A natural question about Dedekind sums is to find conditions on the integers
a1,a2, and b such that s(a1,b)=s(a2,b). We prove that if the
former equality holds then b∣(a1a2−1)(a1−a2). 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 a1 and a2 that are relatively prime to b,
the hypothesis rn(a1,b)=rn(a2,b) implies that
b∣(6n2+1−a1a2)(a2−a1).Comment: 6 page