Estimates for lattice points of quadratic forms with integral coefficients modulo a prime number square

Abstract

Let be a nonsingular quadratic form with integer coefficients, n be even. Let $V=V_Q=V_{p^2}$ denote the set of zeros of $Q(\mathbf{x})$ in ${\mathbb{Z}}_{p^2}$, p be an odd prime, and $|V|$ denote the cardinality of V. In this paper, we are interested in giving an upper bound of the number of integer solutions of the congruence $Q(\mathbf{x})\equiv 0(mod{p}^2)$ in small boxes of the type $\{\mathbf{x}\in {\mathbb{Z}}_{p^2}^n|a_i⩽x_i<a_i+m_i,1⩽i⩽n\}$ centered about the origin, where $a_i,m_i\in \mathbb{Z}$, and $0<m_i<p^2$ for $1⩽i⩽n$.

MSC:11E04, 11E08, 11E12, 11P21.

1 Introduction

Let $Q(\mathbf{x})=Q(x_1,x_2,\dots ,x_n)={\sum }_{1⩽i⩽j⩽n}{a}_{ij}{x}_i{x}_j$ be a quadratic form with integer coefficients in n-variables, and $V=V_{p^2}(Q)$ the algebraic subset of ${\mathbb{Z}}_{p^2}^n$ defined by the equation $Q(\mathbf{x})=0$. When n is even, we let ${\mathrm{\Delta }}_p(Q)=({(-1)}^{n/2}det{A}_Q/p)$ if $p\nmid det{A}_Q$ and ${\mathrm{\Delta }}_p(Q)=0$ if $p|det{A}_Q$, where $(\cdot /p)$ denotes the Legendre-Jacobi symbol and $A_Q$ is the $n\times n$ defining matrix for $Q(\mathbf{x})$. Our interest in this paper is in the problem of finding points in V with the variables restricted to a box of the type

$$\mathcal{B}=\{\mathbf{x}\in {\mathbb{Z}}_{p^2}^n|a_i⩽x_i<a_i+m_i,1⩽i⩽n\},$$

(1)

where $a_i,m_i\in \mathbb{Z}$, and $0<m_i<p^2$ for $1⩽i⩽n$. Consider the congruence

$$Q(\mathbf{x})\equiv 0(mod{p}^2).$$

(2)

The final result of this paper is stated in the following theorem.

Theorem 1 Suppose n is even, Q is nonsingular $(modp)$, and $V_{p^2,\mathbb{Z}}=V_{p^2,\mathbb{Z}}(Q)$ is the set of integer solutions of the congruence (2). Then for any box ℬ of type (1) centered about the origin, if ${\mathrm{\Delta }}_p=\pm 1$,

$$|\mathcal{B}\cap V_{p^2}|⩽{\gamma }_n(\frac{|\mathcal{B}|}{p^2}+p^n),$$

(3)

where the brackets $||$ are used to denote the cardinality of the set inside the brackets, and

$${\gamma }_n=\{\begin{array}{cc}{2}^n(1+\frac{2^{(n/2)+1}}{p}),\hfill & \mathrm{\Delta }=-1,\hfill \\ 2^n(1+2^{(n/2)+1}),\hfill & \mathrm{\Delta }=+1.\hfill \end{array}$$

We shall devote the rest of Section 4 to the proof of Theorem 1. If V is the set of zeros of a ‘nonsingular’ quadratic form $Q(\mathbf{x})(modp)$, then one can show that

$$|V\cap \mathcal{B}|=\frac{|\mathcal{B}|}{p}+O\left(p^{n/2}{(logp)}^{2n}\right),$$

(4)

for any box ℬ (see [1]). It is apparent from (4) that $|V\cap \mathcal{B}|$ is nonempty provided

$$|\mathcal{B}|\gg p^{(n/2)+1}{(logp)}^{2n}.$$

For any x, y in ${\mathbb{Z}}_{p^2}^n$, we let $\mathbf{x}\cdot \mathbf{y}$ denote the ordinary dot product, $\mathbf{x}\cdot \mathbf{y}={\sum }_{i=1}^n{x}_i{y}_i$. For any $x\in {\mathbb{Z}}_{p^2}$, let $e_{p^2}(x)=e^{2\pi ix/p^2}$. We use the abbreviation ${\sum }_{\mathbf{x}}={\sum }_{\mathbf{x}\in {\mathbb{Z}}_{p^2}^n}$ for complete sums. The key ingredient in obtaining the identity in (4) is a uniform upper bound on the function

$$\varphi (V,\mathbf{y})=\{\begin{array}{cc}{\sum }_{\mathbf{x}\in V}{e}_{p^2}(\mathbf{x}\cdot \mathbf{y})\hfill & \text{for }\mathbf{y}\ne \mathbf{0},\hfill \\ |V|-p^{2(n-1)}\hfill & \text{for }\mathbf{y}=\mathbf{0}.\hfill \end{array}$$

