The plane can be tiled without gaps by congruent two-dimensional regular simplices (i.e., equilateral triangles). The three-dimensional Euclidean space cannot be tiled by congruent three-dimensional simplices (i.e., equilateral tetrahedra). Can this be done for higher dimensions? I have a vague recollection that I saw a proof that for some dimensions the tiling exists and that the existence is somehow related to the divisibility of the dimension by four, but this may be false. Can anybody please formulate the correct statement and provide either a proof or a reference?
Thanks!



Tiling by regular simplices
Senechal's [survey paper](http://www.jstor.org/stable/10.2307/2689983) has a beautiful historical overview and some standard references. Recently, we proved in [this paper](http://www.math.ucla.edu/~pak/papers/acute24.pdf) that one cannot tile $\Bbb R^d$, $d\ge 4$, with congruent copies of an acute simplex (this answers your question as well). Interestingly, existence of an acute simplex in $\Bbb R^3$ which tiles the space is open since Sommerville's classification does not allow reflections (see e.g. [this](http://www.springerlink.com/content/j1161728640p1025/) paper). Curiously, any polytope which tiles $\Bbb R^3$ or $\Bbb R^4$ must be [scissors congruent](http://en.wikipedia.org/wiki/Scissors_congruence) to a cube, even if the tiling is aperiodic (see [here](http://www.springerlink.com/content/gh57r3q733444808/)). 


