Single Comment

I am now more certain that I understand what you're saying, and that my proof from before works.  (Joseph has the same idea.)  Let me give some more details:

Given a cell, an orthogonal direction, and a diagonal direction at a right
angle to the orthogonal direction, the set of cells reachable from the
given cell in those directions has a rectangular geometry (it is a proper
subset of the entire board, consisting of all cells of two colors in the
4-coloring).

So, given starting and destination cells, there are at most three ways to
write the leap as a combination of orthogonal and diagonal (at right
angles) steps.  These three ways correspond to the three orthogonal (or
diagonal) directions.  But the 4-coloring has the property that these three
rectangular sub-geometries (not using that as a technical term) each
consist of the color of the starting cell and one other color, and these
three other colors are distinct among the three sub-geometries.

Furthermore, the 4-coloring has the following property: the destination
cell's color is the same as the starting cell's color if and only if the
number of diagonal steps and the number of orthogonal steps have the same
parity.  (This can be seen by noting that the 4-coloring induces the usual 2-coloring on any of the rectangular subgeometries.)

So, if the SOLL is odd, the target cell is of different color than the
starting cell, and so there is only one of these subgeometry paths.
If the SOLL is even, then the target cell is of the same color as the
starting cell, and there are three distinct subgeometry paths (except for
SOLL=0).  These three are distinct in that they use different
ortho-diagonal subgeometries, but they may have the same number of
orthogonal and diagonal steps.

For example, m=1,n=5; m=2,n=4; m=3,n=1 all have SOLL 28 and have a common
destination cell.  (Perhaps this is the smallest with all three distinct?)