(5)

In order to show that $\mathcal{B}\cap V$ is nonempty we can proceed as follows. Let $\alpha (\mathbf{x})$ be a complex valued function on ${\mathbb{Z}}_{p^2}^n$ such that $\alpha (\mathbf{x})⩽0$ for all x not in ℬ. If we can show that ${\sum }_{\mathbf{x}\in V}\alpha (\mathbf{x})>0$, then it will follow that $\mathcal{B}\cap V$ is nonempty. Now $\alpha (\mathbf{x})$ has a finite Fourier expansion

$$\alpha (\mathbf{x})=\sum _{\mathbf{y}}a(\mathbf{y})e_{p^2}(\mathbf{y}\cdot \mathbf{x}),$$

where

$$a(\mathbf{y})=p^{-2n}\sum _{\mathbf{x}}\alpha (\mathbf{x})e_{p^2}(-\mathbf{y}\cdot \mathbf{x}),$$

for all $\mathbf{y}\in {\mathbb{Z}}_{p^2}^n$. Thus

$$\begin{array}{rl}\sum _{\mathbf{x}\in V}\alpha (\mathbf{x})& =\sum _{\mathbf{x}\in V}\sum _{\mathbf{y}}a(\mathbf{y})e_{p^2}(\mathbf{y}\cdot \mathbf{x})\\ =\sum _{\mathbf{y}}a(\mathbf{y})\sum _{\mathbf{x}\in V}{e}_{p^2}(\mathbf{y}\cdot \mathbf{x})\\ =a(\mathbf{0})|V|+\sum _{\mathbf{y}\ne \mathbf{0}}a(\mathbf{y})\sum _{\mathbf{x}\in V}{e}_{p^2}(\mathbf{y}\cdot \mathbf{x}).\end{array}$$

Since $a(\mathbf{0})=p^{-2n}{\sum }_{\mathbf{x}}\alpha (\mathbf{x})$, we obtain

$$\sum _{\mathbf{x}\in V}\alpha (\mathbf{x})=p^{-2n}|V|\sum _{\mathbf{x}}\alpha (\mathbf{x})+\sum _{\mathbf{y}\ne \mathbf{0}}a(\mathbf{y})\varphi (V,\mathbf{y}),$$

(6)

where $\varphi (V,\mathbf{y})$ is defined by (5). A variation of (6) that is sometimes more useful is

$$\sum _{\mathbf{x}\in V}\alpha (\mathbf{x})=p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})+\sum _{\mathbf{y}}a(\mathbf{y})\varphi (V,\mathbf{y}),$$

(7)

which is obtained from (6) by noticing that $|V|=\varphi (V,\mathbf{0})+p^{2(n-1)}$, whence

$$\begin{array}{rl}\sum _{\mathbf{x}\in V}\alpha (\mathbf{x})& =a(\mathbf{0})[\varphi (V,\mathbf{0})+p^{2(n-1)}]+\sum _{\mathbf{y}\ne \mathbf{0}}a(\mathbf{y})\varphi (V,\mathbf{y})\\ =p^{2n-2}a(\mathbf{0})+\sum _{\mathbf{y}}a(\mathbf{y})\varphi (V,\mathbf{y}).\end{array}$$

Equations (6) and (7) express the ‘incomplete’ sum ${\sum }_{\mathbf{x}\in V}\alpha (\mathbf{x})$ as a fraction of the ‘complete’ sum ${\sum }_{\mathbf{x}}\alpha (\mathbf{x})$ plus an error term. In general $|V|\approx p^{2(n-1)}$ so that the fractions in the two equations are about the same. In fact, if V is defined by a ‘nonsingular’ quadratic form $Q(\mathbf{x})$ then $|V|=p^{2(n-1)}+O(p^n)$. (That is, $|\varphi (V,\mathbf{0})|\ll p^n$.)

To show that ${\sum }_{\mathbf{x}\in V}\alpha (\mathbf{x})$ is positive, it suffices to show that the error term is smaller in absolute value than the (positive) main term on the right-hand side of (6) or (7). One tries to make an optimal choice of $\alpha (\mathbf{x})$ in order to minimize the error term. Special cases of (6) and (7) have appeared a number of times in the literature for different types of algebraic sets V; see Chalk [2], Tietäväinen [3], and Myerson [4]. The first case treated was to let $\alpha (\mathbf{x})$ be the characteristic function ${\chi }_S(\mathbf{x})$ of a subset S of ${\mathbb{Z}}_{p^2}^n$, whence (7) gives rise to formulas of the type

$$|V\cap S|=p^{-2}|S|+\text{Error}.$$

Equation (4) is obtained in this manner. Particular attention has been given to the case where $S=\mathcal{B}$, a box of points in ${\mathbb{Z}}_{p^2}^n$. Another popular choice for α is to let it be a convolution of two characteristic functions, $\alpha ={\chi }_S\ast {\chi }_T$ for $S,T\subseteq {\mathbb{Z}}_{p^2}^n$. We recall that if $\alpha (\mathbf{x})$, $\beta (\mathbf{x})$ are complex valued functions defined on ${\mathbb{Z}}_{p^2}^n$, then the convolution of $\alpha (\mathbf{x})$, $\beta (\mathbf{x})$, written $\alpha \ast \beta (\mathbf{x})$, is defined by

$$\alpha \ast \beta (\mathbf{x})=\sum _{\mathbf{u}}\alpha (\mathbf{u})\beta (\mathbf{x}-\mathbf{u})=\sum _{\mathbf{u}+\mathbf{v}=\mathbf{x}}\alpha (\mathbf{u})\beta (\mathbf{v}),$$

for $\mathbf{x}\in {\mathbb{Z}}_{p^2}^n$. If we take $\alpha (\mathbf{x})={\chi }_S\ast {\chi }_T(\mathbf{x})$ then it is clear from the definition that $\alpha (\mathbf{x})$ is the number of ways of expressing x as a sum $\mathbf{s}+\mathbf{t}$ with $\mathbf{s}\in S$ and $\mathbf{t}\in T$. Moreover, $(S+T)\cap V$ is nonempty if and only if ${\sum }_{\mathbf{x}\in V}\alpha (\mathbf{x})>0$.

We make use of a number of basic properties of finite Fourier series, which are listed below. They are based on the orthogonality relationship,

$$\sum _{\mathbf{x}\in {\mathbb{Z}}_{p^2}^n}{e}_{p^2}(\mathbf{x}\cdot \mathbf{y})=\{\begin{array}{cc}{p}^{2n}\hfill & \text{if }\mathbf{y}=\mathbf{0},\hfill \\ 0\hfill & \text{if }\mathbf{y}\ne \mathbf{0},\hfill \end{array}$$

and they can be routinely checked. By viewing ${\mathbb{Z}}_{p^2}^n$ as a ℤ module, the Gauss sum

$$S_p(Q,\mathbf{y})=\sum _{\mathbf{x}\in {\mathbb{Z}}_{p^2}^n}{e}_{p^2}(Q(\mathbf{x})+\mathbf{y}\cdot \mathbf{x}),$$

is well defined whether we take $\mathbf{y}\in {\mathbb{Z}}^n$ or $\mathbf{y}\in {\mathbb{Z}}_{p^2}^n$. Let $\alpha (\mathbf{x})$, $\beta (\mathbf{x})$ be complex valued functions on ${\mathbb{Z}}_{p^2}^n$ with Fourier expansions

$$\alpha (\mathbf{x})=\sum _{\mathbf{y}}a(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y}),\beta (\mathbf{x})=\sum _{\mathbf{y}}b(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y}).$$

Then

$$\alpha \ast \beta (\mathbf{x})=\sum _{\mathbf{y}}{p}^{2n}a(\mathbf{y})b(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y}),$$

(8)

$$\alpha \beta (\mathbf{x})=\alpha (\mathbf{x})\beta (\mathbf{x})=\sum _{\mathbf{y}}(a\ast b)(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y}),$$

(9)

$$\sum _{\mathbf{x}}(\alpha \ast \beta )(\mathbf{x})=(\sum _{\mathbf{x}}\alpha (\mathbf{x}))(\sum _{\mathbf{x}}\beta (\mathbf{x})),$$

(10)

$$\sum _{\mathbf{x}}|(\alpha \ast \beta )(\mathbf{x})|⩽\left(\sum _{\mathbf{x}}\right|\alpha (\mathbf{x})\left|\right)\left(\sum _{\mathbf{x}}\right|\beta (\mathbf{x})\left|\right),$$

(11)

$$\sum _{\mathbf{y}}|a(\mathbf{y}){|}^2=p^{-2n}\sum _{\mathbf{x}}|\alpha (\mathbf{x}){|}^2.$$

(12)

The last identity is Parseval’s equality.

2 Fundamental identity

Let $Q(\mathbf{x})=Q(x_1,\dots ,x_n)$ be a quadratic form with integer coefficients and p be an odd prime. Consider the congruence (2):

$$Q(\mathbf{x})\equiv 0(mod{p}^2).$$

Using identities for the Gauss sum $S={\sum }_{x=1}^{p^2}{e}_{p^2}(a{x}^2+bx)$, one obtains the following.

Lemma 1 ([[5], Lemma 2.3])

Suppose n is even, Q is nonsingular modulo p, and $\mathrm{\Delta }={\mathrm{\Delta }}_p(Q)$. For $\mathbf{y}\in {\mathbb{Z}}^n$, put ${\mathbf{y}}^{\prime }=\frac{1}{p}\mathbf{y}$ in case $p|\mathbf{y}$. Then for any y,

$$\varphi (V,\mathbf{y})=\{\begin{array}{cc}{p}^n-p^{n-1}\hfill & \mathit{\text{if }}p\nmid y_i\mathit{\text{ for some }}i\mathit{\text{ and }}{p}^2|Q^{\ast }(\mathbf{y}),\hfill \\ -p^{n-1}\hfill & \mathit{\text{if }}p\nmid y_i\mathit{\text{ for some }}i\mathit{\text{ and }}p\parallel Q^{\ast }(\mathbf{y}),\hfill \\ 0\hfill & \mathit{\text{if }}p\nmid y_i\mathit{\text{ for some }}i\mathit{\text{ and }}p\nmid Q^{\ast }(\mathbf{y}),\hfill \\ -\mathrm{\Delta }{p}^{(3n/2)-2}+p^{n-1}(p-1)\hfill & \mathit{\text{if }}p|y_i\mathit{\text{ for all }}i\mathit{\text{ and }}p\nmid Q^{\ast }({\mathbf{y}}^{\prime }),\hfill \\ \mathrm{\Delta }(p-1)p^{(3n/2)-2}+p^{n-1}(p-1)\hfill & \mathit{\text{if }}p|y_i\mathit{\text{ for all }}i\mathit{\text{ and }}p|Q^{\ast }({\mathbf{y}}^{\prime }),\hfill \end{array}$$

where $Q^{\ast }$ is the quadratic form associated with the inverse of the matrix for Q modp.

Back to (7): we saw the identity

$$\sum _{\mathbf{x}\in V}\alpha (\mathbf{x})=p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})+\sum _{\mathbf{y}\ne \mathbf{0}}a(\mathbf{y})\varphi (V,\mathbf{y}).$$

Inserting the value $\varphi (V,\mathbf{y})$ in Lemma 1 yields (see [6]) the following.

Lemma 2 (The fundamental identity)

For any complex valued $\alpha (\mathbf{x})$ on ${\mathbb{Z}}_{p^2}^n$,

$$\begin{array}{rl}\sum _{\mathbf{x}\in V}\alpha (\mathbf{x})=& p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})+p^n\sum _{p^2|Q^{\ast }(\mathbf{y})}a(\mathbf{y})-p^{n-1}\sum _{p|Q^{\ast }(\mathbf{y})}a(\mathbf{y})\\ -\mathrm{\Delta }{p}^{(3n/2)-2}\sum _{{\mathbf{y}}^{\prime }(modp)}^{p}a\left(p{\mathbf{y}}^{\prime }\right)+\mathrm{\Delta }{p}^{(3n/2)-1}\underset{{\mathbf{y}}^{\prime }(modp)}{\sum _{p|Q^{\ast }({\mathbf{y}}^{\prime })}}a\left(p{\mathbf{y}}^{\prime }\right).\end{array}$$

(13)

3 Auxiliary lemma on estimating the sum ${\sum }_{y_i=1}^{p}a(p\mathbf{y})$

For later reference, we construct the following lemma on estimating the sum ${\sum }_{y_i}^{p}a(p\mathbf{y})$. Let ℬ be a box of points in ${\mathbb{Z}}^n$ as in (1) centered about the origin with all $m_i⩽p^2$, and view this box as a subset of ${\mathbb{Z}}_{p^2}^n$. Let ${\chi }_{\mathcal{B}}$ be its characteristic function with Fourier expansion ${\chi }_{\mathcal{B}}(\mathbf{x})={\sum }_{\mathbf{y}}{a}_{\mathcal{B}}(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y})$. Let $\alpha (\mathbf{x})={\chi }_{\mathcal{B}}\ast {\chi }_{\mathcal{B}}={\sum }_{\mathbf{y}}a(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y})$. Then for any $\mathbf{y}\in {\mathbb{Z}}_{p^2}^n$,

$$a(\mathbf{y})=p^{-2n}\prod _{i=1}^n\frac{{sin}^2\pi m_i{y}_i/p^2}{{sin}^2\pi y_i/p^2},$$

(14)

where the term in the product is taken to be $m_i$ if $y_i=0$. In particular, if we take $|y_i|⩽p^2/2$ for all i, then

$$a(\mathbf{y})⩽p^{-2n}\prod _{i=1}^{n}min\{m_i^2,{\left(\frac{p^2}{2y_i}\right)}^2\}.$$

Lemma 3 Let ℬ be any box of type (1) and $\alpha (\mathbf{x})={\chi }_{\mathcal{B}}\ast {\chi }_{\mathcal{B}}(\mathbf{x})$. Suppose

$$m_1⩽m_2⩽\cdots ⩽m_l<p⩽m_{l+1}⩽\cdots ⩽m_n.$$

(15)

Then we have

$$\sum _{\mathbf{y}\in {\mathbb{Z}}_p^n}a(p\mathbf{y})⩽2^{n-l}{p}^{l-2n}|\mathcal{B}|\prod _{i=l+1}^n{m}_i.$$

Proof We first observe

$$\begin{array}{rl}\sum _{y_i=1}^{p}a(p\mathbf{y})& =\sum _{y_i=1}^p\sum _{x_i=1}^{p^2}\frac{1}{p^{2n}}\alpha (\mathbf{x})e_{p^2}(-\mathbf{x}\cdot p\mathbf{y})\\ =\sum _{x_i=1}^{p^2}\frac{1}{p^{2n}}\alpha (\mathbf{x})\sum _{y_i=1}^p{e}_p(-\mathbf{x}\cdot \mathbf{y})\\ =\underset{\mathbf{x}\equiv \mathbf{0}(modp)}{\overset{p^2}{\sum _{x_i=1}}}\frac{p^n}{p^{2n}}\alpha (\mathbf{x})\\ =\frac{1}{p^n}\sum _{\mathbf{x}\equiv \mathbf{0}(modp)}\alpha (\mathbf{x})\\ =\frac{1}{p^n}\underset{\mathbf{u}+\mathbf{v}\equiv \mathbf{0}(modp)}{\sum _{\mathbf{u}\in \mathcal{B}}\sum _{\mathbf{v}\in \mathcal{B}}1}\\ ⩽\frac{1}{p^n}\prod _{i=1}^n{m}_i(\left[\frac{m_i}{p}\right]+1).\end{array}$$

(16)

To obtain the last inequality in (16) we must count the number of solutions of the congruence

$$\mathbf{u}+\mathbf{v}\equiv \mathbf{0}(modp),$$

with $\mathbf{u},\mathbf{v}\in \mathcal{B}$. For each choice of v, there are at most ${\prod }_{i=1}^n([m_i/p]+1)$ choices for u. So the total number of solutions is less than or equal to

$$\prod _{i=1}^n{m}_i(\left[\frac{m_i}{p}\right]+1).$$

Using the hypothesis (15) then, continuing from (16), we have

$$\begin{array}{rl}\sum _{y_i=1}^{p}a(p\mathbf{y})& ⩽\frac{1}{p^n}\prod _{i=1}^l{m}_i\prod _{i=l+1}^n{m}_i(\frac{m_i}{p}+1)\\ ⩽\frac{|\mathcal{B}|}{p^n}\prod _{i=l+1}^n\left(\frac{2m_i}{p}\right)⩽\frac{2^{n-l}|\mathcal{B}|}{p^{2n-l}}\prod _{i=l+1}^n{m}_i.\end{array}$$

The lemma is established. □

4 Proof of Theorem 1

As we mentioned before our interest in this paper is in determining the number of solutions of the congruence (2):

$$Q(\mathbf{x})\equiv 0(mod{p}^2),$$

with $\mathbf{x}\in \mathcal{B}$, the box of points in ${\mathbb{Z}}^n$ given by (1):

$$\mathcal{B}=\{\mathbf{x}\in {\mathbb{Z}}^n|a_i⩽x_i<a_i+m_i,1⩽i⩽n\},$$

where $a_i,m_i\in \mathbb{Z}$, $1⩽m_i⩽p^2$, $1⩽i⩽n$. Then $|\mathcal{B}|={\prod }_{i=1}^n{m}_i$, the cardinality of ℬ. View the box ℬ as a subset of ${\mathbb{Z}}_{p^2}^n$ and let ${\chi }_{\mathcal{B}}$ be the characteristic function with Fourier expansion

$${\chi }_{\mathcal{B}}(\mathbf{x})=\sum _{\mathbf{y}}{a}_{\mathcal{B}}(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y}).$$

Lemma 4 Let p be an odd prime, $V_{p^2}=V_{p^2}(Q)$ be the set of zeros of (2) in ${\mathbb{Z}}_{p^2}^n$, and ℬ be a box as given in (1) centered at the origin with all $m_i⩽p^2$. If ${\mathrm{\Delta }}_p=-1$, then

$$|\mathcal{B}\cap V_{p^2}|⩽2^n{\gamma }_n^{\prime }(\frac{|\mathcal{B}|}{p^2}+p^n),$$

where

$${\gamma }_n^{\prime }=1+\frac{2^{(n/2)+1}}{p}.$$

Proof We begin by writing (13); we have the fundamental identity $(mod{p}^2)$:

$$\begin{array}{rl}\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})=& p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})+p^n\underset{p^2|Q^{\ast }(\mathbf{y})}{\overset{p^2}{\sum _{y_i=1}}}a(\mathbf{y})-p^{n-1}\underset{p|Q^{\ast }(\mathbf{y})}{\overset{p^2}{\sum _{y_i=1}}}a(\mathbf{y})\\ -\mathrm{\Delta }{p}^{(3n/2)-2}\sum _{y_i^{\prime }=1}^{p}a\left(p{\mathbf{y}}^{\prime }\right)+\mathrm{\Delta }{p}^{(3n/2)-1}\underset{p|Q^{\ast }({\mathbf{y}}^{\prime })}{\overset{p}{\sum _{y_i^{\prime }=1}}}a\left(p{\mathbf{y}}^{\prime }\right).\end{array}$$

Set $\alpha ={\chi }_{\mathcal{B}}\ast {\chi }_{\mathcal{B}}={\sum }_{\mathbf{y}}a(\mathbf{y})e_{p^2}(\mathbf{x}\cdot \mathbf{y})$. Then the Fourier coefficients of $\alpha (\mathbf{x})$ are given by $a(\mathbf{y})=p^{2n}{a}_{\mathcal{B}}^2(\mathbf{y})$ and, since ℬ is centered at the origin, these are positive real numbers. By Parseval’s identity we have

$$\sum _{\mathbf{y}}|a(\mathbf{y})|=p^{2n}\sum _{\mathbf{y}}|a_{\mathcal{B}}(\mathbf{y}){|}^2=\sum _{\mathbf{y}}|{\chi }_{\mathcal{B}}(\mathbf{y}){|}^2=|\mathcal{B}|.$$

(17)

Thus, it follows from (17) that

$$p^n\underset{p^2|Q^{\ast }(\mathbf{y})}{\overset{p^2}{\sum _{y_i=1}}}a(\mathbf{y})⩽p^n\sum _{\mathbf{y}}|a(\mathbf{y})|⩽p^n|\mathcal{B}|.$$

(18)

Notice that the main term in (13) is

$$p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})=p^{-2}\sum _{\mathbf{x}}{\chi }_{\mathcal{B}}\ast {\chi }_{\mathcal{B}}(\mathbf{x})=\frac{|\mathcal{B}{|}^2}{p^2}.$$

(19)

By Lemma 3, we have

$$p^{(3n/2)-2}\sum _{y_i^{\prime }=1}^{p}a\left(p{\mathbf{y}}^{\prime }\right)⩽2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|\prod _{i=l+1}^n{m}_i$$

(20)

and

$$p^{(3n/2)-1}\underset{p|Q^{\ast }({\mathbf{y}}^{\prime })}{\overset{p}{\sum _{y_i^{\prime }=1}}}a\left(p{\mathbf{y}}^{\prime }\right)⩽p^{(3n/2)-1}\sum _{{\mathbf{y}}^{\prime }}a\left(p{\mathbf{y}}^{\prime }\right)⩽2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|\prod _{i=l+1}^n{m}_i,$$

(21)

where l, as defined before, is such that

$$m_1⩽m_2⩽\cdots ⩽m_l<p⩽m_{l+1}⩽\cdots ⩽m_n.$$

Now going back to (13), if $\mathrm{\Delta }=-1$, we have

$$\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})⩽p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})+p^n\underset{p^2|Q^{\ast }(\mathbf{y})}{\overset{p^2}{\sum _{y_i=1}}}a(\mathbf{y})+p^{(3n/2)-2}\sum _{y_i^{\prime }=1}^{p}a\left(p{\mathbf{y}}^{\prime }\right).$$

(22)

Then, by the equality (19) and the inequalities in (18) and (20), we obtain

$$\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})⩽\frac{|\mathcal{B}{|}^2}{p^2}+p^n|\mathcal{B}|+2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|\prod _{i=l+1}^n{m}_i.$$

(23)

We next determine which of the terms $|\mathcal{B}{|}^2/p^2$, $p^n|\mathcal{B}|$, and $2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|{\prod }_{i=l+1}^n{m}_i$ in (23) is the dominant term. We consider two cases:

Case (i): Suppose $l⩽\frac{n}{2}-1$. Then compare

$$\begin{array}{r}\frac{2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|{\prod }_{i=l+1}^n{m}_i}{|\mathcal{B}{|}^2/p^2}\\ =\frac{1}{|\mathcal{B}|}{p}^{l-(n/2)}{2}^{n-l}\prod _{i=l+1}^n{m}_i=\frac{p^{l-(n/2)}{2}^{n-l}}{{\prod }_{i=1}^l{m}_i}\\ ⩽2^{n-l}{p}^{l-(n/2)}=2^n{\left(\frac{p}{2}\right)}^l{p}^{-n/2}⩽2^n{\left(\frac{p}{2}\right)}^{(n/2)-1}{p}^{-n/2}⩽2^{(n/2)+1}\cdot \frac{1}{p},\end{array}$$

which implies that

$$2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|\prod _{i=l+1}^n{m}_i⩽\frac{2^{(n/2)+1}}{p}\frac{|\mathcal{B}{|}^2}{p^2}.$$

Case (ii): Suppose $l⩾\frac{n}{2}$. Then compare

$$\begin{array}{r}\frac{2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|{\prod }_{i=l+1}^n{m}_i}{p^n|\mathcal{B}|}\\ =2^{n-l}{p}^{l-(3n/2)-2}\prod _{i=l+1}^n{m}_i\\ ⩽2^{n-l}{p}^{l-(3n/2)-2}{p}^{2(n-l)}=2^{n-l}{p}^{n/2-2-l}⩽\frac{2^{n/2}}{p^2},\end{array}$$

which leads to

$$2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|\prod _{i=l+1}^n{m}_i⩽\frac{2^{n/2}}{p^2}{p}^n|\mathcal{B}|.$$

So for any l, always we have

$$2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|\prod _{i=l+1}^n{m}_i⩽(\frac{2^{(n/2)+1}}{p}\frac{|\mathcal{B}{|}^2}{p^2}+\frac{2^{n/2}}{p^2}{p}^n|\mathcal{B}|).$$

Returning to (23), we now can write

$$\begin{array}{rl}\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})& ⩽\frac{|\mathcal{B}{|}^2}{p^2}+p^n|\mathcal{B}|+2^{n-l}{p}^{l-(n/2)-2}|\mathcal{B}|\prod _{i=l+1}^n{m}_i\\ ⩽\frac{|\mathcal{B}{|}^2}{p^2}+p^n|\mathcal{B}|+\frac{2^{(n/2)+1}}{p}\frac{|\mathcal{B}{|}^2}{p^2}+\frac{2^{n/2}}{p^2}{p}^n|\mathcal{B}|\\ =(1+\frac{2^{(n/2)+1}}{p})\frac{|\mathcal{B}{|}^2}{p^2}+(1+\frac{2^{n/2}}{p^2})p^n|\mathcal{B}|\\ ⩽{\gamma }_n^{\prime }(\frac{|\mathcal{B}{|}^2}{p^2}+p^n|\mathcal{B}|),\end{array}$$

