For a positive integer $k$ and an arbitrary integer $h$, the classical Dedekind sums $S(h, k)$ is defined by

$S(h,k)=\sum\limits_{a=1}^{k}{((}\frac{a}{k}))((\frac{ah}{k})),$

where

$((x))=\left\{ \begin{array}{*{35}{l}} x-[x]-\frac{1}{2},\text{if}\ x\ \text{is}\ \text{not}\ \text{an}\ \text{integer}; \\ 0,\quad \quad \quad \quad \ \text{if}\ x\ \text{is}\ \text{an}\ \text{integer}. \\ \end{array} \right.$

The various properties of $S(h, k)$ were investigated by many authors, see [2-4, 7-9]. For example, Carlitz [3] obtained a reciprocity theorem of $S(h, k)$. Conrey et al. [4] studied the mean value distribution of $S(h, k)$, and proved the following important and interesting asymptotic formula

$\sum\limits_{h = 1}^k {^\prime |S(h,k){|^{2m}}} = {f_m}(k){(\frac{k}{{12}})^{2m}} + O(({k^{\frac{9}{5}}} + {k^{2m - 1 + \frac{1}{{m + 1}}}}){\log ^3}k),$

(1.1)

where $\displaystyle\mathop{{\sum}'}_h$ denotes the summation over all $h$ such that $(k, h) = 1$, and

$\sum\limits_{n = 1}^\infty {\frac{{{f_m}(n)}}{{{n^s}}}} = 2\frac{{{\zeta ^2}(2m)}}{{\zeta (4m)}} \cdot \frac{{\zeta (s + 4m - 1)}}{{{\zeta ^2}(s + 2m)}}\zeta (s),$

$\zeta(s)$ is the Riemann zeta-function.

Jia [7] improved the error term in (1.1) as $O(k^{2m-1})$, provide $m\geq 2$. Zhang [9] improved the error term of (1.1) for $m = 1$. That is, he proved the following asymptotic formula

$$ \mathop{{\sum}'}^k_{h=1}|S(h, k)|^2 =\frac{5}{144}k\phi(k)\cdot \frac{\displaystyle \prod_{p^{\alpha} \|k }((1+\frac{1}{p})^2-\frac{1}{p^{3\alpha+1}})}{\displaystyle\prod_{p|k}(1+\frac{1}{p} + \frac{1}{p^2})} + O(k \exp(\frac{4\ln k}{\ln\ln k})), $$

where $p^{\alpha}\| k$ denotes that $p^{\alpha}| k$ and $p^{\alpha +1} \dagger k$ and $\exp(y)=e^y$.

Liu and Zhang [11] studied the hybrid mean value involving Dedekind sums and Kloosterman sums $K(m, n; q)$, which defined as follows (see [5] and [6]):

$$K(m,n; q)=\mathop{{\sum}'}^{q}_{b=1}e (\frac{mb+n\bar{b}}{q}),$$

where $e(y)=e^{2\pi iy}$, $\overline{b}$ denotes the solution of the equation $x\cdot b\equiv 1\bmod q$. They proved the following conclusion:

Let $q$ be a square-full number (i.e., $p\mid q$ if and only if $p^{2}\mid q$), then we have

$\begin{eqnarray*} \mathop{{\sum}'}^q_{a=1}\mathop{{\sum}'}^q_{b=1} K(m, a;q)\overline{K(m, b;q)} S(a\overline{b}, q)=\frac{1}{12}\cdot q\cdot \phi^2(q)\prod_{p\mid q}(1+\frac{1}{p}) , \end{eqnarray*}$

where $\displaystyle\mathop{{\sum}'}^q_{a=1}$ denotes the summation over all $1\leq a\leq q$ such that $(a, q)=1$, $\displaystyle \prod_{p|q}$ denotes the product over all distinct prime divisors $p$ of $q$, $\phi(q)$ is the Euler function, and $\overline{f(n)}$ denotes the complex conjugation of $f(n)$.

