array}{c} r=0 \end{array}}^{p-1} e\left( \frac{a_2u_ir^2+a_3v_ir^3}{p}\right) \right| + 1 \le 2\sqrt{p} + 1. \end{aligned}$$

$\square$

Lemma 9.3

$M_t(p) \ge \frac{1}{p^2}\big ((p-1)^t- (p^2-1)(2\sqrt{p}+1)^t\big )$.

Proof

$$\begin{aligned} M_t(p) = \frac{1}{p^2}\sum _{r_2=1}^p \sum _{r_3=1}^p \prod _{i=1}^t W_i(p,r_2,r_3). \end{aligned}$$

We have $W_i(p,p,p)=p-1$ and for $r_2$, $r_3$ not both p, $W_i(p,r_2,r_3) \le 2\sqrt{p}+1$ by Lemma 9.2. Thus

$$\begin{aligned} \left| M_t(p) - \frac{(p-1)^t}{p^2}\right|\le & {} \frac{1}{p^2} \sum _{r_2=1}^p \sum _{\begin{array}{c} r_3=1 \\ \{r_2,r_3\} \ne \{p,p\} \end{array}}^p \prod _{i=1}^t (2\sqrt{p}+1) \\\le & {} \frac{1}{p^2}(p^2-1)(2\sqrt{p}+1)^t. \end{aligned}$$

So we have

$$\begin{aligned} M_t(p) \ge \frac{1}{p^2}\big ((p-1)^t- (p^2-1)(2\sqrt{p}+1)^t\big ). \end{aligned}$$

$\square$

Taking $t=7$, we get

$$\begin{aligned} M_7(p) \ge \frac{1}{p^2}\big ((p-1)^7- (p^2-1)(2\sqrt{p}+1)^7\big ). \end{aligned}$$

This gives that $M_7(p) > 0$ for $p > 40.58$. This means that we now need only check that Lemma 9.1 holds for each prime smaller than 41. This is now a finite number of cases to check and thus can be verified by computer. In the following section, we note several techniques that may be employed to bring the computational difficulty of the task into the realm of feasibility, and in Appendix 1 we provide Sage code for performing the computation.

It is worth noting that $t=7$ appears to only be required for $p=7$. It seems highly probable that $t=5$ will suffice for all other primes; however, reducing t to 5 weakens the bound of Lemma 9.3 to requiring us to check all primes less than 1193, which would require more computation than is feasible, since even after the optimizations of Sect. 10, the algorithm checks $O(p^7)$ distinct forms for solvability to verify Lemma 9.3 for all primes up through p.

10 Computational techniques

First, we note that if every pair U, V modulo p can be represented by the form in $t_0$ variables, then every pair can be represented by t variables for $t>t_0$. So we will start our search with $t=3$ and store the forms that represent all pairs (U, V) of residue classes mod p. We then need only search higher values of t for the forms that failed to represent all pairs of residue classes with a smaller t.

(The methods in this paragraph are closely modeled after those of [10].) By independently substituting $c_ix_i$ for each $x_i$, we can assume each $x_i$ is either 1 or a fixed quadratic nonresidue c modulo p. By rearranging and multiplying by $b^{-1}$ as needed, we can assume that $u_1, \ldots , u_r = 1$, $u_{r+1}, \ldots , u_t = c$ with $r \ge \lceil {t/2}\rceil$. By multiplying the cubic equation by $v_1^{-1}$ and rearranging, we may assume $1 = v_1 \le v_2 \le \cdots \le v_t$. By substituting $-x_i$ for $x_i$ as needed, we can assume $1 \le v_i \le (p-1)/2$ for each $v_i$ without affecting the $u_i$.

As a final optimization, we note that if the system of congruences

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_tx_t^2 \equiv U \pmod p,\\&v_1x_1^3 + \cdots + v_tx_t^3 \equiv V \pmod p \end{aligned} \end{aligned}$$

(75)

represents $p^2-1$ of the possible $p^2$ pairs of residue classes (U, V) modulo p, then

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_{t+1}x_{t+1}^2 \equiv U \pmod p,\\&v_1x_1^3 + \cdots + v_{t+1}x_{t+1}^3 \equiv V \pmod p \end{aligned} \end{aligned}$$

(76)

will necessarily represent all $p^2$ residue classes, since $(u_{t+1}x_{t+1}^2, v_{t+1}x_{t+1}^3)$ must represent at least two distinct pairs of residue classes, so

$$\begin{aligned} \begin{aligned}&u_1x_1^2 + \cdots + u_tx_t^2 = U - u_{t+1}x_{t+1}^2 \pmod p,\\&v_1x_1^3 + \cdots + v_tx_t^3 = V - v_{t+1}x_{t+1}^3 \pmod p \end{aligned} \end{aligned}$$

(77)

will be solvable for some $(u_{t+1}, v_{t+1})$. This turns out to be quite useful: a substantial number of forms represent exactly $p^2-1$ pairs of residue classes modulo p.

Using these techniques to minimize the computation needed, running the Sage code in Appendix 1 verifies that Lemma 9.1 holds for $p < 41$. This allows us to conclude the following unconditional form of Theorem 8.1.

Lemma 10.1

$\mathfrak {S} > 0.$

11 Conclusion

We have that $R(P) = \mathfrak {S}(Q)J(Q) + O(P^{s-5}(\log P)^{-E})$ by Lemma 6.3. Lemma 10.1, in conjunction with Lemma 8.1, shows that $\mathfrak {S}(Q)>0$ uniformly over all $u_i$, $v_i$ satisfying the conditions of Theorem 1.1 or Theorem 1.2.

The singular integral J(Q) is the same as the one Wooley obtains in the corresponding problem over the integers, so by Lemma 7.4 of [12], there exists a positive constant C such that

$$\begin{aligned} J(Q) = CP^{s-5} + O(P^{s-5}Q^{-1/2}). \end{aligned}$$

In addition, we have the asymptotic upper bound $\mathfrak {S}(Q) \ll 1$ from Lemma 7.3. So we have

$$\begin{aligned} R(P) = CP^{s-5} + O(P^{s-5} (\log P)^{-E}) \end{aligned}$$

for $E > 0$, $C > 0$ uniformly.

Thus R(P) is eventually positive. This can only be true if there is a solution of (1) over the primes, so we can conclude Theorems 1.1 and 1.2.

Appendix 1: Sage code

Code: (SageMath 8.6)

Output: