设整数$ q>2 $.对任意与$ q $互素的整数$ a $, 存在唯一的整数$ b $满足$ 1\leq b\leq q $以及$ ab\equiv 1 (\bmod\ q) $. D. H. Lehmer ^[1]建议研究$ a $与$ b $的奇偶性不同的情形.当$ q = p $为奇素数时, 张文鹏^[2]证明了

$ \begin{eqnarray*} \mathop{\mathop{\mathop{\sum\limits_{a = 1}^p} \mathop{\sum\limits_{b = 1}^p}}_{ab\equiv 1(\bmod p)}}_{2\nmid a+b}1 = \frac{1}{2}p+O\left(p^{\frac{1}{2}} \log^2 p\right). \end{eqnarray*} $

随后在文献[3, 4]中, 张文鹏还得到了渐近公式

$ \begin{eqnarray*} \mathop{\mathop{\mathop{\sum\limits_{a = 1}^q}_{(a, q) = 1} \mathop{\sum\limits_{b = 1}^q}_{(b, q) = 1}}_{ab\equiv 1(\bmod q)}}_{2\nmid a+b}1 = \frac{1}{2}\phi(q)+O\left(q^{\frac{1}{2}} d^2(q) \log^2 q\right), \end{eqnarray*} $

其中$ \phi(q) $为Euler函数, $ d(q) $是除数函数.

设$ k $为非负整数.张文鹏^[5]进一步证明了

$ \begin{eqnarray*} \mathop{\mathop{\mathop{\sum\limits_{a = 1}^q}_{(a, q) = 1} \mathop{\sum\limits_{b = 1}^q}_{(b, q) = 1}}_{ab\equiv 1(\bmod q)}}_{2\nmid a+b}\left(a-b\right)^{2k} = \frac{1}{(2k+1)(2k+2)}\phi(q)q^{2k}+O\left(4^k q^{2k+\frac{1}{2}}d^2(q)\log^2 q\right). \end{eqnarray*} $

此外设$ 0\leq x, y \leq 1 $, 文献[5]中还得到了

$ \begin{eqnarray*} \mathop{\mathop{\mathop{\sum\limits_{a\leq xq}}_{(a, q) = 1}\mathop{\sum\limits_{b\leq yq}}_{(b, q) = 1}}_{ab\equiv 1 (\bmod q)}}_{2\nmid a+b}1 = \frac{1}{2}xy\phi(q)+O\left(q^{\frac{1}{2}}d^2(q)\log^2q\right). \end{eqnarray*} $

设实数$ l, \delta $满足$ l\geq0 $与$ 0<\delta\leq 1 $.王晓瑛与赵秋红在文献[6]中给出了渐近公式

$ \begin{eqnarray*} &&\mathop{\mathop{\mathop{\sum\limits_{a\leq xq}}_{(a, q) = 1}\mathop{\sum\limits_{b\leq yq}}_{(b, q) = 1}}_{ab\equiv 1 (\bmod q)}}_{2\nmid a+b}|a-b|^l = \frac{2x^{l+2}}{(l+1)(l+2)}\varphi(q)q^l+O\left(q^{l+\frac{1}{2}}d(q)\log q(\sigma_{-\frac{1}{2}}(q)+\sigma_{-\frac{3}{2}}(q)\log q)\right), \\ &&\mathop{\mathop{\mathop{\mathop{\sum\limits_{a\leq xq}}_{(a, q) = 1}\mathop{\sum\limits_{b\leq yq}}_{(b, q) = 1}}_{ab\equiv 1 (\bmod q)}}_{2\nmid a+b}}_{|a-b|<\delta q}|a-b|^l = \varphi(q)q^l\left(\frac{x\delta^{l+1}}{l+1}-\frac{\delta^{l+2}}{l+2}\right) +O\left(q^{l+\frac{1}{2}}d(q)\log q(\sigma_{-\frac{1}{2}}(q)+\sigma_{-\frac{3}{2}}(q)\log q)\right). \end{eqnarray*} $

设$ q, c, n $为整数, 满足$ n\geq 2 $, $ q\geq 3 $以及$ (n, q) = (c, q) = 1 $.设$ 0<\delta_1, \delta_2\leq 1 $.陆亚明与易媛^[7]给出了D. H. Lemher问题的推广

$ \begin{eqnarray*} \mathop{\mathop{\mathop{\sum\limits_{a\leq \delta_1q}}_{(a, q) = 1}\mathop{\sum\limits_{b\leq \delta_2q}}_{(b, q) = 1}}_{ab\equiv c(\bmod q)}}_{n\nmid a+b}1 = \left(1-\frac{1}{n}\right)\delta_1\delta_2\phi(q)+O\left(q^\frac{1}{2}d^6(q)\log^2q\right). \end{eqnarray*} $

本文进一步考虑D. H. Lemher问题在短区间的并集上的推广.主要结论如下.

定理1.1   设$ p $是奇素数, $ H>0 $, $ K>0 $, 并设$ I_1^{(j)} $, $ I_2^{(j)} $是$ (0, p) $的子区间, $ 1\leq j\leq J $, 满足$ |I_1^{(j)}| = H $, $ |I_2^{(j)}| = K $, 以及$ I_1^{(j)}\bigcap I_1^{(k)} = \emptyset $, 当$ j\neq k $时.设$ c $, $ n $为整数, 满足$ n\geq 2 $以及$ (n, p) = (c, p) = 1 $.则有

$ \begin{eqnarray*} \sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\nmid x+y}1 = \left(1-\frac{1}{n}\right)\frac{JHK}{p}+O\left(J^{\frac{1}{2}}p^{\frac{1}{2}}\log p\log H\right). \end{eqnarray*} $

推论1.1   当$ J\gg \frac{p^{3}\log^{4}p}{H^{2}K^{2}} $时, 存在$ j\in \left\{ 1,2,\cdots ,J \right\} $, 使得方程$ x \in I_1^{(j)}, $ $ y \in I_2^{(j)}, $ $ xy\equiv c (\bmod\ p), $ $ n\nmid(x+y) $有解.

设$ p>2 $为素数, $ m $与$ n $为任意整数.经典的Kloosterman和的定义为

$ \begin{eqnarray*} K(m, n;p) = \sum\limits_{a = 1}^{p-1}e\left(\frac{ma+n\overline{a}}{p}\right), \end{eqnarray*} $

其中$ e(y) = \hbox{e}^{2\pi iy} $, $ \overline{a} $表示$ a $关于模$ p $的逆, 满足$ 1\leq \overline{a}\leq p-1 $以及$ a\overline{a}\equiv 1 (\bmod\ p) $.由文献[8]可得著名的上界估计$ K(m, n;p)\ll p^{\frac{1}{2}}(m, n, p)^{\frac{1}{2}}. $

Browning和Haynes ^[9]给出了短区间的并集上的Kloosterman和的某种估计式.

引理2.1   设$ p $是奇素数, $ H $为正整数, $ I_{1}, \cdots , I_{J} $是$ (0, p) $的互不相交的子区间, 且对任意$ j $满足$ H/2<|I_{j}|\leq H $.设整数$ l $与$ p $互素, 则有

$ \begin{eqnarray*} \sum\limits_{j = 1}^{J}\left|\sum\limits_{n\in I_{j}}e\left(\frac{l\overline{n}}{p}\right)\right|^{2}\leq 2^{12}p\log^{2}H. \end{eqnarray*} $

为了证明本文的定理, 需要进一步考虑短区间的并集上的Kloosterman和的估计.

引理2.2   设$ p $是奇素数, $ H, n, s, l $为整数, 满足$ 1\leq H\leq p $, $ n\geq 2 $以及$ (n, p) = (l, p) = 1 $.则有

$ \begin{eqnarray*} \sum\limits_{r = 1}^{p}\left|\mathop{\sum\limits_{m = r+1}^{r+H}}_{m\not\equiv 0 (\bmod p)}e\left(\frac{sm}{n}+\frac{l\overline{m}}{p}\right)\right|^2\ll pH. \end{eqnarray*} $

证   由剩余系的性质可得

$ \begin{eqnarray*} &&\sum\limits_{r = 1}^{p}\left|\mathop{\sum\limits_{m = r+1}^{r+H}}_{m\not\equiv 0(\bmod p)} e\left(\frac{sm}{n}+\frac{l\overline{m}}{p}\right)\right|^2 = \sum\limits_{r = 1}^{p}\left|\mathop{\sum\limits_{h = 1}^{H}}_{r\not\equiv -h(\bmod p)} e\left(\frac{s(r+h)}{n}+\frac{l\overline{r+h}}{p}\right)\right|^2\\ & = &\sum\limits_{r = 1}^{p}\mathop{\sum\limits_{h_{1} = 1}^{H}}_{r\not\equiv -h_{1}(\bmod p)} e\left(\frac{s(r+h_{1})}{n}+\frac{l\overline{r+h_{1}}}{p}\right) \mathop{\sum\limits_{h_{2} = 1}^{H}}_{r\not\equiv -h_{2}(\bmod p)} e\left(-\frac{s(r+h_{2})}{n}-\frac{l\overline{r+h_{2}}}{p}\right)\\ & = &\sum\limits_{h_{1} = 1}^{H}e\left(\frac{sh_{1}}{n}\right) \sum\limits_{h_{2} = 1}^{H}e\left(-\frac{sh_{2}}{n}\right)\mathop{\sum\limits_{r = 1}^{p}}_{r\not\equiv -h_{1}, -h_{2}(\bmod p)} e\left(\frac{l(\overline{r+h_{1}}-\overline{r+h_{2}})}{p}\right)\\ & = &\sum\limits_{h_{1} = 1}^{H}e\left(\frac{sh_{1}}{n}\right) \sum\limits_{h_{2} = 1}^{H}e\left(-\frac{sh_{2}}{n}\right) \mathop{\sum\limits_{x = 1}^{p-1}\sum\limits_{y = 1}^{p-1}}_{x-y\equiv h_{1}-h_{2}(\bmod p)} e\left(\frac{l(\overline{x}-\overline{y})}{p}\right)\\ & = &\frac{1}{p}\sum\limits_{h_{1} = 1}^{H}e\left(\frac{sh_{1}}{n}\right) \sum\limits_{h_{2} = 1}^{H}e\left(-\frac{sh_{2}}{n}\right) \sum\limits_{x = 1}^{p-1}\sum\limits_{y = 1}^{p-1} e\left(\frac{l(\overline{x}-\overline{y})}{p}\right) \sum\limits_{a = 1}^{p}e\left(\frac{a(x-y)+a(h_{2}-h_{1})}{p}\right)\\ & = &\frac{1}{p}\sum\limits_{h_{1} = 1}^{H}e\left(\frac{sh_{1}}{n}\right) \sum\limits_{h_{2} = 1}^{H}e\left(-\frac{sh_{2}}{n}\right) \sum\limits_{a = 1}^{p}e\left(\frac{a(h_{2}-h_{1})}{p}\right) \sum\limits_{x = 1}^{p-1}e\left(\frac{ax+l\overline{x}}{p}\right) \sum\limits_{y = 1}^{p-1}e\left(-\frac{ay+l\overline{y}}{p}\right)\\ & = &\frac{1}{p}\sum\limits_{a = 1}^{p}\left|\sum\limits_{h = 1}^{H}e\left(\frac{sh}{n}-\frac{ah}{p}\right)\right|^2 \left|\sum\limits_{x = 1}^{p-1}e\left(\frac{ax+l\overline{x}}{p}\right)\right|^{2}. \end{eqnarray*} $

再由Kloosterman和的经典估计, 可得

$ \begin{eqnarray*} &&\sum\limits_{r = 1}^{p}\left|\mathop{\sum\limits_{m = r+1}^{r+H}}_{m\not\equiv 0(\bmod p)} e\left(\frac{sm}{n}+\frac{l\overline{m}}{p}\right)\right|^2 \ll\sum\limits_{a = 1}^{p}\left|\sum\limits_{h = 1}^{H}e\left(\frac{sh}{n}-\frac{ah}{p}\right)\right|^2\\ & = &\sum\limits_{h_{1} = 1}^{H}e\left(\frac{sh_{1}}{n}\right) \sum\limits_{h_{2} = 1}^{H}e\left(-\frac{sh_{2}}{n}\right) \sum\limits_{a = 1}^{p}e\left(\frac{a(h_{2}-h_{1})}{p}\right)\\ & = &pH. \end{eqnarray*} $

引理2.3   设$ p $是奇素数, $ I_1, \cdots, I_J\subseteq (0, p) $是互不相交的子区间, 且对任意的$ j $满足$ \frac{H}{2}<|I_j|\leq H $, 其中$ H\leq p $为正整数.设$ n, s, l $为整数, 满足$ n\geq 2 $以及$ (n, p) = (l, p) = 1 $.则有

$ \begin{eqnarray*} \sum\limits_{j = 1}^J\left|\sum\limits_{r\in I_j}e\left(\frac{sr}{n}+\frac{l\overline{r}}{p}\right)\right|^2\ll p\log^{2}H. \end{eqnarray*} $

证   利用引理2.2以及文献[9]中的方法, 不难证明引理2.3.为了完整起见, 在此给出详细的证明.

设$ R_{j} $为集合$ I_{j} $中的最小正整数, 并假设$ R_{1}<R_{2}<\cdot\cdot\cdot <R_{j} $.显然有$ R_{j+1}-R_{j}\geq \frac{H}{2} $.定义

$ \begin{eqnarray*} S(r, H) = \mathop{\sum\limits_{m = r+1}^{r+H}}_{m\not\equiv 0 (\bmod p)}e\left(\frac{sm}{n}+\frac{l\overline{m}}{p}\right). \end{eqnarray*} $

容易证明

$ \begin{eqnarray} \sum\limits_{j = 1}^J\left|\sum\limits_{r\in I_j}e\left(\frac{sr}{n}+\frac{l\overline{r}}{p}\right)\right|^2\leq\sum\limits_{j = 1}^J \max\limits_{1\leq h\leq H}\left|S(R_{j}, h)\right|^{2}. \end{eqnarray} $

(2.1)

由$ 1\leq h\leq H $与$ R_{j}-H<r\leq R_{j} $, 可得

$ \begin{eqnarray*} \left|S(R_{j}, h)\right| = \left|S(r, R_{j}-r+h)-S(r, R_{j}-r)\right| \leq 2\max\limits_{1\leq k\leq 2H}\left|S(r, k)\right|. \end{eqnarray*} $

因此有

$ \begin{eqnarray*} \left|S(R_{j}, h)\right| &\leq&\frac{2}{H}\sum\limits_{R_{j}-H<r\leq R_{j}}\max\limits_{1\leq k\leq2H}\left|S(r, k)\right|. \end{eqnarray*} $

再由柯西不等式, 可得

$ \begin{eqnarray*} \left|S(R_{j}, h)\right|^{2} &\leq&\frac{4}{H^{2}}\left(\sum\limits_{R_{j}-H<r\leq R_{j}}\max\limits_{1\leq k\leq2H}\left|S(r, k)\right|\right)^{2}\\ &\leq&\frac{4}{H}\sum\limits_{R_{j}-H<r\leq R_{j}}\max\limits_{1\leq k\leq2H}\left|S(r, k)\right|^{2}. \end{eqnarray*} $

对$ h $取最大值, 并对$ j $求和, 有

$ \begin{eqnarray*} &&\sum\limits_{j = 1}^J\max\limits_{1\leq h\leq H}\left|S(R_{j}, h)\right|^{2} \leq\frac{4}{H}\sum\limits_{j = 1}^J\sum\limits_{R_{j}-H<r\leq R_{j}}\max\limits_{1\leq k\leq2H}\left|S(r, k)\right|^{2}\\ & = &\frac{4}{H}\mathop{\sum\limits_{j = 1}^J}_{2|j}\sum\limits_{R_{j}-H<r\leq R_{j}}\max\limits_{1\leq k\leq 2H}\left|S(r, k)\right|^{2} +\frac{4}{H}\mathop{\sum\limits_{j = 1}^J}_{2\nmid j}\sum\limits_{R_{j}-H<r\leq R_{j}}\max\limits_{1\leq k\leq2H}\left|S(r, k)\right|^{2}. \end{eqnarray*} $

注意到$ R_{j+1}-R_{j}\geq \frac{H}{2} $, 易证

$ \begin{eqnarray} \sum\limits_{j = 1}^J\left|\sum\limits_{r\in I_j}e\left(\frac{sr}{n}+\frac{l\overline{r}}{p}\right)\right|^2 \leq\frac{8}{H}\sum\limits_{r = 1}^{p}\max\limits_{1\leq k\leq 2H}\left|S(r, k)\right|^{2}. \end{eqnarray} $

(2.2)

取正整数$ t $, 满足$ 2H\leq2^{t}\leq4H $.此时显然有$ t+1\leq4\log H $.对任意的$ 1\leq r\leq p $, 选择合适的正整数$ k = k(r)\leq2H $, 使得$ \mathop {\max }\limits_{1 \le h \le 2H} \left|S(r, h)\right| = \left|S(r, k)\right| $.记$ k = \sum\limits_{d \in D} {{2^{t - d}}} $, 其中$ D $是$ [0, t] $之间整数的某个集合, 因此

$ \begin{eqnarray*} S(r, k) = \sum\limits_{d\in D}S(r+v_{r, d}2^{t-d}, 2^{t-d}), \end{eqnarray*} $

其中$ {v_{r, d}} = \sum\limits_{\begin{array}{*{20}{c}} {e \in D}\\ {e \in d} \end{array}} {{2^{d - e}} < {2^d}}. $

利用柯西不等式, 有

$ \begin{eqnarray*} &&\max\limits_{1\leq h\leq2H}\left|S(r, h)\right|^{2} = \left|S(r, k)\right|^{2} = \left(\sum\limits_{d\in D}S(r+v_{r, d}2^{t-d}, 2^{t-d})\right)^{2}\\ &\leq&|D|\sum\limits_{d\in D}\left|S(r+v_{r, d}2^{t-d}, 2^{t-d})\right|^{2}\\ &\leq&(t+1)\sum\limits_{0\leq d\leq t}\sum\limits_{0\leq v< 2^{d}}\left|S(r+v2^{t-d}, 2^{t-d})\right|^{2}. \end{eqnarray*} $

上式两边对$ r $求和, 并结合引理2.2, 有

$ \begin{eqnarray} \sum\limits_{r = 1}^{p}\max\limits_{1\leq h\leq2H}\left|S(r, h)\right|^{2} &\leq&(t+1)\sum\limits_{0\leq d\leq t}\sum\limits_{0\leq v<2^{d}}\sum\limits_{r = 1}^{p}\left|S(r+v2^{t-d}, 2^{t-d})\right|^{2}\\ &\ll&(t+1)\sum\limits_{0\leq d\leq t}2^{d}(p2^{t-d})\ll(t+1)^{2}p2^{t}\\ &\ll&pH\log^{2}H. \end{eqnarray} $

(2.3)

$ \begin{eqnarray} \end{eqnarray} $

(2.4)

结合(2.1)–(2.3)式, 立即可得

$ \begin{eqnarray*} \sum\limits_{j = 1}^J\left|\sum\limits_{r \in I_j}e\left(\frac{sr}{n}+\frac{l\overline{r}}{p}\right)\right|^2 &\ll&p\log^{2}H. \end{eqnarray*} $

易证

$ \begin{eqnarray*} \sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\nmid x+y}1 = \sum\limits_{j = 1}^{J}\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}1-\sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\mid x+y}1. \end{eqnarray*} $

由三角恒等式, 有

$ \begin{eqnarray*} &&\sum\limits_{j = 1}^{J}\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}1 = \frac{1}{p}\sum\limits_{j = 1}^{J}\sum\limits_{x \in I_1^{(j)}}\sum\limits_{y \in I_2^{(j)}}\sum\limits_{|l|\leq\frac{p-1}{2}}e\left(\frac{l(c\overline{x}-y)}{p}\right)\\ & = &\frac{1}{p}\sum\limits_{|l|\leq\frac{p-1}{2}}\sum\limits_{j = 1}^{J}\sum\limits_{y \in I_2^{(j)}}e\left(-\frac{ly}{p}\right) \sum\limits_{x \in I_1^{(j)}}e\left(\frac{lc\overline{x}}{p}\right)\\ & = &\frac{JHK}{p}+\frac{1}{p}\mathop{\sum\limits_{|l|\leq\frac{p-1}{2}}}_{l\neq 0}\sum\limits_{j = 1}^{J}\sum\limits_{y \in I_2^{(j)}}e\left(-\frac{ly}{p}\right) \sum\limits_{x \in I_1^{(j)}}e\left(\frac{lc\overline{x}}{p}\right). \end{eqnarray*} $

再由柯西不等式以及引理2.1可得

$ \begin{eqnarray*} &&\frac{1}{p}\mathop{\sum\limits_{|l|\leq\frac{p-1}{2}}}_{l\neq 0}\sum\limits_{j = 1}^{J}\sum\limits_{y \in I_2^{(j)}}e\left(-\frac{ly}{p}\right) \sum\limits_{x \in I_1^{(j)}}e\left(\frac{lc\overline{x}}{p}\right)\\ &\ll&\frac{1}{p}\mathop{\sum\limits_{|l|\leq\frac{p-1}{2}}}_{l\neq 0}\sum\limits_{j = 1}^{J}\left|\sum\limits_{y \in I_2^{(j)}}e\left(-\frac{ly}{p}\right)\right| \cdot\left|\sum\limits_{x \in I_1^{(j)}}e\left(\frac{lc\overline{x}}{p}\right)\right|\\ &\ll&\mathop{\sum\limits_{|l|\leq\frac{p-1}{2}}}_{l\neq 0}\frac{1}{|l|}\sum\limits_{j = 1}^{J}\left|\sum\limits_{x \in I_1^{(j)}}e\left(\frac{lc\overline{x}}{p}\right)\right| \ll\mathop{\sum\limits_{|l|\leq\frac{p-1}{2}}}_{l\neq 0}\frac{1}{|l|}\cdot J^{\frac{1}{2}}\left(\sum\limits_{j = 1}^{J}\left|\sum\limits_{x \in I_1^{(j)}}e\left(\frac{lc\overline{x}}{p}\right)\right|^2\right)^{\frac{1}{2}}\\ &\ll& J^{\frac{1}{2}}p^{\frac{1}{2}}\log p\log H. \end{eqnarray*} $

因此

$ \begin{eqnarray} \sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\nmid x+y}1 = \frac{JHK}{p}-\sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\mid x+y}1+O\left(J^{\frac{1}{2}}p^{\frac{1}{2}}\log p\log H\right). \end{eqnarray} $

(3.1)

另一方面, 由三角恒等式有

$ \begin{eqnarray} &&\sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\mid x+y}1 = \frac{1}{p}\sum\limits_{j = 1}^{J}\mathop{\sum\limits_{x \in I_1^{(j)}}\sum\limits_{y \in I_2^{(j)}}}_{n\mid x+y}\sum\limits_{l = 1}^{p}e\left(\frac{l(c\overline{x}-y)}{p}\right)\\ & = &\frac{1}{p}\sum\limits_{j = 1}^{J}\mathop{\sum\limits_{x \in I_1^{(j)}}\sum\limits_{y \in I_2^{(j)}}}_{n\mid x+y}1+\frac{1}{p}\sum\limits_{l = 1}^{p-1}\sum\limits_{j = 1}^{J}\mathop{\sum\limits_{x \in I_1^{(j)}}\sum\limits_{y \in I_2^{(j)}}}_{n\mid x+y}e\left(\frac{l(c\overline{x}-y)}{p}\right)\\ & = &\frac{JHK}{np}+\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{j = 1}^{J}\sum\limits_{x \in I_1^{(j)}}\sum\limits_{y \in I_2^{(j)}} e\left(\frac{l(c\overline{x}-y)}{p}\right)\sum\limits_{s = 1}^{n}e\left(\frac{s(x+y)}{n}\right)+O\left(1\right)\\ & = &\frac{JHK}{np}+\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\sum\limits_{j = 1}^{J}\sum\limits_{y \in I_2^{(j)}} e\left(\frac{(sp-ln)y}{np}\right)\sum\limits_{x \in I_1^{(j)}}e\left(\frac{sx}{n}+\frac{lc\overline{x}}{p}\right)+O\left(1\right). \end{eqnarray} $

(3.2)

再由柯西不等式以及引理2.3可得

$ \begin{eqnarray} &&\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\sum\limits_{j = 1}^{J}\sum\limits_{y \in I_2^{(j)}} e\left(\frac{(sp-ln)y}{np}\right)\sum\limits_{x \in I_1^{(j)}}e\left(\frac{sx}{n}+\frac{lc\overline{x}}{p}\right)\\ &\ll&\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\sum\limits_{j = 1}^{J}\left|\sum\limits_{y \in I_2^{(j)}} e\left(\frac{(sp-ln)y}{np}\right)\right|\cdot\left|\sum\limits_{x \in I_1^{(j)}}e\left(\frac{sx}{n}+\frac{lc\overline{x}}{p}\right)\right|\\ &\ll&\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\frac{1}{\left<\frac{sp-ln}{np}\right>} \sum\limits_{j = 1}^{J}\left|\sum\limits_{x \in I_1^{(j)}}e\left(\frac{sx}{n}+\frac{lc\overline{x}}{p}\right)\right|\\ &\ll&\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\frac{1}{\left<\frac{sp-ln}{np}\right>}\cdot J^{\frac{1}{2}}\left( \sum\limits_{j = 1}^{J}\left|\sum\limits_{x \in I_1^{(j)}} e\left(\frac{sx}{n}+\frac{lc\overline{x}}{p}\right)\right|^2\right)^{\frac{1}{2}}\\ &\ll&\frac{1}{np}\sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\frac{1}{\left<\frac{sp-ln}{np}\right>}\cdot J^{\frac{1}{2}}\cdot(p\log^{2}H)^{\frac{1}{2}}, \end{eqnarray} $

(3.3)

其中$ \left<\alpha\right> = \min\left(\{\alpha\}, 1-\{\alpha\}\right) $.记$ m = sp-ln $.则当$ s $取遍模$ n $的完全剩余系, $ l $取遍模$ p $的简化剩余系时, $ m $取遍模$ np $的完全剩余系中与$ p $互素的整数.因此

$ \begin{eqnarray} \sum\limits_{l = 1}^{p-1}\sum\limits_{s = 1}^{n}\frac{1}{\left<\frac{sp-ln}{np}\right>} = \mathop{\sum\limits_{m = 1}^{np}}_{(m, p) = 1} \frac{1}{\left<\frac{m}{np}\right>}\ll np\log(np). \end{eqnarray} $

(3.4)

$ \begin{eqnarray} \end{eqnarray} $

(3.5)

结合(3.1)–(3.4)式, 立即可得

$ \begin{eqnarray*} \sum\limits_{j = 1}^{J}\mathop{\mathop{{\mathop{{\sum}}_{x \in I_1^{(j)}}\mathop{{\sum}}_{y \in I_2^{(j)}}}}_{xy\equiv c (\bmod p)}}_{n\nmid x+y}1 = \left(1-\frac{1}{n}\right)\frac{JHK}{p}+O\left(J^{\frac{1}{2}}p^{\frac{1}{2}}\log p\log H\right). \end{eqnarray*} $

定理1.1证毕.