When are two Dedekind sums equal?

TL;DR

This paper establishes new divisibility conditions on integers for when two Dedekind sums are equal, providing the first such results for classical and Dedekind-Rademacher sums.

Contribution

It proves novel divisibility criteria for the equality of Dedekind sums and extends these results to Dedekind-Rademacher sums, filling a gap in the literature.

Findings

If s(a_1,b) = s(a_2,b), then b divides (a_1a_2 - 1)(a_1 - a_2).

For Dedekind-Rademacher sums, equality implies b divides (6n^2 + 1 - a_1a_2)(a_2 - a_1).

These are the first known conditions of their kind in the literature.

Abstract

A natural question about Dedekind sums is to find conditions on the integers $a_1, a_2$, and $b$ such that $s(a_1,b) = s(a_2, b)$. We prove that if the former equality holds then $ b \ | \ (a_1a_2-1)(a_1-a_2)$. Surprisingly, to the best of our knowledge such statements have not appeared in the literature. A similar theorem is proved for the more general Dedekind-Rademacher sums as well, namely that for any fixed non-negative integer $n$, a positive integer modulus $b$, and two integers $a_1$ and $a_2$ that are relatively prime to $b$, the hypothesis $r_n (a_1,b)= r_n (a_2,b)$ implies that $b \ | \ (6n^2+1-a_1a_2)(a_2-a_1)$.