设整数$h$, $q$满足$q>0$.经典的Dedekind和的定义为

$ S(h, q)=\sum\limits_{r=1}^q\left(\left(\frac{r}{q}\right)\right)\left(\left(\frac{hr}{q}\right)\right), $

其中

$ \begin{eqnarray*} ((x))&=&\left\{\begin{array}{ll} x-[x]-\frac{1}{2},&\text{当}x\text{不是整数}, \\ 0,&\text{当}x\text{是整数}. \end{array}\right. \end{eqnarray*} $

Dedekind和$S(h, q)$在Dedekind $\eta$函数的研究中起着重要作用, 详见文献[1, 2]或者[3]的第3部分.

Berndt^[4]引入了如下的Hardy和:

$ H(h, q)=\sum\limits_{j=1}^{q-1}(-1)^{j+1+\left[\frac{hj}{q}\right]}, $

并研究了其性质. Sitaramachandrarao ^[5]将Hardy和表示成Dedekind和的如下形式:

$ H(h, q)=-8S(h+q, 2q)+4S(h, q). $

在文献[6]中, 徐哲峰与张文鹏研究了短区间上的Hardy和的均值, 并得到了如下渐近公式.

命题1.1 设$p\geq 5$为素数, $\overline{b}$为$b$关于模$p$的乘法逆, 则有

$ \sum\limits_{a\leq\frac{p}{4}}\sum\limits_{b\leq\frac{p}{4}}H(2a\overline{b}, p)=\frac{3}{16}p^2+O(p^{1+\epsilon}), $

其中$\epsilon$为任意小的正数.

Liu^[7]也类似的研究了短区间上的Hardy和, 并得到如下命题.

命题1.2 设$p\geq 5$为素数, 则有

$ \begin{eqnarray*} &&\sum\limits_{a\leq\frac{p}{3}}\sum\limits_{a\leq\frac{p}{3}}H(2a\overline{b}, p)=\frac{1}{5}p^2+O(p^{1+\epsilon}), \\ &&\sum\limits_{a\leq\frac{p}{3}}\sum\limits_{a\leq\frac{p}{4}}H(2a\overline{b}, p)=\frac{27}{320}p^2+O(p^{1+\epsilon}). \end{eqnarray*} $

本文将进一步研究合数模上Hardy和的均值.主要结果如下.

定理1.1 设$q\geq 5$为奇数.则有

$ \sum\limits_{a\leq\frac{q}{4}}\sum\limits_{b\leq\frac{q}{4}}H(2a\overline{b}, q)=\frac{3}{16}q^2\prod\limits_{p^\alpha\parallel q}\frac{\left(1-\frac{1}{p^2}\right)\left(1-\frac{1}{p^{3\alpha}}-\left(1+\frac{1}{p}+\frac{1}{p^2}\right) \left(\frac{1}{p^{2\alpha}}-\frac{1}{p^{3\alpha}}\right)\right)}{\left(1+\frac{1}{p^2}\right)\left(1+\frac{1}{p}+\frac{1}{p^2}\right)} +O\left(q^{1+\epsilon}\right), $

其中$\epsilon$为任意小的正数.

首先将在第2节中把Hardy和的均值表为Dirichlet $L$-函数的均值, 然后在第3节中计算相应的Dirichlet $L$-函数的均值, 最后在第4节中证明定理1.1.

引理2.1 设整数$q, h$满足$q\geq 3$与$(h, q)=1$.则有

$ S(h,q)=\frac{1}{{{\pi }^{2}}q}\sum\limits_{d|q}{\frac{{{d}^{2}}}{\phi (d)}}\sum\limits_{\begin{smallmatrix} \chi \ {\text{mod}}\ d \\ \chi \text{(-1)=-1} \end{smallmatrix}}{\chi (h){{\left| L(1,\chi ) \right|}^{2}}.} $

证  见文献[8]中的引理2.

引理2.2 设$q\geq 3$为奇数, $h$为任意整数, 满足$(h, q)=1$.则有

$ \begin{eqnarray*} H(h, q)=\left\{\begin{array}{ll} \displaystyle{-\frac{16}{\pi ^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d} \mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(h) |L(1, \chi\chi^0_{2d})|^2},&2|h, \\ 0,&2\nmid h, \end{array}\right. \end{eqnarray*} $

其中$\displaystyle\mathop{\sideset{}{^*}\sum_{\chi\bmod m}}_{\chi(-1)=-1}$表示对模$m$的奇原特征求和.

证  由引理2.1有

$ \begin{eqnarray*} H(h, q)&=&-8S(h+q, 2q)+4S(h, q)\\ &=&-\frac{4}{\pi^2q}\sum\limits_{d|{2q}}\frac{d^2}{\phi(d)}\mathop{\sum\limits_{\chi\bmod d}}_{\chi(-1)=-1} \chi(h+q)|L(1, \chi)|^2+\frac{4}{\pi^2q}\sum\limits_{d|{q}}\frac{d^2}{\phi(d)}\mathop{\sum\limits_{\chi\bmod d}}_{\chi(-1)=-1}\chi(h)|L(1, \chi)|^2\\ &=&-\frac{16}{\pi ^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop{\sum\limits_{\chi\bmod d}}_{\chi(-1) =-1}\chi(h+q)\chi_2^0(h+q)|L(1, \chi\chi^0_2)|^2\\ &=&\left\{\begin{array}{ll} \displaystyle{-\frac{16}{\pi ^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop{\sum\limits_{\chi\bmod d}}_{\chi(-1) =-1}\chi(h)|L(1, \chi\chi^0_2)|^2},&2|h, \\ 0,&2\nmid h. \end{array}\right.\\ &=&\left\{\begin{array}{ll} \displaystyle{-\frac{16}{\pi ^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d}\mathop{\sideset{}{^*} \sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(h) |L(1, \chi\chi^0_{2d})|^2},&2|h, \\ 0,&2\nmid h. \end{array}\right. \end{eqnarray*} $

定理2.1设$q\geq 5$为奇数, 则有

$ \begin{eqnarray*} &&\sum\limits_{a\leq\frac{q}{4}}\sum\limits_{b\leq\frac{q}{4}}H(2a\overline{b}, q)\\ &&=\frac{16}{\pi^4q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d}m\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\left(1+2\chi(2)+\chi(4)\right)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2. \end{eqnarray*} $

证  由引理2.2可得

$ \begin{eqnarray*} \sum\limits_{a\leq\frac{q}{4}}\sum\limits_{b\leq\frac{q}{4}}H(2a\overline{b}, q)&=&-\frac{16}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d} \mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2)|L(1, \chi\chi_{2d}^0)|^2 \sum\limits_{a\leq\frac{q}{4}}\chi(a)\sum\limits_{b\leq\frac{q}{4}}\overline{\chi}(b). \end{eqnarray*} $

另一方面, 设$\chi$为模$m$的原特征, 整数$r\geq 1$, $\lambda \in [0, 1)$且$\lambda\neq \frac{r}{m}$.由文献[9]可得特征和的Fourier展式如下:

$ \begin{eqnarray*} \sum\limits_{a\leq \lambda m}\chi(a)=\left\{\begin{array}{ll}\displaystyle \frac{\tau(\chi)}{\pi}\sum\limits_{n=1}^{+\infty}\frac{\overline{\chi}(n)\sin(2\pi n\lambda)}{n}, &\text{如果}\chi(-1)=1;\\ \displaystyle\frac{\tau(\chi)}{\pi i}\sum\limits_{n=1}^{+\infty}\frac{\overline{\chi}(n)(1-\cos(2\pi n\lambda))}{n}, &\text{如果}\chi(-1)=-1, \end{array}\right. \end{eqnarray*} $

其中

$ \displaystyle\tau(\chi)=\sum\limits_{a=1}^m\chi(a)\text{e}\left(\frac{a}{m}\right) $

是Gauss和, $\displaystyle\text{e}(y)=\text{e}^{2\pi iy}$.由此可得

$ \begin{eqnarray*} \sum\limits_{a\leq\frac{q}{4}}\chi(a)&=&\sum\limits_{a\leq m\cdot\frac{q}{m4}}\chi(a)=\sum\limits_{a\leq m\cdot\left\{\frac{q}{4m}\right\}}\chi(a)=\frac{\tau(\chi)}{\pi i}\sum\limits_{n=1}^{+\infty}\frac{\overline{\chi}(n)\left(1-\cos(2\pi n\frac{q}{4m})\right)}{n}\\ &=&\frac{\tau(\chi)}{\pi i}\left(\mathop{\sum\limits_{n=1}^{+\infty}}_{n\equiv 1 (\bmod 2)}\frac{\overline{\chi}(n)}{n}+\mathop{\sum\limits_{n=1}^{+\infty}}_{n\equiv 2 (\bmod 4)}\frac{2\overline{\chi}(n)}{n}\right)\\ &=&\frac{\tau(\chi)}{\pi i}\left((1+\overline{\chi}(2))\sum\limits_{n=1}^{+\infty}\frac{\overline{\chi}\chi_2^0(n)}{n}\right) =\frac{\tau(\chi)}{\pi i}\left(1+\overline{\chi}(2)\right)L(1, \overline{\chi}\chi_2^0). \end{eqnarray*} $

