For a,b,c∈N, let a′=gcd, 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,

\begin{align}

12 a b c \ \mathfrak{S}(a,b,c) \equiv (ab)^{2} + (bc)^{2} + (ca)^{2} + d^{2} \ \pmod{a b c}.

\end{align}

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

\mathfrak{S}(a_{1},\dots, a_{n}) = \sum_{i = 1}^{n} a_{i}^{\prime} \, \mathfrak{s}\left( \tfrac{a_{1} \cdots \hat{a}_{i} \cdots a_{n}}{a_{n+1}}, \tfrac{a_{i}}{a_{1}^{\prime} \cdots \hat{a}_{i}^{\prime} \cdots a_{n}^{\prime}} \right).

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):

12 a_{1} \cdots a_{n} \, \mathfrak{S}(a_{1}, \dots, a_{n}) \equiv \left( \sum_{i = 1}^{n} (a_{1} \cdots \hat{a}_{i} \cdots a_{n})^{2} \right) + 1 \ \pmod{a_{1} \cdots a_{n}}.

Is this more general congruence known?

**Answer**

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