"And, the orientation of the hypotenuse is determined by the choice of right-angled orthogonal-diagonal pair."
That is certainly part of what I was hoping someone could prove or disprove. It is self-evident for two orthogonals at right angles on a square-cell board, or even three on a cubic one. It is self-evident for a diagonal and an orhogonal with a 45° turn on a square-cell board. It is even self-evident for two orhogonals with a 60° turn on a hex board. It is however not only not self-evident for a diagonal and an orhogonal at right angles on a hex board, but untrue in the case of an even SOLL. Again taking m diagonal and orthogonal steps, m=1, n=1 gives the same destinations, not just leaps of the same length as by m=0, n=2. Likewise m=1, n=5 gives the same destinations as m=3, n=1. For odd-SOLL leaps I suspect from lack of counterexamples, and for prime-SOLL leaps I strongly suspect, that such duplicate vaules of m and n for the same destination are impossible, but cannot yet prove either. I have now checked all values of m up to 20 and n up to 40.
That is certainly part of what I was hoping someone could prove or disprove. It is self-evident for two orthogonals at right angles on a square-cell board, or even three on a cubic one. It is self-evident for a diagonal and an orhogonal with a 45° turn on a square-cell board. It is even self-evident for two orhogonals with a 60° turn on a hex board. It is however not only not self-evident for a diagonal and an orhogonal at right angles on a hex board, but untrue in the case of an even SOLL. Again taking m diagonal and orthogonal steps, m=1, n=1 gives the same destinations, not just leaps of the same length as by m=0, n=2. Likewise m=1, n=5 gives the same destinations as m=3, n=1. For odd-SOLL leaps I suspect from lack of counterexamples, and for prime-SOLL leaps I strongly suspect, that such duplicate vaules of m and n for the same destination are impossible, but cannot yet prove either. I have now checked all values of m up to 20 and n up to 40.