从而

$ \begin{eqnarray*} &&\sum\limits_{a\leq\frac{q}{4}}\sum\limits_{b\leq\frac{q}{4}}H(2a\overline{b}, q)\\ &&=\frac{16}{\pi^4q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d}m\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\left(1+2\chi(2)+\chi(4)\right)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2. \end{eqnarray*} $

引理3.1 设整数$m$, $r$满足$m\ge 2$与$(r, m)=1$, $\chi$为模$m$的Dirichlet特征.则有恒等式

$ \displaystyle\mathop{{\sum}^*}_{\chi \bmod m}\chi(r) =\displaystyle\sum\limits_{d|(m, r - 1)}\mu\left(\frac{m}{d}\right)\phi(d) $

和$J(m)=\displaystyle\sum_{d|m}\mu(d)\phi\left(\frac{m}{d}\right), $这里$\displaystyle\mathop{{\sum}^*}_{\chi \bmod m}$表示对模$m$的所有原特征求和, $J(m)$表示模$m$的原特征的个数.

证  见文献[10]中的引理3.

引理3.2 设$d$为奇数, $m|d$, $\displaystyle{r(n)=\sum_{t|n}\chi_d^0(t)}$.则有

$ \mathop {\mathop \sum \limits_{n = 1}^{ + \infty } }\limits_{(n,2m) = 1} \frac{r^2(n)}{n^2}= \frac{3\pi^4}{128}\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right). $

证  利用Euler乘积公式, 有

$ \begin{eqnarray*} \mathop{\sum\limits_{n=1}^{+\infty}}_{(n, 2m)=1}\frac{r^2(n)}{n^2}&=&\prod\limits_{p\nmid 2m}\left(1+\frac{r^2(p)}{p^2}+\frac{r^2(p^2)}{p^4}+\cdots\right) =\frac{27}{80}\prod\limits_{p\nmid m}\left(1+\frac{r^2(p)}{p^2}+\frac{r^2(p^2)}{p^4}+\cdots\right)\\ &=&\frac{27}{80}\prod\limits_{p\nmid d}\left(1+\frac{r^2(p)}{p^2}+\frac{r^2(p^2)}{p^4}+\cdots\right) \mathop{\prod\limits_{p\mid d}}_{p\nmid m}\left(1+\frac{r^2(p)}{p^2}+\frac{r^2(p^2)}{p^4}+\cdots\right)\\ &=&\frac{27}{80}\prod\limits_{p\nmid d}\left(1+\frac{2^2}{p^2}+\frac{3^2}{p^4}+\cdots\right) \mathop{\prod\limits_{p\mid d}}_{p\nmid m}\left(1+\frac{1}{p^2}+\frac{1}{p^4}+\cdots\right)\\ &=&\frac{27}{80}\prod\limits_{p\nmid d}\frac{1+\frac{1}{p^2}}{\left(1-\frac{1}{p^2}\right)^3} \mathop{\prod\limits_{p\mid d}}_{p\nmid m}\frac{1}{1-\frac{1}{p^2}}=\frac{27\zeta^4(2)}{80\zeta(4)}\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right)\\ &=&\frac{3\pi^4}{128}\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right), \end{eqnarray*} $

其中$\zeta(s)$是Riemann zeta函数, 满足$\zeta(2)=\frac{1}{6}\pi^2$, $\zeta(4)=\frac{1}{90}\pi^4$.

定理3.1 设$d$为奇数, $m|d$, $k$为给定的正整数.则有

$ \begin{eqnarray*} &&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2=\frac{3\pi^4}{256}J(m)\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right)+O\left(m^\epsilon\right), \\ &&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2\ll m^\epsilon. \end{eqnarray*} $

证  考虑非负整数$k$.令$A(y,\chi ) = \sum\limits_{N < n \le y} {\chi (n)r(n),} $其中$N$为满足$m\leq N\leq m^4$的参数, 且$\displaystyle{r(n)=\sum_{t|n}\chi_d^0(t)}$.由Abel恒等式可得

$ L(1, \chi\chi_{2d}^0)L(1, \chi\chi_2^0)=\sum\limits_{n=1}^{+\infty}\frac{\chi\chi_2^0(n)r(n)}{n} =\sum\limits_{1\leq n\leq N}\frac{\chi\chi_2^0(n)r(n)}{n}+\int_N^\infty\frac{A(y, \chi\chi_2^0)}{y^2}dy. $

因此有

$ \begin{eqnarray} &&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2\nonumber\\ &&=\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)\left(\sum\limits_{1\leq n_1\leq N} \frac{\overline{\chi}\chi_2^0(n_1)r(n_1)}{n_1}+\int_N^\infty \frac{A(y, \overline{\chi}\chi_2^0)}{y^2}dy\right)\nonumber\\ &&\qquad\times \left(\sum\limits_{1\leq n_2\leq N}\frac{\chi\chi_2^0(n_2)r(n_2)}{n_2}+\int_N^\infty\frac{A(y, \chi\chi_2^0)}{y^2}dy \right)\nonumber\\ &&=\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\sum\limits_{1\leq n_1\leq N}\frac{\overline{\chi}\chi_2^0(n_1)r(n_1)}{n_1}\right) \left(\sum\limits_{1\leq n_2\leq N}\frac{\chi\chi_2^0(n_2)r(n_2)}{n_2}\right)\nonumber\\ &&\qquad+\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\sum\limits_{1\leq n_1\leq N}\frac{\overline{\chi}\chi_2^0(n_1)r(n_1)}{n_1}\right)\left( \int_N^\infty\frac{A(y, \chi\chi_2^0)}{y^2}dy \right)\nonumber\\ &&\qquad+\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\sum\limits_{1\leq n_2\leq N}\frac{\chi\chi_2^0(n_2)r(n_2)}{n_2}\right)\left(\int_N^\infty\frac{A(y, \overline{\chi}\chi_2^0)}{y^2}dy \right)\nonumber\\ &&\qquad+\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\int_N^\infty\frac{A(y, \overline{\chi}\chi_2^0)}{y^2}dy\right)\left(\int_N^\infty\frac{A(y, \chi\chi_2^0)}{y^2}dy\right)\nonumber\\ &&:=M_1+M_2+M_3+M_4. \end{eqnarray} $

(3.1)

首先估计$M_2$, $M_3$与$M_4$.注意到分拆恒等式

$ \begin{eqnarray*} A(y, \chi\chi_2^0)&=&\sum\limits_{n\leq \sqrt{y}}\chi\chi_2^0(n)\sum\limits_{s\leq y/n} \chi\chi_{2d}^0(s)+\sum\limits_{s\leq\sqrt{y}}\chi\chi_{2d}^0(s)\sum\limits_{n\leq y/s}\chi\chi_2^0(n)\\ &&-\sum\limits_{n\leq \sqrt{N}}\chi\chi_2^0(n)\sum\limits_{s\leq N/n}\chi\chi_{2d}^0(s)-\sum\limits_{s\leq\sqrt{N}}\chi\chi_{2d}^0(s)\sum\limits_{n\leq N/s}\chi\chi_2^0(n)\\ &&-\left(\sum\limits_{n\leq \sqrt{y}}\chi\chi_2^0(n)\right) \left(\sum\limits_{n\leq\sqrt{y}}\chi\chi_{2d}^0(n)\right) +\left(\sum\limits_{n\leq \sqrt{N}}\chi\chi_2^0(n)\right)\left(\sum\limits_{n\leq\sqrt{N}}\chi\chi_{2d}^0(n)\right), \\ \end{eqnarray*} $

则有

$ \begin{eqnarray*} \mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}|A(y, \chi\chi_2^0)|\ll y^{\frac{1}{2}}m^{\frac{3}{2}+\epsilon}, \qquad\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}|A(y, \chi\chi_2^0)|^2\ll ym^{2+\epsilon}. \end{eqnarray*} $

因此

$ \begin{eqnarray} M_2&=&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)\left(\sum\limits_{1\leq n_1\leq N}\frac{\overline{\chi}\chi_2^0(n_1)r(n_1)}{n_1}\right)\left( \int_N^\infty\frac{A(y, \chi\chi_2^0)}{y^2}dy \right)\nonumber\\ &\ll&\sum\limits_{1\leq n_1\leq N}n_1^{\epsilon-1}\int_N^\infty\frac{1}{y^2}\left(\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}|A(y, \chi \chi_2^0)|\right)dy\nonumber\\ &\ll&N^\epsilon\int_N^\infty\frac{m^{\frac{3}{2}+\epsilon}}{y^{\frac{3}{2}}}dy \ll\frac{m^{\frac{3}{2}+\epsilon}}{N^{\frac{1}{2}}}. \end{eqnarray} $

(3.2)

同理可得

$ \begin{eqnarray} M_3\ll\frac{m^{\frac{3}{2}+\epsilon}}{N^{\frac{1}{2}}}. \end{eqnarray} $

(3.3)

此外利用Cauchy不等式, 有

$ \begin{eqnarray} M_4&=&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\int_N^\infty\frac{A(y, \overline{\chi}\chi_2^0)}{y^2}dy\right) \left(\int_N^\infty\frac{A(y, \chi\chi_2^0)}{y^2}dy\right)\nonumber\\ &\leq&\int_N^\infty\int_N^\infty\frac{1}{y^2z^2}\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1} |A(y, \overline{\chi}\chi_2^0)||A(z, \chi\chi_2^0)|dydz\nonumber\\ &\ll&\int_N^\infty\int_N^\infty\frac{1}{y^2z^2}\left(\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1} |A(y, \overline{\chi}\chi_2^0)|^2\right)^{\frac{1}{2}} \left(\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}|A(z, \chi\chi_2^0)|^2\right)^{\frac{1}{2}}dydz\nonumber\\ &\ll&\left(\int_N^\infty\frac{1}{y^2}\left(\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1} |A(y, \overline{\chi}\chi_2^0)|^2\right)^{\frac{1}{2}}dy\right)^2\nonumber\\ &\ll&\left(\int_N^\infty \frac{m^{1+\epsilon}}{y^{\frac{3}{2}}}dy\right)^2\ll\frac{m^{2+\epsilon}}{N}. \end{eqnarray} $

(3.4)

因此结合(3.1)-(3.4)可得

$ \begin{eqnarray} &&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2\nonumber\\ &&=\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\sum\limits_{1\leq n_1\leq N}\frac{\overline{\chi}\chi_2^0(n_1)r(n_1)}{n_1}\right) \left(\sum\limits_{1\leq n_2\leq N}\frac{\chi\chi_2^0(n_2)r(n_2)}{n_2}\right)\nonumber\\ &&\qquad+O\left(\frac{m^{\frac{3}{2}+\epsilon}}{N^{\frac{1}{2}}}\right)+ O\left(\frac{m^{2+\epsilon}}{N}\right)\nonumber\\ &&=M_1+O\left(\frac{m^{\frac{3}{2}+\epsilon}}{N^{\frac{1}{2}}}\right)+ O\left(\frac{m^{2+\epsilon}}{N}\right). \end{eqnarray} $

(3.5)

设$(a, m)=1$.由引理3.1有

$ \begin{eqnarray*} \mathop{\mathop{{\sum}^*}_{\chi \bmod m}}_{\chi(-1) =-1}\chi(a)&=& \frac{1}{2}\mathop{{\sum}^*}_{\chi\bmod m}\left(1-\chi(-1)\right)\chi(a)=\frac{1}{2}\mathop{{\sum}^*}_{\chi \bmod m}\chi(a)-\frac{1}{2}\mathop{{\sum}^*}_{\chi \bmod m}\chi(-a)\\ &=&\frac{1}{2}\sum\limits_{s|(m, a-1)}\mu\left(\frac{m}{s}\right)\phi(s) -\frac{1}{2}\sum\limits_{s|(m, a+1)}\mu\left(\frac{m}{s}\right)\phi(s). \end{eqnarray*} $

