"So is your conjecture that in the hex-grid (or triangular grid depending on how you look at it), there are no two right triangles that share a hypotenuse with odd squared length... where the legs of the triangles are along the orthogonal and diagonal directions?"
No, as I have already cited a counterexample to that conjecture if you include triangles of side zero. The orientation of the hypotenuse on the board matters as well as its length. I will have a think to see whether I can formulate the conjecture with purely numeric variables.
"I think the 4-coloring of the hex board can help to prove this. See
wikipedia's 4-coloring image. If a leap has odd SOLL, then in any right angle diagonal-orthogonal path--say m diagonal and n orthogonal as before--we must have that m and n have different parities (else 3m^2+n^2 is even). Then the starting and landing cells have different colors in the 4-coloring. But each orthogonal-diagonal pair of directions at right angles involve exactly two colors, one of which is the starting cell's color. So traveling along distinct orthogonal-diagonal directions lands at distinct color cells."
I will have to think about this offline and reply at a later date.
No, as I have already cited a counterexample to that conjecture if you include triangles of side zero. The orientation of the hypotenuse on the board matters as well as its length. I will have a think to see whether I can formulate the conjecture with purely numeric variables.
"I think the 4-coloring of the hex board can help to prove this. See wikipedia's 4-coloring image. If a leap has odd SOLL, then in any right angle diagonal-orthogonal path--say m diagonal and n orthogonal as before--we must have that m and n have different parities (else 3m^2+n^2 is even). Then the starting and landing cells have different colors in the 4-coloring. But each orthogonal-diagonal pair of directions at right angles involve exactly two colors, one of which is the starting cell's color. So traveling along distinct orthogonal-diagonal directions lands at distinct color cells."
I will have to think about this offline and reply at a later date.