On a recent reciprocity formula for Dedekind sums

TL;DR

This paper explores a family of reciprocity formulas for Dedekind sums, extending recent results by using the three-term relation rather than L-series connections, revealing deeper structural properties.

Contribution

It demonstrates that Du and Zhang's reciprocity formula is a special case within a broader class of formulas derived via the three-term relation.

Findings

Generalizes reciprocity formulas for Dedekind sums

Shows the formulas form a series of related identities

Uses the three-term relation as the main analytical tool

Abstract

Let $s(a,b)$ denote the classical Dedekind sum and $S(a,b)=12s(a,b)$. Recently, Du and Zhang proved the following reciprocity formula. If $a$ and $b$ are odd natural numbers, $(a,b)=1$, then $$ S(2a^*,b)+S(2b^*,a)=\frac{a^2+b^2+4}{2ab}-3, $$ where $aa^*\equiv 1\mod b$ and $bb^* \equiv 1 \mod a$. In this paper we show that this formula is a special case of a series of similar reciprocity formulas. Whereas Du and Zhang worked with the connection of Dedekind sums and values of $L$-series, our main tool is the three-term relation for Dedekind sums.

Equations

$$S(2a^{\ast },b)+S(2b^{\ast },a)=\frac{a^2+b^2+4}{2ab}-3,$$

$$s(a,b)=\sum _{k=1}^b((k/b))((ak/b)).$$

$$((x))=\{\begin{array}{ll}x-\lfloor x\rfloor -1/2& \text{mbox}ifx\in \mathbb{R}\setminus \mathbb{Z};\\ 0& \text{mbox}ifx\in \mathbb{Z}\end{array}$$

$$S(a,b)=12s(a,b)$$

$$S(a,b)+S(b,a)=\frac{a^2+b^2+1}{ab}-3.$$

$$S(2a^{\ast },b)+S(2b^{\ast },a)=\frac{a^2+b^2+4}{2ab}-3,$$

$$S(t{a}^{\ast },b)+S(t{b}^{\ast },a)=\frac{a^2+b^2+t^2}{tab}-3+S(ab,t).$$

$$S(a^{\ast },b)=S(a,b)$$

$$S(t{a}^{\ast },b)+S(t{b}^{\ast },a)=\frac{a^2+b^2+t^2}{tab}-3.$$

$$S(t{a}^{\ast },b)+S(t{b}^{\ast },a)=\frac{a^2+b^2+t^2}{tab}+\{\begin{array}{ll}-t-2/t,& \text{mbox}ifb\equiv a\mathrm{mod}t;\\ t+2/t-6,& \text{mbox}ifb\equiv -a\mathrm{mod}t.\end{array}$$

$$S(5a^{\ast },b)+S(5b^{\ast },a)=\frac{a^2+b^2+25}{5ab}-3.$$

$$S(5a^{\ast },b)+S(5b^{\ast },a)=\frac{a^2+b^2+25}{5ab}-27/5.$$

$$S(5a^{\ast },b)+S(5b^{\ast },a)=\frac{a^2+b^2+25}{5ab}-3/5.$$

$$S(a,c)=\frac{(b^2-1)a}{tb}-S(ab,t)+S(a{t}^{\ast },b).$$

$$S(a,c)=-S(c,a)+\frac{a^2+c^2+1}{ac}-3.$$

$$S(a,b)=S(c,d)+\epsilon S(r,|q|)+\frac{b^2+d^2+q^2}{bdq}-3\epsilon$$

Full text

[TABLE]

where $aa^{*}\equiv 1\mod b$ and $bb^{*}\equiv 1\mod a$. In this paper we show that this formula is a special case of a series of similar reciprocity formulas. Whereas Du and Zhang worked with the connection of Dedekind sums and values of $L$-series, our main tool is the three-term relation for Dedekind sums.

1. Introduction and Result

Let $a$ be an integer, $b$ a natural number, and $(a,b)=1$. The classical Dedekind sum $s(a,b)$ is defined by

[TABLE]

Here

[TABLE]

(see [7, p. 1]). It is often more convenient to work with

[TABLE]

instead. We call $S(a,b)$ a normalized Dedekind sum.