从而

$ \begin{eqnarray} M_1&=&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k) \left(\sum\limits_{1\leq n_1\leq N}\frac{\overline{\chi}\chi_2^0(n_1)r(n_1)}{n_1}\right) \left(\sum\limits_{1\leq n_2\leq N}\frac{\chi\chi_2^0(n_2)r(n_2)}{n_2}\right)\nonumber\\ &=&\left(\sum\limits_{1\leq n_1\leq N}\frac{\chi_2^0(n_1)r(n_1)}{n_1}\right)\left(\sum\limits_{1\leq n_2\leq N}\frac{\chi_2^0(n_2)r(n_2)}{n_2}\right)\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^kn_2\overline{n}_1)\nonumber\\ &=&\frac{1}{2}\mathop{\sum\limits_{1\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2 \leq N}}_{(n_2, 2m)=1}\frac{r(n_1)r(n_2)}{n_1n_2} \sum\limits_{s|(m, 2^kn_2\overline{n}_1-1)}\mu\left(\frac{m}{s}\right)\phi(s)\nonumber\\ &\qquad&-\frac{1}{2}\mathop{\sum\limits_{1\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2 \leq N}}_{(n_2, 2m)=1}\frac{r(n_1)r(n_2)}{n_1n_2} \sum\limits_{s|(m, 2^kn_2\overline{n}_1+1)}\mu\left(\frac{m}{s}\right)\phi(s)\nonumber\\ &=&\frac{1}{2}\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\mathop{\sum\limits_{1\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2 \leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{r(n_1)r(n_2)}{n_1n_2}\nonumber\\ &\qquad&-\frac{1}{2}\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\mathop{\sum\limits_{1\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2 \leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv -n_1(\bmod s)} \frac{r(n_1)r(n_2)}{n_1n_2}\nonumber\\ &:=&M_{11}-M_{12}. \end{eqnarray} $

(3.6)

注意到估计式$r(n)\ll n^\epsilon$.利用剩余系的性质可得

$ \begin{eqnarray} M_{12}&\ll&N^{\epsilon}\sum\limits_{s|m}\phi(s)\mathop{\mathop{\sum\limits_{1\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2 \leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv -n_1(\bmod s)}\frac{1}{n_1\cdot 2^kn_2}\nonumber\\ &\ll&N^{\epsilon}\sum\limits_{s|m}\phi(s)\mathop{\sum\limits_{0\leq r_1\leq \frac{N}{s}-1}\sum\limits_{l_1=1}^{s-1}\sum\limits_{0\leq r_2\leq \frac{2^kN}{s}-1}\sum\limits_{l_2=1}^{s-1}}_{r_2s+l_2\equiv -(r_1s+l_1)(\bmod s)}\frac{1}{(r_1s+l_1)(r_2s+l_2)}\nonumber\\ &=&N^{\epsilon}\sum\limits_{s|m}\phi(s)\sum\limits_{0\leq r_1\leq \frac{N}{s}-1} \sum\limits_{0\leq r_2\leq \frac{2^kN}{s}-1}\sum\limits_{l=1}^{s-1}\frac{1}{(r_1s+l)(r_2s+s-l)}\nonumber\\ &=&N^{\epsilon}\sum\limits_{s|m}\phi(s)\sum\limits_{0\leq r_1\leq \frac{N}{s}-1} \sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\sum\limits_{l=1}^{s-1}\frac{1}{(r_1s+l)(r_2s-l)}\nonumber\\ &=&N^{\epsilon}\sum\limits_{s|m}\phi(s)\sum\limits_{0\leq r_1\leq \frac{N}{s}-1} \sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\sum\limits_{l=1}^{s-1}\frac{1}{(r_1+r_2)s}\left(\frac{1}{r_1s+l}+\frac{1}{r_2s-l}\right)\nonumber\\ &=&N^{\epsilon}\sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{0\leq r_1\leq \frac{N}{s}-1}\sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\frac{1}{r_1+r_2}\sum\limits_{l=1}^{s-1}\frac{1}{r_1s+l}\nonumber\\ &&+N^{\epsilon}\sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{0\leq r_1\leq \frac{N}{s}-1}\sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\frac{1}{r_1+r_2}\sum\limits_{l=1}^{s-1}\frac{1}{r_2s-l}\nonumber\\ &\ll&N^{\epsilon}\sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\frac{1}{r_2}\sum\limits_{0\leq r_1\leq \frac{N}{s}-1}\sum\limits_{l=1}^{s-1}\frac{1}{r_1s+l}\nonumber\\ &&+N^{\epsilon}\sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{0\leq r_1\leq \frac{N}{s}-1}\frac{1}{r_1+1}\sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\sum\limits_{l=1}^{s-1}\frac{1}{r_2s-l}\nonumber\\ &\ll&N^{\epsilon}\sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{1\leq r_2\leq \frac{2^kN}{s}}\frac{1}{r_2}\sum\limits_{1\leq n_1\leq N}\frac{1}{n_1} +N^{\epsilon}\sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{0\leq r_1\leq \frac{N}{s}-1}\frac{1}{r_1+1}\sum\limits_{1\leq n_2\leq 2^kN}\frac{1}{n_2}\nonumber\\ &\ll&N^{\epsilon}. \end{eqnarray} $