In this paper, we use the analytic methods and mean value theorem of Dirichlet $L$-functions to study the hybrid mean value properties involving Dedekind sums and Legendre’s symbol, and prove some new identities and asymptotic formulae. That is, we shall prove the following several conclusions:

Theorem 1.1 Let $p$ be an odd prime with $p\equiv 3\bmod 4$, then we have the identity

$\sum\limits_{a = 1}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {(\frac{{a + 1}}{p})} } (\frac{{b + 1}}{p})S(a\bar b,p) = \frac{{(p - 1)(p - 2)}}{{12}} - h_p^2,$

where $h_p$ denotes the class number of the quadratic field $\mathbf{Q}$$(\sqrt{-p})$.

Theorem 1.2 Let $p$ be an odd prime with $p\equiv 1\bmod 4$, then for any positive integer $k$, we have the identity

$\sum\limits_{a = 1}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {(\frac{{a + 1}}{p})} } (\frac{{b + 1}}{p}){S^{2k - 1}}(a\bar b,p) = \frac{{p \cdot {{(p - 1)}^{2k - 1}} \cdot {{(p - 2)}^{2k - 1}}}}{{{{(12p)}^{2k - 1}}}}.$

Theorem 1.3 Let $p$ be an odd prime, then we have the asymptotic formulae:

(A) $\sum^{p-1}\limits_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S^2(a\overline{b}, p)=\frac{p^2\cdot (p-11)}{144}+ O(p\cdot \exp(\frac{4\ln p }{\ln\ln p })), $ if $p\equiv 3\bmod 4$.

(B) $\sum^{p-1}\limits_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S^2(a\overline{b}, p)=\frac{p^2\cdot (p-11-C_1)}{144}+ O(p\cdot \exp(\frac{4\ln p }{\ln\ln p })), $ if $p\equiv 1\bmod 4$, where $C_1= \frac{4\pi^4}{5}\displaystyle \mathop{{\prod}^*}_q \frac{(q^2+1)^2}{(q^2-1)^2}$ is a constant, $\displaystyle \mathop{{\prod}^*}_q$ denotes the product over all primes $q$ such that $(\frac{q}{p})=1$, and $\exp(y)=e^{y}$.

As an application of Theorem 1.1 (of course, one can also give a proof directly), we can give an interesting computational formula for $h_p$, which we described as following:

Corollary 1.4 Let $p$ be an odd prime with $p\equiv 3\bmod 4$, then we have the computational formula

${h_p} = {\left\{ {\sum\limits_{a = 1}^{p - 1} S ({a^2}, p)} \right\}^{\frac{1}{2}}} = 2\mathop \sum \limits_{a = 1}^{\frac{{p - 1}}{2}} [\frac{{{a^2}}}{p}] - \frac{{(p - 1)(p - 5)}}{{12}} = \frac{{p - 1}}{2} - \frac{2}{p}\sum\limits_{i = 1}^{\frac{{p - 1}}{2}} {{r_i}}, $

where $[x]$ denotes the greatest integer $\leq x$, $r_i$ ($i=1, \ 2, \ \cdots, \ \frac{p-1}{2})$ denotes all quadratic residues $\bmod\ p$ in the interval $[1, \ p-1]$.

In this section, we shall give some lemmas which are necessary in the proof of our theorems. First we have the following:

Lemma 2.1 Let $q>2 $ be an integer, then for any integer $a$ with $(a, \ q)=1$, we have the identity

$S(a, q){\rm{ = }}\frac{1}{{{\pi ^2}q}}\sum\limits_{d|q} {\frac{{{d^2}}}{{\phi (d)}}\sum\limits_{\begin{array}{*{20}{c}} {\chi \;{\rm{mod}}\;d}\\ {\chi ( - 1) = - 1} \end{array}} {\chi (a)} } |L(1, \chi ){|^2}, $

where $\phi(n)$ is the Euler function, $\displaystyle \mathop{\sum_{\chi \ \mathrm{mod} \ d}}_{\chi(-1)=-1}$ denotes the summation over all odd character modulo $ d$, $L(s, \chi)$ denotes the Dirichlet $L$-function corresponding to $\chi$ modulo $d$.

