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White has proved (White, G. K. Lattice tetrahedra -- Canad. J. Math. 16 1964 389–396.) the following theorem:

If $T$ is a closed tetrahedron and $\Lambda$ is a lattice which contains the vertices of $T$, then the two following conditions are equivalent:

(1) The only points of $\Lambda$ in $T$ other than the vertices lie on a pair of opposite edges of $T$;

(2) There is a pair of parallel lattice planes of $\Lambda$ through a pair of opposite edges of $T$ such that no points of $\Lambda$ lie between these planes.

His proof is rather complicated. Has anybody proved this theorem in a simpler way? What proof is the simplest one?

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Perhaps I misunderstaand. (2) seems to imply a version of (1) which does not insist on there being any points besides vertices in the intersection of tetrahedron and lattice. Also, why do you want another proof? Gerhard "Ask Me About System Design" Paseman, 2012.08.12 –  Gerhard Paseman Aug 13 '12 at 1:00
    
Also, I can imagine a true 2 dimensional analogue of this (once I understand the term "lattice plane" thoroughly), but I intuit (perhaps wrongly) that a 4d version might fail. A lot hinges on the strictness of the term "lattice plane" Perhaps you could also include White's theorem verbatim? Gerhard "Sometimes Needs The Fine Print" Paseman, 2012.08.12 –  Gerhard Paseman Aug 13 '12 at 1:06
    
Yes, we can assume that our $T$ has only 4 lattice points. Then White's theorem will describe empty simplices. There is a multi-dimensional generalization at arxiv.org/abs/1004.3411 –  Alexey Ustinov Aug 13 '12 at 5:25
    
I always look for a simpler ways. –  Alexey Ustinov Aug 13 '12 at 5:29

2 Answers 2

A simple illustration of this interesting theorem:
          Empty Lattice Tetrahedron

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I think an improvement would show the lattice points on the two planes that are near but not on the tetrahedron (Colored in a different color.) Gerhard "Ask Me About System Design" Paseman, 2012.08.12 –  Gerhard Paseman Aug 13 '12 at 1:18
    
Also, something that would suggest more generality would be to have the two red edges not be perpendicular (if transported to the same plane). Gerhard "Ask Me About System Design" Paseman, 2012.08.12 –  Gerhard Paseman Aug 13 '12 at 1:29

I have not looked at the literature, so the following may fail for some geometric reason of which I am unaware; this reflects my thinking, and I do not yet see how else to get the results.

It should be clear that (2) implies (1), so let's think about not (2) implying not (1). Then for each pair of parallel planes going through opposite edges of T, if they are lattice planes then there is another lattice plane between them that goes through T and has "as many" points of the lattice as do the parallel planes. I am hoping there is enough regularity so that the middles plane looks like one of the parallel planes shifted; perhaps I am wrong. Now we have for each pair of opposite edges of T a lattice plane going through T between the pair.

Following Sidney Harris, "And then a miracle occurs...", which means I do not know how to take the three planes intersecting the middle of T and produce a point lying in T and on the three planes, and make it a lattice point. If I were looking for a simple (and natural) proof of the result, however, this is what I would try. Perhaps I would need a map to pull the argument into Z^3 and use some analytic geometry to finish things, but I hope not.

Gerhard "Not Copying Someone Else's Catchphrase" Paseman, 2012.08.14

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