(3.7)

现在考虑$M_{11}$.把对$n_1$和$n_2$的求和式分成下列四种情况讨论:

(ⅰ) $ s\leq n_1\leq N$, $\frac{s-1}{2^k}+1\leq n_2\leq N$;

(ⅱ) $s\leq n_1\leq N$, $1\leq n_2\leq \frac{s-1}{2^k}$;

(ⅲ) $1\leq n_1\leq s-1$, $\frac{s-1}{2^k}+1\leq n_2\leq N$;

(ⅳ) $1\leq n_1\leq s-1$, $1\leq n_2\leq \frac{s-1}{2^k}$.

不难证明

$ \begin{eqnarray} &&\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\mathop{\sum\limits_{s\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{\frac{s-1}{2^k}+1\leq n_2\leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{r(n_1)r(n_2)}{n_1n_2}\ll N^{\epsilon} \sum\limits_{s|m}\phi(s) \mathop{\mathop{\sum\limits_{s\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{\frac{s-1}{2^k}+1\leq n_2\leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{1}{n_1\cdot 2^kn_2}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\phi(s) \mathop{\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\sum\limits_{l_1=1}^{s-1}\sum\limits_{1\leq r_2\leq\frac{2^kN}{s}-1}\sum\limits_{l_2=1}^{s-1}}_{r_2s+l_2\equiv r_1s+l_1 (\bmod s)}\frac{1}{(r_1s+l_1)(r_2s+l_2)}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\phi(s)\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\sum\limits_{1\leq r_2\leq\frac{2^kN}{s}-1}\sum\limits_{l=1}^{s-1}\frac{1}{(r_1s+l)(r_2s+l)}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\phi(s)\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\frac{1}{r_1s}\sum\limits_{1\leq r_2\leq\frac{2^kN}{s}-1}\sum\limits_{l=1}^{s-1}\frac{1}{r_2s+l}\ll N^{\epsilon} \sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\frac{1}{r_1}\sum\limits_{1\leq n_2\leq 2^kN}\frac{1}{n_2}\nonumber\\ &&\ll N^{\epsilon}. \end{eqnarray} $

(3.8)

此外还有

$ \begin{eqnarray} &&\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\mathop{\sum\limits_{s\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2\leq \frac{s-1}{2^k}}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{r(n_1)r(n_2)}{n_1n_2}\ll N^{\epsilon} \sum\limits_{s|m}\phi(s)\mathop{\mathop{\sum\limits_{s\leq n_1 \leq N}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2\leq \frac{s-1}{2^k}}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{1}{n_1\cdot 2^kn_2}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\phi(s) \mathop{\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\sum\limits_{l_1=1}^{s-1}\sum\limits_{1\leq n_2\leq\frac{s-1}{2^k}}}_{2^kn_2\equiv r_1s+l_1 (\bmod s)}\frac{1}{(r_1s+l_1)\cdot 2^kn_2}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\phi(s)\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\sum\limits_{l=1}^{s-1}\frac{1}{(r_1s+l)l}\ll N^{\epsilon} \sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{1\leq r_1\leq\frac{N}{s}-1}\frac{1}{r_1}\sum\limits_{l=1}^{s-1}\frac{1}{l}\nonumber\\ &&\ll N^{\epsilon}, \end{eqnarray} $

(3.9)