Lemma 2.2 Let $p $ be an odd prime, then we have the asymptotic formulae

(I) $\displaystyle \mathop{\sum\limits_{\chi \ \mathrm{mod} \ p}}_{\chi(-1)=-1}|L(1, \chi)|^4= \frac{5\pi^4}{144} \cdot p + O(\exp(\frac{4\ln p}{\ln\ln p}));$

(II) $\displaystyle \mathop{\sum\limits_{\chi \ \mathrm{mod} \ p}}_{\chi(-1)=-1}|L(1, \chi)|^2\cdot |L(1, \chi\chi_2)|^2= C_1 \cdot p + O(\exp(\frac{4\ln p}{\ln\ln p})), $

where $C_1= \frac{\pi^4}{180}\displaystyle \mathop{{\prod}^*}_q \frac{(q^2+1)^2}{(q^2-1)^2}$ is a constant, $\displaystyle \mathop{{\prod}^*}_q$ denotes the product over all primes $q$ such that $(\frac{q}{p})=1$.

Proof In fact Lemma 2.1 and Lemma 2.2 are two early results of the second author, their proof can be find in references [8] and [10].

Lemma 2.3 Let $p$ be an odd prime, then for any non-real character $\chi \bmod p$, we have the identity $ |\sum\limits_{a=1}^{p-1}(\frac{a+1}{p})\chi(a)|=\sqrt{p}, $ where $(\frac{*}{p})$ is the Legendre’s symbol.

Proof Since $\chi_2$ is a primitive character $\bmod\ p$, so from the properties of Gauss sums $\tau(\chi)$ we have

$\begin{array}{l} \sum\limits_{a = 1}^{p - 1} {(\frac{{a + 1}}{p})} \chi (a) = \frac{1}{{\tau ({\chi _2})}}\sum\limits_{a = 1}^{p - 1} \chi (a)\sum\limits_{b = 1}^{p - 1} {{\chi _2}} (b)\;e(\frac{{b(a + 1)}}{p})\\ = \frac{1}{{\tau ({\chi _2})}}\sum\limits_{b = 1}^{p - 1} {{\chi _2}} (b)\bar \chi (b)\;e(\frac{b}{p})\sum\limits_{a = 1}^{p - 1} \chi (ab)\;e(\frac{{ab}}{p})\\ = \frac{1}{{\tau ({\chi _2})}}\sum\limits_{b = 1}^{p - 1} {{\chi _2}} (b)\bar \chi (b)\;e(\frac{b}{p})\sum\limits_{a = 1}^{p - 1} \chi (a)\;e(\frac{a}{p}) = \frac{{\tau (\chi ) \cdot \tau (\bar \chi {\chi _2})}}{{\tau ({\chi _2})}}, \end{array}$

(2.1)

where $\chi_2=(\frac{*}{p}) $ is the Legendre’s symbol.

Now Lemma 2.3 follows from (2.1) and the identity $\mid\tau(\chi)\mid= \sqrt{p}$, if $\chi$ is not a principal character $\bmod\ p$.

In this section, we shall complete the proof of our theorems. First we prove Theorem 1.1. For odd prime $p$, from Lemma 2.1 and the definition of $S(a, p)$ we have

$S(a, p) = \frac{p}{{{\pi ^2}(p - 1)}}\sum\limits_{\begin{array}{*{20}{c}} {\chi {\kern 1pt} \, \bmod \, {\kern 1pt} p}\\ {\chi ( - 1) = - 1} \end{array}} {\chi (a)|L(1, \chi ){|^2}} $

(3.1)

and

$\sum\limits_{\begin{array}{*{20}{c}} {\chi {\kern 1pt} \, \bmod \, {\kern 1pt} p}\\ {\chi ( - 1) = - 1} \end{array}} {|L(1, \chi ){|^2}} = \frac{{{\pi ^2}{{(p - 1)}^2}(p - 2)}}{{12{p^2}}}$

(3.2)

Then from (3.1) and Lemma 2.3 we have

$\begin{array}{*{20}{l}} {\sum\limits_{a = 1}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {(\frac{{a + 1}}{p})} } (\frac{{b + 1}}{p})S(a\bar b,p)}\\ { = \frac{p}{{{\pi ^2}(p - 1)}}\sum\limits_{\begin{array}{*{20}{c}} {\chi {\kern 1pt} \,\bmod \,{\kern 1pt} p}\\ {\chi ( - 1) = - 1} \end{array}} {} |\sum\limits_{a = 1}^{p - 1} {(\frac{{a + 1}}{p})} \chi (a){|^2} \cdot |L(1,\chi ){|^2}}\\ { = \frac{p}{{{\pi ^2}(p - 1)}}\sum\limits_{\begin{array}{*{20}{c}} {\chi {\kern 1pt} \,\bmod \,{\kern 1pt} p}\\ {\chi ( - 1) = - 1} \end{array}} {} |\sum\limits_{a = 1}^{p - 1} {(\frac{{\tau (\chi ) \cdot \tau (\bar \chi {\chi _2})}}{{\tau ({\chi _2})}})} {|^2} \cdot |L(1,\chi ){|^2}} \end{array}$

