The (classical) Dedekind sum $s(h,k)$ is defined as
$$s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\bigg(\frac{hr}{k}-\Big[\frac{hr}{k}\Big]-\frac{1}{2}\bigg)$$
for $\gcd(h,k)=1$.
A natural question is, when will we have $s(h_1,k)=s(h_2,k)$?
P.S. Of course a sufficient condition is that $h_1\equiv h_2\pmod k$ or $h_1h_2\equiv 1\pmod k$. Moreover, if $k$ is a prime number it is easy to see that the above is also a necessary condition.
You can find a patial answer in the article On a criterion for the equality of Dedekind Sums $s(h_1,k)=s_2(h_2,k)\pmod{\mathbb Z}$ iff $(h_1-h_2)(h_1h_2-1)=0\pmod k$. The key tool is a formula for Dedekind sum $s(k,h)$ in terms of continued fraction $h/k=[a_0;a_1,\ldots,a_n].$ In the proof the sum $$\tag{*}\sum_{j=1}^n(-1)^{j-1}a_j$$ vanishes modulo $\mathbb{Z}$. If we want exact equality $s(h_1,k)=s_2(h_2,k)$ then we have to add restrictions on the sums $(*)$ for fractions $h_1/k$ and $h_2/k$.
More precise necessary conditions are given in Equality of Dedekind sums mod ℤ, 2ℤ and 4ℤ (Tsukerman, 2014) and Equality of Dedekind sums mod 8ℤ (Tsukerman, 2015). Probably it is our best up to date knowledge.
Another good sourse of formulae is the article Hall, R. R. & Huxley, M. N. Dedekind sums and continued fractions Acta Arith., 1993, 63, 79-90.
