  If L_1 and L_2 are two Brunnian links with all pairwise linking numbers 0,
then we show that L_1 and L_2 are equivalent if and only if they have
homeomorphic complements. In particular, this holds for all Brunnian links with
at least three components. If L_1 is a Brunnian link with all pairwise linking
numbers 0, and the complement of L_2 is homeomorphic to the complement of L_1,
then we show that L_2 may be obtained from L_1 by a sequence of twists around
unknotted components. Finally, we show that for any positive integer n, an
algorithm for detecting an n-component unlink leads immediately to an algorithm
for detecting an unlink of any number of components. This algorithmic
generalization is conceptually simple, but probably computationally
impractical.