(3.3)

If $p\equiv 3\bmod 4$, then $\chi_2(-1)=-1$, so for $\overline{\chi}=\chi_2$, $\overline{\chi}\chi_2$ is the principal character $\bmod\ p$ and $|\tau(\overline{\chi}\chi_2)|=1$. Note that $L(1, \chi_2)= \pi h_p/\sqrt{p}$, from (3.2), (3.3) and Lemma 2.3 we have

$\begin{eqnarray*} \sum^{p-1}_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S(a\overline{b}, p) =\frac{p}{\pi^2(p-1)}[\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1}p|L(1, \chi)|^2-(p-1)|L(1, \chi_2)|^2]\nonumber\\ =\frac{(p-1)(p-2)}{12}-\frac{p}{\pi^2}|L(1, \chi_2)|^2=\frac{(p-1)(p-2)}{12}- h^2_p. \end{eqnarray*}$

This proves Theorem 1.1.

Now we prove Theorem 1.2. If $p\equiv 1 \bmod 4$, then $\chi_2(-1)=1$, so for any odd character $\chi$, $\chi\chi_2$ is not the principal character $\bmod\ p$. This time, from the properties of Gauss sums, (3.1), (3.2) and Lemma 2.3 we have

$\begin{eqnarray*} \sum^{p-1}_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S^{2k-1}(a\overline{b}, p) =(\frac{p}{\pi^2(p-1)})^{2k-1} \cdot p\cdot( \mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1}|L(1, \chi)|^2)^{2k-1}\\ =(\frac{p}{\pi^2(p-1)})^{2k-1} \cdot p\cdot(\frac{\pi^2 (p-1)^2\cdot (p-2)}{12p^2} )^{2k-1} = \frac{p\cdot (p-1)^{2k-1}\cdot (p-2)^{2k-1}}{(12p)^{2k-1}}. \end{eqnarray*}$

This proves Theorem 1.2.

To prove Theorem 1.3. First from (3.1) and Lemma 2.3 we have

$\begin{eqnarray} \sum^{p-1}_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S^2(a\overline{b}, p)\nonumber\\ =\frac{p^2}{\pi^4(p-1)^2}\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1}\mathop{\sum\limits_{\lambda \bmod p}}_{\lambda(-1)=-1}|\sum\limits_{a=1}^{p-1}(\frac{a+1}{p})\chi(a)\lambda(a)|^2\cdot |L(1, \chi)|^2|L(1, \lambda)|^2\nonumber\\ =\frac{p^2}{\pi^4(p-1)^2}\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1}\mathop{\sum\limits_{\lambda \bmod p}}_{\lambda(-1)=-1}|\frac{\tau(\chi\lambda)\cdot \tau(\overline{\chi\lambda}\chi_2 )}{\tau(\chi_2)}|^2\cdot |L(1, \chi)|^2\cdot |L(1, \lambda)|^2. \end{eqnarray}$

(3.4)

If $p\equiv 3\bmod 4$, then $\chi_2$ is an odd character $\bmod\ p$, and $\chi\lambda$ is an even character $\bmod\ p$. So $\overline{\chi\lambda}\chi_2$ is not a principal character $\bmod\ p$. From (3.4), Lemma 2.2 and the properties of Gauss sums we have

$\begin{eqnarray} \sum^{p-1}_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S^2(a\overline{b}, p)\nonumber\\ =\frac{p^2}{\pi^4(p-1)^2}[p\cdot(\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1} \mid L(1, \chi) \mid^2)^2-\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1}(p-1) |L(1, \chi)|^4]\nonumber\\ =\frac{(p-1)^2(p-2)^2}{144p}-\frac{p^2}{\pi^4(p-1)}\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1} |L(1, \chi)|^4 =\frac{p^2\cdot (p-11)}{144}+ O(p\cdot \exp(\frac{4\ln p }{\ln\ln p })).\nonumber\\ \end{eqnarray}$

