“…This is actually stronger than we need. Without reproducing the arguments of [1] and [11] The first factor on the right becomes Q Continuing this procedure with the obvious definitions and using Theorem 7b) to multiply the Jacobi symbols, we finally get The Gauss sum on the right can be evaluated using Theorems 3, 4, and 2 for any known (C, D), In particular, if άetD is odd squarefree, then the Gauss sum on the left equals (2 m det F/det D). Πpidetz?…”