Probably the most important elementary result concerning Dedekind sums is reciprocity law. If $a$ and $b$ are coprime natural numbers, then

[TABLE]

Recently, Du and Zhang have found the following hitherto unknown reciprocity law (see [3]). If $a$ and $b$ are coprime odd natural numbers, then

[TABLE]

where $aa^{*}\equiv 1\mod b$ and $bb^{*}\equiv 1\mod a$.

The proof given in [3] is based on the connection of Dedekind sums and values of $L$-series. The authors of the said paper ask for an elementary proof of their result. Here we give such an elementary proof based on the tree-term-relation of Dedekind sums. Moreover, we show that (2) is a special case of a series of similar reciprocity formulas. Indeed, we have the following.

Theorem 1

Let $a$ and $b$ be coprime natural numbers and $t$ a natural number such that $a^{2}+1\equiv 0\mod t$. Further, let $(b,t)=1$. Then

[TABLE]

As to the case $t=1$, we note

[TABLE]

(see [7, p. 26]) and $S(ab,1)=0$. In the case $t=2$, $a$ and $b$ are odd and $S(ab,2)=0$. Hence we obtain the following.

Corollary 1

The formulas (1) and (2) are immediate consequences of Theorem 1 in the cases $t=1$ and $t=2$.

Corollary 2

Suppose, in the setting of Theorem 1, that $b\equiv\pm 1\mod t$. Then

[TABLE]

Suppose, on the other hand, that $b\equiv\pm a\mod t$. Then

[TABLE]

As to (5), note that $(ab)^{2}\equiv-1\mod t$, which shows that $S(ab,t)=0$ (see [7, p. 28]). In the case of (6), we use $S(1,t)=t+2/t-3$ (which is an immediate consequence of (1)) and $S(-a,b)=-S(a,b)$ (see [7, p. 26]).

Remark. The natural numbers $t$ such that there is a natural number $a$ with $a^{2}+1\equiv 0\mod t$ can be characterized as follows: $t=m$ or $t=2m$, where $m$ is a natural number whose prime divisors are all $\equiv 1\mod 4$ (this includes $m=1$).

Example. Let $t=5$, $a$ such that $a^{2}+1\equiv 0\mod 5$, and $(a,b)=(b,5)=1$. This implies $a\equiv\pm 2\mod 5$. If $b\equiv\pm 1\mod 5$, then

[TABLE]

In the remaining case, we have $b\equiv\pm a\mod 5$. If $b\equiv a\mod 5$, then (6) reads

[TABLE]

If $b\equiv-a\mod 5$, we have

[TABLE]

Proof of Theorem 1

Let $a,b,t$ be natural numbers, $(a,b)=(b,t)=1$, such that $a^{2}+1\equiv 0\mod t$. Put $c=b(a^{2}+1)/t$. Obviously, $(a,c)=1$. Then [5, Th.4 ] says

[TABLE]

where $tt^{*}\equiv 1\mod b$. By the reciprocity law (1),

[TABLE]

However, $c\equiv bt^{*}\mod a$, with $tt^{*}\equiv 1\mod a$. Hence $S(c,a)=S(bt^{*},a)$. We replace $at^{*}$ by $ta^{*}$ and $bt^{*}$ by $tb^{*}$ in the respective normalized Dedekind sums (see (4)). Then a short calculation proves Theorem 1.

We still have to make clear that this remarkably simple proof is based on elementary results. Indeed, (7) follows from the three-term relation

[TABLE]

(see [5]). Here $b,d,$ are natural numbers, $a,c$ integers, $(a,b)=(c,d)=1$, $a/b\neq c/d$. Further, $q=ad-bc$ and $\varepsilon$ is the sign of $q$. Finally, $r=aj-bk$, where $j,k$ are integers such that $-cj+dk=1$. The three-term relation, in turn, can be deduced from the composition rule of the logarithm of Dedekind’s $\eta$-function (see [2, 4]). An elementary proof of this composition rule is given in [6, §4].