(24)

where ${\gamma }_n^{\prime }=1+(2^{(n/2)+1}/p)$. On the other hand,

$$\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})⩾\frac{1}{2^n}|\mathcal{B}||V_{p^2}\cap \mathcal{B}|.$$

(25)

Hence it follows by combining (24) and (25) we find that

$$|\mathcal{B}\cap V_{p^2}|⩽2^n{\gamma }_n^{\prime }(\frac{|\mathcal{B}|}{p^2}+p^n).$$

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Lemma 5 Let p be an odd prime, $V_{p^2}=V_{p^2}(Q)$ be the set of zeros of (2) in ${\mathbb{Z}}_{p^2}^n$, and ℬ be a box as given in (1) centered at the origin with all $m_i⩽p^2$. If ${\mathrm{\Delta }}_p=+1$, then

$$|\mathcal{B}\cap V_{p^2}|⩽2^n{\gamma }_n^{″}(\frac{|\mathcal{B}|}{p^2}+p^n),$$

where

$${\gamma }_n^{″}=1+2^{(n/2)+1}.$$

Proof If ${\mathrm{\Delta }}_p=+1$, again by (13), we have

$$\begin{array}{rl}\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})& ⩽p^{-2}\sum _{\mathbf{x}}\alpha (\mathbf{x})+p^n\sum _{\mathbf{y}}|a(\mathbf{y})|+p^{(3n/2)-1}\sum _{\mathbf{y}(modp)}a(p\mathbf{y})\\ ⩽\frac{|\mathcal{B}{|}^2}{p^2}+p^n|\mathcal{B}|+2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|\prod _{i=l+1}^n{m}_i.\end{array}$$

(26)

We do a similar investigation (as before) to determine which of the terms $|\mathcal{B}{|}^2/p^2$, $p^n|\mathcal{B}|$, and $2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|{\prod }_{i=l+1}^n{m}_i$ of the inequality (26) is the dominant term. In case (i) we find

$$\frac{2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|{\prod }_{i=l+1}^n{m}_i}{|\mathcal{B}{|}^2/p^2}⩽2^{(n/2)+1},$$

which means that

$$2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|\prod _{i=l+1}^n{m}_i⩽2^{(n/2)+1}\frac{|\mathcal{B}{|}^2}{p^2}.$$

And in case (ii) we find

$$\frac{2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|{\prod }_{i=l+1}^n{m}_i}{p^n|\mathcal{B}|}⩽2^{n/2}/p,$$

which gives us that

$$2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|\prod _{i=l+1}^n{m}_i⩽(2^{n/2}/p)p^n|\mathcal{B}|.$$

Hence for any l, we always have

$$2^{n-l}{p}^{l-(n/2)-1}|\mathcal{B}|\prod _{i=l+1}^n{m}_i⩽(2^{(n/2)+1}\frac{|\mathcal{B}{|}^2}{p^2}+\frac{2^{n/2}}{p}{p}^n|\mathcal{B}|).$$

Now on looking at (26), one easily deduces

$$\begin{array}{rl}\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})& ⩽(1+2^{(n/2)+1})\frac{|\mathcal{B}{|}^2}{p^2}+(1+\frac{2^{n/2}}{p})p^n|\mathcal{B}|\\ ⩽{\gamma }_n^{″}(\frac{|\mathcal{B}{|}^2}{p^2}+p^n|\mathcal{B}|),\end{array}$$

(27)

where ${\gamma }_n^{″}=1+2^{(n/2)+1}$. Thus by (27),

$$|\mathcal{B}\cap V_{p^2}|⩽\frac{2^n}{|\mathcal{B}|}\sum _{\mathbf{x}\in V_{p^2}}\alpha (\mathbf{x})⩽{\gamma }_n^{″}{2}^n(\frac{|\mathcal{B}|}{p^2}+p^n).$$

This leads to the proof of the lemma. □

Proof of Theorem 1 This theorem follows immediately from Lemma 4 and Lemma 5 by letting ${\gamma }_n=2^n{\gamma }_n^{\prime }$ if $\mathrm{\Delta }=-1$ and ${\gamma }_n=2^n{\gamma }_n^{″}$ if $\mathrm{\Delta }=+1$. Thus we see from (24) and (27) that for $\mathrm{\Delta }=\pm 1$, one always has

$$|\mathcal{B}\cap V_{p^2}|⩽{\gamma }_n(\frac{|\mathcal{B}|}{p^2}+p^n).$$

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