以及

$ \begin{eqnarray} &&\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\mathop{\sum\limits_{1\leq n_1 \leq s-1}}_{(n_1, 2m)=1}\mathop{\sum\limits_{\frac{s-1}{2^k}+1\leq n_2\leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{r(n_1)r(n_2)}{n_1n_2}\ll N^{\epsilon} \sum\limits_{s|m}\phi(s) \mathop{\mathop{\sum\limits_{1\leq n_1 \leq s-1}}_{(n_1, 2m)=1}\mathop{\sum\limits_{\frac{s-1}{2^k}+1\leq n_2\leq N}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{1}{n_1\cdot 2^kn_2}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\phi(s) \mathop{\sum\limits_{1\leq n_1\leq s-1}\sum\limits_{1\leq r_2\leq\frac{2^kN}{s}-1}\sum\limits_{l_2=1}^{s-1}}_{r_2s+l_2\equiv n_1 (\bmod s)}\frac{1}{n_1(r_2s+l_2)}\ll N^{\epsilon} \sum\limits_{s|m}\phi(s)\sum\limits_{1\leq r_2\leq\frac{2^kN}{s}-1}\sum\limits_{l=1}^{s-1}\frac{1}{l(r_2s+l)}\nonumber\\ &&\ll N^{\epsilon} \sum\limits_{s|m}\frac{\phi(s)}{s}\sum\limits_{1\leq r_2\leq\frac{2^kN}{s}-1}\frac{1}{r_2}\sum\limits_{l=1}^{s-1}\frac{1}{l}\ll N^{\epsilon}. \end{eqnarray} $

(3.10)

结合(3.5)-(3.10)可得

$ \begin{eqnarray*} &&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2\\ &&=\frac{1}{2}\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\mathop{\sum\limits_{1\leq n_1 \leq s-1}}_{(n_1, 2m)=1}\mathop{\sum\limits_{1\leq n_2 \leq \frac{s-1}{2^k}}}_{(n_2, 2m)=1}}_{2^kn_2\equiv n_1(\bmod s)} \frac{r(n_1)r(n_2)}{n_1n_2}+O\left(\frac{m^{\frac{3}{2}+\epsilon}}{N^{\frac{1}{2}}}\right)+ O\left(\frac{m^{2+\epsilon}}{N}\right)+O\left(N^{\epsilon}\right). \end{eqnarray*} $

当$k\geq 1$时, 取$N=m^3$有

$ \begin{eqnarray*} \mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\chi(2^k)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2\ll m^{\epsilon}. \end{eqnarray*} $

而当$k=0$时, 取$N=m^3$, 并利用引理3.2可得

$ \begin{eqnarray*} &&\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2 =\frac{1}{2}\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\sum\limits_{1\leq n \leq s-1}}_{(n, 2m)=1}\frac{r^2(n)}{n^2}+O\left(m^{\epsilon}\right)\\ &&=\frac{1}{2}\sum\limits_{s|m}\mu\left(\frac{m}{s}\right)\phi(s) \mathop{\sum\limits_{n=1}^{+\infty}}_{(n, 2m)=1}\frac{r^2(n)}{n^2}+O\left(m^{\epsilon}\right)\\ &&=\frac{1}{2}J(m)\cdot\frac{3\pi^4}{128}\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right)+O\left(m^{\epsilon}\right)\\ &&=\frac{3\pi^4}{256}J(m)\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right)+O\left(m^\epsilon\right). \end{eqnarray*} $

由定理2.1与定理3.1有

$ \begin{eqnarray*} &&\sum\limits_{a\leq\frac{q}{4}}\sum\limits_{b\leq\frac{q}{4}}H(2a\overline{b}, q)\\ &&=\frac{16}{\pi^4q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d}m\mathop{\sideset{}{^*}\sum\limits_{\chi\bmod m}}_{\chi(-1)=-1}\left(1+2\chi(2)+\chi(4)\right)|L(1, \chi\chi_{2d}^0)|^2|L(1, \chi\chi_2^0)|^2\\ &&=\frac{16}{\pi^4q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\sum\limits_{m|d}m\left( \frac{3\pi^4}{256}J(m)\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right)+O\left(m^\epsilon\right)\right)\\ &&=\frac{3}{16q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\prod\limits_{p\mid d}\frac{\left(1-\frac{1}{p^2}\right)^2}{1+\frac{1}{p^2}}\sum\limits_{m|d}mJ(m)\prod\limits_{p\mid m}\left(1-\frac{1}{p^2}\right)+O\left(q^{1+\epsilon}\right)\\ &&=\frac{3}{16}q^2\prod\limits_{p^\alpha\parallel q}\frac{\left(1-\frac{1}{p^2}\right)\left(1-\frac{1}{p^{3\alpha}}-\left(1+\frac{1}{p}+\frac{1}{p^2}\right) \left(\frac{1}{p^{2\alpha}}-\frac{1}{p^{3\alpha}}\right)\right)}{\left(1+\frac{1}{p^2}\right)\left(1+\frac{1}{p}+\frac{1}{p^2}\right)} +O\left(q^{1+\epsilon}\right). \end{eqnarray*} $

这就证明了定理1.1.