# Dedekind Sum Congruences

For $a,b,c \in \mathbb{N}$, let $a^{\prime} = \gcd(b,c)$, $b^{\prime} = \gcd(a,c)$, $c^{\prime} = \gcd(a,b)$ and $d = a^{\prime} b^{\prime} c^{\prime}$. Define $\mathfrak{S}(a,b,c) = a^{\prime} \mathfrak{s}( \tfrac{bc}{d}, \tfrac{a}{b^{\prime} c^{\prime}}) + b^{\prime} \mathfrak{s}( \tfrac{ac}{d}, \tfrac{b}{a^{\prime} c^{\prime}}) + c^{\prime} \mathfrak{s}( \tfrac{ab}{d}, \tfrac{c }{a^{\prime} b^{\prime}} )$, where $\mathfrak{s}$ is the Dedekind sum. I can prove the following: For $a, b, c \in \mathbb{N}$ with $\gcd(a,b,c) = 1$,

Is this congruence in the literature? I’m aware of the 3-term generalization of Rademacher and Pommersheim and of the work of Beck on the Carlitz-Dedekind Sums, but these don’t seem to include this particular congruence as a special case.

I generalize. For $a_{1}, \dots, a_{n} \in \mathbb{N}$ such that $\gcd(a_1, \dots, a_{n}) = 1$, let $a_{i}^{\prime} = \gcd(a_{1}, \dots, \hat{a}_{i}, \dots, a_{n})$. Define the symmetric summation

where $\hat{}$ denotes omission. For pairwise coprime $a_{1}, \dots, a_{n} \in \mathbb{N}$ (all primed variables equal $1$), I conjecture the following congruence holds for $n = 4$ (and I have a hunch that it continues to hold for $n > 4$):

Is this more general congruence known?

## Answer

Attribution
Source : Link , Question Author : user02138 , Answer Author : Community