The 4x4x4 cube does indeed have a "parity" problem. It may be described
Look at one face of the cube. Number the cubies across one edge as 1, 2,
3, 4. It can be shown that an edge cubie of type two can be either in
a type 2 position unflipped or in a type 3 position flipped.
Next, assuming that the top three planes (or slices) are correct, that
the corners of the last face are correct, and that the proper color is up for
all of the edge cubies, then the edge cubies in group 2 and those in group 3
either have the same "parity" or opposite parity. The parity groups for the
edge cubies are defined as
b d b c
and any of the twelve variations of each obtainable by one or more rotations
of three elements at a time. (Each letter represents a different color.)
When solving the cube, the cubies of group 2 and group 3 will either end up
with the same or opposite parity. If the parity is the same, the cube can be
solved straightforwardly. (I assume people have discovered the transforms that
invert the parity of both groups at the same time.)
However, when the parity is opposing, there is only one "transform" that
will correct the problem. People accustomed to thinking of a useful transform
as one which performs some limited rearrangement of cubies while leaving
everything basically unchanged will find that none of their transforms is
sufficient to solve the problem.
Credit goes to Bill Mann for discovering the "transform" that solves the
problem. I leave it up to him to describe the solution if he wishes to.