(3.5)

If $p\equiv 1\bmod 4$, then $\chi_2$ is an even character $\bmod\ p$, this time we have

$\begin{eqnarray} \sum^{p-1}_{a=1}\sum\limits_{b=1}^{p-1}(\frac{a+1}{p})(\frac{b+1}{p}) S^2(a\overline{b}, p)\nonumber\\ =\frac{p^2}{\pi^4(p-1)^2}[p\cdot(\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1} \mid L(1, \chi) \mid^2)^2-(p-1) \mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1} |L(1, \chi)|^4]\nonumber\\ \qquad-\frac{p^2}{\pi^4(p-1)}\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1} |L(1, \chi)|^2\cdot |L(1, \chi\chi_2)|^2\nonumber\\ =\frac{p^2\cdot (p-11)}{144}- C\cdot p^2+ O(p\cdot \exp(\frac{4\ln p }{\ln\ln p })), \end{eqnarray}$

(3.6)

where $C= \frac{\pi^4}{180}\displaystyle \mathop{{\prod}^*}_q \frac{(q^2+1)^2}{(q^2-1)^2}$ is a constant, $\displaystyle \mathop{{\prod}^*}_q$ denotes the product over all primes $q$ such that $(\frac{q}{p})=1$.

Now Theorem 1.3 follows from asymptotic formulae (3.5) and (3.6).

Using Lemma 2.1 we can also give a direct proof of Corollary 1.4. In fact if $p\equiv 3\bmod 4$, then $(\frac{-1}{p})=-1$, so from Lemma 2.1 and note that the orthogonality of characters $\bmod\ p$, we have

$\begin{eqnarray} \sum\limits_{a=1}^{p-1}S(a^2, p)=\frac{p}{\pi^2(p-1)}\mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1)=-1} \sum\limits_{a=1}^{p-1}\chi^2(a)|L(1, \chi)|^2\nonumber\\ =\frac{p}{\pi^2(p-1)}\cdot (p-1)\cdot |L(1, \chi_2)|^2=h_p^2\ . \end{eqnarray}$

(3.7)

On the other hand, note that $((-x))= -((x))$ and the set $\{1^2, \ 2^2, \ \cdots, \ \frac{(p-1)^2}{4}$, $ -1^2, \ -2^2, \ \cdots, \ -\frac{(p-1)^2}{4}\}$ is a reduced residue system $\bmod\ p$, so from the definition of $S(a, p)$ we have

$\begin{eqnarray} \sum\limits_{a=1}^{p-1}S(a^2, p)= \sum\limits_{a=1}^{p-1} \{\sum\limits_{b=1}^{\frac{p-1}{2}} ((\frac{b^2}{p})) (\frac{b^2a^2}{p})) + \sum\limits_{b=1}^{\frac{p-1}{2}} ((\frac{-b^2}{p})) ((\frac{-b^2a^2}{p})) \}\nonumber\\ =2\sum\limits_{b=1}^{\frac{p-1}{2}} ((\frac{b^2}{p}))\sum\limits_{a=1}^{p-1} ((\frac{b^2a^2}{p}))=4(\sum\limits_{b=1}^{\frac{p-1}{2}} ((\frac{b^2}{p})))^2\nonumber\\ =4(\sum\limits_{b=1}^{\frac{p-1}{2}}\frac{b^2}{p}-\frac{1}{2}\sum\limits_{b=1}^{\frac{p-1}{2}}1- \sum\limits_{b=1}^{\frac{p-1}{2}}[\frac{b^2}{p}])^2=4(\frac{(p-1)(p-5)}{24}- \sum\limits_{b=1}^{\frac{p-1}{2}}[\frac{b^2}{p}])^2. \end{eqnarray}$

(3.8)

Combining (3.7) and (3.8) we may immediately deduce the identity

$ h_p=\{\sum\limits_{a=1}^{p-1}S(a^2, p)\}^{\frac{1}{2}}= 2\sum\limits_{a=1}^{\frac{p-1}{2}}[\frac{a^2}{p}]- \frac{(p-1)(p-5)}{12}. $

This completes the proof of our conclusions.