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 $\gcd(a,b,c) = 1$, \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?

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